Derived Torelli theorem for K3 surfaces
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01:00:26
Diagram
Transcript: English(auto-generated)
00:16
Thank you very much. Thank you to the organizers. And it's really a great pleasure to be here and
00:22
to speak. And I want to revisit some old questions about Torelli theorems that Arthur, of course, have very influential papers, especially with the super singular K3s. And I want to try to sort of rethink a little bit how we might phrase Torelli theorems in positive characteristics.
00:44
I'll talk about joint work with Max Lieblich. It came up in the previous talk. Just to get ideas fixed, let me begin by reviewing some basic Torelli theorems
01:03
over the complex numbers. I have in mind other Torelli theorems, but let me talk about K3 surfaces for this lecture. So let's say we have a K3 surface.
01:33
And so, well, what do you consider? You first consider the the lattice, which is the singular cohomology with z coefficients, which has an inner product.
01:49
And so there's sort of two bits of data. There's this, which is sort of the analytic part, and then there's the Hodge filtration, which is somehow the algebraic part. So for the K3,
02:05
of course, it just looks, well, there's really just one piece. So really you just need to know this, which is the H0 of x kx. From that and the inner product,
02:30
you can get the f1. And so the Torelli theorem in its standard formulation
02:43
says the following. So two K3 surfaces x and y are isomorphic if and only if there exists an isomorphism lambda x,
03:16
so that means including the inner product, preserving the filtrations.
03:31
But it doesn't mean that you can leave the isomorphism. No, no. So this is a delicate question. So as people were,
03:41
I think some of, yeah, so as Ofer says, so let me remark. And this is, actually, I'm glad you asked because I want to discuss this point in positive characteristics. So if, in general, you needed to take a Kähler class to a Kähler class. So if
04:01
let me give it a name, let's say sigma. If sigma takes a Kähler class to a Kähler class,
04:23
then sigma lifts to an isomorphism, put it sigma tilde x to y. This is what you were asking, I think. And also it is unique.
04:47
If it's unique, let me not say anything on the spot. I'm not sure on my feet.
05:01
Yeah, yeah, yeah. That's the question, right? Is it injective, right? No, that's the question. Yeah, I don't, maybe not say something wrong. Okay, and so, right, and so in general you can arrange this
05:26
after, can arrange this by reflecting across minus two curves.
05:43
This is somehow a very classical story. And so, I mean, you see in the very formulation, it's not clear what you should do in positive characteristic because of this integral lattice, right? So of course, August proved a Torelli theorem for super singular K3s,
06:15
which is really the only result I know which uses cohomology as sort of the, you know, you
06:26
some structure, then you have isomorphic varieties. Okay, so sort of the rough idea. So when you say you can arrange this, I think you also need maybe to apply minus one, or I'm not sure because the reflection along minus two curves
06:46
cannot be enough if there are no minus. So you could have the automorphism minus one, minus the identity. Let's see. Yes, I'm sure you're right. Let me, let me,
07:09
so yeah, let me put it, you just put minus one, minus one. Let me just put it like that. Okay, so the rough idea here is to try to make a
07:30
a formulation even in positive characteristic. We'll replace lambda x here by the derived category of, category of coherent sheaves on x. And the Hodge filtration will get replaced by,
08:03
so what filtration can you put on something coming out of the derived category? Well, you can go to the Grothendieck group, tensor Q, which by Riemann-Roch is the chow group tensor Q. And there you have a filtration, which is filtration,
08:24
I mean it's graded, so gets replaced by the co-dimension filtration here. Now this is a very, you know, loose sort of analogy, but I'll explain a Torelli theorem for K3s in this context and
08:43
deduce consequences. Okay, so but in general I don't know to what extent, I mean it could be that this is a much stronger kind of result or condition that you could have Torelli theorems sort of in the derived sense when it might not hold there. So the result you're trying to
09:19
explain, give a new proof over the complex numbers? No, in fact this is, I'll explain,
09:27
sketch the argument for the proof which will reduce to the complex case, unfortunately. All right, so let me give a precise statement. Okay, so K let's say is algebraically closed
09:52
and characteristic not two, for some technical reasons. And so x over K is a K3 surface and so
10:11
there's, some of this came up in the previous talk, but let me get it on the board. So x and y are called Furi-Mukai partners if we have an equivalence of triangulated categories
10:38
between the bounded derived categories of coherent sheaves.
10:44
And so, and as I said, so there's one point which is, which I'll come back to later, but if you take the growth and de-group of the triangulated category, you have to choose this isomorphism in the right way, but you have an isomorphism here
11:06
with chow groups. And actually for the theorems I'm going to discuss, I could pass to numerical equivalence, but anyway that's not too important. So I view this as being filtered. This would be F2, F1, F0. So just in the simple way. And then the theorem
11:44
is that if x and y are K3 surfaces and there exists an equivalence, triangulated categories
12:15
d of x, d of y, preserving the filtration, then x is isomorphic to y. So it is,
12:35
at least with this analogy between viewing the derived category as an integral structure
12:42
plus the hot filtration, then you have isomorphism. And the equivalence of triangulated categories is supposed to be K linear or? Ah, yes. Otherwise you could twist by automorphism of K? Yes, yes, yes. Thank you.
13:02
And if it's just 100 categories in the old sense, not in the any infinity? No, no infinity, no. Yeah, just derived category and coherence views. So what about the previous question, a map from the isomorphism x to y to the equivalence? So this I don't know the answer to. This is a very interesting question.
13:28
I'll sketch the argument because it involves deformation theory and a lot of the classical tools of deformation theory of K3 surfaces, but the proof does not in any way tell how
13:42
the isomorphism you produce is related to the equivalence. Could you remind us what's known for services of other kinds, is there's a whole class of services where isomorphism in the right category? So if either the canonical sheaf is ample or an anti-ample, then you know it already. So then you don't need the filtration.
14:01
What's interesting here, maybe I'll put it here, remark is that, and I'll explain exactly what the examples are, d of x equivalent to d of y is not enough.
14:22
So this is sort of very classical Mukai oral of many people. You can't just ask for equivalence of triangulated categories. This is not enough to imply that x and y are isomorphic. But actually this result, as I'll explain, will tell you exactly to what extent that fails.
14:46
Do you need to assume that y is a K3 surface?
15:21
No, that follows from having the equivalence of triangulated categories. So let me review, there's sort of two ingredients that came up in Francois' talk, but let me review just a few basics.
15:43
So for Mukai transforms, so that's right, so x, y are smooth, projective over K.
16:03
If I have an object of the derived category of the product, I get such a functor by sending K to RP2 lower star P.
16:28
So that's sort of the standard transform, pull back tensor and push down. And a theorem of Orlov, which is really the sort of key thing to get things off the ground,
16:42
is that if I have an equivalence of K linear is of this form, phi p, for some p.
17:19
So this tells you not only, well it gives you some geometry,
17:27
it also tells you the candidates, I mean how you produce as came up in Francois' talk. What you do is you, I mean how could you have a complex on the product, the most natural thing is to take y to be a modulated space of sheaves on x,
17:42
and then p to be the universal family. And d x is d b coherent? Yes, all the derived categories will be the derived categories of coherent sheaves on smooth projective varieties, so there's no issue about finite, I mean they're regular. Yeah, yeah, found it.
18:02
Okay, so what are exactly the moduli spaces? Yeah, so I want to write down, so x over K is a K3 surface,
18:26
and so then let me fix normalization. So for a complex you have the mochai vector, which is the churn character times the Todd class of x square root. Okay, and I view this as being an element of, and for K3 it's defined integrally,
18:49
in general you would have to put some denominators, but for K3 it's there. And it just, I mean it has a very simple description, it's the rank
19:01
c1 of E, and then the rank plus c1 of E squared over 2 minus c2 of E, that's the formula. And then we also fix a polarization on x, so an ample class, and then you can consider the stack
19:33
of, so Giesecker semi-stable sheaves on x with a mochai vector, this vector,
19:55
a given, okay, so you fix your nu and then you make these moduli spaces.
20:05
And then, so for good V, as Francois mentioned, what do you have?
20:26
We have A, every semi-stable sheave is stable, which means that this thing here
20:44
is a GM gerb over some algebraic space, and in fact is a trivial gerb over a K3 surface, which I'll denote in roman font, like that.
21:11
And so, I mean, where the universal sheaf lives on the product of x with a stack, but since it's a trivial gerb, you can descend the universal sheaf to,
21:26
so I'll put universal sheaf E on x cross, so now you have two K3 surfaces and a sheaf on it
21:41
induces an equivalence pi E d of x, d of, so this gives a supply of
22:04
4-mochai partners for a K3 surface. Let's see, oh there's another board, so let me state a theorem which is that every 4-mochai partner, meaning K3 surface with equivalent derived category
22:34
of x, is of this form. Does it mean that the equivalence is after isomorphism
22:46
given by the universal sheaf or that there is another equivalent? Another equivalence, yeah, so I'll explain actually how this follows from the other theorem where already that ambiguity is present. I'm not saying anything about, I don't know how to relate the equivalence to,
23:10
I don't know how to lift the equivalence on the derived category to an isomorphism, yeah, in either case. So what do you say of this form, you mean for some P in the corresponding
23:21
mean? Let me put it here, is a MHV for some VNH, so I should say over C, this is somehow classical. Let's see, there are many names here I should put there.
23:47
But I want to sketch in the spirit of these minus two curves how I view this in relation to the filtered result, so that's by the following proposition. So let's suppose I have two
24:45
K3 surfaces and phi from D of x to D of y an equivalence
25:06
of triangulated categories, of derived categories, then there exists
25:20
a choice, then there exists MHV and, sorry let me say it the way it's written here, then after
25:43
precomposing phi with an autoequivalence of D of x and an equivalence
26:07
D MHV of x, so arising in that way, you can arrange for the equivalence to be filtered.
26:31
So you start out with something which is not a filtered equivalence, but then you have these basic steps you can do. You can apply autoequivalence of D of x and you can do this
26:42
game of replacing x by moduli space of sheaves. And so once you do that, then you can make, and this is just playing games with the sort of invariance of the on the child groups, and so then the corollary is this result. So that implies this
27:02
this theorem here. So once you have the result that two filtered, if you have a filtered equivalence, then they're isomorphic, then from this proposition you get that every Fourier-Mochai partner is a moduli space of sheaves. And this MHV is this on the x side, right?
27:20
Yeah, I mean I could do it on the other side too, but I have to, yeah, I want y to be MHV, so I pre-composed to make it filtered. Okay, all right, so let me state another theorem, which is more subtle, and
27:43
which is there are only finitely many for a given x, only finitely many
28:00
Fourier-Mochai partners. And two, if x is supersingular, so from positive characteristic,
28:21
no non-trivial Fourier-Mochai. So even though I think this condition about preserving the co-dimension filtration looks very strong, in the case of K3 surfaces you can recover
28:47
what's known about Fourier-Mochai partners of K3 surfaces from that one sort of Torelli theorem. Okay, so I want to discuss the idea of the proofs slash
29:51
deformation theory of K3s and K3s with the Fourier-Mochai transforms and so on.
30:06
So let me give sort of the idea over C. I think it's the filtered derived equivalence gives an isomorphism, follows very easily from the classical Torelli theorem. So why is that?
30:26
The point is that if you have x and y, then you get this phi p which is given by pullback tensor and pushdown, so that passes to any cohomology theory.
30:41
And so phi p induces an isomorphism on this what French Swach denoted h tilde, and I think my normalization is slightly different than his.
31:04
I think if I change it now I'll get confused, so let me put it, I think you had a shift by one. Okay, so you do this mochai lattice and so h tilde, so this is, sorry my
31:25
board work is not good, so I define it in this way. So you get an isomorphism on these mochai lattices and the filtered condition tells you that it induces an isomorphism
31:42
on h2. So filtered implies you get an isomorphism on h2 preserving the hot filtration,
32:05
the hot filtration, and so we're done by the usual Torelli. So in some sense this is a much
32:22
stronger, we're imposing a much stronger condition than the usual Torelli condition. Let me just remark over here, I should have said this before, in this case,
32:41
so if you think about it, if I have a point here, let's say k of z, I look at the skyscraper sheaf of a point z in this moduli space, the corresponding sheaf over here is the sheaf corresponding to the point z,
33:02
so the ranks in general get messed up. So here, when you have this kind of equivalence, it's not going to preserve the filtration. I should have remarked that, so in general these things get scrambled up, but once you preserve the filtration, then you're in the usual Torelli theorem setting.
33:21
Okay, so what do you do in characteristic p? So now we're in a bit of a delicate deformation theory problem of k3 surfaces. We want to think about how to deform x, y, and p,
33:46
the triple consisting of two k3 surfaces, an object in the derived category, and the product, and so how do you do that? Well, you consider this very sort of strange stack,
34:06
which is the stack of perfect complexes x cross y, sorry, on x, which are simple
34:27
and I'll put it universally gluable. This p is some object of the derived category, so simple means that the automorphism functor is just GM, so it looks like a simple sheaf,
34:49
and this universally gluable means that the negative x sort of functors p p
35:02
zero for i less than zero. Okay, so you impose some condition, and if you impose these conditions, then this is an old paper of Lieblich, which is that this is an algebraic stack.
35:24
So you have some geometric object to work with, because it's big and non-separated and all that kind of thing, but well in general if you don't put the the simple condition,
35:47
but the simple condition actually gives us a reasonable space here. So the simple condition implies that you have, that this thing dx is a GM-gerb
36:15
over an algebraic space, which maybe I'll roman font dx, so you get the following, so given
36:34
a Fourier-Mochai equivalence p, so phi p from d of x to d of y,
36:55
so you have this p in the product, you can view it sort of asymmetrically as being,
37:05
if you think about y here, and then here's x cross y, you can view it as a family of complexes on x parameterized by y, so here's p, and what so if you look at a point you have to verify things like you basically have to compute x to i of p y p y,
37:28
and so for these conditions you have the x zero and the negative x, and this is happening in the derived category of x, but because this is a Fourier-Mochai equivalence, this is the same as
37:42
x i d of y k of y, so you compute the, you can bring it to the triangulated category on the other side, okay, so these conditions are actually easy to verify when you have a
38:02
Fourier-Mochai equivalence, and so what you end up with is a map from y to the stack, which is this mu given, let's say mu p, so it gives you this map, and call this mu p bar,
38:23
and the condition that phi p is an equivalence, actually is fully faithful,
38:45
is equivalent to this mu p bar is an open immersion, okay, so you somehow translate
39:00
the condition to have a Fourier-Mochai equivalence to sort of a map from this algebraic space. All right, so what can you do with this? So let me state a proposition. Well, let's consider now the deformation functor of x, so we have a k3 surface,
39:24
we have its deformation functor, which we know everything about, so a goes to the liftings. dx isn't separated, is that the point? dx is not separated? The Roman font? Yeah,
39:45
nothing is separated. I mean, I'm not saying anything about global geometry. Yeah, liftings of x to a. Okay, so we have the deformation functor of x, and I'll put it over
40:01
here. Proposition, there's an isomorphism, there's an isomorphism of functors, I mean,
40:22
these are pro-representable, so this, of deformation functors, so that's not such a big deal, but such that
40:44
for every L in the Picard group of x, so I'd say x and y are k3 surfaces, k3, and I have a Fourier-Mochai equivalence dx to dy.
41:05
And, ah, sorry, you're right, and let me further assume, ah, okay, sorry, I'm getting ahead of myself. Sorry, thank you. So once, when you're in this situation,
41:24
so you can, after changing your choice, changing the choice of p, so you always have to do these kind of basic operations, you can arrange
41:43
that 1, 5p100 is 100, and so I'm assuming this is filtered also. So I'm putting a lot of conditions, but once you have that, you can arrange that, and that it takes the ample cone
42:01
of x to the plus or minus of the ample cone of y. So now you see why there's really not much control over, you start with some general equivalence, make it filtered, and then you
42:21
do more operations to make further, so there's not much control. But once you do this, the proposition is that, yeah, reflection, it's sort of like the, I mean, I view it by very rough analogy with the classical theory as making reflections and things like that.
42:46
So then you have an isomorphism of deformation functors, delta, from the deformation functor of
43:04
x to the deformation functor of y, such that for every l in the Picard group, so we're looking at k3, so the deformation functor of x with a line bundle is a subfunctor,
43:22
and so you can ask what it does. So delta of this deformations of the pair xl maps isomorphically to the deformations of the pair y5l. Okay, so, and because of the way I set it up, this is the class of a line bundle. Okay, so
43:45
in other words, you can use the Fourier-Mochai transform to identify the deformation spaces in a way that matches up the line bundle. So I'll explain, I'll explain. Maybe I should say
44:08
there's the basic problem we encounter, which I think I'll have time to mention, in the singular case, you run into the, so here, so in the classical literature, like in your papers,
44:25
you consider deformations of x and an l, but you could also consider deformations of x with a bunch of line bundles, and that gets more and more complicated the more line bundles you consider. But this is an important problem, I think, because there's a number of situations
44:42
where you might try to deform, if you have a high rank k3, you might want to deform the k3 plus, well, of course, if it's super singular, you can't deform the whole lattice, or whole neuron severity, but you might do want a big lattice. Okay,
45:01
big Picard of the generic fiber. Okay, right, so how is this going to work? So the construction of delta, yeah, so well, I think it's basically this picture, so you take,
45:24
let's say I have A as an artinian local W algebra, and I have XA over A, all right, so then I'd get a lifting of the dx and the dy, so I get, let me do it on, I get dxA,
45:46
and then here I have the dx that I started with, and then here's my y, which maps here, this is the mu p bar, and this is open, all right, and so, well, open subsets lift uniquely, so, and this is open,
46:09
so there it is, that's the deformation, okay, so this defines delta. No, it doesn't matter, I just, I have an open subset, and I have a new potent thick,
46:24
new potent thickening of, dx could be as bad as you want, yeah. It is nice in the sense that it is locally a finite type, on the stuff over A, yeah, so there's a subtle point which is special to the K3, which is that I need to know this is flat over A, yeah, locally a finite type and so on,
46:47
yeah, that's right, I'm sweeping a lot under the rug, but the main subtle point is, I mean, why couldn't this just be dx, right, maybe it was, you need the flatness of this over it doesn't satisfy any separation, no, I don't believe so, no,
47:03
but there are several white spaces, certainly it is quasi-separated, yes, but not locally separated, I'd have to sit down and think about it, but, yeah, yeah, but it, but I mean, it's an important point that it doesn't actually, I mean,
47:21
you have to verify that it has some reasonable properties, but not, it doesn't have to be a projective variety or anything, you're lifting open subsets. Okay, now, in fact, there's a little bit more here, which is you can lift the p, so you see there's an annoying thing, which is that the Roman font dx is something like
47:41
a coarse modulized space, and you really want to lift the Fermi-Kai transform too, so in fact, this takes a little more work, you can lift the p, which is really the, I mean, this is this map mu p to p a over
48:06
in d of x a cross over a, okay, so this takes some more work, but let me not dwell on that. Okay, and so now, how do you proceed? You use
48:27
this here, because we want to lift to characteristic zero, you better preserve some ample line bundle, so you can algebraize and get the whole thing, so now, so the conclusion, so what do you get? You get a, let's say, a mixed characteristic DVR, v, and deformations x, y, p over v,
48:59
lifting x, y, p, and then
49:10
yeah, I know, yes, absolutely, so I need, I have at least one map to a mixed characteristic DVR, right, so and that'll, I'll mention a related point, right, and so then generic fibers,
49:32
and actually, I mean, there's a lot to check, but this deformation here, actually, if I start out with a filtered equivalent satisfying those conditions, this deformation is also
49:42
going to be a filtered equivalence, okay? Is it unique, the p a, or the script p? This thing, this thing here, let's see, so I have to calculate, no, I don't need it to be unique,
50:10
yeah, I mean, it has to do with this sort of germ there, and whether there could be a homological, I mean, a torus, or the, well, okay, so what I do, I choose my,
50:31
an ample line bundle on x, and then I look at the versatile, the deformation space of x, l, and then it gets paired up here, and then the only thing I have not sort of
50:43
discussed is why you actually get this, an actual p, because you only have a map to the Roman font, right? So I have a map to the course moduli space, I have to trivialize a germ over that to get the actual p. Okay, all right, so then generic fibers admit
51:05
a filtered form of chi-equivalence, and so then you have to replace v by another extension to get the generic fibers isomorphic, and then you go back down because using some minimality of K3 surfaces, okay, and then use characteristic zero result,
51:26
zero result, and then go back back to closed fiber, okay? So there's a lot of technical points here, but I think the key idea is this kind of,
51:40
sorry, yeah, I should make a complete DVR. So the second feature literally preserves the filtration automatically in this case, it's not even close. No, it's not something, yeah, it just follows from the, it comes from this big stack here. Okay, all right, so let me make a remark.
52:05
So that gives that if you have a filtered equivalence and they're isomorphic, the finiteness result, how do you get the finiteness result and the, I think the super-singular case is maybe the most interesting because how could you prove that
52:28
in fact that if you have a super-singular K3, then any form of chi partner has to be itself. This argument that I just sketched does not lend itself very well to that, but the
52:48
problem you encounter, so for a super-singular K3, so people who do lattice theory know that if you have a
53:01
neuron severity which has rank at least three in characteristic zero, then you don't have non-trivial form of chi partners. Okay, so now the game becomes deform K3 surface plus a rank three
53:29
subgroup inside neuron severity of X, and there's a lot to do, but that's the basic deformation theory problem you encounter.
53:43
So in this case you can calculate using the deformation theory of polarized K3s that you do still have a characteristic zero point of the deformation space, but maybe I'll put it as a question in general. I don't know in general, I don't know how to describe
54:19
deformations. Let me write it this way, deformations of X, E. So where you fix the groups on
54:28
Arthur's paper, there's discussion of if you have a, you can get a quadratic, you know, make a quadratic extension in Deline's article and so on, but in general,
54:41
what does the deformation space, something like that, look like? I don't know. All right, it might not have any characteristic zero point. Yeah, yeah, for a super, yeah, for, I mean, take the super singular case, yeah. When you look at this thing about this P and PA, so there was a GM job there, and
55:04
when you have this, the opinion, the local ring deformation that it could visit, that you're talking about characteristic physics, so it could be that you don't know that the H2 class is zero. So the point is that you do some trick with the determinant
55:23
of the P and relate that to the obstruction class. So there is a serious argument there. I mean, it's not, but you have to do, you're right, there is an obstruction in general, but if you calculate what that obstruction is, you can relate it to the determinant
55:43
of P, and so if you do it, if you set it up right, you can arrange for that obstruction to vanish. Because it's a one-dimensional obstruction space, and you have to relate
56:03
what that obstruction is to something having to do with P. So it's a non-trivial point, yeah. Okay, let's see, yeah, two or three minutes. Okay, so maybe I'll just mention sort of
56:21
a couple more questions here. So I don't know how to do this in general. The other point is, which was raised several times, is the way this is set up, this, you know, the isomorphism, construct the isomorphism, right, I mean, from d of x.
56:47
I don't know how to just take the derived equivalence and then construct the map, right, this seems, I don't know where to do it. It's tempting to think about the group of zero cycles and sort of, if it's filtered, you get a map on zero cycles,
57:05
but that sort of looks like you should just see the morphism, but I don't know how to do that. And I guess the other question, I won't write it, is I don't know in general whether filtered equivalence is a much stronger condition, derived equivalence is much stronger
57:24
than sort of classical Torelli. There seems to be promising progress in the hypersurface case, but I think that's about where we're at right now. Thank you.
57:47
So is the last problem in the problem in generalizing to no algebra, high cost space theory? No, the problem is that sort of the way I've presented, the way I understand it, I don't know,
58:01
I mean, in the honest Torelli situation, you know a lot about the isomorphism in relationship to what you do in cohomology, right? So you have some transformations that you understand, and then you can lift it to an isomorphism if it takes the Kähler class to the Kähler class,
58:21
and so on. But here I think it's a total mess. What is the relationship between the isomorphism and the equivalence of triangulated categories? Right, yeah, so given, you take this thing, how could you even, you know,
58:42
maybe you could impose some condition that says then there's an actual map, right? I don't know. Very nice question. So you discussed the super-singular case. Yes. So in the object you have the ordinary case. Yeah. And then form a modular, you know, group structure, form a total.
59:02
Ah, yes. So then what is, how does the formation theory work? I don't see it because either it's local or global, and I'm not too sure. Could you repeat the question? What is the story for ordinary?
59:23
So I guess in the ordinary case we know dx is really clear, or def x of dx. Oh, def x, yeah, yeah, yeah, def x, yes, yes. Ah, yeah, so what have, and what happens under this delta? Well... What is the story then? Yeah, I don't know because I don't understand sort of the general phi p, right? So, I mean, somehow...
59:44
The basic ingredient was that the tangent space to woman dx was x1. That's right, yeah, yeah. So that was the basic ingredient to get your... Yeah, so then I have to see what these churn classes do as it moves over to the other side.
01:00:02
Yes. It was the ordinary case. Ah, I see. I see. Actually, Francois suggested, and I think this is another promising, it may work, to prove it using Arthur's theorem, deformed to the super-singular K3.