Thank you so much for inviting me. It's a real joy to come here for such a wonderful occasion. Also, since I'm the last one, probably should thank the organizers for their wonderful job. And it was absolutely great to be here. Well, anyway. So let me, I will talk about a typology situation, but let me remind first the theory that was developed by Kashiwara and Shepera already 30 years ago and their book, Shifts on Manifolds. So here is the situation. Let X be complex manifold. Of dimension n. And suppose that I have a constructible sheaf, or a sheaf for me will be mean complex of sheafs, all this, well, with some coefficients. And then here is the definition. The singular support of f, it is a closed subset in the Katanjian bundle to x. And it is defined as follows. It is the smallest subset, which decides the following condition. If we have any pair u and f, where u is an open subset in x and f is a holomorphic function on u, such that if I consider the differential of f and consider its graph, then it does not intersect my closed subset f equals. So if I have such a pair which satisfies this condition, then f is locally a cyclic relative to f. Locally a cyclic means that if, well, there is, well, I assume that the definition is known, but this means that if I will compute vanishing cycle, cycle there will be no, there will be with coefficients in f, there will be trivial. So that's the definition. And it's more or less, so this is that, well, it's easy to see that this is a conical complex subvariety in t star x. And moreover, it's easy to check that the following two completely dielectric properties are that the singular support is empty. This is the same as f equals to 0. And singular support equals x. And when I write x as a subset of Cartesian bundle, this means that it's just a 0 section. Then this means that f is a locally constant and non-zero. Now, a theorem, a theorem they prove is the following that it is that, well, it's complex subvariety. We can look at its irreducible components. And the claim is that all irreducible components have a dimension n. So it's evident in these two examples, but it's true in general. And in fact, they show that the singular support is Lagrangian. Well, it's subvariety, it can have singularities, but on its open part, it is Lagrangian. And since it's also a conical subset, this means that actually it is every irreducible component of singular support is the conormal bundle to some closed subvariety in x. By conormal bundle, I mean that on smooth part you take the usual conormal bundle, and then you take the closure. Well, now, so that is their first main theorem in this situation. And here is another theorem that let me write it maybe on the second board. It is now supposed that my coefficient, well, is a field. Well, then the claim is that, well, you have a collection of subvarieties of middle dimensions, the Cartesian bundle, and the claim is that one can naturally assign to every irreducible component a number in integer, which will make it a cycle, which is called characteristic cycle, denoted in this manner. So that the following properties will hold. So the first property is that suppose that we have a pair u, f, as before. So we have an open subset and a homomorphic function on it. But before we considered situation when differential of f did not take value in the singular support. But now suppose that it intersects the singular support at a single point. Well, then we know that by the definition of singular support, it's true that our function f outside of x is locally acyclic, so there is no vanishing cycles. And so this means that if I will compute the vanishing cycles for f, then it will be skyscraper at x. And so I can compute its dimension. Well, dimension earlier it's complex, so it will be earlier characteristic, but let me write dimension. Well, and the claim is that the following formula for this dimension holds. I should put here the sine minus. And then here will be the local intersection index. We have df, which is a section, so we have df of u. It's a sub-variety in the Cartesian bundle. And here, so we consider it as a cycle, certainly with multiplicity 1. And here we put the characteristic cycle. And so we have two cycles which intersect by one point, so there is the notion of local intersection index. And that point and the claim that it will be exactly the dimension of vanishing cycles. You don't claim that the multiplicities are, they could be positive, negative, or zero. You don't claim anything. I don't claim anything. Well, I will claim anything, but in a short while. It's equality. It's equality. Well, so the second assertion is global. So assume that x is compact. Then let us compute, let's consider the earlier characteristic of x with k-efficient in f. So this means that it's, we compute the homology of x and compute the earlier characteristic. And the formula says the following that it's just the intersection index of x and the characteristic cycle. So here is the situation. The Cartesian bundle itself, it's certainly non-compact. But since x is, you have zero section and x itself is compact, then certainly you can intersect x with cycle of complementary dimensions to this well-defined integer in the claim that you have this equality. Well, now the last property is the answer to Arthur's question. And let me put it here, that if f is perverse, if f is perverse shift, then characteristic cycle is effective. Find it for a usual shift. So the same works for a bounded complex? Well, yes. What do you mean the same holds for both? You consider the condition that the diversity of locally acyclic in the sense of the vanishing cycle Yes, in the sense of vanishing cycle. OK, but you can do it for either f to be a single shift or an object in the derived category. For me, shift means object in the derived category, OK? But you don't claim that what you get is OK. Singular support is just subset, no coefficients. Characteristic cycle has coefficients. And you can have singular support huge, but characteristic cycle being zero. For example, take any f shifted by 1 and take direct sum. Singular support will not change, but vanishing cycle will disappear. Sorry, characteristic cycle will disappear. You want to say that this cycle is uniquely determined by f? Yes, yes, uniquely determined by f. So that I will say in a moment, OK? So this first condition actually completely determines characteristic cycle. Well, it's clear that if you have any components, then you can choose, you can find function which is differential in per-sex as the thing at its smooth part. And then this formula will produce you, immediately produce you the multiplicities. And not only effective, but with strictly positive? Yes, that I did not. Yes, and moreover, it's strictly positive. So the support of Cc equals the singular support. So one comment is that certainly characteristic cycle depends, well, it depends on the shift in additive manner. So it's a homomorphism from the k-zero group of constructible shifts to the group of cycles. On the Cartesian bundle, this follows immediately from this formula 1. Well, the second thing is that let's consider an example when my shift is constant. Then, as I told you, the singular support, it is a zero section. And by the last property, you see that, well, equals x. And maybe let me write that characteristic cycle equals minus 1 power n times x. Well, and then let us see. So this first formula, then, it is exactly Milner's formula. So what stands to the right? It is exactly the Milner's number of the singularity of f. So this x means that it's critical point. So if it's intersex zero, and here stands the Milner's number, and here stands the dimensional vanishing cycle. And minus has to do with it because I can normalize it to be positive for perversive and not usual one that you can forget. And similarly, the second thing is the formula of the Euler characteristic. Well, here it tells you that it's self-intersection of the zero section of the Cartesian bundle. And again, because of this sine thing, this is equivalent to the standard formula that's Euler characteristic of many of all the self intersection number of the diagram. So at least in this situation, everything is perfectly fine. So one last remark about Kashiwara-Shipra is that their proofs are very transcendental. For example, the theory they develop, it works on real analytic manifolds. And if you consider real analytic manifolds, then you can refine any real analytic stratification to just decomposition by simplices. And in such situations, constructible shifts are easy, and so you can work it. But it's extremely transcendental operation, and you cannot just push it to a tight situation. Just one question about formula two. The right hand side is the hard collage. You're right, that's the wrong collage on the right. Well, if you can't x the intersection number of x with a zero cycle with itself in the cotangent bundle, that gives you the hard ones, right? Well, it gives you one number. It's a total hard one. Pardon? No, it's just usually a characteristic. What's the problem with demonule? Yes, before, I said the first time I heard about this was Walensky. Yes, yes, yes. I should certainly have told that their story was completely motivated. It came not from topology at all, and not from complex geometry, but from the theory of demodels. And in theory of demodels, the notion of singular support of demodels, that's one of the very first notion and basic technical notion, how you just develop the theory of demodels from the very beginning. And the notion, well, there is standard. The basic fact is that singular support, you can define for any demodel, but it will have dimension larger than n. And those demodels which have, for which singular support, has dimension, and it's the basic geometric objects that are polynomial demodels. And the formula multiplicities, again, are also defined just from the very definition, like multiplicities and commutative algebra. And the formula, this globally characteristic formula in the RAM setting, so for demodels, it is, I think it is due to Brilinsky-Dupson and Kashiwara if I'm not mistaken. But I should stress that in this definition, its proof is extremely simple. It is one line theorem compared to quite complicated proof that you do for constructible shifts. Both definitions actually prove us. But again, it's one line proof, but it's extremely non-motivic. It's just the structure that you deal with demodels and not with geometry. And they also prove that when the coefficients are characteristic 0, then it corresponds to the demodal characteristic cycle. Yes, they prove that it corresponds to the demodal characteristic cycle. Sorry, I should not have omitted the history, but anyway, first on, I will omit also other parts. So, well, I'm not so much of time. Well, now the basic question that the story I will be talking about is the question if you can just if similar assertion holds in a pulse situation. And if you look at this definition, everything, well, suppose that we work now with an algebraic variety over some field, then you can see that everything in principle makes sense, so the definition there makes sense, and you can also state myself the theorem make perfect sense. And in fact, so what I will be talking about is, yes, it's OK, but with minor modifications. So let me do the minor modifications first, and then I will discuss the story. So here is the minor modification. So let's consider, first, the case of constant sheath. And then certainly then the Milner's formula, well, it's known that it's true, but you should, in case of it's true, in case when the base field has zero characteristic, but if it has finite characteristic, it should be slightly modified. And this, namely, here instead of dimension, there should stand total dimension. And the word total means that, well, it's a vector space in which the Galois group of the disk down below, for punctured disk down below acts. And in case of finite characteristic, you can have a wider wealth ramification. And total means that you add to dimension the term, this one conductor. And the corresponding formula for constant sheath, it was approved by Dalin in the second volume of SJ7. There is his talk, which is exactly called Milner's formula. So, well, so that's one modification that should be made. And another modification I should make on the left board. And this is, OK, so this assertion is false. And as for the rest assertions, yes, they're all true. And so what is done on the left blackboard, well, there is my note on archive that proves the theorem. And what is on the right blackboard is proved in a preprint of Takya Shasaita, which is probably also on the archive, or it will be on archive in the nearest future. OK. Now, maybe before I will continue the talk, let me comment about this situation. Why the tank is not Lagrangian. And the fact that you really just the Lagrangian property must disappear in characteristic B, I think it was, well, at least I first heard that from Dalin long ago back in Moscow. But somehow at that moment you think that there is no theory. And stop thinking about this. Well, anyway, so let me produce an example. Well, maybe first I need notation since it's convenient and I will also use it afterwards. So suppose that you have a map between smooth algebraic varieties, which is proper. And suppose that I have a conic subvariety inside of the Cartesian bundle to y. And then it yields in a pretty standard way a conical subvariety in the Cartesian bundle to x, which I will denote in this manner. And by definition, it consists of all points in the Cartesian bundle. So what will be x and covector nu at x, such that there exists y with the properties that it leaves in the fiber, and such that df at point y applied to nu will lie in C. There is a standard way to push forward the conical subset by the proper map. And a small remark that it follows directly from the definition that if I have a shift g on y, well, let me denote by d of y the category of. Pardon? Yes, again. Oh, we are. Sorry, sorry, sorry. It's R. Thank you. Thank you so much. But if I have a shift on y, then if I consider its direct image, this will be a shift on x. And if I compute its singular support, then it lies inside of the image in that sense of the singular support of g. So this thing gives you an upper estimate for the singular support. One small remark that this upper estimate can be clever in the sense that if R is closed in bedding, then you have a quality.  For example, if R is Frobenius' map, then this produces you the whole Cartesian bundle. And so it just tells you nothing. In the case of characteristic 0, it tells you always tells you something. But in characteristic, we know. Well, now, for example, let's consider a map. Just R, y and x will be for us just the coordinate planes. And I want that the map actually depends only that the second coordinate would not change. So it will be given by a formula. Well, let's denote the coordinates here as t and y. And here is x and y. And this will be, say, g of dy. So let's consider such a transformation. And my shift will be just a direct image of the constant shift on the A2. Just consider just the shift here. Now, in such a situation, you've produced all possible things that cannot happen in characteristic 0. So let's consider the stupidest example where g of dy equals t power p plus dy square. Just absolutely is this example. Now, what do you see? So first, you see that if y, now what holds? So first, what properties? So first, certainly R is finite. Second property, if y is not 0, so outside of y equals 0, the thing is a tile. Now let's look at what happens over the axis y equals to 0. Yeah, you're assuming p is now equal to? p is corrected. I don't care. I don't care. You mean for a tile thing? Well, you compute the differential. And the differential will be this you can forget about. And here will be y squared times dt plus something. And there will be dy. And so it's invertible. OK, but now what happens on the axis? On the axis, it is very strange thing happens. You have dr. You have the map dr. And this map does the following. If I consider the vector along the axis dt, then it sends a tangent vector. dt is sent to 0. You differentiate. And then dy will be sent to dy. So it is a very strange map on this axis. Again, it's differential along the axis itself equals to 0. But in the normal direction, it is a density. Now, if you apply this estimate, certainly f itself, it's not local. It's not local system. It's not smooth error certification, and so on. So therefore, it's singular support. We'll look as follows. So there will be 0 section. Well, because it's local system outside of the open thing. And plus something. And there must be something else because it's not smooth. And there's something else come exactly if we apply this estimate. And if you apply this estimate and look at this formula, you immediately see that it equals x, so the 0 section, times c, where c is a cone over the axis y equals to 0, generated by dt, dx. And so you see that it's absolutely not Lagrangian, OK? So that's all for this example. Well, now maybe I should let me formulate a little extension of this theorem, which is due to the link. And that's a assumption that if we consider many faults of dimension 2 surfaces, then absolutely any conical subset of dimension 2 in the Cartesian bundle can be realized as singular support of some constructible sheath. It's a reducible component of singular support of some constructible sheath. So it's a theorem of non-integrability of characteristics. Well. Now, I want to discuss proofs. So proofs here are not difficult at all. But I want to show a part of the story. And this is a part of the story that explains how you see what singular support is. Certainly, singular support, for example, it's an interesting invariant. Well, you have some conical things, but you don't know how to see them basically because, well, testing by functions, it's not a pleasant thing to do. But now I will describe how one can actually see singular support. And for example, to see that it has right dimension. Well, so certainly, singular support has local origin. So I can assume that, and also, as I told you, that if you embed something by closed embedding, then it transforms in an evident manner. So I can assume that I live on the projective space. And I will use two in order to show how the thing looks like. I will use two tools. The one is Brinsky Radon Transform, and the second is Veronesi Embedding. So let me just recall momentarily what the Radon Transform is. So we have my P. We have the dual projective space. And we have the standard correspondence Q. Well, and the Radon Transform, it is, so maybe before Radon Transform, then you see that in this standard diagram, there is a canonical identification of the projectivizations of the Cartesian bundles to P and to P check. Namely, both of them canonically identified with Q. That's a very classical thing and essentially evident, because what is a point in, say, in projectivization of the Cartesian bundle to P? Well, it is a point in P and a hyperplane in the tangent space to this point. And certainly, since we live on projective space, the hyperplane in the tangent space extends uniquely to a hyperplane in the whole space, passing to point. And so we get a point, projective space, and the hyperplane passing through it. And that's an element of Q. That is this identification. Well, same manner here. Well, this thing is called, by the way, Legendre Transform. And now you have the Radon Transform. And the function R from the category of shifts on P to those in P check, which is given by this correspondence. Well, it has some wonderful, easy standard properties, but I will not discuss them. Well, they used in the proof, but I will not discuss them now. But one thing that is very easy to check is that this identification, in some sense, it's classical approximation to the Radon Transform, which means the following, that if I have a shift F here, then I can compute, let's take a singular support. That's a cone here. So I can consider the corresponding projectivized. It's projectivization, which will be sub-variety here. And on the other hand, I can do the same thing for the Radon Transform of F. And the claim is that they're the same. That's a simple fact, but I do not have time to describe it. Instead, let me pass to the story that I want to have. Now, certainly just playing with Radon Transform help you nothing. Well, if you don't know what singular support of a shift is, then you don't know what just Radon Transform does not help immediately. But it helps after the very nice embedding. So let's do the following. Let's consider an embedding of my projective space. Let's call it now a small projective space, a very nice embedding of some degree more than one, any degree. And then I will do Radon Transform on this larger projective space. So I have P. Let's embed it. So this will be very nice embedding. And then I consider the larger projective space and the Radon Transform on this larger space. Well, now notation. For suppose that I have a cone C inside of T star P, well, then I can consider its image by I extended to a cone here. Well, and then I would like to take its projectivization and just notation will be that I will denote it by C in square braided. And this thing leaves here, so here and here. OK. Well, that's the first notation. And second, so we have F now, which leaves here. And what I want to do is to apply all these functions. So I consider the Radon Transform of I star F. So this is a sheave here. And let's look at its ramification divisor. And ramification divisor means the following that I just restricted to the generic point. So there I have local system. And then local system extends as a local system while whatever it can extend and where it can at its ramification divisor. Well, so I consider it. So to define it, I need to know the thing at the generic point of my projective space. Good. Now the theorem. It's when you see divisor you mean multiplicity. Oh, no, no, no, no, just plain subset. So it's subset of dimension one with no multiplicities. F is any sheave on p, on small p. No, but do you take the ramification divisor of the Radon Transform? Yes. So I is not derived, are you OK? I take I law star F, then I take R being Radon and not write derived. I'm sorry. There is no right derived function in this notation. F is a complex of sheaves. Well, now the theorem. So in formula, it is the way how you reconstruct from D, which is somehow visible in variant of F, that you can reconstruct the singular support of F. And this in particular tells you that it has right dimension. OK, so the first assumption is that D itself can be recovered from this thing. Namely, it's just the image of C in square brackets. Oh, sorry, sorry, sorry. So now from now on, let's put C equal the singular support of F. Then D equals image of F. Excuse me. I think this slide is the third slide from the bottom. That is the problem. So there are some symbols. So it's I not C, C, P star, P tilde. No, it's continuous. Subset of T star. You can see it. Subset sign. Subset sign. And then that is P upper star. No, the next. The next is projectivization. It's projectivization of the cone. So I have C which leaves over small p. The next standard and the standard way of the cone are now projectivized. Maybe there are two tilde there on the right. Oh, yes. Absolutely. Thank you. Well, here these notations were without tilde, but okay. Now, so that is the first thing. Second, the second assertion is, moreover, so my D has different reducible components and C has different reducible components. The same method. In this manner, they correspond one to another. Namely that for every reducible component, D alpha of D, there is a unique reducible component, C alpha of C, such that D alpha equals... So basically, this is another thing and then you do Legendre transform and somehow spreads components that could... Here, they could project to something the same position in P, but when you do the thing, they will project to absolutely different... Yes, that's the effect of veronizing bedding. That's exactly... Pardon? Any of the three more than one. Identity is not allowed. And of course, you shouldn't allow the zero-dimensional objective space. Then I already pointed out that this... Then you don't tell... Probably... You don't tell me much, but... Zero... Yes. Okay. Number three is that actually, this condition two, it uniquely defines C alpha. So that C alpha... In fact, C alpha is a unique column of... in T star P of dimension... of dimension N with that property, with property two. Okay, and maybe I should say also... Well... Maybe I should say also property four, that the map, this projection from C alpha to D alpha is generically radical. You are in the small project space on the big one. Pardon? C alpha is in the small one? No, no, no, no, no. I'm sorry, so here should be bracket. I'm sorry. And the... No, here is... Thank you. Okay. So the thing is generically radical. In a... Classical situation... Sorry, in characteristics hero situation, in fact, the map is birational, well, certainly is birational, and also this thing is... Well, it sits in the Cartesian bundle, projectivization of the Cartesian bundle to P tilde, yes, and D is divisor there, and this is projectivization of the conormal to the alpha. But in case of characteristic P, in case of finite characteristics, this absolutely does not need to be... So, since it's needn't be true, so since it's needn't be Lagrangian, the singular support, well, but somehow you can recover it from the alpha. What about multiplicities? What about multiplicities? About multiplicities a little bit later. I... Okay, so I will... Well, maybe the question is that I would very much want to know how to recover the thing as geometrically as possible. Even in case of projection of surfaces, finite projection of one surface to another, direct image of a constant sheath, so here we were lucky that we could recover it by this stupid situation, but even just iteration of those two things of degree P, it will lend you to a situation that you just cannot recover it from geometry. I cannot, but probably some deeper geometry. Veronaise, just to avoid linear something, the things which are linear, which will break the... No... If your sheath at the beginning has nothing, no locus of ramification, which is linear... No, no, no, no, no. You see, if you have identity map, then the thing is just indices by direction between quadratic cons. So if you have absolutely any quadratic cons, it need not have... Its image need not be divisor, it need not be radiation over its image, and so on. You can just take anything and then go back and produce the corresponding sheath. So Veronaise does something very drastical. Okay, now let's pass to characteristic cycle. Well, at Takeshi's work, it is... Well, it's subtle and it uses many other inputs. So this story is pretty rough and elementary. And I cannot discuss it just because of the absence of time. But instead of it, I will try... Well, there is something to be done there yet. It's not all the story. And one thing that comes in Takeshi's story is that characteristic cycle in what he can do, it has not integral coefficients, but there are... Could be powers of pin, the denominator. And that's for the reason that components of C can be purely inseparable over its image. And the multiplicity in the intersection in Milner's formula will have unavoidable powers of pin. And so that's one thing that one would like very much to do. Now, what I would like to say is sort of a... Well, maybe hoped for formulas that would explain the story. Also, I hope it will provide understanding of things like you can consider for global intersection formula for Zeolier characteristic. You can ask for finer things. For example, to compute the determinant of cohomology. And I would like to have just the story simultaneously and to have definition of characteristic... Some finer definition of characteristic cycle, which would answer also the second question. And that would not involve in itself this... That would be this proof that it does not depend on the choice of functions in Milner formula and so on, but somehow Milner formula will be just corollary. So let me try to put this in the remaining minutes. Let me try to put it on the blackboard. So first, there is an ocean of... The moment you have the notion of singular support, you have the notion of micro-local shifts. Well, usual shifts, they leave on our space and the category of shifts, the triangulated category, they form a shift of triangulated categories on my space. I can consider for every open and the corresponding category. And this will be... So we have D, shift, triangulated categories on X. Well, the moment we have the notions of singular support, we can do the following thing. I can consider the Katanjan bundle. Can you write that again? I can't read what you wrote. We have our manifold X and D is... Well, it's a shift of triangulated categories on X. So I signed to open set, the category of shifts, the triangulated category of shifts on it. But it is not a shift... Well, it's... In modern parlance, the triangulated category means whatever infinity index you will put there. Okay. Now, the moment I have the notion of singular support, you can micro-localize D over the Katanjan bundle. Well, so consider the Katanjan bundle, but I will consider it not with planes, but only those open subsets, which are conical. Well, how do you define it? Well, so if you have an open conical subset in T star X, then we can put D mu of u. This will be the quotient of D of X, monologue the thick subcategory of those shifts, whose singular support lies in the complement of u. So that is... Well, this is a pre-shift of triangulated categories. It has naturalty structure, which has to do with perverse structure here, and you can ask about things like co-dimension, reconjecture, but that I don't want to discuss, but just let's consider this data. Now, what I want to consider, what sort of a question I want to try to ask is the following. So suppose that my X, I will assume that X is compact from now on, and then we have considered the functor Ergam from D of X to just lambda modules. So that's the point K be algebraically closed. And then I can pass to the corresponding map between K theory spectra. And what I want to do to know is to find this map, this homotopy map of spectra. No, K is K theory, it's a Quilin's K spectrum. So in particular, if I have a sheath, then I have actual sheath, then it defines your point here. And so if I know the thing, I will know the homotopy its image, it will be a homotopy point here.  It defines your characteristic if you pass to connected components. If you look at this element in Poincaré group poet, it will define you that are gamma and all the things. So that's actually what we want to have. Well, now let me try to put on the blackboard what I want to... Oh, yes, I will, I will. So I want to... Basically, I want to have a localization of... So we have this map. And I want to localize it twice. I want to localize it with respect to X. And then I want to microlocalize it to the Cartesian bundle. And the claim is that it's all that is needed for the theory of characteristic cycle. So let's see. So we have point X and the Cartesian bundle X. And let me denote this by pi and this will be P. Well, and here on this zero level, we have this story, this map that I want to understand. What does it mean to localize the thing to X? Well, as I told you, D itself, it forms a shift of categories, of triangulated categories of on X. And then I can apply to it K. And it's easy to see that I will get a shift of spectra. Let's call it K of D. That's shift of spectra over X. Okay. Now, what I want to do is to find first a map. Now it's a map of shift of spectra to the following things. So this is, again, this is a spectrum over the point. And I want to consider its upper shriek pullback to X. So I will say in a moment what it means. Well, such things, well, let's consider usual spectra as part of a native spectra of others. Is that kind of sense of topology? Yes, yes, yes, yes. Spectra for me is always in the sense of topology. A native spectra in the sense of a one of homotopy theory. So the thing is, the thing is the shift of native spectra over X. And it looks as follows. So if instead of K of lambda, this will be Z. And for example, if lambda is a field, you have map to Z. Then upper shriek pullback looks as follows. So we should take Tate motif. And then we should shift it by to N. And then put it to X. So that will be situation in case of Z. So this will be the thing. Well, now, so what I want to have is to get this basically, basically it should come by a junction. So what sits here is essentially the map from direct image. So X is compact from direct image of this. This is a part of direct image. And you have a map to K of lambda. So I want to have this to get such a map by junction. By the way, it is not... Well, such thing exists in classical usual topology, but in algebraic geometry, it's much more interesting. Maybe I will give... Do I have two minutes? Yes. Okay. Okay. So we have this picture. So here I will put small question mark. And here there will be larger question mark. Okay. The larger question mark is this. So let's consider this arrow. So I assume that it exists. And now consider it's just plain pullback by P. So here we'll get... Okay. Now, well, so this shift... Well, this shift of spectra, it has natural map. I recall that there was this D mu. And there is the corresponding K shift. Well, there is a map from pullback of D to D mu. And that's a map on the corresponding K spectra. And now... We are on T star X now. Yes. We are living now here. So here we live over the point. Here we live over X. And here we live over T star X. Well, maybe let me put question mark on the right. So small. And here there will be larger. Okay. And the larger is that there is an absolutely canonical map here. Well, now... What I know is that if you live in classical... In the situation of Kashiwara and Shippara and work with classical topology instead of motifs, then such a construction exists. Well, now... I believe that if... Well, that it should come if you... If one understands how this story with singular support actually related to the story with vanishing cycles over multi-dimensional basis, then this map should come by itself. Just for the reason that in Kashiwara-Shippara situation, it's essentially playing with vanishing cycles over multi-dimensional basis. Well, but some version of it for usual topology. Well, now I would... Let me just... So the claim is that whenever you have such a formula, then you have all things that you wish to have. So for example, you have... So let me just say why vanishing cycles come as a rough, rough, rough... Well, how it comes from this picture. And it comes like this. Characteristic cycle. Characteristic cycle, sorry. So we have a shift. So we have a point here. So we have a section here. And then it comes from a section of the story. And what does it mean to have a section of the thing? So if we have a shift that... So the thing and this element, if I restrict it to the complement to the singular support of my shift, it's trivialized. The section just vanishes as an element of the quotient category. And so given such an arrow, it produces you for every shift. So consider the corresponding section here and it produces your trivialization of the section. When you pull it back to the Cartesian bundle and then restrict to the complement of the singular support. Now let's project the story from K theory to Z just by the other characteristic map. Then here we'll have Z of n and then we'll pull it back here. And then you know that sections of such things are the same as cycles of co-dimension. And actually the Chao group. But if we consider the section supported on some sub-variety, this will be the Chao group of n cycles on this sub-variety. And if the sub-variety, our singular support, has dimension n, this means exactly that we know multiplicity is at the generic point. So in this manner, this thing immediately, if you replace K theory by Z, it produces you in particular the cycle. And believe me, it also produces you all formulas, the global earlier characteristic formula, and so on. So again, that is my hope, but I need to stop now. You said it's very hard to recognize the singular support. How many I give you a closed irreducible sub-variety? Can you tell me what's the singular support and what's the character of the cycle? I need to have a sheet. I'll take the constant, the trivial constant sheet. Well, if it's... Well, nobody... If your sub-variety is smooth, then it is a conormal bundle. Nothing that... If it's not smooth, then nobody knows. It depends on the singularity. What about the singularity? No, I cannot. I cannot and it's... Well, if it's not complete intersection, then even you don't know if it would be perversive or not. So... Hello, Alina. So, Alina, would you briefly explain this construction of the characteristic cycle? Would you like to check your formula for it? What do you mean? You see, are you clear? Yes. From this triangle? From this triangle? Okay. You can check your formula for it. Okay, so if I have... If I have a sheaf on X, so I have a point here, so I will have a section here and just consider its image, its image here. Then since it is the same as if I pass first here to here, this will be a section of the sheaf of spectrum which is trivialized on the complement of the singular support. Just because the image of the section here is trivialized on the complement of the singular support because microlocal sheaf just vanishes outside. Okay? And so you will have a section of this fellow which is equipped... So you replace this by Z, yes? Equipped with trivialization on the complement. This means that you have a Camology class with coefficient in this thing with support in the singular support. And this is absolutely the same as giving multiplicities. So you do actually need the trivialization on the complement to get that? Yes. No, no, to get... Otherwise it will be an element of the Chao group, of the Chao group of the cycles on. And if you consider a Camology of the thing supported on subvariety, it is the same as the Chao group of the subvariety. And since it has the same co-dimension, and that this is just a cycle of the generic points, nothing else. Another thing, I think, one thing is clear on this question that if you take F and DF, so you see the same. If I take AF and dual... Because phi is... Okay. Wonderful, yes. And maybe you can see it here also. Yes, and actually look somehow very... I think that I should say that there should be a relative picture for morphisms of varieties which is very much possible. And I would love to know at least how to spell it out. At least can check, actually. But again, the support for this is that you have absolutely canonical picture in the setting of real analytic varieties. Which is even nice in case of circle. That means that in some sense you are looking for some complex of shift in the cotangent space supported by the... By the singular support. Which reflects the vanishing cycle. It's not complex of shifts. It is a section of the spectrum. No, no, okay, but you are looking for a substitute. Yes, for a substitute, yes, exactly. But on the smooth part of the... No, on the whole thing. On the smooth part, it is just the local system that you can in some sense see with looking the vanishing cycle. Yes, K-class. It's K-class, very probably, yes. But with the micro-local thing, the Japanese team in Kashiwara, they could not produce something. Some module micro-local, they don't know. Probably they produce, but I believe they don't understand it at singular points. And I would like that the theory will be as rough as you don't... Yes, yes, I wanted to leave everywhere since probably you need to know it everywhere to have actual picture on the level of whole K-theory. Probably. For earlier characteristics, you need to know it at the generic points only. But for subtler... Any more questions? So when you consider this... On the Cortangent, on T-Sparx mode GM, this was for the Zariski topology. So several related questions. One thing is that you use the word shift of triangular categories. So I imagine that this means that there is a patching result when you work in some higher context with you. Yes, it's an stable infinity. Then it's a shift, yes. And then in the analytic case, what do you do? You don't have... If you just do analytic topology, how... what do you... Same, absolutely same thing. But you don't have enough... you're not seeing the singular support in the... that is... Do you do it for complex analytic or real analytic? How do you do it for real analytic thing? And then you have enough things with... Then you have enough, yes. I mean, all of them sit in some conormals. And it's sort of a funny shift. Every section, it's supported on... Well, it's supported in codimension N. So... And what was the conjecture that you mentioned there when you spoke about this? This structure and some conjecture if it didn't stay? Ah, okay. So that's just two words. So as I told you that every object here, just if you start with a shift, then it's a section of D-mule. It's a microlocally. It's supported on a singular support, yes. And now... What... Now suppose that you play not with triangulated categories, but with perverse shift with the heart. So let's consider a perverse shift. Now... The following... So the thing, again, so it's supported in codimension N. Now, what you want to do is that the function of restriction outside of codimension N plus one of these categories will be faithful. Restriction to codimension... Yes, N plus two, it will be fully faithful. And then one dimension less, it's an equivalence of categories. So this means that if I have a microlocal perverse shift, then on the whole contingent bundle or on some domain, that it is the same as microlocal perverse shift on its open subset, obtained by removing as many points of codimension N plus three as you want. N plus three, I think, or N plus four, I don't... This is for the analytic topology? Yes, now this is a very old conjecture in the context of demodules with regular singularities. That was proved fairly recently, maybe three years ago by Kashiwara and Villanin. They have... In the archive, it's called something called Dimension Three Conjecture and so on. But amazingly, they cannot prove... They prove it actually for regular demodules or perverse shift, but with coefficients in field of characteristics zero. For perverse shift with Z-Matell coefficients, they don't know how to prove it. The proof is analytic. No, but what is the statement? Because you define this, what you call shift from primary category. So you don't take the quotient of dx by something, and since the singular support is also purely of dimension something, it does not make sense. I mean... Look, so what happened? So suppose that you have perverse shift, yes? But it's at the generic point of the singular support. There you have some data, some categories that it would be nice to describe. Now, your shift... If your shift is non-zero, then this data is non-zero, definitely, that you know. Now you want to reconstruct your shift from this data. What you should do, you should add to it some data, add to the dimension one more, some sort of a gluing data. If two components intersect something of dimension n minus one, then you should add there some sort of a gluing data.  But in order to reconstruct the shift just from this gluing data, you should have compatibility, which will sit in one co-dimension more, a generic point on one co-dimension more. And the moment you have it, you don't need to go any deeper. No, but once you define this d mu of u, a dx-modulo, the six-up category of singular supported the complement of u, when the complement of u has large co-dimension, more than the middle dimension. No, no. The thing is that you shiftify. You shiftify the story. So, when you define it using these equations, you get a pre-shift of categories there. Now you consider the true shift. And for this true shift, in principle, it could have sort of a gluing data, which lives deeper and deeper. Ah, you didn't say that this is... Ah, you have to shiftify. Yes. So, that was over precision. So, I'm saying so vague things.


Hi everyone, Joe for jasmizcasebreaks.com coming at you with the brand new 2020 2021 Panini immaculate basketball Five box full case pick your team number one The next one the next pick your team. I think only has four or five teams left So if you want to do that straight up, we can do it without a without a filler or we're gonna add a filler We'll take a look at that after the break. Now. I appreciate everybody who bought their team straight up. Thank you so much Thanks for spending of your new spending a bit of your New Year's Eve Friday with us We're usually not on on New Year's Eve, but there's some big new releases. So we had to So we started a little early and end a little early now before we put took out all the teams in the number block Mark ended up with last bot mojo with the Atlanta Hawks And here are the number block results If you're looking for the randomizer videos, they're in separate videos. There's one for this group right here. Here it is Oh See what we got Jason jasper's on Instagram live and he's been pulling some at jasper's breaks It's been pulling some nice stuff. So hopefully here on the group break side We'll find some nice stuff here as well. Good luck to everybody. There it is great stuff here Six cards per box on average Looks like we have a thin card in the middle click a redemption in the middle Uses blank card here to slow play some of the hits a little bit Well, I'd love to find a logo man out of here Shondi that's 40 out of 99 For the heat that is Jimmy Butler and that'll be for Miami zero And that'll be for Peter Miami zero We have Marvin Bagley the third standout memorabilia 43 out of 99 for Sacramento It'll be for Clayton and the Kings 24 out 25 Jalen Brown for the Celtics two-color patch That'll be for David B and Boston The redemption is patch autographs red buddy healed For the Kings and once again, that's Clayton with the Kings and back here signature moves George Gervin the Iceman 76 out of 99. That's pretty sweet. I like that Font that they're using there and that's a good-looking card for Danny and the Spurs And we got shadow box signatures Deandre Hunter 63 out of 99. That's Mark and the Hawks last spot mojo strikes again They say around these parts they say that 70% of the time last spot mojo hits a hundred percent of time Especially in group breaks. I don't know if that holds at I don't know if that stat holds as as strongly You know when you're talking about like fillers or stuff like that, but For the main breaks definitely seems like it Especially full case for it. All right next box Some of our biggest polls have been out of immaculate basketball Resale dollars wise so keep getting into it ladies and gentlemen 10 out of 99 coal Anthony Orlando magic David B. He's been playing some good basketball this season. Hopefully keep it up coal Kelly Obrey jr. Squatch is 76 out of 99 That's usually with the Warriors Lamella ball relic 30 out of 99. That is for sunny and the Charlotte Hornets Joachim Noah 3 out of 10 three-color patch and autograph big bold auto there from Joachim Noah That is part of the bulls random number block and that goes to 3 That goes to Chicago 3 Which will be for Jeremy 33 and 3 plus 10 is 13 Jersey number at all counts We got Niko Manion 40 out of 99 Initiation Inc For Eugene and the Golden State Warriors and over here We have a 21 out of 32 Carl Anthony towns three-color patch and autograph a wild cat has appeared for sunny and the T Wolves Maybe we'll find his teammate in here somewhere. All right three boxes to go Got Donovan Mitchell 63 out of 99 for David B in the jazz Subjack what's going on and there's a manual quickly 17 out of 99 For the Knicks that'll be for David B 25 out of 34 Yanis onto the compote two-color patch Hans with the bucks Christian Latiner 55 out of 75 Subject of a grade a grade 30 for 30 documentary Herald and signatures for a sunny and the T Wolves Shadowbox anchors Rick Fox nice 16 out of 99 Michael with the Lakers and a one-of-one Julius Randall modern Marks autograph David B with the New York Knicks And Julius Randall is really starting to come into his own in the last season or so Nice and out of fives and under get the world-famous train whistle all aboard the big hit express I like that. That's nice To a boxes to go right for the piscins. It's Isaiah Stewart 71 out of 99 That'll be for David B We got the standard relic Clippers Paul George 49 out of 99 Clippers are part of that Denver number block group Clippers 9 is going to be for Adam Clippers 9 Danny of Vizier 95 out of 99 It's also part of the number block group that Denver number block group So wizards 5 is Robert Next is modern marks Terence man 32 out of 99 Clippers 2 Goes to last spot mojo Mike with number 2 You got Eric Bledsoe 42 out of 99 past and present signatures Pelicans Paul with the Pells and Scorer's Club Kevin Durant on card autograph 2 out of 25 for the Nets. I feel like we don't see a lot of Kevin Durant on card autos Feel like he always ends up being a redemption Or it's a sticker auto, but here is a live on card auto Kevin Durant 2 out of 25 It's a strong hit Brooklyn It's part of that Chicago number block group. So that goes to Brooklyn too And that's Ryan Cason. There you go Ryan. All right final box We got a we have another pick your team in the store right now ladies and gentlemen no filler yet and It's down to five teams left if you want to get that knocked out straight up It's your last chance to because after we finish this break And wrap everything up and put the break to bed Probably pop those teams into a number block. It's the last chance to get those teams straight up at 51 out of 99 Keras Leverg Pacers Kyle Suke with that one Patrick Beverly 82 out of 99 for the Clippers Clippers 2 It's gonna be from Mike. Nice team logo patch. That's pretty cool 19 out of 25 Robert Woodard the second for the Kings Clayton With the Sacktown Bay area and back down California love for the Pacers past and present signatures Danny Granger 24 out of 99 for Kyle Suke and The Pacers a couple more to go. We got Devin Dotson rookie introductions autograph 32 out of 99 for the Bulls Bulls to Ryan Cason and the last one here marks of greatness For my rivals the Celtics Larry bird on-card autograph seven out of ten And that'll be for David B and Boston It looks like no randomizer is nothing like that pretty strong break a lot of nice stuff a lot of great hits Brand new release just dropped today You got a chance of getting your team straight up right now. We'll see a little I'm Joe We'll see you a little bit later tonight for the next immaculate basketball break. Jasper's case breaks calm


The House of Representatives has erred the federal government to find a lasting solution to power generation and supply challenges in the country. This followed the adoption of a motion of public importance presented by Honourable Unimei Idem on Tuesday. In his presentation, Idem expressed worry about the incessant interruption of power supply across the country in the past weeks. He lamented that the poor supply has disrupted businesses and lives of Nigerians who are equally burdened with several tariff increments. The lawmaker informed his colleagues that the federal government through the Transmission Company of Nigeria and the market operators has begun the disconnection of electricity and not complying with the operation codes and guidelines. The House subsequently directed its committee on power to oversee the implementation of resolution and report back in one week. Millions of electricity consumers in Nigerian homes and communities may be plunged into darkness in the coming weeks as the federal government through the Transmission Company of Nigeria TCN and the market operator MO have begun massive disconnection of digital electricity distribution companies disco from the national grid. Note that an extension of the grace period of 60 days given to the defaulters disco wherewith they are expected to comply with the provisions of the market rules. Respect to payment of the outstanding embossies, posting of adequate bank guarantees and forwarding their active power purchase agreement to the MO slash TCN is almost over.