 Okay, so So it's my pleasure to introduce the speaker today We speak on the cycle class map for zero cycles for local things Thank you very much Okay, I'm sorry No 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 of ramified extension of a PID fields And so especially these will be results about surfaces with positive geometrogens 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 projective variety overfield K capital K and as usual, I will denote the chart group of zero cycles by CH0 of X So we call this is the group of Cycles, so free loop on the close points Modular rational equivalence, so rational equivalence is generated by linear equivalence on the curves lying on X Because X is projective, there is a degree map and I will denote the kernel by a lot of X Just one second we check for the sound Okay, so So in dimension one, this is nothing but the peak of group. Yes shift François This is just a peak of group and if Yeah, there is a rational point Then the degree zero part is nothing 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 need geometrically connected X. Yes. Yes Comparatively really soon For the whole of the talk So Yeah, in dimension greater than one. This is one of the simplest invariance that one can associate to X But it's In many cases, it's very difficult to determine the structure of this group or to compute it in any given example It affects the dimension greater than one and this even over an algebraically closed field So to illustrate this let me recall the situation surfaces over the complex numbers Well, we have a conjecture box conjecture So in dimension one here, you see that this group is the group of points of an evident variety So it's represented by an evident variety and dimension greater than one say mentioned to It's conjectured that This group may not have X should be representable in a suitable sense By the points of some variety if only if Each two of x-y's is zero. So what is known is that If you have it, so this is the geometric genus and what is known is that 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 Okay, however, there is one simple thing that one can say Over over the complex numbers and more generally over algebraically closed fields in any dimension Is the fact that this group is divisible In the case of curves, it's clear from the above description And in general you can reduce to the case of curves by finding curves like on X going through specified points Okay, so This is the situation of the complex numbers There is one situation in which This group is well understood. It's when the field is the finite field And everything is understood because of a higher dimensional class field theory But let me recall the relevant theorem Which is due to Cato and Shouji Saito 1983 So in this case The child group of zero cycles of degree zero is always finite, but it's more precise than this one cannot in fact compute it in examples It's very controlled. So maybe there is a map natural map going from the child group of zero cycles to the Ibibianized fundamental group Defined by Provinus substitutions. It's in the close points the correct to the class of the corresponding Provinus and Cato and Saito proved that this map is injective So to detect rational equivalence you just need to compute the image in this group and the It's in its image of this injection is very well understood. It was determined by 9 So this gives complete control over this this group and for example It's the right is simply connected from this you can deduce that the child group of zero cycles of degrees zero is trivial So the image are just the elements whose degree is integer. Yes, okay Okay, so But it's really the only situation where So from now on I will consider local and strictly local tunes so capital K will be the question R R will be a conciliant discrete valuation ring the residue field is K It's okay for simplicity I will assume that capital K has characteristics zero, but it's really not an important assumption and I'm going to 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 okay local is when it's okay, it's finite And capital K strictly local When small k is a separate So in this situation, yeah, any examples of course are PID fields here to be a finite extension and Here for strictly local there are Natural examples the maximum of ramified extension of the KD field but also See the blocker and this is an interesting example. So in this situation what has a good structure theorem for the chocolate of zero cycle of degree zero which was When you consider the ramified extension, you can consider it either algebraically or complete and I guess there is a general rigidity fact that it Yes I will assume hence again Everything to go So Okay, so the structure theorem my title and subtle in 2010 So here is the 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 trying to pee So precisely You can write as D plus F where F is its finite The ball of prime to P and D is is a group which is divisible So this means For N prime to P You don't say So if P is zero, it's divisible plus finite Was prime to zero is not Okay, so Right, so as I mentioned to D is a bit analogous to to the group over the complex numbers As you've seen it's extremely hard to understand so pure it's Probably even harder So I mean there's little really little known about it. It's only known that you can contain torsion Even for simply connected branches, so this is a due to as a grand site, so But This talk will focus on the finite finite group app, which is pretty analogous to the group over finite fields here so So in this sense this theorem is analogous to the theorem of that point sites Oh, which says that for 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 on its structure so For this so we want to study F F is finite over prime to P. 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 the Classes of cycles is the cycle class map to et alchromology. So when you consider the cycle class map So Which goes to C H 0 of X to So if D is the dimension of the variety goes to H to D So here So Just for beginning interview someone is calling the room so we don't know who is We will don't we will continue Okay, so we can see Integral erratic homology So this is a finicky generated data So because of this any element which is infinity divisible here So this map kills D so only F survived and Another way to phrase this is to consider this book So this kills again this kills the divisible thoughts and really the analogous question To the theorem over finite fields Yes, well is this map injective? So what's about the In fact over a finite field if you consider this over a finite field you really get exactly The map here for being a substitution you can identify this homology group For a fixed L with the pro L pounds of by one up So really it's a generalization and Okay, and if you can answer it and I mean when there is a positive answer then you can hope to to compute F By computing just its image in a talk homology So this this question has received a lot of attention From the 70s to the 90s After initial work by block who initiated a method for studying cycles torsion cycles using a theory and By combining this block August theory and more career associate here It was recognized that one could get a good hold of Co-dimension two cycles on the three varieties. Oh, sorry on the property dimension so So that many names here As we make of course the cattle and site to which is already mentioned So wrote a whole series of papers about this problem using key theory for Cycle of the dimension to 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 geometrical zero so An example of a theorem obtained in this series, which I think it should be attributed. Oh, I should not just write Ruskin so this theorem is Should be attributed to polyurethane and Ruskin 1985 So about this question if so in a strictly local case capital K strictly local maybe I did not Stress this enough the theorem of Saito and Sato 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 So the goal of the talk would be to try to answer To give some answers to 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 to mention is about strictly local fields and Actually over C level parentheses is C. So if you want to To to consider surfaces with non-zero geometric genus the first Passive surfaces you look at this K3 surfaces and In this respect we can prove the following X is a K3 surface Over C level parentheses T With semi-stable reduction Then okay then and the answer is positive So what does it mean? The cycle class map is injective but here The strict so we're in the strictly local situation So it's a field of common logical dimension one and you can try to understand what this common g-group is in this case and Because the surface is simply connected It really means that there's no f in other words The job group of zero circles of degrees zero is divisible. Okay, so let me Let me explain the strategy for proof of this so we will consider a model of our The ring R. So here are in C level bracket T and I will consider script x over R projective Regular model Let me denounce the special fiber by a and it's irreducible components by AI So since we're in a semi-stable case a here is reduced And I will assume in addition that it says it's a simple normal crossings device or on X So this is not needed though in the result of I Mean you can always choose a model like this And we work over a C double parenthesis T. Yes, so I mean I'm explaining the strategy for the proof So we choose such a mobile Okay, no, but if you have a model result simple lower closing. Yes Ah, you mean if it's still some stable in this weaker sense. Yeah. Yes. Okay. Yeah You're right. If you're impose with some stable in the 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 semi-stable So in this situation Let me use a new ball Regular it is not enough to have a log smooth right So in this situation consider the the cycle cost map psi so and so Not to H4 X And we consider what happens on the model on the model We have one cycles on the model and the cycle cost map To the top of the model. They are vertically you can write localization exact sequences. You can restrict to the generic fiber and Before that you have One cycle on a special fiber and here you have probably with support in the special fiber There's a cycle cost map again, which we'll call psi one a One cycles on the special fiber Okay, and now the point is that Saito and Saito Not only prove the finiteness theorem here In fact, they put something much stronger which underlies this theorem They proved that the cycle cost map for one cycles on the model is an isomorphism so this is Saito and Saito and they prove this in in the generality of their theorem arbitrary dimension local or strictly local field But they don't have regular model. Yes, they have regular models. They assume they have one They assume they start with one Then they claim. Oh, you need to prove this. Okay, but to prove this you can reduce to the I mean using your Okay, so Now if you look at this and there's one point here, which is zero cycles extend to one cycles on the model So this map is subjective 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 surjected but in fact, it's never the case for For for the generations of K3 surfaces One can 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 Truly in terms of the special fiber. So I'm just giving you the results. It's free 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 human then What's the human that to the power I? three direct sum of I copies of human then and then it goes to the direction And off the most of the groups of the components That's our human then so this complex Is exact that's the problem. Okay, what's on 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 J's component, it's sent to If you go to the ice factor you look at the class of a I intersection a J That's a lambda the ice factor or I different from G J if I equals J you can anyway Modify by an element 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 now that's the second part of the strategy is exactly in the middle. Yes, exactly. Yeah, just this So now we 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 was a whole literature about this starting from the work of Fulikov person can come 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 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 reduction Then using these two things you can translate the problem into a combinatorial one so You obtain a combinatorial problem So mainly so it's about the combinatorics of the of the special fiber They need the biggest case of the classification tells you that so so in many cases you have a K3 surface which degenerates to Rational surfaces in such a way that the dual complex triangulates this here, so then you you You obtain questions about triangulations of this here in fact the negatively curved triangulations of this here 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 you can solve it that's not solve it and I'm sweeping Lots of details and the rug I should mention that these the generations are not as you break 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 What do you mean? You don't get skipped you get algebraic spaces or you get so I think you get algebraic spaces But it's not written in the literature So they use an elastic manifold In your results that you mentioned you always the same projective Yes, do you realize the results hold the proof use projective for example that Do the the proof of cut cut inside? Yes In many places you just hydroplane sections You use about any theorems and things like that so So in other so even in the last thing about the generation of case you really need projective No, I mean you need the Burtini theorems to prove sorry to prove for example This is one reason 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 classification that they have yes analytic the generation will apply in particular to the projective ones. Yes So it's but they but they give you know, but what they say is if you start with a 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 To get the condition that K is trivial on the total space And this is what you need to To analyze the special fabric 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? Is it does it change the problem? Sure. Yeah, you lose you lose the question. I mean it's a question about a finite group. I mean, okay Okay, so that's about Look up theorem one. So a few comments So 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 As the exact as the exact as the exactness sorry of the certain complex which looks like this Except it's a bit more complicated in general and 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 fiber Um Okay, the second comment is that if we had the results of Kulikov person can come here in the Morrison in mixed characteristic then the whole proof will go through over 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 a note of X is point to be divisible They work analytically, but then they must use not formal series but convergent series I don't know. Sure. So the first indeed the first step is to reduce from C double parenthesis T to Convergence So Okay, so again if we had these classification results in mixed characteristic We would get practical visibility over Qp and R and the third comment is that's For K3 surfaces, which are not so miss 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 instability is that it's only for some stable K3s That there is a classification Okay, so now I turn to local fields and in the remote and 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? Some you said it's not always the case that there is such or they show the research No, they show that there is such but it's not always a scheme With reduced special. Yes. Yes. I mean you start with the assumption that it's semi stable as a scheme You assume to start with that X is 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 I Okay, so I turn to local fields Okay is local field and so I will say to theorem so this So as you will see this will be a much stronger theorem than this one So I guess it's the main theorem of the talk. 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 I keep the notation X a regular predictive model Again a is the special fiber And I assume so maybe there are multiplicities now, I'm not assuming some instability, but I assume that a red Is a simple normal crossings divisor Okay, so then theorem two so they have to make two assumptions So the first assumption is about the AI's so The AI's 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 albin is a variety of X So I assume that's out X That's potentially good reduction. Of course, this is the case for example if X is simply connected So the theorem is that again the side is injective. So the cycle map to H4ZL2 Is injective for any other form of P. So we get a positive answer in this situation So again, this was known for surfaces with geometry genus zero. It's it was proved by Trudicito in the 90s and the point so the point is that So it applies to surfaces with non-zero geometric genus and as I mentioned these are those for which over the complex numbers the Chao group is Is not representable So a few remarks first we're not without 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 K3 surfaces, but P3 surfaces X For which You have a model and the special and the special type or 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 Kind of many examples where this can be applied second remark is that this the conclusion can be formulated and in an equivalent way Which also appears in the literature So equivalently So there's a natural pairing between the Chao group and the Rauer group You evaluate classes of the bra group at points and you take the local invariance of a classical theory and so this goes to humans And the conclusion is equivalent to saying that the left kernel here is divisible by there So So this can be seen as a higher dimensional generalization of Liechten-Montaic duality Liechten-Montaic duality says that for curves this is a perfect pairing on both sides and and clearly The maximal divisible subgroup has to be in the kernel because the bra rule is torsion Okay, and the third remark is about the other assumption About the Albanese in fact if you remove it's then the theorem is wrong this there are cancer examples due to parimalism in the 90s wrong without Without any assumption on the Albanese so you have to put some assumptions Okay, so in principle This should apply to all simply connected surfaces So now I will explain what the ingredients of the proof are ingredients, so you start in the same way as as for theorem one you you write Yeah, you look at the diagram on the top you writes the Cycle class map for one cycles on the model and on the special fiber and You use the theorem of Saito and Sator that the map in the middle is Injective is an isomorphism, sorry, but then Here in this situation, we will actually prove that the map on the top is surjective so here so We'll prove Cy one a So from each One a To you will support so really that's it's our homology each for a Thanks By the theorem of Saito and Sator this is enough, but it's Perhaps unexpected is growing in theorem one. This was not true, but so Okay here, that's what that's what we prove Okay, so we start by Reformulating this a bit as we did in theorem one so the first thing is second If and only if so here again you do lies and So you write the dual of these groups by combining compatibility and utility for the Galois homology of So you find that the dual of this group is h3 Here h3 Two elements that are one 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 out by the certain subgroup which I did not see each one a i orthogonal and I will 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 It's just the group of those classes Which die when you restrict them to curves the set of classes alpha in h3 a i one of the other one Such that the restriction of alpha to any curve is zero Okay, so then I will explain how the various hypothesis Play a role and how you how you prove this I mean I would just give A picture Okay, the first thing is this You have to to use the hypothesis on the albanese So the albanese is is related to h1 of x of the geometric generic fiber And because x is a surface it's related to h3 priority Okay, 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 the upper potency to the Another way is conjecture for surfaces in mixed structure stick So here you have to argue I'm not doing it But these two ingredients Will allow you to conclude That's h3 Of any bar so bar means you go to an algebra closure In a ql It's cure of weight three You get the weights of the action of Frobenius on this and because a bar is proper But maybe not smooth a free re it's it's the weights three and possibly lower, but in fact Using this hypothesis you define it as cure of weight three and because it's pure of weight three It has to inject into the direction of the same groups over the components This is because you can write what the kernel is and you find the kernel is pure of weight I'm sorry, maybe not sure, but it's I mean there's no way three in the kernel Okay, so that's one point Then the second point Is to use the tape conjecture in Not in an essential way but Okay, it's a fact that the tape conjecture implies that The action of Frobenius on the h2 Of a i bar is semi-simple for the eigenvalue one So, I mean sorry Yeah, it's partially so let me say it's partially semi-simple So it's this is a classical fact and if you use this semi-simplicity First some weight arguments From from this You can deduce You can deduce that h1 of the finite field with values in h2 of a bar 2l of one Inject into the direction of the same groups over the components. I'm sorry. What is partially? So it means just for the eigenvalue one. It's semi-simple for the eigenvalue one Ah, this is part of that conjecture. Yeah, I mean it's a consequence of it. Okay, okay Because of normality. Okay, I'm putting these two things together You deduce that h3 of a by the ocean-southpacal sequence h3 of a q1 of one Inject into the direction of the same groups Over the components a miracle that you can deduce from this the same thing with portion coefficients h3 of a q1 of z of one Inject into the same same thing and the proof here is very simple. I mean there's nothing deep But it uses crucially the fact that we're looking at surfaces and it's the only place where it's really crucial You have to use some biogets of respect for sequence and you want some groups to be torsion-free and Well, it works in dimension two and not in higher dimension Okay, so you're almost done. You've proved that the first marker is injected But the second map is no problem at all. That's where we use the take conjecture So in fact the take conjecture here Implies that these groups c1 of ai or thermal law are zero if you could all of this together Okay, so that's how the proof looks So a few comments First of all, it applies. So as I mentioned to k3 surfaces, even in the non-semestable case And also in the semestable case as you see there's no need for any classification results In the sense, it's much easier. There's no combinatorics in both But you assume that it is strict normal crossing by before you did not assume. You're right. You're right But I mean conjecturally But I did I did not assume that the multiplicities were one Okay, so, um, so that's that's one point and um in particular So this Do you see if we had the classification in mixed characteristic? We would get the injectivity For k3 surfaces over the maximal and ramified extension of chaotic fuel Now this theorem Gives it for free assuming take conjecture And an existence of models For k3 surfaces over over chaotic fields But it's a bit strange that if you want to prove the same results for the maximum ramified extension then Apparently you need You need very strong geometric results Okay, and I want to finish by Coming back to strictly locomotives All right, can I ask a question? Yes Yeah, so you You don't assume that the cross-fibers are reduced But you said you use the latest spectrum signals. What what does this mean? so, I mean So the point is To use the Local invariant cycle theorem, which tells you that h i of a say a bar Well surjects H i of s bar Invariance under the initial group But to prove this you can reduce By using the Gavardian answerations you can reduce to To the semi-stable case Yeah, thank you. Okay, so I want to to Come back to strictly local fields. Okay, so We have a positive answer for Piedi fields at least for simply connected surfaces. We have a positive answer for for Piedi fields as you mean to conjecture 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 this here in three and this works over both types of strictly local fields. So there exists x a simply connected smooth projective surface of a syllable plant as a stream Such that so again What is the second map here? We're in the strictly local case. So the the ecology group h2d Only contains information about the degree and the pi one a billion pi one But because it's simply connected, but really nothing else than the degree So here it comes for example is an example where the group is not divisible and here We have a not of x not two Is z not two? So the side is not injected and also so that's one point Another point is the same Over the maximum of the normified extension Of the periodic fields for infinity many prime p So, uh, so in particular This shows that the hope for an analog of the catocyto theorem Which was about finite fields. You could hope for an analog over this field, which is quasi-finites Well, this hope does not hold. I mean the A billion fundamental group does not control The finite quotients of a not of x finds maximum divisible for group And um and another thing which I think well for me it was really unexpected Is that and 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 Caliphial 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 fields for which the cycle map is injected and the same And in this example to take contractory satisfied. So that's okay And the same over any finite extension of the chaotic field you can or at least any finite unromified because then you You see that the trajectory is satisfied You can still apply theorem too, but when you go to the limit You get to go to example, so I think that that was a It's a strange phenomenon Okay, so There are many Many things that we don't understand. I mean for chaotic fields, there's a rather complete result except when the urbanized variety Is not trivial or does not have potentially good reduction Then we don't have a good answer a good understanding what happens Also over strictly local fields We don't have it would be good to have a better understanding of what goes on what to expect and Also in higher dimension Most of what I've said works exactly in 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 the only place where dimension two was really used Is this one and here we don't I mean 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 a of an open question simple open question in high dimension is is this if you take x A smooth projective and a rationally connected variety Overseasable parameters t Well Is the chalkbook trivial For such a variety it's known that it has finite exponents. So Trudul is equivalent to being divisible here and That's that's something we don't know in dimension greater than two But you know it in dimension two because it is rational Yes, it's rational. Okay, that's zero Okay, so I'll stop here. Thank you very much. Thank you So Could you maybe start from tokyo be gene and then come back to first? Okay, so we start from tokyo Okay, in fact, we don't see you very well. There is no light You Yeah So I Not completely sure. I think I think the geometry genius is three All right. I mean I've computed this but I've forgotten It's color I mentioned one this I can tell you I can tell you also it The degeneration in this case is very simple. It degenerates into the union of two surfaces which meet the long one curve And each surface each of these surfaces I think has geometry genius one The curve is an elliptic curve And they have color right dimension one two Okay, I think that's all I can say So you you just use the property of the Modifiable or you You mean to to to obtain this Yeah, yes. So actually, uh, 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 case three 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 at ch1 of ai or sub at all But in the situation of case three surfaces But this group was was trivial So, yeah, that's again a 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 if you take the the curve here the intersection curve Say c Then you should have that the intersection number of c with any divisor in In either surface Is even always even for any i and any d Divisor on ai That's what you need to do Thank you Okay, so we do we have some questions from begin Okay, so any questions? No There's no questions for me We have a question You said something about I did not Register exactly so about the some passage to the limit for the strict Zalization of the periodic field there was some comment that you made Oh here at the end. Yes. No. So here I was saying that So there is an example like this Over the maximum of ramified extension of some chaotic field. Yes. Now the surface is defined over some chaotic field And to this surface over the chaotic field you can apply theorem too Yes, so you have this situation with the surface of a chaotic field and On on this chaotic field the cycle cost map is injective and it's still injected if you go to any Unramified extension or maybe any finite extension if you believe it's a conjecture But if you go to the maximum and ramified extension, it's not injected anymore That's the count for example here. Ah, so the point is the chromology doesn't come here Because it's the Zal coefficients. Ah, okay, not finite coefficients. But yeah, still it's maybe a bit unexpected Other questions I think that um at some places you mentioned that uh To use gabbers l prime alterations here and yes, could you give a specific example? Because here it's not too convincing because we work with ql. So Oh, I'm sorry. I'm sorry. I'm sorry. You're right here. The long is enough. I'm sorry. No, you're right No, no, it's really used uh, just to prove the theorem of sito and sato The isomorphism there I mean if you want to prove this you can do any alteration of degree prime as well and and uh And reducing this case to Can you work with finite coefficients? Yeah, you're right. You're right. I'm sorry. Yeah, this theorem is true for finite coefficients. And that's how you prove it Yes, thank you Other questions Let's thank the speaker again And then we say goodbye to to regime and Oh, it's not maybe I can uh, yes, so we have to get a question. So I'm not uh, I didn't know Where is it used? So you there was a question From tokyo. So you said that you have this uh, local monodrome theorem 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 weight three Yes, so what you do is that in fact it's in an exact sequence. So it surjax onto h3 So you have this 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 height expansion with support in a bar And so this uh, 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 here. It's not completely clear Uh, 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 urbanism and rapoport think assume that you have semi stable Yes, you assume regular model. No, yes, right. But again, we can use alterations to to prove the Wait on the drawing spectral Conjecture we can use alterations to prove it Okay, okay Still no questions So if not, then we say goodbye Goodbye