 It's a very great pleasure and I'm very honored to be invited to give a talk in honor of Mazaki. When I started studying mathematics in the early 80s, I learned a lot from Eskeika, which I have to say was not so easy to read. Also, I learned a lot from the much easier paper on the rationality of roots of B functions and also many things in studying the lecture notes from the Wilternus Lectures. When I was proposed to give a talk, I had to think about what I could talk about because my problem is that in recent years I was mostly involved in model theory, non-archive geometry, things like that, which are very nice but maybe do not fit so well with this conference. So I have decided to give a lecture on arc spaces and I will mostly concentrate on work by other people. So I have to start by reminding what arc spaces are. Historically, the study of arc spaces started in the 60s by early work by John Nash, which was unpublished until the 90s. And then there was a big comeback when Maxime introduced the Motivic Interaction. So we will work of our field K. We will take the variety of K, the variety, I mean a scheme of finite type, reduced and separated. And for any integer n, we have ln of ik, which is the space of n-jet, which represents the function from K-scheme to set, sending something called z. Okay. So ln of ik is a K-scheme. We have truncation maps. One may define the arc space as the inverse limit of these guys. And so these morphisms are very easy to understand and so this is a natural way of scheme. So it has some universal properties. So if you have any field extension, an l-point of this space is just a l-t point of x. And in fact, this is very easy to prove, but in fact there is a theorem of bat that the same holds for any K-algebra l. But surprisingly, this is very hard to prove. It uses derived geometry. I mean, it's strange. I mean, it says no direct proof of it. Okay. But we won't need that. So first example is a fine space, ad, which, okay, let's choose coordinates. And l of ad is just, you have an infinite number of variables. Okay. Just parameterizing this power series. And if x is a closed in ad defined by polynomial equation fl, say, it's easy to get equations for the corresponding. Oh, it's not. It's a bad one. It's a good one? Okay. If x is, for simplicity, if x is alpha in a quasi-projective, I suppose it's easy to verify this for K-algebras by direct. Like for alpha and K, it's trivial. But that is for any scheme. Yeah, yeah. Okay. Yeah. And so how do you get equations from elix? Okay. So you have a naive way to write down equations. You plug in power series xit into the equation f equals 0. You develop. And you have the coefficient of all the t to some power. But you have a, this provides your equation, but you have a more advanced way, a more conceptual way. In fact, you have some Hasse-Schmidt derivation sending for di. Okay. It's not good. I mean, okay. I have switched the indices. It's still correct, even if it's disturbing. So these are Hasse-Schmidt derivation. And then elix is just defined by this guy. It's completely equivalent to the naive description, but it's a bit more conceptual. And in general, if x is a fine, then, okay, these spaces are also a fine. And this we know. And infinity are just the Hasse-Schmidt derivation of okay. And here you just have to truncate the Hasse-Schmidt derivation up to the very end. Okay. So I will concentrate on the case where x is a singular variety. So we have truncation morphism from elix to ln of x. And when x is a singular, of course, in general, pyn will not be subjective. This means you have equations and you have solutions, you have a power theory solution up to some degree and you are asking if they live to actual solution. And there is an observation to do that when you are not smooth. Okay. But it follows from a result of Greenberg that pyn of elix is constructible. Okay. I should have said that these ln are finite type. So this as a set is a constructible set. Why is it? Because Greenberg says that, okay, to know that you are liftable to an actual solution, it's enough to know that you are liftable high enough. And this means that pyn of this guy is also pyn of some ln for m big enough. And then you apply Chevalet theorem. Okay. Now, this leads to a very old result. So we consider the gothic ring of varieties with a class of the affine line inverted. And we have this theorem that now is quite old by Deneff and myself. So it was published in 99. So if you consider, pyn of elix is constructible. So the liftable arcs is form a constructible set. So they have a class in this ring. With Deneff, we prove that this is rational. So this shows a strong regularity of this pyn. And of course, the statement and the proof came by analogy with the periodic integration. And to prove this, we use the motivic integration. And what I would like to concentrate on today is a small lemma in this paper by Jan and myself. And many things I will talk about today are variations or elaborations of this small lemma. Of course, I don't claim at all that the theorem by other people are included in this lemma. But this lemma can be seen as kind of an ancestor for such a result. So what is this lemma saying? Do you have an expression for this rational function or maybe you're going to discuss it? Oh, it's a good point. You have different kind of theories and sometimes it's easy to prove the rational. Sometimes it's not so. So it's easy when, for instance, you can compute things explicitly on a resolution. And for instance, this series is also rational. What does it mean, the rational series? Wait, wait, wait. I can only answer one question at a time. So this series is rational, too, but this is easy because you can view it on a resolution. Thank you. Thank you. Thank you so much. This is not known in statistic, not zero. Okay? And so to answer Valérie's question, the point is that it's quite hard to compute this even for, in low dimension, in small dimension and so on. It's not, I mean, it's not computable on a resolution so you don't have many, many things you can do to compute it. No, but it would not help to know, it would not help. It would not help because it's not enough to know resolution and it's not needed to either. Because in a resolution, okay, perhaps it's proven in statistic P, perhaps not, I don't know, but at least it's conjectured. But things we use here, specific to statistic zero, there is no sensible conjecture on statistic P. So situation is worse. And Offer was asking a question, what does it mean to be rational? In fact, the statement is more precise, so the denominator has a very precise type and so if I would have written the more precise statement on the blackboard, it would be clear what is meant by rational, but I don't want to do it here. And why the sum of lnx is using resolution? Because, because it's, okay. I mean, the only way to prove something easy is to do it, I think. Okay. Okay, so the whole story here is the fact that x is singular, and so we have to, some have to, okay, the game is you have some singular, you have the singular part of x. Okay, I picture arcs like that. So if the arc is completely outside, it's nothing important, nothing serious happens. But, okay, if your arc has its origin on a singular set, but the remaining of the arc is outside, some interesting behavior appears and you have to measure this. Okay. So, I will assume now that x is of pure dimension d, just for simplicity. And I have the Jacobian ideal of x, which is the, some fitting ideal of, of the Kepler differential. So the fitting ideal is, you take the Jacobian matrix and it's the ideal generated by certain minors. So it's, I think it's a minor order, the number of variables, a minor number of equations, something like that. But, okay, it's not important for this talk. And now I will consider arcs on, so this is an ideal. And I can consider the order of this ideal along t and I ask it to be e. Okay, when I pull back the ideal on the arc, it's just t to the e, generated by t to the e. Okay. And I will, later I will also introduce this notation. This is just a set. Okay. Yes. So, it will be important later to view elix as a scheme. But these are, okay, this is viewed as a set, yes. So the lemma, which we, at the time, we didn't realize it would, it was of any, anything important. So fix, you fix the first natural number e and then you take any n, at least e. Then you have two things. First is, okay, in fact, I haven't not, I haven't introduced this notation. But you take the same definition because n plus e is larger than e, so it makes sense. And so this is a precise form of, of, of liftability. And so in particular, you get that when you truncate this guy, it's also, you get also a constable set. I'm sorry. Is it n plus e n? No. Okay. Oh, the right conception. Oh, here, I'm sorry. Okay, thank you. Okay. And then you can just look at one step truncation and then it's a piecewise trivial vibration. I just say what I mean by this. With fiber, a fine space of dimension d, where d was the dimension of x. So piecewise trivial vibration, of course, is risky. And it just means that you have a partition into locally closed subset on which you have trivial fabrication. Okay. Now, so this space, Alex, is a problem, of course, is that it's not a finite type. So we cannot apply a usual tool of algebraic geometry. And now I will, I will present two kinds of results which have some relations, but we, which are different, showing that for certain aspects, you have some finite, some finiteness. So the first question I want to present, which is a question of notarianity of completed local rings. This is true in any characteristic. Maybe K should be perfect. So now I want K to be perfect. And so I recall a subset of X cylinder, if it's of the form, if it's an inverse image of some C, with C constructible, and N integer. K looks ad hoc. It seems quite natural. But in fact, it's relatively easy to check that it's, it's just another name for something which is present in, in the, in the EGS. It's just the same as being a constructible in Alex. But the notion of constructability when you are not a finite type is not, not so well known. So you claim that any constructible on Alex comes from a constructible on a certain level. This is what I'm saying. I will say right now. So what is a constructible in general? So we, we, we, we usually it's just a Boolean combination of finite Boolean combination of open and closed subset. But in a modernized situation, it's defined as follow. You have the notion of being retro compact. So being retro compact means it's time you enter, you intersect with some open, open quasi-compact, you're still quasi-compact. And then the general notion of constructible set is that just finite Boolean combination of open retro compact subset. And so with this definition, you, you have this more conceptual description of cylinders. Yeah. Very stupid question. I'm trying to process the second statement. Is it like a sort of implicit function theorem? You're in the second part of it. I mean, inflating, very. Yeah, absolutely. Yes. Yes. Okay. Now I will introduce certain points on the workspace. So this notion was developed by Reguera. I mean these cylinders, they were very essential in constructing motivic integration when the story started. And so the definition of Reguera is a point of elix, is stable. I'm sorry for, okay, so too many things are already called stable. I mean, but I'm not responsible for this definition. If I is a generic point of some irreducible cylinder in elix, and what do I mean by a cylinder being irreducible? It just means that it's the ice enclosure is irreducible. And then you have this very nice theorem. Take, if, take fire knock, which is not completely, there is a functionality of L. And so I can view the arcs on the singular part as being a subset of elix. So I take fire knock, which is not completely inside. Of course, it may have its origin inside. And fire is stable. Then if you take the local ring, you complete it, you get a nuttering ring. Now I just want to, okay. I will present a way to prove this later in my talk. But right now I will just explain what the use of this. So it's used via a curve. Selection lemma is the proofs, there are two known proofs of the Nash conjecture for surfaces. So the proofs are by Deboubert, Dier and Pereira on one hand, and then later on another proof by Defenex and Ducampo. Okay, so let me explain this curve selection lemma. I won't write a statement. I will draw a picture, okay, to save time. So I'm in the space of arcs. This is C cylinder. I, or constructible set, I have C prime another one, which is bigger, or C lies in the closure of C prime depending on, okay. And I have a point here, which is generic, which is a stable arc. So here the picture is on the arc space, so a point corresponds to an arc, and we will consider arcs in the arc space. That's life. Okay, so if you were in a nice finite-dimensional situation, then things would be, you have a curve selection lemma saying that's okay. You have a formal arc, which is the generic point line in C prime, and in the spatial point. And here, having that this local ring at phi is notarian is just enough for the proof to work. And so you have this. And also I should explain what is this Nash conjecture. So this was the main content of the paper by Nash. It was just, after Ionaka proved his resolution theorem, so Ionaka's resolution was completely new at the time. And what Nash wanted to understand was what he called the essential components of resolutions. So to speak, those who are there in any resolution, okay, one can't give a precise definition. I won't do. But let's concentrate on the case of surfaces. For surfaces, you have the notion of minimal resolution, and then you can take the irreducible component of the exceptional divisor. Now you take with my notation, with the notation we have considered we can take this. So this just means the arcs in Lx with origin lies on the singular locus. And so Nash was concerned with irreducible components of this guy. And the conjecture for surfaces was that there is a 1-1 correspondence with this set and the set of the irreducible components of exceptional divisor of minimal resolution. So x is a surface. What do you mean by minus 1 of what? Of L or L0? Of L of a singular. Oh, no, there's no L, I'm sorry. I'm very sorry. Okay, so this conjecture was proved by these two group of people. First by the first group and then a few years later by the second group. And they use this very strong notoriety property. You can try to extend the conjecture to any dimension, but you have, for example, starting in dimension 3. So it's a bit mysterious. I know that there is no minimal resolution. No, no, but you can devise a notion of essential divisorial evaluation and so on. Okay, but there is no meaning in trying to state it because it's a contract example. Now I would like to discuss some more recent work, or quite a very recent because it was on archive this year, by Defernex and Docampo that shares new light on this regular work. So it's not their work on the Nash problem, it's something else. They are doing something where that, essentially no one did be considered before, is computing, understanding the color differential on the workspace. For this, they do the following. This is not their notation. Okay, a point of Elix corresponds to a knack. The only natural morphine from Elix to X is just assigning to an arc its origin. You cannot assign to an arc itself as a sub-object, but you can do this by considering, so to speak, the universal family of arcs over Elix. I don't have to define it, I guess. So it's a universal family of arcs. And you have something similar with ln, okay? Here you said you use a back view? No, there is no back view, I'm here. X is anything or affine? What is X? It's not a skin? No, but I can, okay. In the affine case, if Elix is just this guy, a spectrum of this guy, then the other one is just this guy. Yeah, but if it's not affine, or here you assume it's affine, X is affine. I mean, to give this definition? But then you can glue it. I mean, I don't think they don't need a back view. And so now the theorem is the following. They have, and you have something similar with ends. So of course I have to explain what is P. P is a small disturbance, but not a big thing. So it's a locally free sheave of one coin. And so in affine, so if I'm in the affine setting, it's just this guy. So it's a negative part, so to speak. So there's a kind of duality going on there, and a similar expression in level A. Okay, so this is very nice computation. It's not harder. You have to do the right thing, but it's just another formula. But it helps a lot. From this formula, they derive, okay, for any pointer. It's an expression for the embedding dimension. So embedding dimension of a local ring is just the dimension of M over M2. Okay? What is infinite? The paper? Yeah, because LX is infinite. No, but the point is the scheme. No, but it could be infinite. Okay. Hi, this is your question? Yeah. Yeah, it can be infinite. Okay, okay. Because in five minutes, we will say something in terms of finiteness. Okay? But it's true whenever in any case. So you take the image. So take the dimension of the LISC closure. And so you have this as a limit. And this kind of expression already occurred in the Motivic Indication. But since this is closely related to omega, they are able to do this computation. And then, of course, you have that one side is finite, if and only if the other side is finite. And so this implies not without too much difficulty. The following COM. Do you take a point which doesn't lie in L of the singular locus? No, it's always true. But I guess if it lies in the singular locus, it's infinite on both sides. I don't think there is any condition on five for this to be true. So the embedding dimension is finite. If and only if phi doesn't lie in the... Okay, this answers of a question. And phi is stable. Embedding dimension is... Dimension of M over M2. It's finite. It's finite. Oh, okay. Okay. So you get easily this because you may express this condition in terms of this. And then you get the notoriety for free. This is incredible. Because the regular sprue was quite hard. To read something generally in EGA, I think that you take a limit of local ring. The total local ring of M over M2 is the same somehow in the limit you get in the area. Osunagata did something like this, so there is a... Yeah, but here you need, I mean, you don't... At a certain point, you are not Italian. At a certain point, you are not. Okay, wait one second, maybe a bit more, but you will see the situation is not nice. And so what to do... Okay, then you have the lemma or remark in any generality. If you have local ring, I mean, it's... So this is a small thing. The embedded dimension is finite if and only if the completion is not Italian. This is an easy exercise in commutative algebra. And then you get regular sprue M for free and you get even a bonus because you get the characterization of stability in terms of the tinnity of the... In the case where the embedded dimension is finite, is it also the case that the maximal ideal is finitely generated? I mean, I'm not sure. Okay, and so let me make two remarks. Let me make two remarks. So remark even if phi is stable, this guy may not be... may not be not Italian. There are easy examples. And even if phi is stable and this guy is not Italian, it may not be excellent. In fact, it may be reduced, but the completion is not reduced. So it's interesting because you see that part of the pathology of commutative algebra, you can already observe it when you're working on arc spaces. Okay, so... And so this lemma I wrote here on the blackboard, we did not see it because I did not provide any details on proof, but in fact it plays a role again and again. In fact, there are some people who divide the general brochure on who they vote to paper, where they divide the general principle from which many different... This was a corollary and other statement, like the Drindfeld, Greenberg, Kasdan, I will introduce now. Okay, so there is a general framework, but that subsumes this kind of statement, okay, it's not my aim to present this. So, my last part. And so there is also a COM of Greenberg, another Greenberg with an I, not two E's, and Kasdan, which was later provided another proof by Drindfeld, which stays following. And so here you take arc, which is not completely in the singular lockers, then you take this completion, which may or may not be notarian, but anyhow it's... Okay, so what does it say? It says that as formal schemes, so to speak, five-dimensional, but this is far from... This is different from being notarian. Okay, it says that you are the local ring of a localization of a finite-dimensional, finite-type object, and here you have just the affine space in infinite number of variables, and you take, say, the point zero and you complete. And this is what it says. This is very useful, for instance, in paper by Boutier. In infinite number, how it gets built in special cases? This is... One has to think, at least at first one thing, you can see contradiction, but there is no contradiction. But if you take this completed tens of order with infinite number of variables, then I think that there the thing is not material. There is no contradiction. I mean, the two statements are two, I mean... What do you want? Do you trust you are just saying this? Yes, because I mean... Do you explain? No, but it's not... There is no contradiction, no. No, maybe you mean that you localize it with another point of infinite dimensional space or not closed, not the origin, or... Yes, otherwise it's not finite dimensional. What do you mean by finite dimensional? You mean a k-point, or a scratch point. Yes. You have to remove infinite many points, many... Okay, the point... Okay. Okay, let me... Okay, let me... ... ... ... ... ... ... ... ... ... ... I'm sorry, I'm afraid that the theorem of Rindfeld and Greenberg and Casdan... I will provide you another... I will provide you another... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... Okay, the point is that you would like to have perverse sheaves on LX, intersectional converging. But it doesn't make sense at this stage. But once you have this kind of decomposition, at least formally you have something of finite type. And you can speak of at least some kind of gem of intersectional converging sheaves. And if you work over a finite field, say, you may take trace of Frobenius, and they showed in some special cases that this trace of Frobenius was computing the real thing, what you were testing. But the problem was to define an intersectional converging complex on LX. And this is what I want to present in maybe the five last minutes. So this is a very nice work of Butier and Kasdan. Okay. So Butier and Kasdan provide a kind of global version of the DGK theorem you did not like. And so fix some E. I remind you that this notation, so it was just, okay, an inequality instead of some equality. So this means that you are not too close from the singular locus. But of course E may move. So it's not such a big deal. So they prove the following. So there exists some scheme Z with a morphism to this guy and another morphism. Okay. So with G post-smooth, okay, just a projectively, a field-hand projectively of smooth morphism. Subjective. And do you think, okay, there isn't some hypothesis which I forgot. So it's K, algebraically closed. Okay. And do you think an isomorphism on a formal neighborhood at each point, at each K point of Z? And U is a poetal. And Y is a finite type. Yeah, but maybe then it's the zero. Yeah, maybe it's the zero, yeah. I'm sorry for that yet. So how can it be if it is pro... So you don't want this to be poetal, the G. You said it's an isomorphism on the formal neighborhood. U is poetal. No, no, but the G. The G looks like poetal because it's the complete... Well, if you have an inverse limit of smooth things, of course you don't. They can change the dimensions. Okay, that's a statement. No, no, but I'm sorry about this. This is extremely easy to make. I mean, you make epsilon changes and then something which was true becomes false. I probably made it by writing this zero. And so I want to stick to what... Okay. And so the rules, so it's a global version. So as someone said, I mean, the original lemma is kind of implicit function. And so you do these jobs to prove the GK... Drinfeld, Greenberg, Cajan, CRIM, for instance. But you have some choices. And here the idea for proving that is, okay, you don't care about the choices. I mean, you put all the choices, so to speak, in this scheme Z, which is very big, which is endowed with a global map to something like that. So they call this an atlas. I mean, this is very justified. And I don't think... I think I'm already over time, so I won't explain more. I will just explain in two words what I'm doing. So they introduce a topology, which they call the post-prosthopology. And which is related with... For a topology, but you... Okay. And this allows them to build a nice category of eladic sheaves on this kind of people with the six operations, the duality operator. And of course it's compatible if you may increase E. So you could take the co-limit of these categories. And they are also able to introduce a notion of perversity and to define intersection complexes. So they define a global version of IC complexes and for which they check that they correspond exactly to what they did, I mean, ECC, that was done previously just in the formal level. I think it's time to stop. Before potential questions, I think it's time to thank the organizers for the wonderful job they did. And also to thank Mazaki for the wonderful mathematics he gave us and for also what he will continue to give us for many years. It's just that your fee or file whatever there is care-rational. That's why there's no contradiction. Because in the case where it's not care-rational. Yes, to remove zero. It's not only a infinity zero. No, but the fee or file there is care-rational. It's not... you have to add the off-pay. Oh, yeah, yeah, yeah, yeah, yeah. Thank you. Yeah, yeah. Yeah. Okay, I'm sorry. You have a file? No, so I want to produce a counter-example to one of your... Okay, good. I'm the only... I will be the only person responsible for that and not the authors of the papers. No, okay, so let us... I want to look at a non-normal surface obtained by pinching, that is, you take a line in A2, let's say the x-axis, and you get x to x-square, and so you pinch using this. Yes. Then you have now a singular locus. Yes. And such that if you take a curve on the singular locus, if you take a formal curve, which is non-zero tangent, it doesn't live to the normalization. So the idea is that in the art space in the neighborhood, you get just the art space of the singular locus in order to do the full thing. And then you have written a formula for the dimension of the... embedded, the... for the dimension of m over m squared. Yes. Ah, but it is... maybe it is... I'm just looking at k-points and not... I'm looking... Okay. Okay. Good for me. Thank you.