 Yeah, so thank you very much. So it is the first time of the lectures, of the series of lectures, and the first three are given by Pierre Chapeau. And I think he explained how it works at least in the regular case. And I think I'll start a little bit what's going on in the regular case, and then I'll shift to irregular case. OK, so we consider a complex manifold, X, and we consider the OXT. So that is in the sheaf, so all the tempered functions. And that behave very well in some sense in the following way, if there is a morphism of complex manifolds. Then first, there are two big, two major factorial properties, one is you start OX, tempered one on X, and you tensor with the transfer by module. So that is a dy dx module, and dx is a ring of differential operators on X. And that is got to white. So more or less the pullback of the tempered distribution on white can be described on X. And the next one. So this paper strip is just in the sense of the relative or that is without the paper strip on the. Ah, yeah, yeah, that is, yes, that is in the version of the, yes, yes. So that is in the sheaf of the, no, no, no, no, I'm sorry, F inverse divide module. So that is in the, in the sheaf sense, that is also in the sheaf sense. And another one is if F is coherent OX module, and if the support of F is proper over Y, then F tensor OX to F star is isomorphic to star F tensor, that is OX. So that is stability. So that is again a coherent sheaves, coherent sheaves converges, and so that commutes. So that's an actual one. And so the other consequence that holds in the derived sense, in the demodule case, or good, I don't know, good or case I good, DX. And we are assuming that support M is proper over Y, then so that is a demodule sense D M F M tensor D OX to E's tensor D OX. So that is in I divide. So both are true. So those are the very factorial properties of the in the sheaf of tempered distributions and tempered homework functions. And by using this, the main theorem for that one is if M is regular holonomic in the derived category of DX modules with regular holonomic converging groups, then first is a drum. Of M, tempered drum. So that is by the definition M, where OX T is OX tensor OX, OX T, and omega X is the highest degree difference of forms sheaf. And so that is isomorphic to the usual drum. So in the regular case, the dramatic homology calculated in the tempered sense is the same as the usual. And the second one is OX T tensor M, that is solution sheaf, and here M is solution sheaf. And so that is in I DX. And if you go to DX modules, then M is isomorphic to R home. I think by his notation, I DX for M OX. So those are three properties. And as a query, there is an equivalence of categories DX is isomorphic to constructively sheaves CX by drum factor. So that is the main theorem. And I think he explained a little bit how to prove it. And main ingredient is to reduce this theorem to have very simple sheaves by devisage. And it means in a foreign sense. So at PM, we have a statement, some statement on M, so that is DX. And assume that it satisfies the foreign condition that I'm going to write. Then P is true for any. So it means in order to prove it, we prove something, some properties on M. So the first one is locality. So PM is true if P UI for open covering. So that is locality. So that properties as PX to make sure that it concerns with X. So there is an implicit condition, P of M depends only on the isomorphism class of M. Yes, but that is a corollary property. So that is a first property. So I will give another property. So I think that, yeah, I think it's, for example, I think that in private this. So if that is the distinct trial and PXM prime and PXM double prime is true, then PXM is true. Yeah, let me see. I think, yeah, that is a corollary of the other one. So it is not the end, so, yeah, and C is PXM is true, then PXM, the shifted one is true. And M double prime is true. So that's what M is true. And I think it is noted up here that F is projective and M is the electronomic on X. And we assume that the PXM. Then it's directly imagines. Do you assume that it has a good filtration condition? Ah, yes. Yes, better to assume it. But in the regular case, that is automatically true. Yeah, but better to assume it. Yeah. I think that. And is it, and the good means that on each relatively compact there is a filtration? Yes, that's right. And projective means that globally it is in PNCOSI or it is locally, what does it mean? Yeah, of course, because that property is local. So you can assume that that is projective locally on Y, but on forever. So assume that that is projective globally. So there is an ample bundle on AXM. And that is the last one. If M has a normal form, a normal crossing divisor, then PXM is true. So I think it depends on the definition. For example, if you assume that M, the zero has a normal form, then PXM is true. The definition of normal form or something like for rank one system, so for. Yeah, yeah. First, in his definition you have to assume that PXM is true. Yes. For rank one, then C is direct sum. Yeah, yeah, yeah. Yes, direct sum is true. So the purpose is we want to generalize it to the irregular case. So there are several parts we have to change. Of course, for those kind of things, you have to consider the statement for autonomic modules and the problem is what is the normal form. And in that regular case, that is rather simple. But unfortunately, in the irregular case, that is rather complicated. So that I'm going to explain now. Let's say, so that is a classical case, classical ordinary differential equations. The theory of linear with ordinary differential equations. So you take a singular point, say, in a complex, in an open set of a complex domain and you take for the sake of simplicity, dx over dxp, where p is a zero z, dzm plus mz, where ajz is a homomorphic one. And a zero has a zero is a singular point. And so at least m is isomorphic to 4xm outside at a local area outside zero as a homodule. And then the classical theorem says the following thing. As I said, I will not be very precise. So here, I think if you want to cover all the cases, it's a little bit more complicated. But I think to understand what's going on, I think it's enough. That is not the exact theorem, but approximated to, with something. And so I assume those conditions. Then there is fjz and or write all the things. So that's a finite sum and k new is, that is finite sum, say, k new and q. So that is a place of g, what is the function. And sum, there are m solutions. And there are lambda j, you see. And another function, ajz. So that is z log z, new sum number. So that is a polynomial in log z. And aj new is formal policies in a functional power of, say, a for some integer. So that uj, say, z is exponential phi jz, say, lambda j, ajz is a formal solution of pu. So they are nearly independent solutions. So that is the first one. And the second one is, so that is a formal solution. And that helps to get a real solution. So you have 0. And you take an arbitrary direction, say, data 0. Then you can find some angular neighborhood, say, u. And ujz, that is a homomorphic on this angular neighborhood. And it is a solution, homomorphic solution defined on u. And uj is asymptotically equal to uj. So that is the one. So that is j equal to 1 to m. So as you know that there are outside singularity, there are m linearly independent solutions of pu equal to 0. And uj gives certain linearly independent solutions. And so the meaning of asymptotically means uj minus uj. I will explain. It satisfies some condition. So that is the third lambda j. Where ujz is, so here that is a formal sound and you truncate. So ujz is this part. And here there is aj, but you truncate. So that is, say, aj nu. Yes, yes, arbitrary integer. k is smaller than s and aj nu is, it doesn't matter so much. So that's a satisfied condition. So if you truncate its form of c's up to some degrees, then that is a plus mutary, ujz, ujz, s. So in this way we can get the solution. But the problem is those uj's are not unique. So uj heart is, if you more or less you can specify, but for uj heart, for uj, even if uj heart is determined, uj is not unique. Because there's some rule. For example, real fj prime is smaller than, say, very much smaller than real fj on u. Then uj prime plus uj or uj is asymptotically u heart j prime. Or let me see, it's bigger. Yeah, uj prime is u heart j. So it means if that is true, then prime is very small. So even if you add it, this asymptotic expansion doesn't change because here it takes a big time. And uj prime is very small compared to this. You stated this for solutions of an n-soldered equation. Yes. Can you, one can pass through the associated first-order for a vector. Is there a difference in the statement? No, no, no, same, same. You have to adjust it. Yeah. So it just says, if those are, let me see. Yeah. So if that is not in this form, let's say dz to u is dz to u. And so those are quantum vectors of functions. Then you assume that this part is the same. So those are quantum vectors. And this part you don't change. And not exact, what does it mean? You said, tell them not exact. I think if I say it like that, it is exact. I'm sorry. Maybe you should assume maybe k nu are strictly negative. What? It's an exponent for it. On the first part, no, fij is equal to, no, on the left. Yes. It's still on the left here. This one? No, not still on the left. Yes. Just on the left. First column. Just columns. Sir, k nu should be strictly negative. Yeah. You can add it. But it doesn't change. Yeah. Yeah. So as Maximil said, you can change it. Yes. So fij is not unique. So that is one part. One important part. So that is the question. And so how to interpret it? So there is a work of Drine and at the same my branch to how to interpret it. So you consider x and you take a real flow. So recall that it is more or less our data. And the data is so that is a complex number with absolute. So it is like that. And so that is x tilde. And you go. That is 0. And you blow up by square. So it contains s. So this part is s. So that is s1. OK. So you consider a x tilde. So that is x minus s is 0. And that is contained tilde. So that is a formal function outside s. And f is template. So here, more or less, f is smaller than some constant. So that is a shift of rings on x tilde. And you have m. And you now what? Pi inverse. Pi inverse. Pi inverse. U. So you drift the ring from OX to AX tilde. Temperature. OK. So that is a DA module. And that is, in fact, isomorphic to DA expansion. So this fact is a little rather easy. Because if you write that theorem, uj, exponential minus phi j. So that is, by the condition, that is tempered. And moreover, that is invertible. So that is AX tilde, invertible one. So by this one, you can give a explicit isomorphism between them. So that is isomorphic. So that is one part. And what they, to really consider is the following thing. You consider the solution here. And you consider it, say, u is in j star over x theta s, 0. And you restrict s. So that is a shift on s. And you consider a filtration of s. So, namely, you consider phi. So that is u. Such that u is, say, this part is not, say, exponential, real part. Where phi z is, those pieces, say, or something. Some pieces. Pardon? How big that n is there? Some n. Some constant. Some constant, positive constant, and some n, and exponential. So, for example, those ujz, uj, uj belongs to scj. And if it satisfies this condition, uj prime will also be on it. So that is, as psi, if, for example, real part of phi is real part of phi. Of course, on some region. So, of course, this relation changes according to s. So which point you consider? So that is a filtration, but it moves on s. So you consider this. And then the theorem says, more or less, those filtered objects. So that gives a filtration of s. And more or less, the minimum connection, s, x, with singularity, at the tail, is corresponding to s towards filtered objects. So that is the theorem. OK. So that is one dimensional case. And, of course, that is to naive to extend to an arbitrary dimension and for arbitrary, holonomic demodges. So how to interpret this one in a higher dimensional case? And the method is to use so-called enhanced shift. So it means, so we consider the following object. Consider r plus s plus r. s is that boundary of x tilde. And you consider s, or first, you consider on this. So say p, p plus s. And you consider s, p, and so c is real boundary. OK. And you sum up. So that is a constant shift on this space. So that is a sub-shift. And then, so to give a filtration is more or less equivalent to give such an object. It is more or less equivalent. So that is one. But so that is not enough because as Maxim said, phi is not unique. For example, s phi is s phi plus constant. Because the condition of the modulation estimation does not change. So that is not enough. Because if you interpret, because that you can put c here, then it becomes all the constant shift. So if you walk in a usual shift, it doesn't work. So the idea is you don't consider this one, but you consider it in the shift. In fact, it is not exactly on r, but you have to compress it on. So that is a shift. So p bar inverse, p bar is, so you consider compactification of r to s. And you consider this one. It doesn't matter so much. I will explain it. Say infinity over. It doesn't matter so much. After that, I'm going to explain it. So you consider this one, then that is one. So that is in the shift. And so now you consider not on s, but you can consider on x now. If such an object can be considered as in the shift on x. So that is, so we call that enhanced efficiency. Enhanced means we add one variable here. And so as you see that to add one variable is more or less to give a moving iteration, iteration that moves. So here there are, so we add one variable and we have to compress it on r. In fact, so in the shift itself, if you consider on r, then that has a local property. So that is stuck on s times r. So if she goes to infinity, more or less that becomes a constant shift. If you consider it on s times r, but if you compactify it, then it is not. So that is I'm going to explain now. So I repeat it. So in an ordinary differential equation case, more or less, if you go to x tilde and if you consider tempered, that is tempered form function defined on angular neighborhood, then that becomes to a very simple one. And that is one point. So because of that, so the solution have some iteration and that can be interpreted as in the shift on the base space times r, one dimension. Mindfully add it. So the next, I want to explain how we realize those kind of fields. So one is, as I explained, we have to compactify it. So that I'm going to explain now. So in the general case, you have some manifold, but you add an infinite point. So that is hopefully getting. So that is a kind of a compactification or plus a compactification. So here, and the complement is closed set. So that is closed set. So you have the hat by I star or the same. So that is a full subcategory of, there is a category of sheaves on m hat. And in this case, you can divide it, but it does not give a new thing. So that is nothing but Km. So in this case, in a usual shift case, we don't need such a maneuver, but in the shift case, we do the same thing. It is not the IKM. They are not the asthma. So we have to store the compactification is rather important. So the definition is, so m and bar is a border space if m is open. And we define this category, they have the category of in the sheaves or the border space. So that's the definition. And I think I will not, that is a little rather routine. So I will not so much develop the theory. So we say that there are two border space and we want to define a category of border spaces. And that is a morphism. It means f is a continuous map and gamma f bar and bar is proper. Where gamma f is a graph and gamma f bar is the closure in the closure of the graph. And we assume that is proper. So then it is easy to see that that becomes a category. So you can consider the category of border spaces. In each border space you can define the category of in the sheaves and you have a factorial one. So you have the operations of one. For example, if f morphism then you can define so direct image and direct image with compass support and f inverse. That's the same way. So here there are no similarity assumptions on the spaces. Just arbitrary typological space? I think better to assume that. So we call that the space, the typological space is good. It means it has local compact and the number of the number is countable at infinity and perhaps finite of the termination. I think that is easy. It's not that you get confused if m hat and m hat are compact but this gives a morphism. If m hat and m hat are compact. Yes, yes, yes. So, yes, that's right. Yeah. So what is the last line? Is f upper plus or f upper? I don't read. This one? So that is direct. And f inverse? Yes. Yeah. I think that definition is like that. I think that is not important. So you have two projections. Say Q1 Q2 and this one is Q2 and so on. I think they are joint. So you can get it easily. And then use your properties of direct image or home or also you can define tensor product and so on. So it is exactly the same as the usual in the sheaves. So where the finite lovely dimension enters in the sheaves? I think it is not so important but I think that is here I was not so precise but so if you consider bounded one then it needs. So I think you can remove it but then you have to be careful on the dimension. So they are so bounded complex goes to infinite boundary. So that is a bit technical one. I think you don't need it but then you have to be very careful. Because in the case of in the sheaves there is no injective objects. In the category of in the sheaves there is no injective objects and I think the injective dimension is infinite. So you have to be a little bit careful. But I think that is a very technical assumption. So the example is assuming that m hat is a border of space then so you consider that that is a border of space m m and hat that is border of space m hat. Then those are the morphisms of border space. So if you write j so you have j star no no no j inverse so j inverse so that is in fact isometric to j star so that is inverse image and that is nothing but the quotient map. So by the definition that is a quotient of this one this one so it quotient of a function so that is a quotient of a function. So they are not equal and for example this one k is k m k and j star k r i home k m so that is I think because it behaves exactly similar to the usual direct image function and so on so it I will not explain more ok so I think I will take up 5 minutes or 10 minutes ok we shall start ok so now I explain those I will explain those adding one variable case so now you consider the as I explained not r but it is a so one way is you consider r bar so that is so that is a topological space so that is as one to close the interval and all p1 r is a line and then both are at so r bar or so those are both of the space and they are isomorphic as both of the space so we shall write it r is infinity so that is both of the space yes so that is one point but so the complement is two points and then one but morphism is symmetrical on the complement morphism is symmetrical of open parts you remember that m bar and morphism is only a continuous map from m to n satisfying some condition such that gamma f bar to n bar that is proper so so that is isomorphic so those are isomorphic are the both of the space so basically yeah it should be otherwise you can't work so I think it should be it's not so many difference but you I think that is a little bit easier to think about but anyway so you take a real manifold so good manifold and are you thinking so that is a both of the space say are or p bar are or that is both of the spaces ok so now you consider the infinity and so the idea is to how to interpret them in terms of in the shift theoretically ok so we here we define the convolution so you have you consider the two so that is it you have the projection to so that is the first projection you take the first component and the second component and you add up infinity so mu is t1 say t2 is x t1 so they are morphisms of both of the spaces and then define the plus the convolution I like it in this way so that is the and another one I home plus so that is the I think strict dimension I think here perhaps p is better but I think it's minus plus is better but I think I will not write it those are no but not so relevant and it's rather tedious so I will not write it and definition is so that is the p1p2 your definition I home plus p1k2 is p2 or p1 or p1 you better to change it doesn't matter so that is the definition and the theorem is this here that one I don't know it is commutative monoidal category or what tensor category so that is the tensor function and you need to object is t equals 0 so that means m so that is definition so when you take t1 plus p2 what is it what is plus infinity plus no it depends on the open part yeah that is the open part so so that is one of the more important space yeah so that is already defined it's a map open space with certain condition and so that is a unit object means of course k is with some functional property and I home is a I don't know the word the word in my home for the tensor category it means it means home k1 tensor plus k2k3 is home it is canonical as more home k1 I home plus so that is infinity and that is so that is okay so so that is the one so there are many important so that part is problem so of course this one is important in the sense that if you tensor to this one that is kt equals 0 so that is a unit object so it must be like that and this one also important and negative also that I don't write it and similar similarly this one is important there are many important and this one negative one is it orthogonal so you remember that it is important we are called it orthogonal if you want to 0 and so that is as 0 okay so and other one is infinity one so that is also important and so there are many important they are related in a foreign way so for example you add two things so that is important that is important and they are also similar to each other so the sum is important and you have kt equals 0 and as you got it I think so so that is this thing triangle or k or which is no no no I am sorry I am sorry kt which is better so there are things to plan so this means that the morphisms are compatible with the identipotent the morphisms are compatible with the isomorphism so an identipotent is there an object plus there is a morphism yes that's right and the maps are compatible and the boundary maps you don't say anything about two maps are compatible and the boundary maps let me see I don't know I haven't checked it maybe there is some relation when you shoot bay one it is no longer in the quadrant here no it makes no sense because with the boundary map it makes the object shoot at bay one it makes no sense I don't know I haven't checked it but there can be some relation but I don't know anyway so so perhaps so so if it is identipotent then you can have a subcategory so that is the one so I will not explain why I write like that so that is K 0 and of course that is because of this I am sorry plus yes and in fact that is can be written I think similarly that is true also we need something I think and you can define this positive just a sign and you have the intersection that is so that is in fact K such that K is as much to Pi inverse L for L I KM where Pi is a projection so that is constant along the 5 so that is this so you have here D infinity so there are several ones so here the intersection is 0 and because of this if you so that one is isomorphic to the I KM and usually in a similar way the quotient of this one is equal to the quotient of this one and this quotient is asomorphic to equivalent to the quotient category and that one plus that one is that one so that is the usual way so we have the following so we define the shift of enhanced shift is so this one you divide with that one and you have you define plus minus I KM is so that's the definition so that is E plus plus that is minus minus and this one is the quotient those are the quotient yeah so so that is isomorphic to the diagonal so that is the intersection this one so explicitly that is given by K so that is this component and K so that is the one okay so all the terms of categories here all the symmetric monolithic categories here is this the plus one symmetric yes yes yeah the monolithic categories symmetric or commutative you mean symmetric means commutative okay yes it is and I think there is a one theorem where it is I should state yeah so for example I said that this one is equal to this one and that is related to this one and so so here there are two operators plus and plus and they are in fact related in the wrong way okay so this map is so that is functional so this map because there is a map from morphism from this one to this one and that is K because that is identity element and that is at home plus 0 K because that is unit element unit object so that is a composition map and so that is embedded in this thing and the third term is an inverse image of some shift on M in the shift on M and L is there are many ways to write it I think for example so the point is so those two are isomorphic in the enhanced in the shift category so we call it the category of of enhanced in the shifts and in this category those two are isomorphic so there are two ways of expressions and those are projection to E plus and E plus so that is a relation and so remark is so you can answer the direct image and this one is in fact 0 and so it did a part remains and that is it so that is the category we need so for example we consider those ones in that category so for example for example K is isomorphic to K 1 M because I because I think that is clear you have this sequence and statically is 0 so it is equal to 2 and we write this one is so that is original that is in the shift that is in the shift K M and you consider that is in the shift on the border space M times infinity and so you go to here so that is you can consider it as enhanced shift also and so that is the so that is similar one so that is was very similar to that and then of course in a similar way K in so now this category has very similar property for example they are monoidal categories symmetrical monoidal category plus operators so that is also well defined so that is also well defined and this one is also well defined so that is well defined and that that is a tensor category and similar property let me see I don't think let me see I think I think so that's right 0 when you have one positive and one negative you get 0 or 1 still I think that maximum question plus so it plus so this one is something and that one is something and so it's commutative and that tensor product of this one this one is equal to so that must be also well known for this one too so in fact in fact E is a direct sum of two categories but we only use this category so from the beginning it's enough to concept this one but I think I don't know that is a problem of formalism and it is not essential so you have a constant map so this has left and right so left left left left left left left left left left left left left so those are right adjoint left adjoint from going here to this one so so that is a so of course this one and this one is isomorphic so I mean you have M there are two map RE and two ITM and the quotient map and this one is isomorphic to identity both for LE and R so those two objects are isomorphic so that I explained by this one that is clear so the next step is so this category also behaves similarly to the usual category it means assume that there is a map M to N then you can define the pull back and push forward for the enhanced initiatives so that is rather easy because you have a product map and you have 50 to 50 so you have for example RF weak and R so you have the quotient map and so that gives two functions that I denote F, weak and E F star so this so you take the quotient then I mean this composition factor in this way similarly you have inverse image also so the IK and 50 you have inverse image and upper shake and it factors also and they satisfy usual relation I don't write it anymore but rather usual for example E, F, E star I think G, F star or the other way around anyway those kind of there are many main properties that is similar to the usual so I will not write it for example for PEPS or IHOME E, F star K E, F star IHOME plus E, F inverse L where K is on M and L is on N so that is a if you replace home then that is a usual relation and for the other one also IHOME plus F shake KL so that is F star IHOME plus K upper shake so those are the usual relations and so basically so it satisfies all the relations similar properties so to manipulate those one it is very similar to the usual shift the category of shift so so basically with other those kind of techniques in shift theory can work for enhanced engines so now a little bit complicated object so here we consider very special kind of objects and that is what we call stable objects so let K is plus IKM so it means K is then the following conditions are equivalent so that is isometric to K by definition so that is a restriction map or the other way around IHOME plus so that is a isometric K or IKM definition for first K and KME is that KME plus infinity so that is a definition and you answer this one and there are a lot of properties or those are all equivalent so and this one satisfies that condition that is nothing but this one satisfies that condition for which L for some L E IKM for some L no, of course K satisfies that condition but for any we say that in this case K is stable if those conditions I think better to remark the following thing so the problem is in those for the in the sheaves it usually to calculate this one for arbitrary K IKM so that is rather difficult but if K belongs to so is it usually difficult to calculate but so IKM F is in the it's not in the sheaves but sheaves and K is in the sheaves that is rather easy more or less if K is so it is not exactly true because that is in the sheaves category but more or less if it is like that and that is at home K I and that is the limit of the sheaves so for example if you take the homology group that is true so that is it calculable because that is in the sheaves theory but for this one it's not it's it's calculable but you need so as you see that IKM limit something is limit of projective limit of IKM so here there is projective limit and projective limit is not exact so that part causes a problem so usually it is difficult but for stable object it's a little bit easier in the following sense for example R at home plus K is tensor plus F and F is say the M R it's a sheave and K is E so that is R E home plus so you can put it here F E home plus M E K if you assume that K is stable then that is K if K is stable so now that is a sheave so it is more or less calculable otherwise you have to be very careful to calculate those kind of those kind of object so that is the advantage of a stable object and so let me see we have a time 5 minutes so just a definition so assume now perhaps I think I will stop stop here what is the photo of K, T bigger than equal to 0 and K of T bigger than 0 yes I think so yes you mean this one T bigger than 0 I don't know T bigger than 0 so you have this one and you cut this one then that is a common view of the interval so that is 0