 And let me thank Ahmad and Fabrice for this nice invitation and the organization of the conference, and particularly Ahmad for his hospitality during my current stay here at the RHS. Thank you. So I'm going to be discussing today some problems concerning the deformation and variation of algebraic cycles focusing on the equal characteristic P situation. So to begin, let me describe the rough type of problem that we'll be interested in and some existing and new conjectures. So for the most part, we'll consider some smooth projective family of smooth varieties over a field. We'll have some closed point of the base, closed just for simplicity. And to be precise, let's take some codimension I cycle on the fiber over this point. And I'm not going to be to set any torsion questions today, so I'll always be working with cycles tensor q. Then the type of classical problem that one wants to study is the following question. When does this cycle extend to the whole family x? In some sense, there are various more precise formulations you can give. So in characteristic 0, the problem is at least conjecturally reasonably well understood, because we have the variational hodge conjecture of Groten-Dieg, 66, if I remember correctly. So this says that assuming the characteristic of k is 0 and with notation exactly as here, then the following are equivalent. So firstly, the condition that really interests us, the cycle does extend. So the cycle extends from the special fiber to all of x to be precise, mod homological equivalence. So that is the same, or precisely, I can find some z tilde to the codimension I cycle on the whole family x with the same DRAM class after restriction to the original fiber. And this extension, mod homological equivalence, is really the best one can ever hope for. So that's what we'll study throughout. Although it's slightly misrepresent for Groten-Dieg wrote, the expectation is that a sufficient condition is the obviously necessary. So you have the quity, what is the DRAM homology? You can take 1 over q over k. So which one is this? That doesn't matter. And if I've taken a closed point, it doesn't matter. No, but k is the field of characteristic 0. So it's the field of the rational. Oh, k, k, k. Yeah, DRAM homology over k. Over k. Yeah, of course, of course. I exclude it from the notation because DRAM homology is not going to appear very much. Yeah, absolutely over k. Yeah. So the expectation is that the obvious necessary condition is sufficient. That is to say, it's enough that the DRAM cycle class, living in the DRAM homology of the special fiber, lifts to the DRAM homology of the whole family x, DRAM homology over k. Or do you assume that the point is a resin field k? Well, I take it to be, in fact, very soon, k is going to be algebraically closed. But in the cases, as long as it's a closed point that doesn't matter. And if I compute DRAM homology using the DRAM complex, it doesn't matter whether I compute it over k or over finite extension of k. No, but I have a similar condition, too, that the class for k could sink, but it's not there. In any case, I only care about the case k algebraically closed. So all we can impose the condition that the class is flat. That is to say, it lifts to a global section on the base of this DRAM local system. So what we know about this conjecture, in the sense that there are two primary results, what I'll call theorem a. So this is part of Delinia's DRAM de la Parti fix. So it's the condition 2 and 3 are equivalent. So I can't reach up there any longer. But it says that the obvious map from the group in 2 to the group in 3 is surjective. And from that surjectivity, one sees that the two conditions are equivalent. And then we have the result of left shats, which in the absolute case goes all the way back to 1924, stating, of course, he didn't have the variational hodge conjecture. He was considering the absolute case, but it follows from left shats result that the variational hodge conjecture is true for divisors. That is to say, in code I mentioned 1. And apart from that, OK, we have some special cases where we know the hodge conjecture, and therefore we can deduce variational statements. But essentially, apart from that, we don't know much else. But as I said, it's not characteristic 0 that interests us today. So from now on, OK is going to be a perfect field of positive characteristic. So in this setup, we can still consider the problem. And Groten-Diek, in the same footnote where this variational hodge conjecture appears, also proposed, I say proposed, there's about five words on the topic, proposed a variational take conjecture, which is completely analogous. But very little seems to be known about. It seems not to have even been very widely studied. And so instead, I'd like to, in fact, follow the same philosophy that Le Fou mentioned at the very end of his talk yesterday, that to study defamation of problems in characteristic p, we should work periodically. And so the main aim today, instead of trying to formulate such variational questions using allatic co-amology, we're going to form an analog in crystalline co-amology. And let me remark immediately that I'll make comments later about rigid co-amology. So let me fix some standard notation for working with crystalline co-amology. W is W of k, the ring of width vectors of our perfect field k. Big k is its field of fractions. For any reasonable scheme over k, we'll be more clear later about what reasonable means. I can consider the crystalline co-amology. And I will continue to ignore torsion questions. So I'll always work with crystalline co-amology with p inverted. And I'll, hoping this doesn't upset you, Professor Kerr, I'm going to continue to omit the k. Everything's taken over k. We have the absolute Frobenius, which acts on the crystalline co-amology groups. And so I can consider the eigenspaces, which I'll denote with standard superscript notation. And finally, I recall that analogous to the Dirac cycle class maps that we see in the variational Hodge conjecture of p to the i. So it's the eigenspace of the Frobenius with eigenvalue p to the i. So we have in this crystalline set up crystalline cycle class maps, which associate, to any code I mentioned, i cycle on our variety, its corresponding value in the 2i crystalline co-amology. And in fact, we know that it lands in the middle eigenspace of the Frobenius. In the cases in which I apply the crystalline class map, it is indeed going to be proper. If it's not proper, then you're right that there's been some subtlety regarding the crystalline cycle maps. But you can certainly define it now using some technique in k-theory. I mean, if you use, let's say, Geisel-Lavine's computation of k-theory and characteristic p, one can easily define the crystalline cycle class maps for any finite type k-ski. Well, I got quite worried about this at one point. When I started looking back over the classical literature, this restriction of the cycle class map to the proper case. And although we don't need it, it's maybe helpful to recall at this point just to put ourselves in the right frame of mind, at least part of it's very large. If x is not proper, then it's very large. It's very big, yes. This is why I'll make some comments about rigid co-amology later. So as I say, just to put ourselves in the right frame of mind, let me recall part of the crystalline formulation of the Tate conjecture, which is normally formulated erratically. If x is smooth, projective over a finite field, then the surjectivity part of the conjecture will state that after tentering the cycle class map up by qp, we get a surjection onto the eigenspace. So I think it can now go here. So now we have enough to formulate an analog and crystalline co-amology of the variational hot conjecture. So k continues to be perfect of characteristic p and f, s, and z are as in the formulation of the general problem at the beginning. And to save me writing it so many times, I'm going to let c be the crystalline class of my cycle z. Then the conjecture states that the following four conditions are all equivalent. And I'm going to give them names so that I can talk about them. So firstly, the condition of interest is that the cycle z extends to the whole family mod-homological equivalent. So that says the cycle extends mod-homological equivalent. And one suspects that a sufficient condition is the obvious necessary one, which are called the crystalline condition, namely that the crystalline cycle class in the crystalline co-amology of the special fiber lifts to the crystalline co-amology of the whole family. Now since we know what eigenspace that should lie in, we could impose the slightly stronger condition instead that the cycle class lifts to the expected eigenspace. And finally, since we would like to have a nice analog of the situation of the variational hot conjecture, we also consider a flatness condition stating that the crystalline cycle class of our cycle is flat. That is to say, it lifts. In fact, if the lift exists, it's unique to a global section over the base of Augus's convergent F-isocrystal. Now I'm not going to use the flatness condition really at all today. So if you are not an expert on F-isocrystals, you can more or less ignore this final part. But now again, the same problem about non-closed, I mean, the residue field of the point, because the crystalline co-amology we use is over the field factor of the bit vector of k. But when you have a closed point, you have to consider the bit vector of the extension. Then you have vector space on a different field. So when you say the class lifts, I mean, so it is also not clear that what you are doing is well that your kind of conjecture is enough to consider it for algebraically closed field, because it's something missing in, I don't know. How about I take it k to be algebraically closed? No, but the question is, do you prove that your k to be algebraically closed field implies the general case, or did you see it? Well, if you construct a cycle at the level of the algebraic closure, it necessarily comes from some finite level. And a standard transfer argument then will push it down to whatever finite level you want. So it was enough to construct the cycle on the whole family x after base changing to the algebraic closure. No, but the point is, suppose you have a small s to find out the quadratic extension. So over the algebraic closure, you get two points. So suppose you can lift each one of them, but you don't mean that the lifting are the same. So you can check the lift. It does, because it's flat. It's determined by its value at any one of the points. No, but one of your conditions is that the homology class extends just to h to i or something like that. Then the non-flat condition too. And with this condition, it is not clear that if this condition is verified, then suppose you know the conjecture of the algebraic closure, then it is not clear that the cycle you construct is the sense to, it's clear that it is over the quadratic extension, but it's not the same. But then I can simply transfer it down to the original extension. If I can sort of cycle over some quadratic extension, see I can sort of line bundle over the quadratic extension, I can then just take its norm down. Because it doesn't satisfy the condition that you want on the quadratic extension. It satisfies only at one of the conjugate points, but possibly not at the other. It does, because I assume the existence, I assume the flatness. I assume the existence of a flat class, which hits it at the special point. No, in the conjecture that you said the following are equivalent, yes? You wrote there the same question as before. So the content of the conjecture, this is the crystalline analog. For simplicity, we can look at the original one. So the question is whether it is reasonable to put this conjecture when there is this change of field, the closed point is not rational. And so I'm saying that it doesn't trivially follow from knowing the conjecture over the practically closed field that condition 2 is. No, I agree. I agree. In order to reduce the algebraic closed case, you need to work with condition 3. No, but to prove that the same problem will be improving the 23R equivalent. Because the way you would do it is you use, again, you use delinso axiope to work on the algebraic closure. And there you know that you can lift one of them to a flat. Again, you will have the same problem in descending from the quadratic extension. It will not be a. Can you assume to date? Yeah, we can assume it's algebraically closed then. And we can discuss afterwards. But once you've got the equivalence of the flatness condition with the others, I believe this problem can be avoided. It could be a problem in your ethnicity. So let me then take it to be algebraically closed if that will settle the issues for today. I can't even see what the assumption is now. He's happy to have questions, I think. So I make some immediate remarks. So firstly, we can certainly see that the deformability condition, the crystalline condition, and the flatness condition are exact analogs of what we see in the variational Hodge conjecture. Secondly, regarding some obvious implications, deformability implies the crystalline phi condition, which in turn implies the crystalline condition, which in turn implies the flatness condition. And finally, so this rigid cohomology comment I've been alluding to, we could consider some obvious analogs of the conditions in which we replace crystalline cohomology everywhere by rigid cohomology and the convergent isochrystall by an overconvergent isochrystall. But it turns out that these conditions are completely redundant in that you get an equivalent conjecture. So if you can prove the conjecture for crystalline cohomology, you automatically get it for rigid cohomology by virtue of the fact that the crystalline cycle class map factors through rigid cohomology. And conversely, if you can prove it for rigid cohomology, then because rigid cohomology maps the crystalline cohomology, you get the conjecture for crystalline cohomology. But it turns out that working directly with rigid cohomology is a lot less convenient than working directly with crystalline cohomology. So we follow the crystalline route. Although from some philosophical point of view, the conjecture might be more natural with rigid cohomology. Let me state the main results then. Maybe I'll try to stick to algebraically closed field k to avoid any further problems. So the first theorem, which is an analog of theorem A up there, states that the conditions Chris phi and flat are, in fact, equivalent. I'm assuming k is algebraically closed. And the second theorem, which provides an analog of left sets result for line bundles, states that this crystalline variational take conjecture is true for divisors. So in the flat condition, the group A0, Chris, is a finite dimension? It is finite dimensional, yes. It embeds into the crystalline cohomology the special fiber. And my claim is that because it's an embedding, these problems that you mentioned don't matter. OK. And so in the proof, you just work with crystalline with out rigid. That's correct. That's correct. I mean, an alternative approach. So by Kedlaya's fully faithfulness result, the global sections of the convergent isochrystal agree with the convergent sections of its over-convergent extension if the over-convergent extension exists, which is essentially a conjecture of Bertelot. And so then you can then play the whole game in rigid cohomology instead. So as an application, we can consider the following situation where x is smooth and projective over a finite field, which unfortunately is not algebraically closed. y is a smooth hyperplane section. And we can consider the question, let me take this to be dimension at least 3. We can consider the question of extending line bundles on the hyperplane section to the ambient variety. And the type of corollary that comes from a simple pencil argument, so I've replaced divisors by line bundles now, if I take this line bundle on the hyperplane section, then it extends to a line bundle on the ambient variety x. If and only if its first crystalline churn character extends to the crystalline cohomology of the ambient variety. And what one can do with that is show that if you want to prove the take conjecture for divisors, where I now even mean the allatic etal formulation of the take conjecture over some finite field, then it's enough to consider the case of surfaces. So roughly, how does this go? You want to prove it, let's say, for your three fold once you know it for surfaces. You pick some hyperplane section carefully. Specifically, that means that you want to ensure that the hyperplane section doesn't have too many more, doesn't have a much larger Frobenius eigenspace than the ambient variety. And then you simply apply the previous corollary to start constructing line bundles on the ambient variety. And I should mention explicitly that I recently learned this corollary does exist in some unpublished short manuscript of De Jong from a few years ago. So it does seem to be known to a handful of experts. So let me turn out away from the applications again and back to the results. So the key to establishing theorem two is going to be a local formulation of it, in which I definitely want K to be algebraically closed though it's not true. So I'll take X to be smooth and proper over the spectrum of some power series ring. Y to be the special fiber and L to be some line bundle on the special fiber. Then the result states that L extends to a line bundle on the whole family X if and only if its first crystalline churn class has the analogous extension property. Except as opposed to previously, I'm now going to be careful as to which eigenspace I work in. So I am going to insist that the lift of the churn class does line the expected eigenspace. That can be eliminated, but it turns out to be much more natural to work with the eigenspace condition. So again, it says that if you've got some line bundle on the special fiber, then the obvious necessary crystalline obstruction is in fact sufficient for extending the line bundle. For example, if you take X to be a smooth, proper family of K3 surfaces or abelian varieties over KT, or maybe I emphasize again that definitely K is now able to breakly closed, excuse me, smooth, proper, super singular family. To be precise, I just require H2 to be isoclinic of slope 1. Then it follows from theorem 3 that the co-kernel of the specialization map from the Navon-Severi group of the geometric generic fiber to the Navon-Severi group of the special fiber is killed by a power of p, which for smooth, proper families of K3 surfaces is an old result of my garden. But I feel that this somehow gives the conceptual explanation behind it. You have surjectivity of Picard groups because you have surjectivity in crystalline coloramology. I've got to use this fishing device now. So for the final 20 minutes, I'll discuss some ideas behind the proofs, maybe main ideas. I hope. I'd like to think there's more than one. So I guess all the main theorems are up. So theorem 1, which is the equivalence of these conditions, at least over an abelian closed field, is proved by modifying Delinier's original argument in characteristic 0. Delinier's original axiomatic argument is no restriction on the characteristic at all. And so that immediately gives some of the implications. And the hard part is incorporating the Frobenius to get a handle on the crystalline phi condition that imposes some restriction on the eigenspace. That's what doesn't appear anywhere in Delinier's original situation, to generation of Ray spectral sequences. Objective OK. Yeah. So then once you've got theorem 1, it's enough to prove theorem 3 to get theorem 2. So theorem 3 is the local formulation of theorem 1, theorem 2, excuse me. And if you can prove the result in the local formulation, then by some relatively standard type of art and approximation argument, you can prove the result in the global setting. So everything reduces to theorem 3. And so from now on, we put ourselves in that type of local situation. A will be power series in some number of variables over k, which again, from now on, is definitely algebraically closed for these results. X over spec A is smooth and proper. Y is the special fiber of this family. Y sub R is the infinitesimal thickening of the special fiber Y. Then theorem 3 is going to be deduced from a considerably stronger result, theorem 4, as follows. So we suppose we're going to deal with all co-dimensions at once. So I'll take some element in k0 of y. Everything's more torsion, so I can identify this with the sum of the child groups of y. Though it's probably better to think of z as being the class of some vector bundle on the special fiber y. And the theorem gives conditions under which the class of this vector bundle admits infinitesimal extensions of all orders. And we'll see that these infinitesimal extensions to all order exist if and only if the crystalline churn class of our class z, which sits in the crystalline co-demology of the special fiber, lifts churn, sorry, sorry, sorry. Keep doing this. Thank you. If and only if it's crystalline churn character lifts to the expected piece of the crystalline co-demology of x. So several comments I should make about the theorem. Firstly, it does imply theorem 3 since the Picard group is a direct sum earned of k0 on which the only non-vanishing part of the churn character is the first churn class. And this is key. Theorem 4 tells us about infinitesimal extensions of vector bundles. But if you have a line bundle which admits infinitesimal extensions of all orders, it's the same as a line bundle on the whole Schemax. So if you want to extend the line bundle up to x, it's enough to extend it infinitesimally to all orders. And that's how theorem 4 gives you theorem 3. Secondly, I'd like to stress that theorem 4 should be viewed as an infinitesimal version of this crystalline form of the variational Tate conjecture I'm discussing, which is valid in all co-dimensions as opposed to most of the results today which have only been for divisors or line bundles. And finally, it's an analog of the result of Bloch, Anno, and Kertz, for those of you who know this, in that 2012 inventionist paper in mixed characteristic. So they consider the situation of having some smooth proper x over zp. And they start with a k0 class on the special fiber. And they prove that it admits infinitesimal extensions to all orders if and only if it's crystalline Schoen character. And now the test is to see how the crystalline Schoen character looks in terms of the Hodge filtration on the special fiber. And they prove that if the necessary condition is satisfied, then indeed you have infinitesimal extensions to all orders. And this is a perfectly analogous statement but in equal characteristic p. And so in the end, everything reduces to theorem 4. And so I'd like to finish by giving a brief sketch of the proof of that theorem. So although the theorem is inherently only a statement about k0, the proof is based on recent results in higher algebraic k theory and topological cyclic homology. So I can't give them, I certainly can't give a definition of topological cyclic homology. That takes a two hour talk. But you should think of it, it's not completely wrong to think of topological cyclic homology as some form of derived crystalline cohomology or maybe derived log-Hodge-Witt cohomology would be better. But let me say derived crystalline cohomology. That already gives you some idea of what it's doing. So then the proof goes as follows. We need to understand the difference between vector bundles on y and vector bundles on the r-thin fantastical thickening of y. And the general machinery of topological cyclic homology gives you a comparison map called the tracemap. Let me omit the zeroes, because as I mentioned, it's really a statement in higher algebraic k theory. Gives you a comparison between the difference in k theory to the difference in topological cyclic homology, which as I've said, it's on derived form of crystalline cohomology. So we can almost believe that the result is going to work now. It tells us that the obstruction for k0 is some derived obstruction in crystalline cohomology. Because we have a result of McCarthy from 97 telling us that this tracemap is an isomorphism in essentially full generality. You take any nilpotent thickening, even of non-commutative rings, and the theory produces an isomorphism for you. So that's the very general statement that starts us off. Can you call that step one? I'll try to break. And the k theory question involves, is the positive or also negative or the negative is zero? Also negative. The negative of y is zero, but the negative yr might not be. OK. And the difference is in the sense of spectra? Yeah, yeah. So I mean the homotopy fibers are weekly equivalent. Is this the after-tensoring with q? This is on the nose. On the nose for nilpotent thickening in any characteristic? Let's say characteristic p. Or at least where p is in the nilpotent ideal. Otherwise, I would need to p-complete both sides. The statement will be true in full generality after p-completing both Tc and k theory. Taking the p-completion of the spectrum. No, but you don't have a p in general. So I don't need to do it here. I don't need to. I mean if you're in characteristic zero, it's a different. Then it's true ration. No, then it's just not true. Then you should replace the right-hand side by classical cyclicomology. In characteristic zero, it is classical. Cyclicomology. And then you can play the same game actually to prove deframational Hodge statements in characteristic zero. The comparison with cyclicomology is older. That goes back to Goodwillian in the early 90s. Maybe in late 80s. So a consequence of this more or less tautologically is that our vector bundle, class Z, admits infinitesimal extensions to all orders if and only if the same can be said of its value in topological cyclicomology. So it makes it compute some topological cyclicomology. And the key second result, which is a result of Geisler and Heselholt from 99, tells you what topological cyclicomology of a smooth variety looks like. Emphasis on smooth. So because Y is smooth, they calculate the topological cyclicomology as pieces of logarithmic Hodge-Wittcormology. So already at this stage, we see that the class of the vector bundle admits infinitesimal extensions of all orders if and only if some property is satisfied by something awfully crystalline-looking. Thank you. And so the main difficulty is then to describe the topological cyclicomology of all these nilpotent thickenings, which are manifestly non-smooth. So we compare it just by functoriality to the topological cyclicomology. I should answer this also by q. Of x itself. And the third key step, which is recent joint work of myself and Bjorn Dundas, is the formal functions theorem for topological cyclicomology, telling us that this is, in fact, an isomorphism. So in contrast to k0, where a limit of k0 classes on the thickenings does not give you a k0 class on x. In fact, in topological cyclicomology, you have a limit of Tc classes on all the thickenings of the special fiber, and it algebraizes to a Tc class of x itself. And set p is in catecp. All mixed characteristic, as long as you p complete. It's a very general statement that if you work over any, say any Zp algebra, which mod p has a finite p basis, then the result goes through. Any Zp? So the result works if you have a smooth proper scheme over the spectrum of a complete ring, complete netherian ring, a complete netherian Zp algebra, complete for some i-addict topology, any ideal i, with the property that the ring mod p is finitely generated over its p-th powers. Yeah, yeah, that's finite p basis. No, no. And nor are there any smoothness conditions necessary. So very general, very general formal functions there. And i is contains p. Wait, i doesn't contain. No, n name is complete for the zero i-addict topology, so. Well, then I get a trivial statement that Tcx equals Tcx. It's relative to i, so I take the thickenings of the special fiber over i relative to powers of i. It's the exact analog of Groten-Tix formal functions for coherent cohomology. Sequeling of, do you use also special fiber for p? No. No, so once you'd probably assume. Yes, we'd better assume p is in i. And here also, when you take limit of Tc0, yr, isn't it more natural to do homotopy? Yeah, so I'm being sloppy about limbs and homolims. In fact, in this case, I can kill the limb one term. But the natural statement is indeed that Tcx identifies with the homotopy limit of the Tcyr's. But in this particular case, I can kill the limb one term. Though that doesn't play any role in the proof. You can ignore the limb one term anyway, so it's not necessary to kill it. So what's so great about this is that we've now reduced the problem to understand topological cyclic homology of x, which might not be a smooth variety, but it's still regular. And that's enough to apply the Geyser-Hessel result again. So we apply step two again, which describes Tc of regular things to identify this with the log-hodge-wit co-homology of the whole family x. It's bad, but it turns out to work. Thank you, I keep forgetting the Q. I mean, so many of the snaps hold without putting Q, but. Of course, the usual references for there are V, it is not in this generality. So fortunately, either one looks at Professor Ilouzi's paper and checks that many of the results hold in greater generality using some now on the Popescu trick. Or you look at a paper by Atsushi Shihou, who has already checked many of these results. Yeah, yeah. No, I highly recommend that. He has a paper in which he exactly considers hodge-wit co-homology of arbitrary regular F.P. schemes and checks what are the minimum necessary conditions so that everything works. So it's not so bad after all. So finally then, step four, we want to identify this logarithmic hodge-wit co-homology with the expected eigenspaces in crystalline co-homology, which is, in this case, a completely classical comparison, since y is a smooth popper variety. But it turns out that the analogous comparison map for x itself is not quite an isomorphism, but you can still show that it's surjective using a very similar arguments to the smooth popper case. And so the final conclusion is that your vector bundle class on the special fiber admits infinitesimal extensions to all orders if and only if its value in the crystalline co-homology of y lifts to the crystalline co-homology of x with the restriction, which you can eliminate by, if you're projective, you can eliminate one of these left-shed arguments. But somehow, for the statement of the theorem, it's most natural to impose the eigenspaces restriction. Yeah. So that completes the proof of the sketch and of the talk. Thank you very much. Is there any questions? Yes. I have a question here on one. All right. Remember, the new key spot here and the proof is the weight of the number. So when you have your egg, which is on parter, then it's going to come back to take an egg bar. And the image of the HN of the egg bar into the HN of the egg in the bottom part is the weight inflation. So do you use some other option? Delinio uses this to extend the degeneration of the Lorentz spectral sequence from the projective case to the proper case. And I deliberately imposed the assumption that my family was projective to avoid that. But actually, I think that's a very interesting question. And I think it, I've given it some thought, and it seems to reveal some unknown aspect, either of crystalline or rigid coromology. We can consider precisely the situation of having some closed embedding of smooth, proper varieties. And we can suppose that I have some open, which contains the image of y. And then what Delinio does is he compares the image of the coromology of u in y with the image of the coromology of x in y. And he shows that the same. And it would be great to have an analog in crystalline or rigid coromology, but it doesn't seem to exist at present. And definitely, if I'm eliminating these projective assumptions or for extending the results, I think that's a necessary step. So I've talked to some experts in piatic coromology about this, but I get some general feeling that the theory of piatic coromology is still not at an optimal state. So here's the second question. You said that Tc is how it's close to that dirac crystalline coromology. Yes. You have the unheat coromology, so could you make the statement that you have to be precise? So I'd make the statement firstly that, as you see, it coincides on smooth things with logarithmic Gram-wit. It's very hard to say anything precise. But I think Bargab would agree with me. It's very interesting that this was coming. Yeah, I mean, I know he's thinking about this. I know he's thinking about the relationship of his theory with groups that you also see on the topological cyclochomology side. Well, let's discuss afterwards. Let's discuss afterwards. Yeah, yeah. I'm going to continue after the line. Thank you.