 Thank you very much. Thank you to the organizers. And it's really a great pleasure to be here and 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. So I'll talk about joint work with Max Lieblich. I came up in the previous talk. And just to get ideas fixed, let me begin by reviewing some basic Torelli theorems over the complex numbers. And I'll say, I have in mind other Torelli theorems, but let me talk about k3 surfaces for this lecture. But so let's say we have a k3 surface. And so what do you consider? You first consider the lattice, which is the singular comology z coefficients, which has an inner product. 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, of course, it just looks like, yeah. Well, there's really just one piece. Really, you just need to know this, which is the h0 of x kx from that and the inner product. You can get the f1. And so the Torelli theorem in its standard formulation says the following. So two k3 surfaces, x and y, are isomorphic, if and only if there exists an isomorphism lambda x. So that means including the inner product, preserving the filtrations. But it doesn't mean that you can leave the isomorphism. No, no. So this is a delicate question. So as people were going to, I think some of, yeah. So as Hofer says, so let me remark. 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 scalar class to a scalar class. So let me give it a name. Let's say sigma. If sigma takes a scalar class to a scalar class, 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, the lifting. Ah, if it's unique, let me not say anything on the spot. I'm not sure on my feet. More precisely, it's putting a harm of x to y, goes to harm of x to y, what do you say about the back sets? Oh, I saw the x to y. Yeah, yeah, yeah. That's the question, right? Is it injective, right? Yeah. No, that's the question. Yeah, I don't, I, maybe not say something wrong, so. OK, and so, right. And so in general, you can arrange this after, you can arrange this by reflecting across minus two curves. So 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 k threes, which is really the only result I know which uses comology as sort of the, if you have an isomorphism of comology respecting some structure, then you have isomorphic varieties, OK? 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 not. I'm not sure because the reflection along minus two curves cannot be enough if there are no minus two. So you could have the automorphism minus one, minus the identity, certainly. Ah, ah, yes. And minus one is for a reason. Let's see. Yes, I'm sure you're right. Let me, let me, so, yeah, let me put it. You just put minus one, minus one. Well, let me just put it like that, OK? So, OK, so the rough idea here is to try to make 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 hot filtration will get replaced by, so what filtration can you put on something coming out of the derived category? Well, you can go to the growth and deep group, tensor q, which by Riemann-Roch is the Chao group tensor q. And there you have a filtration, which is a filtration. I mean, it's graded. So it 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 deduce consequences, OK? 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 of when it might not hold there. Will the result you're trying to explain give a new proof of the complex numbers? No. In fact, this is, I'll explain, sketch the argument for the proof which will reduce to the complex case, unfortunately, yeah. All right, so let me give a precise statement, OK? So k, let's say it's algebraically closed and characteristic, not 2, for some technical reasons. And so x over k is a k3 surface. And so some of this came up in the previous talk, but let me get it on the board. So x and y are called Fui-Muqai partners if we have an equivalence of triangulated categories between the bounded derived categories of coherent sheaves. And as I said, so there's one point which I'll come back to later, but if you take the growth and degroup of the triangulated category, you have to choose this isomorphism in the right way. But you have an isomorphism here with Chao 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 is the F2, F1, F0, so just in a simple way. And then the theorem is that if x and y are k3 surfaces over k and there exists an equivalence of triangulated categories, d of x, d of y, preserving the filtration, then x is isomorphic to y. So at least with this analogy, between viewing the derived category as an integral structure 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 the morphs of k. Yes, yes, yes. Thank you. And it is just 100 categories in the old sense, not in the any infinity. No, no infinity. No. Just derived category of coherent shoes. So what about the previous question? An isomorphism is x to y to the equivalence of k. So this I don't know the answer to. This is a very interesting question, which is, and yeah. So 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 the isomorphism you produce is related to the equivalence. Could you remind us what's known for surfaces of other canines? Because the whole class of surfaces where isomorphism is a category of bicep isomorphism. So if either the canonical sheaf is ample or anti-ample, then you know it already. So then you don't need the filtration. So 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. So this is sort of very classical Moukai Orlov, many people. So you can't just ask for equivalence of triangulator 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. Do you need to assume that y is a k3 surface? Now that follows from having the equivalence of triangulator categories here. So let me review. There are sort of two ingredients that came up in Francois' talk. But let me review just a few basics. So for Moukai transforms, so that's x, y are smooth, projective over k, 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. So that's sort of the standard transform, pullback, tensor, and pushdown. And a theorem of Orlov, which is really the sort of key thing to get things off the ground, is that if I have an equivalence, an equivalence of triangulator categories, k linear is of this form, phi p for some p. So this tells you not only, well, it gives you some geometry. 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 modular space of sheaves on x, and then p to be the universal family. And dx is db 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. OK, so what are exactly the modular spaces? Yeah, so I want to write down. So x over k is a k3 surface. And so then let me fix normalization. So for a complex, you have the Mokai vector, which is the churn character times the Todd class of x square root. And I view this as being an element of, and for k3, it's defined integrally. In general, you would have to put some denominators. But for k3, it's there. And it just has a very simple description. It's the rank c1 of E. And then the rank plus c1 of E squared over 2 minus c2 of E. That's the formula. And let me also fix a polarization, x, or an ample class. And then you can consider the stack of, so geysiker-sevi-stable sheaves on x with Mokai vector, this vector, a given, so you fix your nu. And then you make these moduli spaces. And then, so for good, as Francois mentioned, what do you have? We have a, every semi-stable sheave is stable, which means that this thing here is a GM gerb over some algebraic space. And in fact, is a trivial GM gerb over a k3 surface, which denoted Roman font like that. And so, I mean, where the universal sheave lives on the product of x with a stack. But since it's a trivial gerb, you can descend the universal sheave to, so I'll put universal sheave E on x cross. So now you have two k3 surfaces and a sheave on it. Induces an equivalence phi E d of x d of. So this gives a supply for Mokai partners for a k3 surface. Let's see how there's another board. So let me state a theorem, which is that every for Mokai partner, or meaning k3 surface with equivalent derived category of x, is of this form. Does it mean that the equivalence is after the morphism given by the universal sheave or that there is another equivalent? Another equivalence, yeah. So I'll explain, actually, how this follows from the other theorem, which already that ambiguity is present. I'm not saying anything about I don't know how to relate the equivalence to I don't know how to lift the equivalence on the derived category to an isomorphism. In either case. So what is that of this form you mean for some phi in the corresponding meaning, you said? Let me put it here. Is a mHv for some v and h? So I should say over c, this is somehow classical. Let's see there are many names here I should put there. 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 forward proposition. So let's suppose I have two k3 surfaces and phi from d of x to d of y, an equivalence of triangulated categories, of derived categories. Then there exists choice. Then there exists mHv and after precomposing phi with an auto equivalence of d of x and an equivalence d mHv of x, so arising in that way, you can arrange for the equivalence to be filtered. 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 the auto equivalence of d of x and you can do this game of replacing x by a moderate space of sheeps. And so once you do that, then you can make. And this is just playing games with the sort of invariance of the Chao groups. And so then the corollary is this result. So that implies this theorem here. So once you have the result that two filtered, if you have a filtered equivalence, and then there isomorphic, then from this proposition you get that every Fourier-Makai partner is a moderate space of sheeps. And this mHv sits on the x side, right? Yeah, I mean, I could do it on the other side too, but I want y to be mHv. So I precompose to make it filtered. OK, all right, so let me state another theorem, which is more subtle, which is there are only finitely many Fourier-Makai partners, and two, if x is super singular, so for a positive characteristic, then no non-trivial. 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 what's known about Fourier-Makai partners of K3 surfaces from that one sort of Torelli theorem. So I want to discuss the idea of the proofs slash deformation theory of K3s and K3s with the Fourier-Makai transforms and so on. So let me give sort of the idea over C. I think the filtered derived equivalence gives an isomorphism follows very easily from the classical Torelli theorem. So why is that? 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 homology theory. And so phi p induces an isomorphism on what Frantz-Swarck deducted h tilde. And I think my normalization is slightly different than his. I think if I change it now, I'll get confused. So let me put it. I think you had a shift by one. OK, so you do this Mokai lattice. So h tilde, sorry, my board work is not good. So I define it in this way. So you get an isomorphism on these Mokai lattices. And the filtered condition tells you that it induces an isomorphism on H2. So filtered implies you get an isomorphism on H2, preserving the hot filtration. And so we're done by the usual Torelli. So in some sense, 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, 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. 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. OK, 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, the triple consisting of two k3 surfaces, an object, and then derived category in the product. And so how do you do that? Well, you'd consider this very strange stack, which is the stack. A perfect complex is x cross y, sorry, on x, which are simple. And I'll put it universally glueable. So what 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 sheath. And this universally glueable means that the negative x, so functors, p, p is 0 for i less than 0. So you impose some condition. And if you impose these conditions, then this is an old paper of Leiblich, which is that this is an algebraic stack. So you have some geometric object to work with, because it's big and non-separated and all that kind of thing. Well, in general, if you don't put the simple condition, but the simple condition actually gives us a reasonable space here. So the simple condition implies that this thing dx is a gm gerb over an algebraic space, which maybe I'll put it on dx. So given a Fourier-Mochai equivalence p, so we'll put phi p from d of x to d of y. So you have this p in the product. You can view it sort of asymmetrically as being, if you think about y here and then here's x cross y, you can view it as a family of complexes on x parametrized by y. So here's p. And 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. So for these conditions, you have to x 0 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 x i d of y, k of y. So you compute the, you can bring it to the triangulated category on the other side. So these conditions are actually easy to verify when you have a Fourier-Mochai equivalence. And so what you end up with is a map from y to this stack, which is this mu given, let's say mu p. So it gives you this map and this mu p bar. And the condition that phi p is an equivalence, actually is fully faithful, is equivalent to this mu p bar is an open immersion. So you somehow translate the condition to have a Fourier-Mochai equivalence to sort of a map from this thing to this algebraic space. So 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. We have its deformation functor, which we know everything about. So a goes to the liftings. dx isn't separated, is that the point? What is your point? dx is not separated? The Roman font? Yeah, nothing is separated. I mean, I'm not saying anything about global geometry. Liftings of x to A. So we have the deformation functor of x. And I'll put it over here. Proposition, there's an isomorphism of functors. I mean, these are pro-representable. So this of deformation functors. So that's not such a big deal. But such that 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-Makai equivalence dx to dy. And ah, sorry, you're right. And let me further assume. OK, sorry, I'm getting ahead of myself. Sorry, thank you. So when you're in this situation, so you can, after changing your choice, changing the choice of p, so you always have to do these kind of basic operations, can arrange 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 of x to the plus or minus of the ample cone of y. So now you see y, there's really not much control over. You start with some general equivalence, make it filtered. And then you do more operations to make further. So there's not much control. But once you do this, the proposition is that you involve some sort of reflection. 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. So then you have an isomorphism of deformation functors, delta from the deformation functor of 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 sub functor. 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 y, 5l. And because of the way I set it up, this is the classical line bundle. So in other words, you can use the form of chi transform to identify the deformation spaces in a way that matches up the line bundle. I'll explain. I'll explain. Maybe I should say there's the basic problem we encounter, which I think I'll have time to mention, in the super singular case, you run into the. So here, in the classical literature, like in your papers, 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 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 neurons vary. But you might do want a big lattice. OK. Big pick hard of the generic fiber. OK. Right. So how's this going to work? So the construction of delta. Yeah. So, well, I think it's basically this picture. So you take, let's say I have a is an Artinian local w algebra, and I have xa over a. So then I get a lifting of the dx and the dy. So I get, let me do it on, I get dxa. 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. And so, well, open subsets lift uniquely. So this is open. So there it is. That's the deformation. OK, so this defines delta. Do you know why it's dancing dx? No. It doesn't matter. I just have an open subset, and I have a nupon thickening of. dx could be as bad as you want, yeah. It is nice in the sense that it is locally of finite time. Sure, yeah. Or the stuff over a is essentially flat 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. Locally of finite type and so on. 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? I mean, it was, you need the flatness of this over here. Generally, it doesn't satisfy any separation. No, I don't believe so, no. Well, there are several of such wide spaces. Certainly it is quite separated. Yes. But not locally separated or not. I'd have to sit down and think about it, but yeah, yeah. But I mean, it's an important point that it doesn't actually, I mean, you have to bear this by that it has some reasonable properties, but it doesn't have to be a projective variety or anything. You're lifting open subsets. Okay, and 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 a course of modular space, and you really want to lift the Pore-Makai transform, too. So in fact, this takes a little more work. Can lift the P, which is really the, I mean, this is this map mu P to P a over in d of x a cross over a. So this takes some more work, but let me not dwell on that. Okay, and so now, how do you proceed? You use this here, because we want to lift a characteristic zero, you better preserve some ample line bundle so you can algebraize and get the whole thing. So now, so what do you get? You get a, let's say a mixed characteristic DVR, V, and deformations x, y, P over V, lifting x, y, P, and then... Can you just, you know, the def-x over l is flat over, you know, the d's and f. Yeah, I know, yes, absolutely. So I need that I have at least one map to a mixed characteristic DVR, right? So, and that'll mention a related point. Right, and so then generic fibers, 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 gonna be a filtered equivalent, okay? Is it unique, or is it PA unique, or what? Is it unique, the PA, or the scripting, or whatever? This thing here, let's see, so I have to calculate. No, I don't need it to be unique, yeah. I mean, it has to do with this sort of germ there and whether there could be a comalogical, I mean, a torsor that... How do you get this to be? Well, okay, so what do I do? I choose my enamel line bundle on X and then I look at the deformation space of XL and then it gets paired up here and then the only thing I have not sort of 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 model space, I have to trivialize a germ over that to get the actual P, okay? All right, so then generic fibers admit a filtered for Makai equivalents 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, zero result and then go back to closed fiber, okay? So there's a lot of technical points here but I think the key idea is this kind of... Sorry, yeah, I should make a complete DVR. So the exact... It preserves the filtration of automatic in this case. Nothing has been posed. No, it's not something, yeah, it just follows from the... It comes from this big stack here, but yeah. Okay, all right, so let me make a remark. So that gives that if you have a filtered equivalents and there is a morphic, the finiteness result, how do you get the finiteness result and I think the super singular case is maybe the most interesting. Because how could you prove that, in fact, 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 problem you encounter so for a super singular K3, so people who do lattice theory know that if you have a neuron severity which has ranked at least three in characteristic zero, then you don't have non-trivial form of Chi partners. So now the game becomes deform K3 surface plus a rank three subgroup E inside neuron severity of X and there's a lot to do but that's the basic deformation theory problem you encounter and 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 deformations, I'll write this way, deformations of X E. So when you fix the groups on Arthur's paper there's discussion of if you have a, you can get a quadratic thing, make a quadratic extension in Delene's article and so on but in general, what is the deformation space? Something like that look like, I don't know. All right. It might not have any characteristic zero point, yeah. For a super, yeah, I mean, take the super singular case, yeah. When you look at this thing about this P and PA, so there was a GM gel there and when you have this, the opinion, the local ring deformation, it could be that you're talking about characteristic pieces so it could be that you don't know that the H2 class is zero over an opinion. So the point is that you do some trick with the determinant 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 of P and so if you do it, if you set it up right, you can arrange for that obstruction to vanish. Determinant of the P, I mean, you can, because it's a one-dimensional obstruction space and you have to relate 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 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. I don't know how to just take the derived equivalents and then construct the map, right? This seems, I don't know where to do it. It's tempting to think about the Chao group of zero cycles and sort of, if it's filtered, you get a map on zero cycles, but that sort of looks like you should just see the morphism, but I don't know how to do that, yeah. And I guess the other question I won't write it is, I don't know in general whether filtered equivalents is a much stronger condition, derived equivalents is much stronger than sort of classical Tarelli. There seems to be promising progress in the hypersurface case, but I think that's about where we're at right now. Thank you. So is the last problem, the problem in generalizing to no algebraic or closed space here? No, the problem is that sort of the way I've presented, the way I understand it, I don't know. I mean, in the honest Tarelli situation, you know a lot about the isomorphism in relationship to what we do in comology, right? So you have some transformations that you understand and then you can lift it to an isomorphism if it takes the Cayley class to the Cayley class and so on. So, but here I think it's a total mess, you know, what is the relationship between the isomorphism and the quillerns of triangular categories, I don't know. Which isomorphism is corresponding to this specifically by morphism? Right, yeah, so given, you take this thing, how could you even, you know, maybe you could impose some condition that says then there's an actual map, right? That gives it, I don't know. I have a question, so we discussed the super-singular case. Yes. So the object you have the ordinary case. Yeah. And then form a modular in a nice group structure, form a tour. 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 that's a true. Could you repeat the question? What is the story for all the names? Well, I guess in the ordinary case we know DX is really clear. From DEF-X, not DX, from DEF-X. 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 there? Yeah, I don't know because I don't understand sort of the general Phi P, right? So, I mean, somehow. The basic ingredient was the dependent space to woman DX, and it was X1. That's right, yeah, yeah. So that was the basic ingredient to get you on. Yeah, so then I have to see what these churn classes do as it moves over to the other side. So the ordinary case was just as in X. Ah, I see, I see. Andrew Francois suggested, and I think this is another promising, it may work, to prove it using Arthur's theorem on super, to form to the super singular K3s.