On the cycle class map for zero-cycles over local fields
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Wittenberg, Olivier
The Chow group of zero-cycles of a smooth and projective variety defined over a field k is an invariant of an arithmetic and geometric nature which is well understood only when k is a finite field (by higher-dimensional class field theory). In this talk, we will discuss the case of local and strictly local fields. We prove in particular the injectivity of the cycle class map to integral l-adic cohomology for a large class of surfaces with positive geometric genus over p-adic fields. The same statement holds for semistable K3 surfaces over C((t)), but does not hold in general for surfaces over C((t)) or over the maximal unramified extension of a p-adic field. This is a joint work with Hélène Esnault.
 OK, so it's my pleasure to introduce the speaker today. Olivier Wittonberg will speak on the Cycler class map for zero cycles for local fields. Thank you very much. Thank you for the introduction and for the invitation to speak in this seminar. So everything I'm going to say today is joined first with the d'Annie de Villeva. Pardon me? Ah, I'm sorry. OK, I'm sorry. Everything I'm going to say is joined first with the NNNO. So the goal will be to explain some new results we obtained about chart groups of zero cycles for surfaces over local fields and strictly local fields. So local fields is the usual sense of the word. Strictly local means, for example, the maximum at running a five-distancing of a PID field. And especially these will be results about surfaces with positive geometrigeness. And this will be the point, as I will explain. But let me start by setting up some notation for the whole of the talk. So capital X will be a smooth, productive variety over a field K, capital K. And as usual, I will denote the chart group of zero cycles by CH0 of X. So recall this is the group of cycles. So free-loop on the closed points, modulo rational equivalence. So rational equivalence is generated by linear equivalence on the curves, pying on X. Because X is projected, there is a degree map. And I will denote the kernel by 8 of X. OK. So just one second, we check for the sound. OK. So recall, so in dimension one, this is nothing but the pick-up group. Yes, shift, François. OK. Thank you. This is just a pick-up group. If there is a rational point, then the degree zero can't be anything but the set of rational points of the Jacobian variety. So this group in dimension one is very well understood. Now in dimension greater than one, you mean geometrically connected X. Yes, yes. OK. So this is the whole of the topic. So in dimension greater than one, this is one of the simplest invariants that one can associate to X. But in many cases, it's very difficult to determine the structure of this group or to compute it in any given example. In fact, this dimension greater than one. And this even over an algebraically closed field. So to illustrate this, let me recall the situation with the circumstances over the complex numbers. Well, we have a conjecture, Locke's conjecture. So in dimension one here, you see that this group is the group of points of an idyllian variety. It's represented by an idyllian variety and in dimension greater than one, say dimension two, it's conjectured that this group may not have X should be representable in a suitable sense by the points of some variety, if and only if h2 of X-y is zero. So what is known is that if you have it, so this is the geometric genus. And what is known is that if you have a surface with non-zero geometric genus, then this group is huge. It's not representable in any sense. And this conjecture in the other direction is not so rich. Okay, however, there is one simple thing that one can say over the complex numbers and not generally over algebraically closed fields in any dimension is the fact that this group is divisible. In the case of curves, it's here from the above description. And in general, you can reduce to the case of curves by finding curves like on X going through the specified points. Okay, so this is the situation of the complex numbers. Okay, there is one situation in which this group is well understood. It's when the field is the finite field. So when k is the finite field at q, then everything is understood because of a higher dimensional class field theory. So let me recall the relevant theorem, which is due to Cato and Shouji's cycle, 1983. So in this case, the childhood of zero cycles of degrees zero is always finite. But it's more precise than this. One can in fact compute it in examples. It's very controlled. So namely, there is a map, a natural map going from the trial group of zero cycles to the abinionized fundamental group defined by Frobenius substitutions. It's in the close points to the class of the corresponding Frobenius. And Cato and Shouji prove that this map is injected. So to detect rational equivalence, you just need to compute the image in this group. And the image of this injection is very well understood. It was determined by nine. So this gives complete control over this group. For example, if the right is simply connected from this, you can deduce that the trial group of zero cycles of degrees zero is trivial. So the image are just the elements whose degree is integer. Yes, exactly. OK. So it's really the only situation where one has a complete hold on this group. So from now on, I will consider local and strictly local cubes. So capital K will be the quotient field of R. R will be a Hanselian discrete-valuation ring. The residue field is K. It's not K. For simplicity, I will assume that capital K has characteristic zero, but it's really not an important assumption. And I will let D denote the characteristic of small k, which may be zero. And so I will consider the two cases when small k is finite or separately close. So capital K local is when small k is finite and capital K is strictly local. When small k is separate of the close. So in this situation, yeah, any examples, of course, are KD fields here, QP and finite extensions. And here, for strictly local, there are two natural examples. The maximum and ramified extension of the KD field, but also C double parenthesis teach. It's an interesting example. So in this situation, one has a good structure theorem for the chart of the zero cycle of degree zero, which was proved by Sainto and Sato. When you consider the ramified extension, you can consider it either algebraically or completely. And I guess there is a general rigidity fact that it doesn't matter. Yes, yes, yes. I will, I will use you, Hanselian, to find everything to go, two cases. So, okay, so the structure theorem, like Sato and Sato in 2010. So here is a field that has an infinite transcendence degree. So you cannot avoid a big divisible subgroup just as in the complex case. But the theorem tells you that apart from this, there's just a finite group, at least prime to P. So precisely, you can write a lot of X as D plus F, where F is finite, of all the prime to P. And D is a group which is divisible prime to P. So this means G mod NB equals zero for N prime to P. So if P is zero, you don't say prime to P. Right, so if P is zero, it's divisible plus finite. Yes, because prime to zero is not. Yeah, okay, yeah, I agree. Okay, so, right, so as I mentioned, so D is a bit analogous to the group over the complex numbers. As you've seen, it's extremely hard to understand. So here it's probably even harder. So, I mean, there's little, really little known about it. It's only known that D can contain torsion even for simply connected varieties. So this is due to a saccharine cytos. But in this talk, we'll focus on the finite group F, which is really analogous to the group over finite fields here. So, in this sense, this theorem is analogous to the theorem of Catoin Cytos, which says that over finite fields, the group is finite. Now, one would like to have an analog for the second part of the theorem, too, which allows you to compute it in examples or to get a hold on its structure. So, for this, we want to study F. F is finite over prime to B, so we fix a prime L different from P. And we look at the L primary torsion of F. And the general tool to study classes of cycles is the cycle class map to et alchromology. So, let me consider the cycle class map, which goes to CH0 of X to... So, if D is the dimension of the variety, it goes to H to D, and probably X to the left. So, here you would be... Can you stop hearing me? François? If someone is calling for an answer... Who is calling for an answer? If someone is calling for an answer from the outside, I don't know. Okay, so... He's going to stop all by himself? I hope so. Okay, so just for beginning, interview someone is calling the room, so we don't know who he is. We will continue. Okay, so we can... Integral and et alchromology. So, this is a finitely generated et alchromology. So, because of this, any element which is infinitely divisible here must die in there. So, this map kills D. So, only F survives. And another way to phrase this is to consider this loop, tensor hat et al. The tensor hat means you consider the inverse limit of C H 0 of X, not 0, L to the N. C H 0 of X. So, again, this kills the divisible thought. And really, the analogous question to the theorem of a finite field is, well, is this map injective? So, what's about the injectivity of... Sorry. In fact, over a finite field, if you consider this over a finite field, you really get exactly the map, your probinus substitution. So, you can identify this homology group, well, for a fixed L with the 4L pounds of pi 1 amp. So, really, it's a generalization. Okay, and if you can answer it, I mean, when there is a positive answer, then you can hope to compute F by computing just its image in et alchromology. And this question has received a lot of attention from the 70s to the 90s after initial work by Bloch, who initiated a method for studying torsion cycles using k-theory. And by combining this and the Bloch-August theory and the Markovian-Sussman theorem, it was recognized that one could get a good hold of co-dimension two cycles on the three varieties, sorry, on the varieties of our country dimension. So, there are many names here to be mentioned. So, Kuliotelen, Sonsjuk, and Suley, of course, the Katow and Saito, which are over dimensions. And also Saito and Sout. So, wrote an old series of papers about this problem using k-theory for cycles of the dimension two, so for surfaces for zero cycles. And one question, really, these methods produced very effective results for rational surfaces and more generally for surfaces with geometric zero. So, an example of a theorem obtained in this series, which I think should be attributed, oh, I should not just write Ruskin. So, this theorem should be attributed to Kuliotelen and Ruskin, 1985. So, about this question, if, so in a strictly local case, capital K, strictly local, yes, maybe I did not stress this enough, the theorem of Saito and Sout is for both cases, local and strictly local. So, here if K is strictly local and X is a surface with geometric zero, then there is a positive answer. Then, Saito is injected. So, the goal of the talk would be to try to answer, to give some answers to this question for surfaces with non-zero positive, sorry, with non-zero geometric genus. So, over local fields and over strictly local fields. So, actually the strongest results will be about local fields, but I also hope to show that the situation is perhaps unexpectedly very different between local fields and strictly local fields. Okay, so maybe I might start here. So, the first theorem, I want to mention, is about strictly local fields. And actually, our C double parenthesis is C. So, if you want to consider surfaces with non-zero geometric genus, the first class of surfaces you look at is K3 surfaces. And in this respect, we can prove the following. So, if X is a K3 surface, our C double parenthesis is C, with semi-stable reduction. Then, okay, then the answer is positive. So, what does it mean? The cycloplasma is injected. But here, so we're in the strictly local situation, so it's a field of chronological dimension one, and you can try to understand what this chronological group is in this case. And because the surface is simply connected, it really means that there's no F. In other words, the child group of zero circles of degrees zero is divisible. Okay, so let me explain the strategy for proof of this. So, we will consider a model of the ring R. So, here R is C double-reffected to T. And I will consider a script X over R, with a projected regular model. Let me denote the special fiber by A, and its irreducible components by A i. So, since we're in the semi-stable case A here is reduced, and I will assume in addition that it's a simple normal process device on X. So, this is not needed in the result? I mean, you can always choose a model like this. I mean, we want a recelible parenthesis T. So, I mean, I explained the strategy for the proof, so we choose such a model. No, but if you have a model without simple normal processing, then... Ah, you mean it's still semi-stable in this weaker sense? Yes. Yes, okay, yeah, you're right. The theorem holds with semi-stable in a weaker sense. That's what's written in the paper, but I focus on this for simplicity. Okay, so, yeah, okay, you're right. I mean, there's an ambiguity in semi-stable. So, in this situation, let me use a new model. And you needed the total space is regular. So, it's nothing like a long, smooth semi-stable. Okay. So, in this situation, we consider the cycle of cos-map psi. So, C of 0 of x, then it's our half zl, not to h4 xzl2, and we consider what happens on the model. So, on the model, we have one cycle on the model, and the cycle cross-map to its outcome of the model. Now, vertically, you can write localization exact sequences. You can restrict to the generic fiber. And before that, you have one cycle on the special fiber. And here, you have cobalogy with supports in the special fiber. And here, there's a cycle cross-map again, which we'll call psi 1a. So, one cycle is on the special fiber. Okay, and now the point is that psi 2 and psi 2 did not only prove the finiteness theorem here. In fact, they proved something much stronger, which underlies this theorem. They proved that the cycle cross-map for one cycle on the model is an isomorphism. So, this is psi 2 and psi 2. And they proved this in the general idea of the theorem, arbitrary dimension, local or strictly local theorem. But they don't have regular models. They have regular models. They assume they have one. They assume they start with one, and then they claim, oh, you need to prove this. You can reduce using your improvement of the theorem. Okay, so now if you look at this, and there's one point here which is zero cycle extends to one cycle on the model, so this map is surjective. If you look at this, you see that you want to prove that the bottom map is injective. Of course, it would be enough for the top map to be surjective. But in fact, it's never the case for the generations of K3 surfaces. You want to check that this is never the case. The top map is never surjective. So, one has to be more precise. But still, it's possible by looking at this diagram and using this isomorphism to translate the injectivity of the bottom map purely in terms of the special fiber. So, I'm just giving you the results. It's really reformulation using the isomorphism. So, that's really step one. So, from this, you deduce that this group is divisible if and only if a certain complex is exact. So, this is a translation you can do in general, but for simplicity, let me use some very simple properties which come from the fact that I'm looking at the K3 surface to give you the translation. So, here we find that a certain complex should be exact. So, let me write it, q mod z plus q mod z to the power i. So, direct sum of i copies of q mod z. And then it goes to the direct sum of the divisible groups of the components, tensor q mod z. So, if this complex is exact, that's the problem. Okay, what are the maps? Here, it's just the diagonal map. And here, well, if the dimension was one, it would be the intersection matrix. So, in general, it sends, if you take lambda on the jth component, it sends to, if you go to the i factor, you look at the class of a i intersection a j tensor lambda on the i factor, or i different from j. If i equals j, you can anyway modify by another man coming from the left to determine the image. So, this describes the map. Okay, so this is a concrete criterion which you now have to check. So, that's the second part of the strategy. It's exactly in the middle, you may. Yes, exactly, yeah, just this. So, now you have transformed the problem into something which is purely on the special fiber. It's completely concrete. So, then you have to use the fact that special fibers of the generations of semi-stable K3 surfaces have been classified. So, there's a whole literature about this, starting from the work of Kulikov, the person of income, and Mirinda and Morrison in the 70s, the 80s. Morrison. So, they classified semi-stable generations of K3 surfaces. So, here one should, I mean one would have to be precise. It means that if you have a semi-stable K3 surface, they give you some nice model for which they can tell you what a special fiber is. And so, there is a classification. And if you use this, and if you know enough about the surfaces AI which appear here, and which in this classification are rational surfaces or ruled surfaces, except in the case of good reduction, then using these two things, you can translate the problem into a combinatorial one. So, you obtain a combinatorial problem. So, it's about the combinatorics of the special fiber. Namely, the biggest case of the classification tells you that, so in many cases, you have a K3 surface which degenerates to rational surfaces in such a way that the dual complex triangulates the sphere. So, then you obtain questions about triangulations of the sphere. In fact, negatively characterized triangulations of the sphere. These questions are in fact hard in general. I mean, these objects are hard in general from a combinatorial point of view. But it turns out that for this particular problem, you can solve it. That's the last point. Solve it. And I'm sweeping lots of details on the rug I should mention, that these degenerations are not algebraic. It's only in the analytic category. You don't get schemes. So, you have to work a bit to glue the pieces together, but it can be not. What do you mean you don't get schemes? You get algebraic spaces or you get... I think you get algebraic spaces, but it's not written in the literature. So, they use analytic manifolds. And in your results that you mentioned, you always say projective. Yes. Do your results hold the proof used projective? For example, do the proof of Catern's site use projective or not proper? In many places, you use hydroplane sections. You use Bertini CRMs and things like that. So, in the last thing about the generation of cases, you really need projective... No, I mean, you need Bertini CRMs to prove, for example, this is isomorphism, but once you have it and you have translated the question in this form, then, I mean, you just look at this complex and you forget everything else, and now you can lose projectiveness. But the specification that they have for analytic degenerations will apply in particular to the projective ones. Yes. So, it... But they give you... But what they say is if you start with the K3 surface, then there is a model which is regular, and such that the... So, the total space is regular and the canonical divisor on the total space is trivial. But this is not a projective scheme. This is only an analytic manifold. I mean, to get the condition that K is trivial on the total space. Ah. And this is what you need to analyze the special fiber. And then you have to go from this to your model. Yes. So, you have to use weak factorization and things like that. You use weak factorization to... and check that the complex doesn't change and so on. And what about ramifying the original local field? Does it change the problem or not? Sure, sure. Yeah, you lose the question. I mean, it's a question about a finite group. I mean... Okay. F, so... No, you lose everything. Okay. Okay, so, that's about......the group of theorem one. So, a few comments. So, the first comment is that the first step of the strategy here is completely general. You can draw this diagram in any dimension and you can always translate the injectivity of psi in... as the exactness, sorry, of a certain complex which looks like this, except it's a bit more complicated in general. In particular, if x is not simply connected, then it's not true or at least not clear at all that it only depends on the special factor. Okay, the second comment is that if we have the results of Kulikov-Person-Pinkham-Mirano-Morison in mixed characteristics, then the whole proof will go through over, say, a maximum of ramified extension of Qp. And you will get in the same way that the group F is trivial, which means here that a0 of x is trying to be divisible. You said that they work analytically, but then they must use not formal series but convergent series, otherwise it doesn't make sense. Sure, sure. So, the first step is to reduce from C double parenthesis T to convergence series. Oh, yeah. And I'm speaking the total number. So, okay, so again, if we have these classification results in mixed characteristics, we will get trying to be divisible over Qp and R. And the third comment is that for K3 surfaces which are not semi-stable, we don't know. I mean, it's an open question for us. We don't know what happens. The only reason why we need some miscibility is that it's only for semi-stable K3s that there is a classification. Okay, so now I turn to local students. And in the remote, have you said that they have the total space regular, but is it with reduced fiber or not? Yes, they always work with reduced fibers. But what do they have to assume to get it? You said that it's not always the case that there is such or that they show that there is such. No, they show that there is such, but it's not always a scheme. With reduced speciality? Yes, yes. I mean, you start with the assumption that it's semi-stable as a scheme. You assume to start with that X has semi-stable reduction as a scheme. And then they tell you you can find another model, which is maybe not a scheme, which is still semi-stable, and for which K is triple. Okay, this is the point. I think it's better to let it go. Yes, you can. I'm sorry. Okay, so I turn to local fields. Okay, so K is a local field. And so I will set the theorem. So as you will see, this will be a much stronger theorem than this one. So I guess it's the main theorem at the top. It's also about surfaces. And I will, again, work with the model. So this time I assume that I have one like this. So I keep the notation X, a regular projective model. Again, A is the special fiber. So I keep more of the notation. So maybe there are multiplicities now, I'm not assuming semi-stability, but I assume that A in red is a simple normal crossings divisor. Okay, so then theorem two. So you have to make two assumptions. So the first assumption is about the AIs. So the AIs are smooth projective surfaces over a finite few, and I assume that they satisfy the take conjecture. And the second assumption has to do with the alvanism variety of X. So I assume that X has potentially good reduction. Of course, this is the case, for example, if X is simply connected. And then, so the theorem is that, again, psi is injective. So the second map to H4ZL2 is injective for any other form of P. So that's a positive answer in this situation. So again, this was known for surfaces with geometry genus zero. It was proved by Trudy Saito in the 90s. And the point is that, so it applies to surfaces with non-zero geometry genus. And as I mentioned, these are those for which, over the complex numbers, the chalk group is not representable. So a few remarks. First, we're not about the take conjecture. Here we assume it for the components of the special fiber. But it's very often the case that it's satisfied for trivial reasons. So there are many, for example, there are many P3 surfaces, P3 surfaces, X for which you have a model. And the special type is the union of rational surfaces or ruled surfaces. But AI is, by rationally, ruled for each eye. And in this case, AI satisfies the take conjecture trivially. So there are many examples where this can be applied. Second remark is that this, the conclusion can be formulated in an equivalent way. We should also take an electric chart. So, equivalently. So there is a natural pairing between the chalk group and the brauer group. You evaluate classes of the bra group at points, and you take the local invariant of our classical theory. And so this goes to q-molz. And the conclusion is equivalent to saying that the left kernel here is divisible by n. So this can be seen as a higher dimensional generalization of Liechten-Monteguality. Liechten-Monteguality says that for curve, this is a perfect terrain on both sides. And clearly, the maximal divisible subgroup has to be in the kernel because the brauer group is torsion. And the third remark is about the other assumption about the albanese. In fact, if you remove it, then the theorem is wrong. There are cancer examples due to Karim Alain's crash in the 90s. So it's a theorem wrong without assumptions. Is there any assumption on the albanese? So you have to put some assumptions. So in principle, this should apply to all simply connected surfaces. So now I will explain what the ingredients of the proof are. So ingredients of the proof. So to start in the same way as for theorem one, you look at the diagram on the top. You write the cycle class map for one cycle on the model and on the special fiber. And you use the theorem of Saito and Sato that the map in the middle is injective. It's an isomorphism, sorry. But then here in this situation, we will actually prove that the map on the top is surjective. So we will prove the surjectivity of psi 1A. So from each 1A turns our head to the left. To a chronology with supports, so really that's it's an alpha homology, each for A x z0 to 2. So again, by the theorem of Saito and Sato, this is enough. But it's perhaps unexpected this problem. In theorem one, this was not true. But so, okay, here that's what we proved. So we start by reformulating this a bit as we did in theorem one. So the first thing is psi 1A is surjective, if and only if. So here again, you do analyze. And so you write the dual of these groups by combining congratulity and duality for the gamma homology of the compil. So you find that the dual of this group is H3. So let me write here H3 of A 12 z1. And then you consider the following map. First of all, you go to the direct sum over each component. And then you mod out by a certain subgroup. So the same group on the top, and I mod out by the certain subgroup which I did not see H1Ai orthogonal. And I'm going to define this group in a minute. But the condition is that the composition here is injective. This map here is injective. So this is a truly formal reformulation. And the group, I'm sorry, I forgot to mention what it is. It's just the group of those classes which die when you restrict them to curves. So the set of classes alpha in H3Ai 12 z1 such that the restriction of alpha to any curve is 0. So then I will explain how there is hypothesis, a payroll, how you prove this. I mean I will just give a picture. The first thing is this. You have to use the hypothesis on the urbanize. So the urbanize is related to H1 of the geometric-generic fiber. And because X is a surface, it's related to H3 priority. And now if you use this, you can try to relate this H3 of the generate geometric fiber to what happens on the special fiber using the results of upper-potency to prove the another main way is conjecture for surfaces in mixed characteristics. So here, I have to argue, I'm not doing this, but these two ingredients will allow you to conclude that H3 of A bar, so bar means you go to an algebraic closure, and QL is Q of weight 3. You get the weight of the action of Frobenius on this, and because A bar is proper, but maybe not smooth, a 3 Re is in a way 3 and possibly lower. But in fact, using this hypothesis, you define it as pure of weight 3. And because it's pure of weight 3, it has to inject into the direction of the same groups of other components. This is because you can write what the kernel is, and you find the kernel is pure of weight. I'm sorry, I'm not sure, but there's no weight 3 in the kernel. So that's one point. Then the second point is to use the tape conjecture, not in an essential way, but it's a fact that the tape conjecture implies that the action of Frobenius on the H2 of A bar is simisimple for the eigenvalue 1. So let me say it's partially simisimple. So this is a classical fact. And if you use this simisimplicity, plus some weight arguments, from this you can deduce that H1 of the finite field with values in H2 of A bar 2L of 1 inject into the direction of the same groups over the components. What is partially simisimple? I'm sorry? What is partially? So it means just for the eigenvalue 1, it's simisimple for the eigenvalue 1. This is part of the tape conjecture. Yeah, I mean it's a consequence of it. Okay, okay, because of the merit. Okay, now putting these two things together, you deduce that H3 of A, by the ocean-southpacal sequence, H3 of A, Q1 of 1, inject into the direction of the same groups over the components. And now there's a little miracle that you can deduce from this same thing with portion coefficients, H3 of A, Q1 of 1, inject into the same thing. And the proof here is very simple, there's nothing deep, but it uses crucially the fact that we're looking at surfaces. So it's a very simple way to deduce where it's really crucially. You have to use some bioregular spectral sequence, and you want some groups to be torsion-free, and it works in dimension 2 and not in a higher dimension. Okay, so you're almost done. You've proved that the first map here is injected, but the second map is no problem at all. That's where we use the tape conjecture. In fact, the tape conjecture here implies that these groups, C1 of A, I, or 7 of O, are 0. So if you could all of this together, it's done. Okay, so that's how the proof looks. So a few comments. First of all, it applies. So as I mentioned, 2K3 surfaces, even in the non-semistable case, and also in the semistable case, as you see, there's no need for any classification results. In the sense it's much easier, there's no combinatorics involved. But you assume that it is strict normal crossing by before you did not assume. You're right. But I mean conjecturally, I did not assume that the multiplicities were 1. Okay, so that's one point. And in particular, so this, you see, if we had the classification in mixed characteristics, we would get the injectivity for K3 surfaces over the maximal and ramified extension of KDQ. Now this theorem gives it for free streaming tape conjecture and existence of models for K3 surfaces over KDQ. But it's a bit strange that if you want to prove the same results with a maximal and ramified extension, then currently you need very strong geometric results. Okay, and I want to finish by coming back to strictly local. Can I ask a question? Yes. So you don't assume that the cross-fibers are reduced. But you said you use the latest spectra systems. What does this mean? So the point is to use the local invariant cycle theorem, which tells you that it's H i of A, say A bar, which will subject onto H i of S bar to an invariant under the inertial group. But to prove this, you can reduce, by using the Gavard-Elyon answerations, you can reduce to the semi-stable case. Okay? Here, thank you. Okay, so I want to come back to strictly local fields. Okay, so we have a positive answer for KD fields, at least for simply connected surfaces. We have a positive answer for KD fields, as you mentioned in the lecture. And for strictly local fields, we have a positive answer for surfaces with geometric genus zero. And then also for K3 surfaces. So what happens in general? Well, so we were quite surprised to find a count, for example, in general. So that's here in three. And this works over both types of strictly local fields. X, that's simply connected, smooth projective surface, over a C-local plane to the C, such that, so again, what is the cycle map here? We're in the strictly local case, so the collage group H2E only contains information about the degree and the P1, the billion P1, because it's simply connected to really nothing else than the degree. So here, for example, is an example where the group is not divisible. And here, we have A0 to X, Z0 to... So the side is not injected. And also, so that's one point. Another point is the same. Over the maximum unromified extension of a K, that deals for infinity to many times P. So in particular, this shows the hope for an analog of the Catocyte theorem, which was about finite fields. You could hope for an analog over this here, which is quasi-finites. Well, this hope does not hold. I mean, the billion fundamental group does not control the finite quotients of A0 to X, maximally divisible for blue. And another thing which I think, well, for me, it was really unexpected, is that, I mean, you also get examples over strictly local fields like this. Now, if you take such an example, well, it's a surface, it's defined over some chaotic field. So if you take this example, you can apply it to its theorem too. So what you get in the end is a surface over a chaotic field for which the cycle map is injected, and the same... And in this example, the deconstructor is satisfied. So that's okay. And the same over any finite extension of the chaotic field, or at least any finite unrungified, because then you see that the deconstructor is satisfied. You can still apply theorem too. But when you go to the limit, you get a counterexample. So I think that was a strange phenomenon. Okay, so there are many things that we don't understand. I mean, for chaotic fields, there are rather complete results, except when the Albanian variety is not trivial or does not have potentially good reduction, then we don't have a good answer, we don't understand what happens. Also over strictly local fields, we don't have... It would be good to have a better understanding of what goes on and what to expect. And also in higher dimension, I want to mention that most of what I've said works exactly the same way in higher dimension, which could be interesting, because the previous techniques were about co-dimension two cycles on varieties of any dimension. So the previous techniques do not say anything about zero cycles in dimension three or more. So the only place where dimension two was really used is this one. And here we just don't know what happens. And because of this problem, we don't know what happens in higher dimension. And an example of an open question, a simple open question in higher dimension, is this, if you take X, a smooth projective, a rationally connected variety, over seedable for anything with T, well, is the chocolate trivial? For such a variety, it's known that it has finite exponents, so trivial is equivalent to being divisible here. And that's something we don't know in dimension greater than two. But you know it in dimension two, because it's Russian, right? Yes, it's Russian, which I'm going to give you a zero. Okay, so I'll stop here. Thank you very much. Thank you. So could you maybe start from Tokyo, Beijing, and then come back to Paris? So we start from Tokyo? Okay, no. In fact, we don't see you very well. There is no light. Ah, really? Really? Yeah. So, so, so, so, so you knew last term, what happened you make a random with the substitute? So before your PZ or a big number, second big number. So I, so I, not completely sure. I think, I think the geometry in this is three. I mean, I've computed this, but I've forgotten. It's color and dimension one. This I can tell you. I can tell you also, the degeneration in this case is very simple. It degenerates into the union of two surfaces, which meet along one curve. And each surface, each of these surfaces, I think has geometriciness one. The curve is an elliptic curve. And they have color and dimension one, too. Okay, I think that's all I can say. So you just mean the property of the, the variable or are you, ah, no. Ah, you mean to obtain this? Yeah. Ah, yes, so actually, yeah, yeah. So thank you for the question. So in, in theorem one, I wrote this complex on the top, which was equivalent to the divisibility of this group. So it's not, it's not correct in, in general. I was looking at K3 surfaces, but in general it's almost this. For a, for a variety which is simply connected, it's exactly this except that on the right, you have to mod out by the same group as here, CH1 of AI or sub-anode. But in, in the situation of K3 surfaces, this group was, was trivial. So, yeah, so that's again the property of the special fiber. And so the exact property you need to get this example is the generation like this, such that the, the generate fiber is simply connected. This is very important. And the second property is that you should take the curve here, the intersection curve, say C. Then you should have the intersection number of C with any divisor in, in either surface. Either surface is even, always even for any i and any d divisor on AI. That's what you need to do. Thank you. Are there further questions? Okay, thank you very much for, for talking. Okay, so we, we have some questions from the dean. Okay, so any questions? No? There's no questions from the dean. You have a question? Yeah, yeah, you said something about, I did not register exactly. So about the, some passage to the limit of the strict visualization of the periodic field, there was some comment that you made. Here at the end. Yes. So here I was saying that, so there is an example like this, over the maximum or unified extension of some chaotic field. Yes. Now the surface is defined over some chaotic field. Yeah. And to this surface, over the chaotic field, you can apply theorem two. Yes. So you have this situation with the surface over a chaotic field, and on, on, on this chaotic field, the surface map is injected, and it's still injected. If you go to any unromified extension, or maybe any finite extension, if you believe that it's a conjecture. But if you go to the maximum unromified extension, it's not injected anymore. That's the counter example here. All right, so the point is the coromology doesn't come out to this limit. Right, because it's the zeta coefficients. Ah, okay. Not finite coefficients. Okay. But yeah, still it's, maybe you did it unexpected. Other questions? What are you, I think that at some places you mentioned that you have to use the Gabbers L prime alterations here. Yes. Could you give a specific example? Because here is not too convincing because you work with QL. So, yeah. Oh, I'm sorry, I'm sorry, I'm sorry. You're right. Here, De Jong is enough. I'm sorry. No, no, you're right. No, no, it's really used just to prove the theorem of Saito and Sato, the isomorphism there. I mean, if you want to prove this, you can do any exploration of the grid prime as well. And, and, and reducing this case to really work with finite coefficients. You're right. You're right. I'm sorry. This theorem is true for finite coefficients. And that's how you prove it. Yes, thank you. Other questions? So, let's bring the speaker again. And then we say good bye to Eugene and to... Yeah, yeah, yeah. I don't know. Maybe I can... Yes. So, we have still a question. No, no, I'm not, I didn't know. Yes. Where is it used? There was a question from Tokyo. So, you said that you have this local monodrome theorem. Yes. So, where does it enter in your, in the... So, here, I mean, if you want to prove that H3 of A bar is pure of weight three. Yes. So, what you do is that, in fact, it's in an exact sequence. So, it surjacks onto H3. We have this. But in fact, this is not the important part. The important part is what comes on the left. You have H3 of the model, say, over the maximum amount of high-file expansion, which is supported in A bar. And so, this local invariant cycle theorem allows you to prove it's not completely obvious, but it allows you to prove that this is an exact sequence. It's exact, pure. It's not completely pure. But once you know it's an exact sequence, you're done because this has weights at least three. This has weights at most three. So, if it's not pure, it must be because of this. And this you can relate to the urbanizing. And Rappaport think assume that you have semi-stable. Yes. You assume regular model, no? Yes, right. But again, we can use alterations to prove the weight on the drawing spectrum. We can use alterations to prove it. Okay. Okay. Still no questions? So, if not, then we say goodbye.
 OK, so it's my pleasure to introduce the speaker today.  Olivier Wittonberg will speak on the Cycler class map for zero cycles for local fields.  Thank you very much.  Thank you for the introduction and for the invitation to speak in this seminar.  So everything I'm going to say today is joined first with the d'Annie de Villeva.  Pardon me?  Ah, I'm sorry.  OK, I'm sorry.  Everything I'm going to say is joined first with the NNNO.  So the goal will be to explain some new results we obtained about chart groups of zero cycles  for surfaces over local fields and strictly local fields.  So local fields is the usual sense of the word.  Strictly local means, for example, the maximum at running a five-distancing of a PID field.  And especially these will be results about surfaces with positive geometrigeness.  And this will be the point, as I will explain.  But let me start by setting up some notation for the whole of the talk.  So capital X will be a smooth, productive variety over a field K, capital K.  And as usual, I will denote the chart group of zero cycles by CH0 of X.  So recall this is the group of cycles.  So free-loop on the closed points, modulo rational equivalence.  So rational equivalence is generated by linear equivalence on the curves, pying on X.  Because X is projected, there is a degree map.  And I will denote the kernel by 8 of X.  OK.  So just one second, we check for the sound.  OK.  So recall, so in dimension one, this is nothing but the pick-up group.  Yes, shift, François.  OK.  Thank you.  This is just a pick-up group.  If there is a rational point, then the degree zero can't be anything but the set of rational points of the Jacobian variety.  So this group in dimension one is very well understood.  Now in dimension greater than one, you mean geometrically connected X.  Yes, yes.  OK.  So this is the whole of the topic.  So in dimension greater than one, this is one of the simplest invariants that one can associate to X.  But in many cases, it's very difficult to determine the structure of this group or to compute it in any given example.  In fact, this dimension greater than one.  And this even over an algebraically closed field.  So to illustrate this, let me recall the situation with the circumstances over the complex numbers.  Well, we have a conjecture, Locke's conjecture.  So in dimension one here, you see that this group is the group of points of an idyllian variety.  It's represented by an idyllian variety and in dimension greater than one, say dimension two,  it's conjectured that this group may not have X should be representable in a suitable sense  by the points of some variety, if and only if h2 of X-y is zero.  So what is known is that if you have it, so this is the geometric genus.  And what is known is that if you have a surface with non-zero geometric genus, then this group is huge.  It's not representable in any sense.  And this conjecture in the other direction is not so rich.  Okay, however, there is one simple thing that one can say over the complex numbers  and not generally over algebraically closed fields in any dimension is the fact that this group is divisible.  In the case of curves, it's here from the above description.  And in general, you can reduce to the case of curves by finding curves like on X going through the specified points.  Okay, so this is the situation of the complex numbers.  Okay, there is one situation in which this group is well understood.  It's when the field is the finite field.  So when k is the finite field at q, then everything is understood because of a higher dimensional class field theory.  So let me recall the relevant theorem, which is due to Cato and Shouji's cycle, 1983.  So in this case, the childhood of zero cycles of degrees zero is always finite.  But it's more precise than this. One can in fact compute it in examples. It's very controlled.  So namely, there is a map, a natural map going from the trial group of zero cycles to the abinionized fundamental group  defined by Frobenius substitutions.  It's in the close points to the class of the corresponding Frobenius.  And Cato and Shouji prove that this map is injected.  So to detect rational equivalence, you just need to compute the image in this group.  And the image of this injection is very well understood. It was determined by nine.  So this gives complete control over this group.  For example, if the right is simply connected from this, you can deduce that the trial group of zero cycles of degrees zero is trivial.  So the image are just the elements whose degree is integer.  Yes, exactly.  OK. So it's really the only situation where one has a complete hold on this group.  So from now on, I will consider local and strictly local cubes.  So capital K will be the quotient field of R.  R will be a Hanselian discrete-valuation ring.  The residue field is K. It's not K.  For simplicity, I will assume that capital K has characteristic zero, but it's really not an important assumption.  And I will let D denote the characteristic of small k, which may be zero.  And so I will consider the two cases when small k is finite or separately close.  So capital K local is when small k is finite and capital K is strictly local.  When small k is separate of the close.  So in this situation, yeah, any examples, of course, are KD fields here, QP and finite extensions.  And here, for strictly local, there are two natural examples.  The maximum and ramified extension of the KD field, but also C double parenthesis teach.  It's an interesting example.  So in this situation, one has a good structure theorem for the chart of the zero cycle of degree zero,  which was proved by Sainto and Sato.  When you consider the ramified extension, you can consider it either algebraically or completely.  And I guess there is a general rigidity fact that it doesn't matter.  Yes, yes, yes.  I will, I will use you, Hanselian, to find everything to go, two cases.  So, okay, so the structure theorem, like Sato and Sato in 2010.  So here is a field that has an infinite transcendence degree.  So you cannot avoid a big divisible subgroup just as in the complex case.  But the theorem tells you that apart from this, there's just a finite group, at least prime to P.  So precisely, you can write a lot of X as D plus F, where F is finite,  of all the prime to P.  And D is a group which is divisible prime to P.  So this means G mod NB equals zero for N prime to P.  So if P is zero, you don't say prime to P.  Right, so if P is zero, it's divisible plus finite.  Yes, because prime to zero is not.  Yeah, okay, yeah, I agree.  Okay, so, right, so as I mentioned, so D is a bit analogous to the group over the complex numbers.  As you've seen, it's extremely hard to understand.  So here it's probably even harder.  So, I mean, there's little, really little known about it.  It's only known that D can contain torsion even for simply connected varieties.  So this is due to a saccharine cytos.  But in this talk, we'll focus on the finite group F, which is really analogous to the group over finite fields here.  So, in this sense, this theorem is analogous to the theorem of Catoin Cytos,  which says that over finite fields, the group is finite.  Now, one would like to have an analog for the second part of the theorem, too,  which allows you to compute it in examples or to get a hold on its structure.  So, for this, we want to study F.  F is finite over prime to B, so we fix a prime L different from P.  And we look at the L primary torsion of F.  And the general tool to study classes of cycles is the cycle class map to et alchromology.  So, let me consider the cycle class map, which goes to CH0 of X to...  So, if D is the dimension of the variety, it goes to H to D, and probably X to the left.  So, here you would be...  Can you stop hearing me?  François?  If someone is calling for an answer...  Who is calling for an answer?  If someone is calling for an answer from the outside, I don't know.  Okay, so...  He's going to stop all by himself?  I hope so.  Okay, so just for beginning, interview someone is calling the room, so we don't know who he is.  We will continue.  Okay, so we can...  Integral and et alchromology.  So, this is a finitely generated et alchromology.  So, because of this, any element which is infinitely divisible here must die in there.  So, this map kills D.  So, only F survives.  And another way to phrase this is to consider this loop, tensor hat et al.  The tensor hat means you consider the inverse limit of C H 0 of X, not 0, L to the N.  C H 0 of X.  So, again, this kills the divisible thought.  And really, the analogous question to the theorem of a finite field is, well, is this map injective?  So, what's about the injectivity of...  Sorry.  In fact, over a finite field, if you consider this over a finite field, you really get exactly the map, your probinus substitution.  So, you can identify this homology group, well, for a fixed L with the 4L pounds of pi 1 amp.  So, really, it's a generalization.  Okay, and if you can answer it, I mean, when there is a positive answer, then you can hope to compute F by computing just its image in et alchromology.  And this question has received a lot of attention from the 70s to the 90s after initial work by Bloch,  who initiated a method for studying torsion cycles using k-theory.  And by combining this and the Bloch-August theory and the Markovian-Sussman theorem,  it was recognized that one could get a good hold of co-dimension two cycles on the three varieties,  sorry, on the varieties of our country dimension.  So, there are many names here to be mentioned.  So, Kuliotelen, Sonsjuk, and Suley, of course, the Katow and Saito, which are over dimensions.  And also Saito and Sout.  So, wrote an old series of papers about this problem using k-theory for cycles of the dimension two,  so for surfaces for zero cycles.  And one question, really, these methods produced very effective results for rational surfaces  and more generally for surfaces with geometric zero.  So, an example of a theorem obtained in this series, which I think should be attributed,  oh, I should not just write Ruskin.  So, this theorem should be attributed to Kuliotelen and Ruskin, 1985.  So, about this question, if, so in a strictly local case, capital K, strictly local,  yes, maybe I did not stress this enough, the theorem of Saito and Sout is for both cases,  local and strictly local.  So, here if K is strictly local and X is a surface with geometric zero,  then there is a positive answer.  Then, Saito is injected.  So, the goal of the talk would be to try to answer, to give some answers to this question  for surfaces with non-zero positive, sorry, with non-zero geometric genus.  So, over local fields and over strictly local fields.  So, actually the strongest results will be about local fields,  but I also hope to show that the situation is perhaps unexpectedly very different between local fields  and strictly local fields.  Okay, so maybe I might start here.  So, the first theorem, I want to mention, is about strictly local fields.  And actually, our C double parenthesis is C.  So, if you want to consider surfaces with non-zero geometric genus,  the first class of surfaces you look at is K3 surfaces.  And in this respect, we can prove the following.  So, if X is a K3 surface, our C double parenthesis is C,  with semi-stable reduction.  Then, okay, then the answer is positive.  So, what does it mean?  The cycloplasma is injected.  But here, so we're in the strictly local situation,  so it's a field of chronological dimension one,  and you can try to understand what this chronological group is in this case.  And because the surface is simply connected,  it really means that there's no F.  In other words, the child group of zero circles of degrees zero is divisible.  Okay, so let me explain the strategy for proof of this.  So, we will consider a model of the ring R.  So, here R is C double-reffected to T.  And I will consider a script X over R,  with a projected regular model.  Let me denote the special fiber by A,  and its irreducible components by A i.  So, since we're in the semi-stable case A here is reduced,  and I will assume in addition that it's a simple normal process device on X.  So, this is not needed in the result?  I mean, you can always choose a model like this.  I mean, we want a recelible parenthesis T.  So, I mean, I explained the strategy for the proof, so we choose such a model.  No, but if you have a model without simple normal processing, then...  Ah, you mean it's still semi-stable in this weaker sense?  Yes.  Yes, okay, yeah, you're right.  The theorem holds with semi-stable in a weaker sense.  That's what's written in the paper, but I focus on this for simplicity.  Okay, so, yeah, okay, you're right.  I mean, there's an ambiguity in semi-stable.  So, in this situation, let me use a new model.  And you needed the total space is regular.  So, it's nothing like a long, smooth semi-stable.  Okay.  So, in this situation, we consider the cycle of cos-map psi.  So, C of 0 of x, then it's our half zl,  not to h4 xzl2, and we consider what happens on the model.  So, on the model, we have one cycle on the model,  and the cycle cross-map to its outcome of the model.  Now, vertically, you can write localization exact sequences.  You can restrict to the generic fiber.  And before that, you have one cycle on the special fiber.  And here, you have cobalogy with supports in the special fiber.  And here, there's a cycle cross-map again,  which we'll call psi 1a.  So, one cycle is on the special fiber.  Okay, and now the point is that  psi 2 and psi 2 did not only prove the finiteness theorem here.  In fact, they proved something much stronger,  which underlies this theorem.  They proved that the cycle cross-map for one cycle on the model  is an isomorphism.  So, this is psi 2 and psi 2.  And they proved this in the general idea of the theorem,  arbitrary dimension, local or strictly local theorem.  But they don't have regular models.  They have regular models.  They assume they have one.  They assume they start with one, and then they claim,  oh, you need to prove this.  You can reduce using your improvement of the theorem.  Okay, so now if you look at this,  and there's one point here which is zero cycle extends  to one cycle on the model, so this map is surjective.  If you look at this, you see that you want to prove  that the bottom map is injective.  Of course, it would be enough for the top map to be surjective.  But in fact, it's never the case for the generations  of K3 surfaces.  You want to check that this is never the case.  The top map is never surjective.  So, one has to be more precise.  But still, it's possible by looking at this diagram  and using this isomorphism to translate the injectivity  of the bottom map purely in terms of the special fiber.  So, I'm just giving you the results.  It's really reformulation using the isomorphism.  So, that's really step one.  So, from this, you deduce that this group is divisible  if and only if a certain complex is exact.  So, this is a translation you can do in general,  but for simplicity, let me use some very simple properties  which come from the fact that I'm looking at the K3 surface  to give you the translation.  So, here we find that a certain complex should be exact.  So, let me write it, q mod z plus q mod z to the power i.  So, direct sum of i copies of q mod z.  And then it goes to the direct sum of the divisible groups  of the components, tensor q mod z.  So, if this complex is exact, that's the problem.  Okay, what are the maps?  Here, it's just the diagonal map.  And here, well, if the dimension was one,  it would be the intersection matrix.  So, in general, it sends, if you take lambda on the jth component,  it sends to, if you go to the i factor,  you look at the class of a i intersection a j tensor lambda  on the i factor, or i different from j.  If i equals j, you can anyway modify by another man  coming from the left to determine the image.  So, this describes the map.  Okay, so this is a concrete criterion  which you now have to check.  So, that's the second part of the strategy.  It's exactly in the middle, you may.  Yes, exactly, yeah, just this.  So, now you have transformed the problem into something  which is purely on the special fiber.  It's completely concrete.  So, then you have to use the fact that special fibers  of the generations of semi-stable K3 surfaces  have been classified.  So, there's a whole literature about this,  starting from the work of Kulikov,  the person of income,  and Mirinda and Morrison in the 70s, the 80s.  Morrison.  So, they classified semi-stable generations  of K3 surfaces.  So, here one should, I mean one would have to be precise.  It means that if you have a semi-stable K3 surface,  they give you some nice model for which they can tell you  what a special fiber is.  And so, there is a classification.  And if you use this, and if you know enough about the surfaces  AI which appear here, and which in this classification  are rational surfaces or ruled surfaces,  except in the case of good reduction,  then using these two things,  you can translate the problem into a combinatorial one.  So, you obtain a combinatorial problem.  So, it's about the combinatorics of the special fiber.  Namely, the biggest case of the classification tells you that,  so in many cases, you have a K3 surface which degenerates  to rational surfaces in such a way that the dual complex  triangulates the sphere.  So, then you obtain questions about triangulations of the sphere.  In fact, negatively characterized triangulations of the sphere.  These questions are in fact hard in general.  I mean, these objects are hard in general  from a combinatorial point of view.  But it turns out that for this particular problem, you can solve it.  That's the last point.  Solve it.  And I'm sweeping lots of details on the rug I should mention,  that these degenerations are not algebraic.  It's only in the analytic category.  You don't get schemes.  So, you have to work a bit to glue the pieces together,  but it can be not.  What do you mean you don't get schemes?  You get algebraic spaces or you get...  I think you get algebraic spaces, but it's not written in the literature.  So, they use analytic manifolds.  And in your results that you mentioned, you always say projective.  Yes.  Do your results hold the proof used projective?  For example, do the proof of Catern's site use projective or not proper?  In many places, you use hydroplane sections.  You use Bertini CRMs and things like that.  So, in the last thing about the generation of cases,  you really need projective...  No, I mean, you need Bertini CRMs to prove,  for example, this is isomorphism,  but once you have it and you have translated the question in this form,  then, I mean, you just look at this complex and you forget everything else,  and now you can lose projectiveness.  But the specification that they have for analytic degenerations  will apply in particular to the projective ones.  Yes.  So, it...  But they give you...  But what they say is if you start with the K3 surface,  then there is a model which is regular,  and such that the...  So, the total space is regular and the canonical divisor on the total space is trivial.  But this is not a projective scheme.  This is only an analytic manifold.  I mean, to get the condition that K is trivial on the total space.  Ah.  And this is what you need to analyze the special fiber.  And then you have to go from this to your model.  Yes.  So, you have to use weak factorization and things like that.  You use weak factorization to...  and check that the complex doesn't change and so on.  And what about ramifying the original local field?  Does it change the problem or not?  Sure, sure.  Yeah, you lose the question.  I mean, it's a question about a finite group.  I mean...  Okay.  F, so...  No, you lose everything.  Okay.  Okay, so, that's about... ...the group of theorem one.  So, a few comments.  So, the first comment is that the first step of the strategy here is completely general.  You can draw this diagram in any dimension and you can always translate the injectivity of psi in...  as the exactness, sorry, of a certain complex which looks like this,  except it's a bit more complicated in general.  In particular, if x is not simply connected, then it's not true or at least not clear at all that it only depends on the special factor.  Okay, the second comment is that if we have the results of Kulikov-Person-Pinkham-Mirano-Morison in mixed characteristics,  then the whole proof will go through over, say, a maximum of ramified extension of Qp.  And you will get in the same way that the group F is trivial,  which means here that a0 of x is trying to be divisible.  You said that they work analytically, but then they must use not formal series but convergent series,  otherwise it doesn't make sense.  Sure, sure.  So, the first step is to reduce from C double parenthesis T to convergence series.  Oh, yeah.  And I'm speaking the total number.  So, okay, so again, if we have these classification results in mixed characteristics,  we will get trying to be divisible over Qp and R.  And the third comment is that for K3 surfaces which are not semi-stable, we don't know.  I mean, it's an open question for us.  We don't know what happens.  The only reason why we need some miscibility is that it's only for semi-stable K3s that there is a classification.  Okay, so now I turn to local students.  And in the remote, have you said that they have the total space regular, but is it with reduced fiber or not?  Yes, they always work with reduced fibers.  But what do they have to assume to get it?  You said that it's not always the case that there is such or that they show that there is such.  No, they show that there is such, but it's not always a scheme.  With reduced speciality?  Yes, yes.  I mean, you start with the assumption that it's semi-stable as a scheme.  You assume to start with that X has semi-stable reduction as a scheme.  And then they tell you you can find another model, which is maybe not a scheme, which is still semi-stable, and for which K is triple.  Okay, this is the point.  I think it's better to let it go.  Yes, you can.  I'm sorry.  Okay, so I turn to local fields.  Okay, so K is a local field.  And so I will set the theorem.  So as you will see, this will be a much stronger theorem than this one.  So I guess it's the main theorem at the top.  It's also about surfaces.  And I will, again, work with the model.  So this time I assume that I have one like this.  So I keep the notation X, a regular projective model.  Again, A is the special fiber.  So I keep more of the notation.  So maybe there are multiplicities now, I'm not assuming semi-stability, but I assume that A in red is a simple normal crossings divisor.  Okay, so then theorem two.  So you have to make two assumptions.  So the first assumption is about the AIs.  So the AIs are smooth projective surfaces over a finite few, and I assume that they satisfy the take conjecture.  And the second assumption has to do with the alvanism variety of X.  So I assume that X has potentially good reduction.  Of course, this is the case, for example, if X is simply connected.  And then, so the theorem is that, again, psi is injective.  So the second map to H4ZL2 is injective for any other form of P.  So that's a positive answer in this situation.  So again, this was known for surfaces with geometry genus zero.  It was proved by Trudy Saito in the 90s.  And the point is that, so it applies to surfaces with non-zero geometry genus.  And as I mentioned, these are those for which, over the complex numbers, the chalk group is not representable.  So a few remarks.  First, we're not about the take conjecture.  Here we assume it for the components of the special fiber.  But it's very often the case that it's satisfied for trivial reasons.  So there are many, for example, there are many P3 surfaces, P3 surfaces, X for which you have a model.  And the special type is the union of rational surfaces or ruled surfaces.  But AI is, by rationally, ruled for each eye.  And in this case, AI satisfies the take conjecture trivially.  So there are many examples where this can be applied.  Second remark is that this, the conclusion can be formulated in an equivalent way.  We should also take an electric chart.  So, equivalently.  So there is a natural pairing between the chalk group and the brauer group.  You evaluate classes of the bra group at points, and you take the local invariant of our classical theory.  And so this goes to q-molz.  And the conclusion is equivalent to saying that the left kernel here is divisible by n.  So this can be seen as a higher dimensional generalization of Liechten-Monteguality.  Liechten-Monteguality says that for curve, this is a perfect terrain on both sides.  And clearly, the maximal divisible subgroup has to be in the kernel because the brauer group is torsion.  And the third remark is about the other assumption about the albanese.  In fact, if you remove it, then the theorem is wrong.  There are cancer examples due to Karim Alain's crash in the 90s.  So it's a theorem wrong without assumptions.  Is there any assumption on the albanese?  So you have to put some assumptions.  So in principle, this should apply to all simply connected surfaces.  So now I will explain what the ingredients of the proof are.  So ingredients of the proof.  So to start in the same way as for theorem one, you look at the diagram on the top.  You write the cycle class map for one cycle on the model and on the special fiber.  And you use the theorem of Saito and Sato that the map in the middle is injective.  It's an isomorphism, sorry.  But then here in this situation, we will actually prove that the map on the top is surjective.  So we will prove the surjectivity of psi 1A.  So from each 1A turns our head to the left.  To a chronology with supports, so really that's it's an alpha homology, each for A x z0 to 2.  So again, by the theorem of Saito and Sato, this is enough.  But it's perhaps unexpected this problem.  In theorem one, this was not true.  But so, okay, here that's what we proved.  So we start by reformulating this a bit as we did in theorem one.  So the first thing is psi 1A is surjective, if and only if.  So here again, you do analyze.  And so you write the dual of these groups by combining congratulity and duality for the gamma homology of the compil.  So you find that the dual of this group is H3.  So let me write here H3 of A 12 z1.  And then you consider the following map.  First of all, you go to the direct sum over each component.  And then you mod out by a certain subgroup.  So the same group on the top, and I mod out by the certain subgroup which I did not see H1Ai orthogonal.  And I'm going to define this group in a minute.  But the condition is that the composition here is injective.  This map here is injective.  So this is a truly formal reformulation.  And the group, I'm sorry, I forgot to mention what it is.  It's just the group of those classes which die when you restrict them to curves.  So the set of classes alpha in H3Ai 12 z1 such that the restriction of alpha to any curve is 0.  So then I will explain how there is hypothesis, a payroll, how you prove this.  I mean I will just give a picture.  The first thing is this. You have to use the hypothesis on the urbanize.  So the urbanize is related to H1 of the geometric-generic fiber.  And because X is a surface, it's related to H3 priority.  And now if you use this, you can try to relate this H3 of the generate geometric fiber to what happens on the special fiber  using the results of upper-potency to prove the another main way is conjecture for surfaces in mixed characteristics.  So here, I have to argue, I'm not doing this, but these two ingredients will allow you to conclude that H3 of A bar,  so bar means you go to an algebraic closure, and QL is Q of weight 3.  You get the weight of the action of Frobenius on this, and because A bar is proper,  but maybe not smooth, a 3 Re is in a way 3 and possibly lower.  But in fact, using this hypothesis, you define it as pure of weight 3.  And because it's pure of weight 3, it has to inject into the direction of the same groups of other components.  This is because you can write what the kernel is, and you find the kernel is pure of weight.  I'm sorry, I'm not sure, but there's no weight 3 in the kernel.  So that's one point.  Then the second point is to use the tape conjecture, not in an essential way,  but it's a fact that the tape conjecture implies that the action of Frobenius on the H2 of A bar  is simisimple for the eigenvalue 1.  So let me say it's partially simisimple.  So this is a classical fact.  And if you use this simisimplicity, plus some weight arguments,  from this you can deduce that H1 of the finite field with values in H2 of A bar 2L of 1  inject into the direction of the same groups over the components.  What is partially simisimple?  I'm sorry?  What is partially?  So it means just for the eigenvalue 1, it's simisimple for the eigenvalue 1.  This is part of the tape conjecture.  Yeah, I mean it's a consequence of it.  Okay, okay, because of the merit.  Okay, now putting these two things together, you deduce that H3 of A,  by the ocean-southpacal sequence, H3 of A, Q1 of 1,  inject into the direction of the same groups over the components.  And now there's a little miracle that you can deduce from this  same thing with portion coefficients, H3 of A, Q1 of 1,  inject into the same thing.  And the proof here is very simple, there's nothing deep,  but it uses crucially the fact that we're looking at surfaces.  So it's a very simple way to deduce where it's really crucially.  You have to use some bioregular spectral sequence,  and you want some groups to be torsion-free,  and it works in dimension 2 and not in a higher dimension.  Okay, so you're almost done.  You've proved that the first map here is injected,  but the second map is no problem at all.  That's where we use the tape conjecture.  In fact, the tape conjecture here implies that these groups,  C1 of A, I, or 7 of O, are 0.  So if you could all of this together, it's done.  Okay, so that's how the proof looks.  So a few comments.  First of all, it applies.  So as I mentioned, 2K3 surfaces, even in the non-semistable case,  and also in the semistable case, as you see,  there's no need for any classification results.  In the sense it's much easier, there's no combinatorics involved.  But you assume that it is strict normal crossing  by before you did not assume.  You're right.  But I mean conjecturally, I did not assume that the multiplicities were 1.  Okay, so that's one point.  And in particular, so this, you see,  if we had the classification in mixed characteristics,  we would get the injectivity for K3 surfaces  over the maximal and ramified extension of KDQ.  Now this theorem gives it for free streaming tape conjecture  and existence of models for K3 surfaces over KDQ.  But it's a bit strange that if you want to prove the same results  with a maximal and ramified extension,  then currently you need very strong geometric results.  Okay, and I want to finish by coming back to strictly local.  Can I ask a question?  Yes.  So you don't assume that the cross-fibers are reduced.  But you said you use the latest spectra systems.  What does this mean?  So the point is to use the local invariant cycle theorem,  which tells you that it's H i of A, say A bar,  which will subject onto H i of S bar to an invariant  under the inertial group.  But to prove this, you can reduce,  by using the Gavard-Elyon answerations,  you can reduce to the semi-stable case.  Okay?  Here, thank you.  Okay, so I want to come back to strictly local fields.  Okay, so we have a positive answer for KD fields,  at least for simply connected surfaces.  We have a positive answer for KD fields,  as you mentioned in the lecture.  And for strictly local fields, we have a positive answer  for surfaces with geometric genus zero.  And then also for K3 surfaces.  So what happens in general?  Well, so we were quite surprised to find a count,  for example, in general.  So that's here in three.  And this works over both types of strictly local fields.  X, that's simply connected,  smooth projective surface,  over a C-local plane to the C, such that,  so again, what is the cycle map here?  We're in the strictly local case,  so the collage group H2E only contains information  about the degree and the P1, the billion P1,  because it's simply connected  to really nothing else than the degree.  So here, for example,  is an example where the group is not divisible.  And here, we have A0 to X,  Z0 to...  So the side is not injected.  And also, so that's one point.  Another point is the same.  Over the maximum unromified extension  of a K, that deals for infinity to many times P.  So in particular,  this shows the hope for an analog of the Catocyte theorem,  which was about finite fields.  You could hope for an analog over this here,  which is quasi-finites.  Well, this hope does not hold.  I mean, the billion fundamental group  does not control the finite quotients  of A0 to X, maximally divisible for blue.  And another thing which I think,  well, for me, it was really unexpected,  is that, I mean, you also get examples  over strictly local fields like this.  Now, if you take such an example,  well, it's a surface,  it's defined over some chaotic field.  So if you take this example,  you can apply it to its theorem too.  So what you get in the end is a surface  over a chaotic field for which the cycle map is injected,  and the same...  And in this example, the deconstructor is satisfied.  So that's okay.  And the same over any finite extension of the chaotic field,  or at least any finite unrungified,  because then you see that the deconstructor is satisfied.  You can still apply theorem too.  But when you go to the limit,  you get a counterexample.  So I think that was a strange phenomenon.  Okay, so there are many things  that we don't understand.  I mean, for chaotic fields,  there are rather complete results,  except when the Albanian variety is not trivial  or does not have potentially good reduction,  then we don't have a good answer,  we don't understand what happens.  Also over strictly local fields,  we don't have...  It would be good to have a better understanding  of what goes on and what to expect.  And also in higher dimension,  I want to mention that most of what I've said  works exactly the same way in higher dimension,  which could be interesting,  because the previous techniques were about  co-dimension two cycles on varieties of any dimension.  So the previous techniques do not say anything  about zero cycles in dimension three or more.  So the only place where dimension two was really used  is this one.  And here we just don't know what happens.  And because of this problem,  we don't know what happens in higher dimension.  And an example of an open question,  a simple open question in higher dimension,  is this, if you take X, a smooth projective,  a rationally connected variety,  over seedable for anything with T,  well, is the chocolate trivial?  For such a variety,  it's known that it has finite exponents,  so trivial is equivalent to being divisible here.  And that's something we don't know in dimension greater than two.  But you know it in dimension two,  because it's Russian, right?  Yes, it's Russian,  which I'm going to give you a zero.  Okay, so I'll stop here.  Thank you very much.  Thank you.  So could you maybe start from Tokyo,  Beijing, and then come back to Paris?  So we start from Tokyo?  Okay, no.  In fact, we don't see you very well.  There is no light.  Ah, really?  Really?  Yeah.  So, so, so, so, so you knew last term,  what happened you make a random with the substitute?  So before your PZ or a big number,  second big number.  So I, so I, not completely sure.  I think, I think the geometry in this is three.  I mean, I've computed this, but I've forgotten.  It's color and dimension one.  This I can tell you.  I can tell you also,  the degeneration in this case is very simple.  It degenerates into the union of two surfaces,  which meet along one curve.  And each surface, each of these surfaces,  I think has geometriciness one.  The curve is an elliptic curve.  And they have color and dimension one, too.  Okay, I think that's all I can say.  So you just mean the property of the,  the variable or are you, ah, no.  Ah, you mean to obtain this?  Yeah.  Ah, yes, so actually, yeah, yeah.  So thank you for the question.  So in, in theorem one, I wrote this complex on the top,  which was equivalent to the divisibility of this group.  So it's not, it's not correct in, in general.  I was looking at K3 surfaces,  but in general it's almost this.  For a, for a variety which is simply connected,  it's exactly this except that on the right,  you have to mod out by the same group as here,  CH1 of AI or sub-anode.  But in, in the situation of K3 surfaces,  this group was, was trivial.  So, yeah, so that's again the property of the special fiber.  And so the exact property you need to get this example  is the generation like this,  such that the, the generate fiber is simply connected.  This is very important.  And the second property is that you should take the curve here,  the intersection curve, say C.  Then you should have the intersection number of C  with any divisor in, in either surface.  Either surface is even, always even  for any i and any d divisor on AI.  That's what you need to do.  Thank you.  Are there further questions?  Okay, thank you very much for, for talking.  Okay, so we, we have some questions from the dean.  Okay, so any questions?  No?  There's no questions from the dean.  You have a question?  Yeah, yeah, you said something about,  I did not register exactly.  So about the, some passage to the limit  of the strict visualization of the periodic field,  there was some comment that you made.  Here at the end.  Yes.  So here I was saying that, so there is an example like this,  over the maximum or unified extension of some chaotic field.  Yes.  Now the surface is defined over some chaotic field.  Yeah.  And to this surface, over the chaotic field,  you can apply theorem two.  Yes.  So you have this situation with the surface over a chaotic field,  and on, on, on this chaotic field,  the surface map is injected, and it's still injected.  If you go to any unromified extension,  or maybe any finite extension,  if you believe that it's a conjecture.  But if you go to the maximum unromified extension,  it's not injected anymore.  That's the counter example here.  All right, so the point is the coromology doesn't come  out to this limit.  Right, because it's the zeta coefficients.  Ah, okay.  Not finite coefficients.  Okay.  But yeah, still it's, maybe you did it unexpected.  Other questions?  What are you, I think that at some places you mentioned  that you have to use the Gabbers L prime alterations here.  Yes.  Could you give a specific example?  Because here is not too convincing because you work with QL.  So, yeah.  Oh, I'm sorry, I'm sorry, I'm sorry.  You're right.  Here, De Jong is enough.  I'm sorry.  No, no, you're right.  No, no, it's really used just to prove the theorem of  Saito and Sato, the isomorphism there.  I mean, if you want to prove this, you can do any  exploration of the grid prime as well.  And, and, and reducing this case to really work with  finite coefficients.  You're right.  You're right.  I'm sorry.  This theorem is true for finite coefficients.  And that's how you prove it.  Yes, thank you.  Other questions?  So, let's bring the speaker again.  And then we say good bye to Eugene and to...  Yeah, yeah, yeah.  I don't know.  Maybe I can...  Yes.  So, we have still a question.  No, no, I'm not, I didn't know.  Yes.  Where is it used?  There was a question from Tokyo.  So, you said that you have this local monodrome theorem.  Yes.  So, where does it enter in your, in the...  So, here, I mean, if you want to prove that H3 of A bar is  pure of weight three.  Yes.  So, what you do is that, in fact, it's in an exact sequence.  So, it surjacks onto H3.  We have this.  But in fact, this is not the important part.  The important part is what comes on the left.  You have H3 of the model, say, over the maximum amount of  high-file expansion, which is supported in A bar.  And so, this local invariant cycle theorem allows you to  prove it's not completely obvious, but it allows you to  prove that this is an exact sequence.  It's exact, pure.  It's not completely pure.  But once you know it's an exact sequence, you're done  because this has weights at least three.  This has weights at most three.  So, if it's not pure, it must be because of this.  And this you can relate to the urbanizing.  And Rappaport think assume that you have semi-stable.  Yes.  You assume regular model, no?  Yes, right.  But again, we can use alterations to prove the weight  on the drawing spectrum.  We can use alterations to prove it.  Okay.  Okay.  Still no questions?  So, if not, then we say goodbye.
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