Higher derived Artin stacks
Porta, Mauro
Definitions. Examples and standard properties. Examples: Eilenberg Mac-Lane stacks, higher stacks related to rationnal homotopy theory.
 Let's start with the engine motivation for the gematic stack. So far, many people use these motions, and after this talk, many people will use their examples. So I would like to motivate in a conceptual way, and lead for gematic stacks. So a basic motivation to consider stacks is that when we work in the category of schemes, we have objects which have a nice gematic distinction, nice gematic algorithm, but in fact they are not like schemes. So I think the object which can be described as a portion by the color and the definition, for example. Number one, natural of the user to consider patterns of the objects is to fully complete the category of schemes, taking the data sheets. So we are considering a unit and that into the data sheets. But I would like to stress that this obligation is quite brutal. So if this destroys completely, on the geometry we already have here. So let me give a simple example. We take X and Y schemes. Inside the category of schemes we get data that you don't union. So here we can consider either the presentable issues associated to X, the ones set to Y, we take the list of the unit and the category of schemes, or we can consider the issues associated to the list of the unit. So what generation? They are quite different in fact. To complete this fact, just evaluate the two of them over two points. In this case we have two points on X or one point on Y. While in this case we can get also the points on X and the point on Y. So they are different. So to recover the geometry here, the way you wrote in the group that did that, the way we do that today is to fix a topology and to think of a topology on schemes, a good topology obviously, as the assignment comes, which should be become or should rename or say, call limits in the category of the schemes. So there is a nice categorical way to describe this. I won't be busy today. So let me just say that we are now in the category of schemes with respect to topology. So which can be the topology. So at this point, this category is quite different. And we can think of objects here as nice gluing of objects in schemes, but the point is that we cannot really control the way we glue objects. So there is an element amount of geometry that we can extend from the category of schemes. So anyway, to solve this problem is to restrict our style to a full subcategory of this category and to specify how to glue the objects in a sort to give you a rule to get objects in here. So this is what I will explain in this talk. The rule I'm talking about is the inductive definition of geometric sets, essentially. So to give the last example and then to move to the main part of the talk, many of you will know the glue line for schemes, classical schemes. This has a simple version, this one, so if we have two schemes and opening motions like this, then we can glue the objects along these opening motions. So here we are considering quite particular maps which are in fact opening motions. So in this exact thing we will change the patching rule. In fact, we will consider multiple types of maps which in fact, in the case of classical, are just to be taken as movement. And this will give many full results. So let me start now for real, so one notation. So since I have to work both in the right and the left side thing, I will just follow the general approach of the IP2, and we will see the connection in a general context and it will be there already. So let me just fix my station. So first of all, we need a category of algebras, so this is done by considering, together with the classical information and that. So this will be a symmetric curve monomial, so this is a category, close. Satisfying some sense, I would like to stress that one of the most important is the fact that commutative objects inside the category C as a transfer of the structure from C. This is not obvious. For example, if you work with this differential gradient algebra, you will see that it is only at 360, and really require that the adjuncts to see, the forgotful and the symmetric, and be used to respect the structure. There are ways to solve the problem, for example, considering a simple algebra, or instead of using commutative objects, using an infinite object, I will just assume this. Next, we consider the category of a fine object in C, which is defined simply as the opposite of commutative other objects. And we need to fix the genetic information we want to preserve, so we need a model, a trait of biology, and we are also satisfying some nice conditions, so model trait of biology. For example, we want this topology to be super-sacanaka, and the end-class commutative case. And finally, we need a class of maps, which play there in my toy example of opening a merge, there will be smooth morphism, so P is a class of morphism, in a fine object, such that it's compatible in some way with tau, such that this means, first of all, that if we have a cover, so if this is a cover, we respect the tau, then ui tau is in P, and next we want that P is stable under equivalences, under for most people back, and under implementation. This is a reasonable assumption, and at last we want that the range in P is local on the plane. And there is a last technical assumption, but I won't need it, so let's see if it works. So I will consider in this talk the get-out side in both the range and the range, so one, a tau, plus tau, plus tau plus tau, with P in both the smooth maps, and second, the tau, I'm going to do a five-step, but to me, again, small steps. The definition of this thing is that the juggernaut is called yesterday. Okay? So now I can give you the definition of the juggernaut. Starts. And I was saying it is an inactive definition. You are going to see that. So, this factor is minus one, dramatic, if it is a juggernaut. Second, the morphism of stacks is minus one, representable. If for every representable stack and for every map from x to g, the base changes over in this equation. Okay? So, this equation is representable for f for all the people back to this representable. And third, f morphism of stacks is minus one, p. If it is minus one, representable. And, moreover, in the diagram before, so, at g, x. This map, as we call it, alpha, is in p. So, this is a map between representable stacks, so, it is already defined. It doesn't mean anything. So, now we can do the reduction stack, so, that n bigger than zero. So, for example, we need the notion of nr plus. So, nr plus of stack f is a family, ui of representable stacks, together with morphism from ui to f, such that each morphism is in n minus one, p. And, moreover, the total morphism from the disdon't human to f is subtractable as morphism of stacks. So, it is subtractable by zero. It is an equivalent. Okay? So, this is the definition of atlas. We say that a stack, alpha, is n geometric. If it has nr plus and the diagram is n minus one representable. Okay? So, now morphism at your map to g is n representable, or you can image it. It's also classical situation at g. So, first, here we have representable stack, representable, and map to base change. And we want this to be an n minus one stack. And then both n geometric. And finally, as before, so f, we call this alpha again, so f to g is in n, p. Okay? So, in this situation, this is an atlas of this stack is nr plus ui of the range. Okay? Subtract each composite, so ui go to this. And then on with alpha and x is n, p. Yeah, in n, how do you think, yeah, it's a morph in between representable stack. So, the last point may seem a little bit bold, but the point is that we want a day in nt to be a local stack. So, I will develop some of the basic properties from the consequences of this. So, all these things implicitly have them for all x, y, g. Yeah, yeah. For every x and for every map. It's the same of the previous situation. I said that for every x and for every map. For example. I thought I said y equals q. So, for map. Not after this. Okay, so, first of all, being n minus 1 representable, or respectively being in n minus 1 p implies that it is n representable. And respectively n p. Okay, for n non-physical. Second of all, n representable and n morphism, okay, Okay, so,reb petty and unrep wealthy and unp росgy and unprom p Kathy involved on the And these two properties, we can already deduce that n-stats is stable under pullbacks, homotopy pullbacks. So n-gimatic, flat, stable under homotopy pullback. And finally, fold property, this is quite useful. So this is the analog of the well-known affinity here of the classical piece. So the factor is a morph between stats, and g is n-gimatic. And moreover, we can find an aphelus of g so that if the state is an n-gimatic, then f is an n-gimatic. And aphelus of g such that if they change, so f, among the people of the circle is n-gimatic, then f is n-gimatic. If you watch about the look of this pointer, this is done as editing by induction. In this case, the hard point is to do the representable case, and it's followed by the general function on the topology. But yeah, it's really more involved the proof with the representable stats than to extend this to the n-gimatic stats. So this is just from being a subcanon. What? I'm saying is it a consequence of the psyche of a subcanon? Yeah, it's essentially that. In fact, with the assumption on topology, subcanonical is not assumed as adoptive topology. And we show subcanonical by this value. And finally, I would like to give you an explanation of stacks as portions. So 2.3 stacks as portions. So this is the construction, let's say, in more or less in one step. What can we say? So in the classical situation, for example, to construct a general space, we consider another equivalence relation. Here we are in an infinity one context, so we should replace equivalence relations with a gupaider or limits of gupaider, a geometric representation of gupaide objects. So what we are going to use is the notion of CW gupaider. So let me call that for you. A simple object in a monogamy model category is a single gupaide. If the following is satisfied, so if 34, the natural map from Xn to X0, X1, this is n times, is a weak equivalence. And second, the natural map from X to X1. This is again a weak equivalence. So here we have considered these five lines. And if X0 star is a single gupaide, we will say in an internal eye context that X star is an NP-CW gupaide. So if it is a single gupaide and moreover, one or two, thank you. And we can write it in the monotonic order. No, no, no, we can write the product of other series or monotonic order. So X0 and X1 are angular. So in this case, C is the capital of X. And for the map, so the face map from X1 to X0 are in X and in X. Okay, so with this notation, we can write the first arrow, so that in your first point you have an arrow and something is written below. This is a weak equivalence. That is correct. Yeah, we can do that. So I can say it in a theorem. So f is static, the following are equivalent. So one, f is n geometric. And second, there exists a single n minus one p gupaide. So n minus one p single gupaide, such that f is weak equivalent to the geneticization of X star. So this by definition is the monotonic limit of Xn. Okay, so you have to do it not hard. On one direction, if f is n geometric, we take an atlas and we form the check nerve of this atlas. So we have the signal coil and we show that the assumption is required by the method. On the other side, we take an n minus one atlas of X0 and we show using product that is also in the top. So the georegulation, but this n minus one atlas becomes an n atlas for us. Maron, should your groupage be in a stack, not in your site? Yeah, I said that from here on this is stack. Okay. So I was saying that we can put this stuff in order to be able to extend geometry on the base site. So extending the geometry. So essentially two different ways to do this. So in one direction, we can look for an enhancement theory. Okay, I won't discuss this. There will be talks about the global transient complex. And on the other side, we can formally extend the notion of certain classes, of being certain classes of both. So for example, in the both, the derived and the right case, we can take flat maps, smooth maps, ethyl maps with respect to the ethyl side with smooth morphisms and extend the notion of being flat for a map between general and flat. So this is really quite formal. We just need to take you, a class of morphisms such that, is stable under conservation, vegidate and equivalences. And the main assumption is that being in Q is local, so local, both on the domain and on the domain. Then the definition of the, what is that essentially. We will say that the morphism between sides is in Q, in NQ. It is unrepresentable and they say that any representable factor is in N-reformity. So F in G is in NQ if it is, so this is the definition. N is representable and for every x is representable, for every morphism. So x is in G, the base say in Q, so x is for G, x is for N-1, 2. So in exercise NQ, the class of morphisms is stable. So this is my equivalence on what we will go back in competition. And second, NQ is local, local on the domain. So this is what I wanted to say about design and theory. So I like to sketch a few problems of geometric sketching that I can use quickly because they don't discuss it essentially. So three, two, one. Okay, so I won't say that the smooth morphism I have talked about is satisfied with the general sense. This is clear in a sense, otherwise we won't be doing the diagonality. So let me recall the contractions of the transition factor even by Peter yesterday. So here we have, I will work with CTGA, so K is the theta characteristic. So we have inclusion of affine schemes into derived affine schemes of K. Here we have T0, calculation. We can extend this to the category of classical stats of K, I of x. And for example, from here, stale and inclusion. And we can extend T0 by the left hand extension. This means completely that we, for the other fine stats, we define explicitly as, and then we require that T0, commuting coordinates. So this gives a functor here, T0, below the began T0. So the capital of the sponsor, now we can extend that nearly. So this is a one with both the number, Alec, and marginal, te также. And that we call, now we will go full table table. I'm not saying that the additive is not paid out. So second, this here stands for angiometric steps to angiometric steps. Third, in visual, together, I'm saying angiometric steps to angiometric steps. I think more is true for those three. I reflect with, if a stack becomes angiometric by... Okay, so another thing I would like to state is a kind of complexity bound. So assume that F is the right stack, which is angiometric. And assume also that A is a CDGA, okay, homologically bounded. So I denote this to this notation. So A is a homology band which is less than M, because of A. Then F of A is N plus M plus 1, okay. The idea to do this is quite... I like it. So we decompose A using the classical tower, and we observe that the more steps. So we first decompose A with the classical tower. Now I will tell you that the more steps appearing in the classical tower are all square-thin extensions. So since it can be shown that the angiometric step has a good intersection theory, a good infinitesimal hypothesis, we can use the intersection theory of A, which will be explained in later talks, I suppose. So use the intersection theory to control the tidal of F at A to... So the M-manifolding theory. And then go by induction, essentially. So the point is that we can show that this fiber is concretely using the intersection theory. So the main idea is that every angle as a positive of the hour, with more fitments, are square-thin extensions, and moreover the section theory allows to control what is the geometric step that allows on the square-thin extensions. So finally we have two four and the intersection. So the last of the three, so the position, always in the right section, so F is the left section, which is angiometric step, then the cohesion of the junction, so this is the presentable. So when I say the presentable, I mean n-presentable for some n, in fact here. So this is first and second for a v-a, which is angle c-v-a, and for every flat morphism from the spectrum, flat them. So we are now going to put that back. And this is essentially saying by definition that, if we are going to use the n-presentable for some n-presentable for some n-presentable for some n-presentable, then we are going to use the n-presentable for some n-presentable. So this is one. So by definition, I've done something like this, instead of pi-0.8, you can do any g-agegraph B, so that the bar from pi-0.8 pi-0.p is subject to. So first of all, let me remark that the n-geometric, and the n-geometric, in the other case, n-geometric, and by n-presentable. So the idea of this proof, the proof is straightforward. The key point is, in fact, the representable facts are zero from n, which is forced into the other case. So this allows to use n-gometric in a function to also, so I can also give this a few. So the base stat is okay, and for n-presentable equal to n-zero, then we fix up and find the representable stat. Can you explain to me the difference between this level and that proposition? What? Sorry? Can you tell me about different contexts for n-zero from the proposition that you stated about? You want to know the definition of this? No, he said that now it's possible to case the proposition. Ah, yeah, yeah, okay. It's a way to put people together. Yeah, yeah, yeah, we're using the input. Yeah, exactly. Yeah, okay. In this case, the proof is much easier. Just use the root space to use n-active flow. So you don't need positive n-optic composition, you don't need the abstraction. You start from the left, huh? And you use the picture theory to see how the proposition is going to be. Okay, there's a lot of further problems. Yeah. No, but okay, I'm discussing on the idea because I want to do the example again. Otherwise, now you've got the position of the other way. Okay, so I will just skip the room for a minute. So the first example I want to discuss is island attack in authors. So we work, for example, in this context with the category of fine schemes, fine schemes with the SVPF probability. Okay, so we fix a sheath of a median use, okay, which goes from commutative calculus to set, sorry, a sheath. So we can define a three sheaths, k with a n, which goes from this time, add to the solution set, setting an island of commutative calculus of r, k of a r and work. This means the island and the space associated with this group, a region group. So a is a... Okay, so yeah, okay, I could... Okay, so this is not the set, but we can specify that. And we can also describe the part of this and the distribution. So I will discuss simply the case of any part one, but it is possible to tonight this construction, just a bit more after. Okay, so using the way to specify definition, k, a, n is the associated stack to k, d, n. So to... So the fact is, basically, there's an associated stack. We use the notion of towards a sheath of a median groups, of groups in fact. So definition, f stack, f is a toaster. Okay, if it has a left a action, okay, m over and class. So for every affine action, I mean action. Yeah, I think it's a median. Okay, because in the case of any part one, we can also pick a monogamy on a sheath. So what I was going to ask is, is it an action in a strict sense or is it a mean sense? I'm the... No, I just want a... So the same thing you mean by action? Yeah, I want a more fit like that. Okay, so for every representable stack, f is a mean action. For every representable stack, f... There exists a cover. Such that f of ui is different from the end set. And second, the natural map... This should be a... Yeah, okay. So with this, we can form a respect in a sheath. So tors A, which goes from commutative algebra to... And here I'm sending r to... So, okay, we have an ocean of tors over r, so we take... This is like a chore that's over the thermal object. Yeah, I'm okay. Sorry. Because you're introducing a... I don't want to... This map, though, should be nothing more than a sheath. Yeah, yeah, thank you. And moreover, this should be a... In my reaction. Okay, also write down the absolute motion and then apply it to the... And then basically... So this will be the final object. So I'm ready to come back to this straight of reactions. This is also an aspect of case of the simplification theorem we've seen. So the group, say... This is a sheath, but it's the same. Take a group, take the... You have a point and a pg, a point and g, so it's another point. Like, we can get the controls from these two central sets the same as pg, from each central set. So this is what we're going to do. So simplification, you need to have time to do your work. But that's the... I mean, the definition of this is the definition. So really this is some logic of reaction. The real thing is the reaction is just some way of accepting it. They've opened up the regions. You write down the regions, you get the question, which is the right one. Okay, so... Yeah, we have the... Do you want to say... Of course, you can say we're going to say that I'm right. But water? Does it matter water? If you place iso-pigopics into... Yeah. I don't know. You want to say it's water. We'll take a look at it. Okay. So... Okay. What can be shown is that this simplification of the sheath is a stack. And it comes in with a more similar of simplification of the sheath. One... What's the reaching uses of the... You're just looking at the nerve. I mean, this is a good point. Obviously, you're really just trying to go around the group of users. But then you get the nerve to get a stack of just... Oh, because the thing is... I don't know if you're really getting the group of users. I mean, this is the group of users. No, what I'm saying is that you're not going to get a higher rate. So this is true. This works, I guess, when A is just a really good group of users. You're going to get a higher rate. Yeah. Yeah, if our presence if A is a sheath of essential group, you need to take the nerve of the equal relationship between the drivers. That would be... And is X a fixed, say, stack? Do you want torsions of A's depend on X? Because you're looking at A torsions over X. A torsions over X means A torsions in the slice......of X. So torsions depend on a transverse. A1? A6. Yeah. X1. Yeah, X is all sides. Yeah, yeah, yeah. That makes me a better... Okay, and... If you can work with the great publishers to do the whole installation, America's back to normal to the top. You're getting a longer break. Yes. I guess you said that we need to do the rest. But do you think this is the existence towards A as a sheath? Yeah, I was thinking about it today. We have an internal map, and this actually tors A as the significant amount of the distribution. So we can identify this with K, A, and Y. Okay. And......Tm......Tm... So if......A is an algebraic space, in fact, is representable by an algebraic space......which is flat......over basically the spectrary......Lan......K, A, and N is NR, in angiomagnetic. Well, you need to look at your flight, I suppose. Look at your flight condition, I suppose. You need to have a great space to look at your flight condition. Who was thinking? Unless it's not you. Because if you take an affine, a very big flat flight, which is a lot of flight types, we may have no more, that K, A, and N, if you start floating on a plane, it won't be a PPL, but it will be flat, so... So you really need to be flat, and look at your flight condition. Okay. I guess that this can be global by induction, so the K is N equal to zero, I think it can be far more than N. N equals one is a......1, and this is a term of 14, already. And then you notice that K1 is a group of objects in all these types. And it's smooth for step K. Okay. So then, for a smooth group of objects, it's pretty clear that the classified space can be too big because there is another one at last. So the only significant case is... And all over. It is......more. Okay. Thank you. I hope for N, sit here. A is for smooth stuff. That's a good thing. The classified stuff for flat groups K of flight orientation is a smooth stuff, for the greatest extent. Thank you. So do I have time on time left or......adjusted? And now......just got a time on this list. So? It was five minutes ago. Okay. Let me try this next couple of statements. So your claim is that this second to five-quad force A isn't at K1. Yes. Okay. And the second is, if A is an algebraic space, it's a group object. Or it'll be in the slide. So......if you want to look at some of the security......I'm not an expert. I won't speak about minimal models. Let me just......explore you that a simply connected space is said to be rational. If it's a amount of news by N, K, with respect to the average point, are......a number of points. So......if you want to look at the number of points that are few vector space for every......another......I'm saying that K is significant. So the parameters......so we'll then......show that the simulation sets can be now in a model structure so that rational spaces are a final object. And we then note this model structure by a point in the rational spaces in this way. And also, Freelance shows that there is an equivalence......freelance equivalence between......simply connected to the other. So......this is a model structure. This is a model structure. This is a equivalence. This is realized by four or five states between an......I don't remember four or five. So what I would like to quickly explain is that in that this thing, Luri, proposes a simplification of this proof using what we call rational types. So... Let me give you the conditions. Sorry, go ahead. To be a two vector space, it's not a property or an interstructure. It's not a... So, a priori, Pi i is a revealing group. Maybe abstractly, it doesn't work in some two vector space, but......is that a property or a structure? It's a......I don't know. I've seen that in the Algos. I don't know. Okay. The point of what we're going to say is exactly to say what's happening over a logical ring. There's an absolute statement because the rational truth is one. Unique, it exists. At least we can find it. But you can replace Q by any ring. Are you going to talk about that one? No. Okay. I would like to talk about it to... explain the......Kanavan and Ian. So, cdh......the visual factor. This is a rational factor. Okay. So, this may be......since the Mr. Algos. We have a client at the axis to create it. Okay. We'll be clear later why we are considering in this exacting, so from the other exacting that we are just restricting ourselves as we look at the object. I mean, extra is equivalent to the IZ or the Z-axis. I've connected. No, okay. And they are. Then we require that for any......community value, not differential, any rhythm, so we'll simply be using zero over zero. X, R is simply an act of that. And... Okay. You're saying that it's really just a......not a random size? And saying that this would be defined over a computer debug. So, Louis said that it's equivalent. He said that this is the kind extension of the easy offense. But it will be clear. I hope why we need to put our stuff in this dynamic. Okay. So, we're going to do a dynamic, but as the dynamic bigger than rule, they......is your K cubed, by the way? What is your K cubed? What is my K? K cubed. Yeah. Thank you. Yeah, so, I'm missing a part. If it is......on the......and this board. So, there's a vector cube, vector face. V, such that R going into pi and X, R is R sent into 3, and so Q, R. So, this by definition, rational type. This is not too far from being a stack on the rules. For example, if all the......a lot of use of X are of an entire......then it can be shown that it is a graphite stack. So, something that could have been presented by a commutative algebra. Commutative address, and data are concentrated in both the two. So, that statement is pi and X are for any case point. What is that? Like, X of R is a point that you're saying with M face point? I can't respond to that. Oh, I didn't see that. Yeah, I had to... But the same is that the point is unique, because of the conditions of the rules. It's simply connected and this pi and X faces. So, the option to have a point is like coherent chronology of spec Q is question directed space. Which is 0. So, what do you think they connected? No, what I mean is that the point is really unique, in the sense that I think the space of... I don't know how to say that. Ah, we mean it's... Okay, you don't need to do a point. A point always exists. You don't need to do a point. If you just mean that the shift pilot is a point, and together with the other conditions, and I feel you can even have I1 as soon as it is like... There are no options, but we have a rule of... Okay. So, we really don't have to have a place one to make. Okay. And... So, what I would like to state, to confuse, is that whenever we have given an irrational type of this kind, so we have a point, we can take the... So, we have Q, going into... I've just seen it as a representable type. Okay. So, we can take the form of condition of X at the point Q. This is... The X is nowhere. What's the X is nowhere? It's an irrational type. What do you mean by... You're for sure F. No, what I mean is what you mean by rational condition of a stack. It's essentially the same thing you do when you associate to a stack. It's form of stack at the point. So, now, okay, I think it is... Let me call this F, Fs. There is a... R is a... F, R over FQ, the point S. This right here. It can be shown that rational type, this has good properties, which makes this into something that looks like a deformation factor. So, in the language of the way this is a formal model, I told them. And we got the... So, this is the result of proving duckpan. And it states essentially that if we start with the formal model, I told them, and we take the tangent spectrum, F, then we realize an equivalence of capitalism. What is proven in duckpan is that this cylinder of functions realizes the equivalent between rational spaces and the general answer. Okay, so this is what I want to state. To design this condition, you probably need an R to give a mathematical answer. I guess it's a little bit of a question. The completion... The question is, does the map from R to FQ, well, do you have a map from R to Q, right? I have... Yeah, because Q is the final object. I can see that. So, sorry, it's the initial object in the category of... I mean, you just need an argumentation so you can talk about the point of R rather than taking the tangent spectrum. Yeah. Yeah, so R is an argument. Sorry. R is an argument to shoot down to R. You're telling us what F hat evaluates on R, and I think R needs to be augmented. I'm going to just give the answer back to R. I think you can respect R to Q, and I think you can respect R to Q, and I'll just respond to what R is. I'll wait for you to do the Q. Wait, I'll do it. I'll do it. You know what point is a subline, or things I'll measure over Q. I mean, I mean, I mean, sorry. Yeah. Yeah, yeah. The key thing is that we're like problems. Rt final, Rt new, CgA, which are kind of augmented. Yeah. Why is the cut cannot be augmented? So, maybe... No, how's it going to... Is it going to look like an extension? Yeah. So, I think space is conflactable, it is. Well... So, in any case, a cut leaves a dust-beam right moment, so it's defined on when it's augmented Q. Yeah. Not anything easy. Yeah, and taking them on. So, if we are working on the capital... on the capital stacks, we have to say they don't combine in parameters, and we get a more efficient problem at R to Fq. No, let's go back to Q. Just because you have such a... you have a more efficient problem. Yeah. All of a sudden. Okay, okay, okay. So, find it should be here. So, we find it here, and in CgA, augmented or augmented. Right? Yes. So, there is a... I mean, there are global versions of this thing. At least there are some people like this. You know, you can always make a... take... you can make a positive image, or no negative image. And there is a spec filter, so, since one of these things would be, say, to map from B dash, you have to have, say, an anti-vape stack, which is your control on rings. Okay? You can show that this control has to be taken from a positive-grade digital response, or K, K is a ring of 30, 0, 2, and an abstract image. Okay. You can compose that to the inclusion, you know. Okay. And you can characterize the image. And isn't the co-development of the stack here on A length? On A length, where A now is inclusive. Yes. And you can show that when K is 2, you can show that this image, I mean, if you are interested to connect it, and simply connect it to G, you can do that with the GD expression on the P-dice. And it's kind of a dysfunction. But in general, you get much more interest. You can get a portion of that find when you include the group schemes. Okay. This is again in the image. So that's why you get, you know, all the fine schemes, like A2 minus 0 is co-affining this sense. Okay. So, I mean, what I mean is, I just don't need it to go to the form of 30, when that first expression of most B30, that's in two steps. Global steps. Okay. Thank you. No, personal comment? So, with the definitions of energy metric, so, I mean, it's inductive. It seems that if you start with the energy metric stack, that, I mean, using the induction and all the addresses you have, you could build a hypercover by appliance. Is the co-skill the functionality of the cybercover related to the end energy metric? So, if you can hand me what you are saying here. Well, I'm saying if I start with, like, we have an encoder of an engine and a stack. Right. To pull a component check, you know, right? Yeah. It starts off at the time that it stops being an appliance, then where it stops, I can then find another cover by appliance and then I can do this and build a hypercover. Yeah. And then you can ask, you know, like, how, how, how, how, how, how can we be truncated if this is cybercover and if it's sort of like, asking, like, at what points can I then restrict to the else we'll see built up and then extend the same thing? And is that, is that, that point will be some positive number? And that's end or close way to end? I, I know I have to think about that. Okay. It sounds very close to me. It can be as close to, as far as, as possible, perhaps, but it sounds very close to me. It sounds very close to me. Yeah. What you can do in cases like, if you have a, if you have an answer, you can always present it as a central scheme. So deciding this post-credits condition plus some, you know, smoothness of the whole extension, where the end is actually this end. So the, the close-coded obvious is again, yeah, but it's a central scheme, so I can find a scheme which is a central scheme. And that's so good, that's really, that's really what I was trying to do. Yeah, I'm going to say that. Okay. This is, this is pretty simple. This is pretty simple. I feel like the central presentation starts. Yeah. And more versions. Are you done? That's it.
 Let's start with the engine motivation for the gematic stack.  So far, many people use these motions, and after this talk, many people will use their examples.  So I would like to motivate in a conceptual way, and lead for gematic stacks.  So a basic motivation to consider stacks is that when we work in the category of schemes,  we have objects which have a nice gematic distinction, nice gematic algorithm,  but in fact they are not like schemes.  So I think the object which can be described as a portion by the color and the definition, for example.  Number one, natural of the user to consider patterns of the objects is to fully complete the category of schemes,  taking the data sheets. So we are considering a unit and that into the data sheets.  But I would like to stress that this obligation is quite brutal.  So if this destroys completely, on the geometry we already have here.  So let me give a simple example.  We take X and Y schemes.  Inside the category of schemes we get data that you don't union.  So here we can consider either the presentable issues associated to X, the ones set to Y,  we take the list of the unit and the category of schemes,  or we can consider the issues associated to the list of the unit.  So what generation? They are quite different in fact.  To complete this fact, just evaluate the two of them over two points.  In this case we have two points on X or one point on Y.  While in this case we can get also the points on X and the point on Y.  So they are different.  So to recover the geometry here, the way you wrote in the group that did that,  the way we do that today is to fix a topology  and to think of a topology on schemes, a good topology obviously,  as the assignment comes, which should be become or should rename or say,  call limits in the category of the schemes.  So there is a nice categorical way to describe this.  I won't be busy today.  So let me just say that we are now in the category of schemes with respect to topology.  So which can be the topology.  So at this point, this category is quite different.  And we can think of objects here as nice gluing of objects in schemes,  but the point is that we cannot really control the way we glue objects.  So there is an element amount of geometry that we can extend from the category of schemes.  So anyway, to solve this problem is to restrict our style to a full subcategory of this category  and to specify how to glue the objects in a sort to give you a rule to get objects in here.  So this is what I will explain in this talk.  The rule I'm talking about is the inductive definition of geometric sets, essentially.  So to give the last example and then to move to the main part of the talk,  many of you will know the glue line for schemes, classical schemes.  This has a simple version, this one, so if we have two schemes and opening motions like this,  then we can glue the objects along these opening motions.  So here we are considering quite particular maps which are in fact opening motions.  So in this exact thing we will change the patching rule.  In fact, we will consider multiple types of maps which in fact, in the case of classical,  are just to be taken as movement.  And this will give many full results.  So let me start now for real, so one notation.  So since I have to work both in the right and the left side thing,  I will just follow the general approach of the IP2,  and we will see the connection in a general context and it will be there already.  So let me just fix my station.  So first of all, we need a category of algebras,  so this is done by considering, together with the classical information and that.  So this will be a symmetric curve monomial,  so this is a category, close.  Satisfying some sense, I would like to stress that one of the most important is the fact that  commutative objects inside the category C  as a transfer of the structure from C.  This is not obvious.  For example, if you work with this differential gradient algebra,  you will see that it is only at 360, and really require that the adjuncts to see,  the forgotful and the symmetric, and be used to respect the structure.  There are ways to solve the problem, for example, considering a simple algebra,  or instead of using commutative objects, using an infinite object,  I will just assume this.  Next, we consider the category of a fine object in C,  which is defined simply as the opposite of commutative other objects.  And we need to fix the genetic information we want to preserve,  so we need a model, a trait of biology,  and we are also satisfying some nice conditions, so model trait of biology.  For example, we want this topology to be super-sacanaka,  and the end-class commutative case.  And finally, we need a class of maps,  which play there in my toy example of opening a merge,  there will be smooth morphism, so P is a class of morphism,  in a fine object, such that it's compatible in some way with tau,  such that this means, first of all, that if we have a cover,  so if this is a cover, we respect the tau,  then ui tau is in P,  and next we want that P is stable under equivalences, under for most people back,  and under implementation.  This is a reasonable assumption,  and at last we want that the range in P is local on the plane.  And there is a last technical assumption, but I won't need it,  so let's see if it works.  So I will consider in this talk the get-out side in both the range and the range,  so one, a tau, plus tau,  plus tau plus tau, with P in both the smooth maps,  and second, the tau,  I'm going to do a five-step, but to me, again, small steps.  The definition of this thing is that the juggernaut is called yesterday.  Okay?  So now I can give you the definition of the juggernaut.  Starts.  And I was saying it is an inactive definition.  You are going to see that.  So, this factor is minus one, dramatic, if it is a juggernaut.  Second, the morphism of stacks is minus one, representable.  If for every representable stack and for every map from x to g,  the base changes over in this equation.  Okay?  So, this equation is representable for f for all the people back to this representable.  And third, f morphism of stacks is minus one, p.  If it is minus one, representable.  And, moreover, in the diagram before, so, at g, x.  This map, as we call it, alpha, is in p.  So, this is a map between representable stacks, so, it is already defined.  It doesn't mean anything.  So, now we can do the reduction stack, so, that n bigger than zero.  So, for example, we need the notion of nr plus.  So, nr plus of stack f is a family, ui of representable stacks,  together with morphism from ui to f, such that each morphism is in n minus one, p.  And, moreover, the total morphism from the disdon't human to f  is subtractable as morphism of stacks.  So, it is subtractable by zero.  It is an equivalent.  Okay?  So, this is the definition of atlas.  We say that a stack, alpha, is n geometric.  If it has nr plus and the diagram is n minus one representable.  Okay?  So, now morphism at your map to g is n representable, or you can image it.  It's also classical situation at g.  So, first, here we have representable stack, representable, and map to base change.  And we want this to be an n minus one stack.  And then both n geometric.  And finally, as before, so f, we call this alpha again, so f to g is in n, p.  Okay?  So, in this situation, this is an atlas of this stack is nr plus ui of the range.  Okay?  Subtract each composite, so ui go to this.  And then on with alpha and x is n, p.  Yeah, in n, how do you think, yeah, it's a morph in between representable stack.  So, the last point may seem a little bit bold, but the point is that we want a day in nt to be a local stack.  So, I will develop some of the basic properties from the consequences of this.  So, all these things implicitly have them for all x, y, g.  Yeah, yeah.  For every x and for every map.  It's the same of the previous situation.  I said that for every x and for every map.  For example.  I thought I said y equals q.  So, for map.  Not after this.  Okay, so, first of all, being n minus 1 representable, or respectively being in n minus 1 p implies that it is n representable.  And respectively n p.  Okay, for n non-physical.  Second of all, n representable and n morphism, okay,  Okay, so,reb petty and unrep wealthy and unp росgy and unprom p Kathy involved on the  And these two properties, we can already  deduce that n-stats is stable under pullbacks,  homotopy pullbacks. So n-gimatic, flat, stable under homotopy pullback.  And finally, fold property, this is quite useful.  So this is the analog of the well-known affinity here  of the classical piece. So the factor is a morph between stats,  and g is n-gimatic. And moreover, we can find an aphelus of g  so that if the state is an n-gimatic, then f is an n-gimatic.  And aphelus of g such that if they change, so f, among the people  of the circle is n-gimatic, then f is n-gimatic.  If you watch about the look of this pointer, this is done as editing by induction.  In this case, the hard point is to do the representable case,  and it's followed by the general function on the topology.  But yeah, it's really more involved the proof with the representable stats  than to extend this to the n-gimatic stats.  So this is just from being a subcanon.  What?  I'm saying is it a consequence of the psyche of a subcanon?  Yeah, it's essentially that.  In fact, with the assumption on topology,  subcanonical is not assumed as adoptive topology.  And we show subcanonical by this value.  And finally, I would like to give you an explanation of stacks as portions.  So 2.3 stacks as portions.  So this is the construction, let's say, in more or less in one step.  What can we say?  So in the classical situation, for example, to construct a general space,  we consider another equivalence relation.  Here we are in an infinity one context,  so we should replace equivalence relations with a gupaider  or limits of gupaider,  a geometric representation of gupaide objects.  So what we are going to use is the notion of CW gupaider.  So let me call that for you.  A simple object in a monogamy model category is a single gupaide.  If the following is satisfied, so if 34, the natural map from Xn to X0,  X1, this is n times, is a weak equivalence.  And second, the natural map from X to X1.  This is again a weak equivalence.  So here we have considered these five lines.  And if X0 star is a single gupaide,  we will say in an internal eye context that X star is an NP-CW gupaide.  So if it is a single gupaide and moreover,  one or two, thank you.  And we can write it in the monotonic order.  No, no, no, we can write the product of other series or monotonic order.  So X0 and X1 are angular.  So in this case, C is the capital of X.  And for the map, so the face map from X1 to X0 are in X and in X.  Okay, so with this notation,  we can write the first arrow, so that in your first point you have an arrow and something is written below.  This is a weak equivalence.  That is correct.  Yeah, we can do that.  So I can say it in a theorem.  So f is static, the following are equivalent.  So one, f is n geometric.  And second, there exists a single n minus one p gupaide.  So n minus one p single gupaide, such that f is weak equivalent to the geneticization of X star.  So this by definition is the monotonic limit of Xn.  Okay, so you have to do it not hard.  On one direction, if f is n geometric, we take an atlas and we form the check nerve of this atlas.  So we have the signal coil and we show that the assumption is required by the method.  On the other side, we take an n minus one atlas of X0 and we show using product that is also in the top.  So the georegulation, but this n minus one atlas becomes an n atlas for us.  Maron, should your groupage be in a stack, not in your site?  Yeah, I said that from here on this is stack.  Okay.  So I was saying that we can put this stuff in order to be able to extend geometry on the base site.  So extending the geometry.  So essentially two different ways to do this.  So in one direction, we can look for an enhancement theory.  Okay, I won't discuss this.  There will be talks about the global transient complex.  And on the other side, we can formally extend the notion of certain classes, of being certain classes of both.  So for example, in the both, the derived and the right case, we can take flat maps,  smooth maps, ethyl maps with respect to the ethyl side with smooth morphisms and extend the notion of being flat for a map between general and flat.  So this is really quite formal.  We just need to take you, a class of morphisms such that,  is stable under conservation, vegidate and equivalences.  And the main assumption is that being in Q is local,  so local, both on the domain and on the domain.  Then the definition of the, what is that essentially.  We will say that the morphism between sides is in Q, in NQ.  It is unrepresentable and they say that any representable factor is in N-reformity.  So F in G is in NQ if it is, so this is the definition.  N is representable and for every x is representable,  for every morphism.  So x is in G, the base say in Q, so x is for G, x is for N-1, 2.  So in exercise NQ, the class of morphisms is stable.  So this is my equivalence on what we will go back in competition.  And second, NQ is local, local on the domain.  So this is what I wanted to say about design and theory.  So I like to sketch a few problems of geometric sketching that I can use quickly  because they don't discuss it essentially.  So three, two, one.  Okay, so I won't say that the smooth morphism I have talked about is satisfied with the general sense.  This is clear in a sense, otherwise we won't be doing the diagonality.  So let me recall the contractions of the transition factor even by Peter yesterday.  So here we have, I will work with CTGA, so K is the theta characteristic.  So we have inclusion of affine schemes into derived affine schemes of K.  Here we have T0, calculation.  We can extend this to the category of classical stats of K, I of x.  And for example, from here, stale and inclusion.  And we can extend T0 by the left hand extension.  This means completely that we, for the other fine stats, we define explicitly as,  and then we require that T0, commuting coordinates.  So this gives a functor here, T0, below the began T0.  So the capital of the sponsor, now we can extend that nearly.  So this is a one with both the number, Alec, and marginal, te также.  And that we call, now we will go full table table.  I'm not saying that the additive is not paid out.  So second, this here stands for angiometric  steps to angiometric steps. Third, in visual, together,  I'm saying angiometric steps to angiometric steps.  I think more is true for those three. I reflect with, if a stack becomes angiometric by...  Okay, so another thing I would like to state is a kind of complexity bound.  So assume that F is the right stack, which is angiometric.  And assume also that A is a CDGA, okay, homologically bounded.  So I denote this to this notation.  So A is a homology band which is less than M, because of A.  Then F of A is N plus M plus 1, okay.  The idea to do this is quite... I like it.  So we decompose A using the classical tower, and we observe that the more steps.  So we first decompose A with the classical tower.  Now I will tell you that the more steps appearing in the classical tower are all square-thin extensions.  So since it can be shown that the angiometric step has a good intersection theory,  a good infinitesimal hypothesis, we can use the intersection theory of A,  which will be explained in later talks, I suppose.  So use the intersection theory to control the tidal of F at A to...  So the M-manifolding theory.  And then go by induction, essentially.  So the point is that we can show that this fiber is concretely using the intersection theory.  So the main idea is that every angle as a positive of the hour,  with more fitments, are square-thin extensions, and moreover the section theory  allows to control what is the geometric step that allows on the square-thin extensions.  So finally we have two four and the intersection.  So the last of the three, so the position, always in the right section, so F is the left section,  which is angiometric step, then the cohesion of the junction, so this is the presentable.  So when I say the presentable, I mean n-presentable for some n, in fact here.  So this is first and second for a v-a, which is angle c-v-a, and for every flat morphism  from the spectrum, flat them.  So we are now going to put that back.  And this is essentially saying by definition that,  if we are going to use the n-presentable for some n-presentable for some n-presentable  for some n-presentable, then we are going to use the n-presentable for some n-presentable.  So this is one.  So by definition, I've done something like this, instead of pi-0.8, you can do any g-agegraph B,  so that the bar from pi-0.8 pi-0.p is subject to.  So first of all, let me remark that the n-geometric,  and the n-geometric, in the other case, n-geometric, and by n-presentable.  So the idea of this proof, the proof is straightforward.  The key point is, in fact, the representable facts are zero from n, which is forced into the other case.  So this allows to use n-gometric in a function to also, so I can also give this a few.  So the base stat is okay, and for n-presentable equal to n-zero, then we fix up and find the representable stat.  Can you explain to me the difference between this level and that proposition?  What? Sorry?  Can you tell me about different contexts for n-zero from the proposition that you stated about?  You want to know the definition of this?  No, he said that now it's possible to case the proposition.  Ah, yeah, yeah, okay.  It's a way to put people together.  Yeah, yeah, yeah, we're using the input.  Yeah, exactly.  Yeah, okay.  In this case, the proof is much easier. Just use the root space to use n-active flow.  So you don't need positive n-optic composition, you don't need the abstraction.  You start from the left, huh?  And you use the picture theory to see how the proposition is going to be.  Okay, there's a lot of further problems.  Yeah.  No, but okay, I'm discussing on the idea because I want to do the example again.  Otherwise, now you've got the position of the other way.  Okay, so I will just skip the room for a minute.  So the first example I want to discuss is island attack in authors.  So we work, for example, in this context with the category of fine schemes,  fine schemes with the SVPF probability.  Okay, so we fix a sheath of a median use, okay, which goes from commutative calculus to set, sorry, a sheath.  So we can define a three sheaths, k with a n, which goes from this time,  add to the solution set, setting an island of commutative calculus of r,  k of a r and work.  This means the island and the space associated with this group, a region group.  So a is a...  Okay, so yeah, okay, I could...  Okay, so this is not the set, but we can specify that.  And we can also describe the part of this and the distribution.  So I will discuss simply the case of any part one, but it is possible to tonight this construction, just a bit more after.  Okay, so using the way to specify definition, k, a, n is the associated stack to k, d, n.  So to...  So the fact is, basically, there's an associated stack.  We use the notion of towards a sheath of a median groups, of groups in fact.  So definition, f stack, f is a toaster.  Okay, if it has a left a action, okay, m over and class.  So for every affine action, I mean action.  Yeah, I think it's a median.  Okay, because in the case of any part one, we can also pick a monogamy on a sheath.  So what I was going to ask is, is it an action in a strict sense or is it a mean sense?  I'm the...  No, I just want a...  So the same thing you mean by action?  Yeah, I want a more fit like that.  Okay, so for every representable stack, f is a mean action.  For every representable stack, f...  There exists a cover.  Such that f of ui is different from the end set.  And second, the natural map...  This should be a...  Yeah, okay.  So with this, we can form a respect in a sheath.  So tors A, which goes from commutative algebra to...  And here I'm sending r to...  So, okay, we have an ocean of tors over r, so we take...  This is like a chore that's over the thermal object.  Yeah, I'm okay.  Sorry.  Because you're introducing a...  I don't want to...  This map, though, should be nothing more than a sheath.  Yeah, yeah, thank you.  And moreover, this should be a...  In my reaction.  Okay, also write down the absolute motion and then apply it to the...  And then basically...  So this will be the final object.  So I'm ready to come back to this straight of reactions.  This is also an aspect of case of the simplification theorem we've seen.  So the group, say...  This is a sheath, but it's the same.  Take a group, take the...  You have a point and a pg, a point and g, so it's another point.  Like, we can get the controls from these two central sets  the same as pg, from each central set.  So this is what we're going to do.  So simplification, you need to have time to do your work.  But that's the...  I mean, the definition of this is the definition.  So really this is some logic of reaction.  The real thing is the reaction is just some way of accepting it.  They've opened up the regions.  You write down the regions, you get the question, which is the right one.  Okay, so...  Yeah, we have the...  Do you want to say...  Of course, you can say we're going to say that I'm right.  But water? Does it matter water?  If you place iso-pigopics into...  Yeah.  I don't know.  You want to say it's water.  We'll take a look at it.  Okay.  So...  Okay.  What can be shown is that this simplification of the sheath is a stack.  And it comes in with a more similar of simplification of the sheath.  One...  What's the reaching uses of the...  You're just looking at the nerve.  I mean, this is a good point.  Obviously, you're really just trying to go around the group of users.  But then you get the nerve to get a stack of just...  Oh, because the thing is...  I don't know if you're really getting the group of users.  I mean, this is the group of users.  No, what I'm saying is that you're not going to get a higher rate.  So this is true. This works, I guess, when A is just a really good group of users.  You're going to get a higher rate.  Yeah.  Yeah, if our presence if A is a sheath of essential group,  you need to take the nerve of the equal relationship between the drivers.  That would be...  And is X a fixed, say, stack?  Do you want torsions of A's depend on X?  Because you're looking at A torsions over X.  A torsions over X means A torsions in the slice... ...of X. So torsions depend on a transverse.  A1?  A6.  Yeah.  X1.  Yeah, X is all sides.  Yeah, yeah, yeah.  That makes me a better...  Okay, and...  If you can work with the great publishers to do the whole installation,  America's back to normal to the top.  You're getting a longer break.  Yes.  I guess you said that we need to do the rest.  But do you think this is the existence towards A as a sheath?  Yeah, I was thinking about it today.  We have an internal map, and this actually tors A as the significant amount of the distribution.  So we can identify this with K, A, and Y.  Okay.  And... ...Tm... ...Tm...  So if... ...A is an algebraic space, in fact,  is representable by an algebraic space... ...which is flat... ...over basically the spectrary... ...Lan... ...K, A, and N is NR, in angiomagnetic.  Well, you need to look at your flight, I suppose.  Look at your flight condition, I suppose.  You need to have a great space to look at your flight condition.  Who was thinking?  Unless it's not you.  Because if you take an affine, a very big flat flight,  which is a lot of flight types, we may have no more,  that K, A, and N, if you start floating on a plane,  it won't be a PPL, but it will be flat, so...  So you really need to be flat, and look at your flight condition.  Okay.  I guess that this can be global by induction,  so the K is N equal to zero, I think it can be far more than N.  N equals one is a... ...1, and this is a term of 14, already.  And then you notice that K1 is a group of objects in all these types.  And it's smooth for step K.  Okay. So then, for a smooth group of objects,  it's pretty clear that the classified space can be too big  because there is another one at last.  So the only significant case is...  And all over.  It is... ...more.  Okay.  Thank you.  I hope for N, sit here.  A is for smooth stuff.  That's a good thing.  The classified stuff for flat groups K of flight orientation  is a smooth stuff, for the greatest extent.  Thank you.  So do I have time on time left or... ...adjusted?  And now... ...just got a time on this list.  So?  It was five minutes ago.  Okay.  Let me try this next couple of statements.  So your claim is that this second to five-quad force A isn't at K1.  Yes.  Okay. And the second is, if A is an algebraic space,  it's a group object.  Or it'll be in the slide.  So... ...if you want to look at some of the security... ...I'm not an expert.  I won't speak about minimal models.  Let me just... ...explore you that a simply connected space  is said to be rational.  If it's a amount of news by N, K,  with respect to the average point,  are... ...a number of points.  So... ...if you want to look at the number of points  that are few vector space  for every... ...another... ...I'm saying that K is significant.  So the parameters... ...so we'll then... ...show that the simulation sets  can be now in a model structure  so that rational spaces are a final object.  And we then note this model structure  by a point in the rational spaces  in this way.  And also, Freelance shows that  there is an equivalence... ...freelance equivalence  between... ...simply connected to the other.  So... ...this is a model structure.  This is a model structure.  This is a equivalence.  This is realized by four or five states  between an... ...I don't remember four or five.  So what I would like to quickly explain  is that in that this thing,  Luri,  proposes a simplification of this proof  using what we call  rational types.  So...  Let me give you the conditions.  Sorry, go ahead.  To be a two vector space,  it's not a property or an interstructure.  It's not a...  So, a priori, Pi i is a revealing group.  Maybe abstractly, it doesn't work  in some two vector space, but... ...is that a property or a structure?  It's a... ...I don't know.  I've seen that in the Algos.  I don't know.  Okay.  The point of what we're going to say is exactly  to say what's happening over a logical ring.  There's an absolute statement  because the rational truth is one.  Unique, it exists.  At least we can find it.  But you can replace Q by any ring.  Are you going to talk about that one?  No.  Okay.  I would like to talk about it to...  explain the... ...Kanavan and Ian.  So, cdh... ...the visual factor.  This is a rational factor.  Okay.  So, this may be... ...since the Mr. Algos.  We have a client at the axis to create it.  Okay.  We'll be clear later why we are considering  in this exacting, so from the other exacting  that we are just restricting ourselves  as we look at the object.  I mean,  extra is equivalent to the IZ or the Z-axis.  I've connected.  No, okay.  And they are.  Then we require that  for any... ...community value, not differential,  any rhythm,  so we'll simply be using zero over zero.  X, R is simply an act of that.  And...  Okay.  You're saying that it's really just a... ...not a random size?  And saying that this would be defined  over a computer debug.  So, Louis said that it's equivalent.  He said that this is the kind extension  of the easy offense.  But it will be clear.  I hope why we need to put our stuff in this dynamic.  Okay.  So, we're going to do a dynamic,  but as the dynamic bigger than rule,  they... ...is your K cubed, by the way?  What is your K cubed?  What is my K?  K cubed.  Yeah.  Thank you.  Yeah, so, I'm missing a part.  If it is... ...on the... ...and this board.  So, there's a vector cube, vector face.  V, such that R  going into pi and X, R  is R sent into 3,  and so Q, R.  So, this by definition,  rational type.  This is not too far from being a stack on the rules.  For example, if all the... ...a lot of use of X  are of an entire... ...then it can be shown that  it is a graphite stack.  So, something that could have been presented by a commutative algebra.  Commutative address,  and data are concentrated in both the two.  So, that statement is pi and X are for any case point.  What is that?  Like, X of R is a point that you're saying with M face point?  I can't respond to that.  Oh, I didn't see that.  Yeah, I had to...  But the same is that the point is unique,  because of the conditions of the rules.  It's simply connected and this pi and X faces.  So, the option to have a point is like  coherent chronology of spec Q is question directed space.  Which is 0.  So, what do you think they connected?  No, what I mean is that the point is really unique,  in the sense that I think the space of...  I don't know how to say that.  Ah, we mean it's...  Okay, you don't need to do a point.  A point always exists.  You don't need to do a point.  If you just mean that the shift pilot is a point,  and together with the other conditions,  and I feel you can even have I1 as soon as it is like...  There are no options, but we have a rule of...  Okay.  So, we really don't have to have a place one to make.  Okay.  And...  So, what I would like to state, to confuse,  is that whenever we have given an irrational type of this kind,  so we have a point, we can take the...  So, we have Q, going into...  I've just seen it as a representable type.  Okay.  So, we can take the form of condition of X at the point Q.  This is...  The X is nowhere.  What's the X is nowhere?  It's an irrational type.  What do you mean by...  You're for sure F.  No, what I mean is what you mean by rational condition of a stack.  It's essentially the same thing you do when you associate to a stack.  It's form of stack at the point.  So, now, okay, I think it is...  Let me call this F, Fs.  There is a...  R is a...  F, R  over FQ,  the point S.  This right here.  It can be shown that rational type, this has good properties,  which makes this into something that looks like a deformation factor.  So, in the language of the way this is a formal model, I told them.  And we got the...  So, this is the result of proving duckpan.  And it states essentially that if we start with the formal model, I told them,  and we take the tangent spectrum, F,  then we realize an equivalence of capitalism.  What is proven in duckpan is that this cylinder of functions  realizes the equivalent between rational spaces and the general answer.  Okay, so this is what I want to state.  To design this condition, you probably need an R to give a mathematical answer.  I guess it's a little bit of a question.  The completion...  The question is, does the map from R to FQ,  well, do you have a map from R to Q, right?  I have...  Yeah, because Q is the final object.  I can see that.  So, sorry, it's the initial object in the category of...  I mean, you just need an argumentation so you can talk about the point of R  rather than taking the tangent spectrum.  Yeah.  Yeah, so R is an argument.  Sorry.  R is an argument to shoot down to R.  You're telling us what F hat evaluates on R,  and I think R needs to be augmented.  I'm going to just give the answer back to R.  I think you can respect R to Q,  and I think you can respect R to Q,  and I'll just respond to what R is.  I'll wait for you to do the Q.  Wait, I'll do it.  I'll do it.  You know what point is a subline,  or things I'll measure over Q.  I mean, I mean, I mean, sorry.  Yeah.  Yeah, yeah.  The key thing is that we're like problems.  Rt final, Rt new, CgA,  which are kind of augmented.  Yeah.  Why is the cut cannot be augmented?  So, maybe...  No, how's it going to...  Is it going to look like an extension?  Yeah.  So, I think space is conflactable, it is.  Well...  So, in any case,  a cut leaves a dust-beam right moment,  so it's defined on when it's augmented Q.  Yeah.  Not anything easy.  Yeah, and taking them on.  So, if we are working on the capital...  on the capital stacks,  we have to say they don't combine in parameters,  and we get a more efficient problem  at R to Fq.  No, let's go back to Q.  Just because you have such a...  you have a more efficient problem.  Yeah.  All of a sudden.  Okay, okay, okay.  So, find it should be here.  So, we find it here,  and in CgA, augmented or augmented.  Right? Yes.  So, there is a...  I mean, there are global versions of this thing.  At least there are some people like this.  You know, you can always make a...  take...  you can make a positive image,  or no negative image.  And there is a spec filter,  so, since one of these things would be, say,  to map from B dash,  you have to have, say, an anti-vape stack,  which is your control on rings.  Okay?  You can show that this control has to be taken  from a positive-grade digital response,  or K, K is a ring of 30, 0,  2, and an abstract image.  Okay.  You can compose that to the inclusion, you know.  Okay.  And you can characterize the image.  And isn't the co-development of the stack here  on A length?  On A length, where A now is inclusive.  Yes.  And you can show that when K is 2,  you can show that this image, I mean,  if you are interested to connect it,  and simply connect it to G,  you can do that with the GD expression on the P-dice.  And it's kind of a dysfunction.  But in general, you get much more interest.  You can get a portion of that find  when you include the group schemes.  Okay.  This is again in the image.  So that's why you get, you know,  all the fine schemes, like A2 minus 0  is co-affining this sense.  Okay.  So, I mean, what I mean is, I just don't need it  to go to the form of 30,  when that first expression of most B30,  that's in two steps.  Global steps.  Okay.  Thank you.  No, personal comment?  So, with the definitions of energy metric,  so, I mean, it's inductive.  It seems that if you start with the energy metric stack,  that, I mean, using the induction  and all the addresses you have,  you could build a hypercover by appliance.  Is the co-skill  the functionality of the cybercover  related to the end energy metric?  So, if you can hand me what you are saying here.  Well, I'm saying if I start with, like,  we have an encoder of an engine and a stack.  Right.  To pull a component check, you know, right?  Yeah.  It starts off at the time that it stops being an appliance,  then where it stops,  I can then find another cover by appliance  and then I can do this and build a hypercover.  Yeah.  And then you can ask, you know,  like, how, how,  how, how, how, how can we be truncated  if this is cybercover  and if it's sort of like, asking,  like, at what points can I then restrict  to the else we'll see built up  and then extend the same thing?  And is that, is that, that point  will be some positive number?  And that's end or close way to end?  I, I know I have to think about that.  Okay.  It sounds very close to me.  It can be as close to, as far as,  as possible, perhaps,  but it sounds very close to me.  It sounds very close to me.  Yeah.  What you can do in cases like,  if you have a,  if you have an answer,  you can always present it as a central scheme.  So deciding this post-credits condition  plus some, you know,  smoothness of the whole extension,  where the end is actually this end.  So the, the close-coded obvious is again,  yeah, but it's a central scheme,  so I can find a scheme which is a central scheme.  And that's so good, that's really,  that's really what I was trying to do.  Yeah, I'm going to say that.  Okay.  This is, this is pretty simple.  This is pretty simple.  I feel like the central presentation starts.  Yeah.  And more versions.  Are you done?  That's it.
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Let's start with the engine motivation for the gematic stack. So far, many people use these motions, and after this talk, many people will use their examples. So I would like to motivate in a conceptual way, and lead for gematic stacks. So a basic motivation to consider stacks is that when we work in the category of schemes, we have objects which have a nice gematic distinction, nice gematic algorithm, but in fact they are not like schemes. So I think the object which can be described as a portion by the color and the definition, for example. Number one, natural of the user to consider patterns of the objects is to fully complete the category of schemes, taking the data sheets. So we are considering a unit and that into the data sheets. But I would like to stress that this obligation is quite brutal. So if this destroys completely, on the geometry we already have here. So let me give a simple example. We take X and Y schemes. Inside the category of schemes we get data that you don't union. So here we can consider either the presentable issues associated to X, the ones set to Y, we take the list of the unit and the category of schemes, or we can consider the issues associated to the list of the unit. So what generation? They are quite different in fact. To complete this fact, just evaluate the two of them over two points. In this case we have two points on X or one point on Y. While in this case we can get also the points on X and the point on Y. So they are different. So to recover the geometry here, the way you wrote in the group that did that, the way we do that today is to fix a topology and to think of a topology on schemes, a good topology obviously, as the assignment comes, which should be become or should rename or say, call limits in the category of the schemes. So there is a nice categorical way to describe this. I won't be busy today. So let me just say that we are now in the category of schemes with respect to topology. So which can be the topology. So at this point, this category is quite different. And we can think of objects here as nice gluing of objects in schemes, but the point is that we cannot really control the way we glue objects. So there is an element amount of geometry that we can extend from the category of schemes. So anyway, to solve this problem is to restrict our style to a full subcategory of this category and to specify how to glue the objects in a sort to give you a rule to get objects in here. So this is what I will explain in this talk. The rule I'm talking about is the inductive definition of geometric sets, essentially. So to give the last example and then to move to the main part of the talk, many of you will know the glue line for schemes, classical schemes. This has a simple version, this one, so if we have two schemes and opening motions like this, then we can glue the objects along these opening motions. So here we are considering quite particular maps which are in fact opening motions. So in this exact thing we will change the patching rule. In fact, we will consider multiple types of maps which in fact, in the case of classical, are just to be taken as movement. And this will give many full results. So let me start now for real, so one notation. So since I have to work both in the right and the left side thing, I will just follow the general approach of the IP2, and we will see the connection in a general context and it will be there already. So let me just fix my station. So first of all, we need a category of algebras, so this is done by considering, together with the classical information and that. So this will be a symmetric curve monomial, so this is a category, close. Satisfying some sense, I would like to stress that one of the most important is the fact that commutative objects inside the category C as a transfer of the structure from C. This is not obvious. For example, if you work with this differential gradient algebra, you will see that it is only at 360, and really require that the adjuncts to see, the forgotful and the symmetric, and be used to respect the structure. There are ways to solve the problem, for example, considering a simple algebra, or instead of using commutative objects, using an infinite object, I will just assume this. Next, we consider the category of a fine object in C, which is defined simply as the opposite of commutative other objects. And we need to fix the genetic information we want to preserve, so we need a model, a trait of biology, and we are also satisfying some nice conditions, so model trait of biology. For example, we want this topology to be super-sacanaka, and the end-class commutative case. And finally, we need a class of maps, which play there in my toy example of opening a merge, there will be smooth morphism, so P is a class of morphism, in a fine object, such that it's compatible in some way with tau, such that this means, first of all, that if we have a cover, so if this is a cover, we respect the tau, then ui tau is in P, and next we want that P is stable under equivalences, under for most people back, and under implementation. This is a reasonable assumption, and at last we want that the range in P is local on the plane. And there is a last technical assumption, but I won't need it, so let's see if it works. So I will consider in this talk the get-out side in both the range and the range, so one, a tau, plus tau, plus tau plus tau, with P in both the smooth maps, and second, the tau, I'm going to do a five-step, but to me, again, small steps. The definition of this thing is that the juggernaut is called yesterday. Okay? So now I can give you the definition of the juggernaut. Starts. And I was saying it is an inactive definition. You are going to see that. So, this factor is minus one, dramatic, if it is a juggernaut. Second, the morphism of stacks is minus one, representable. If for every representable stack and for every map from x to g, the base changes over in this equation. Okay? So, this equation is representable for f for all the people back to this representable. And third, f morphism of stacks is minus one, p. If it is minus one, representable. And, moreover, in the diagram before, so, at g, x. This map, as we call it, alpha, is in p. So, this is a map between representable stacks, so, it is already defined. It doesn't mean anything. So, now we can do the reduction stack, so, that n bigger than zero. So, for example, we need the notion of nr plus. So, nr plus of stack f is a family, ui of representable stacks, together with morphism from ui to f, such that each morphism is in n minus one, p. And, moreover, the total morphism from the disdon't human to f is subtractable as morphism of stacks. So, it is subtractable by zero. It is an equivalent. Okay? So, this is the definition of atlas. We say that a stack, alpha, is n geometric. If it has nr plus and the diagram is n minus one representable. Okay? So, now morphism at your map to g is n representable, or you can image it. It's also classical situation at g. So, first, here we have representable stack, representable, and map to base change. And we want this to be an n minus one stack. And then both n geometric. And finally, as before, so f, we call this alpha again, so f to g is in n, p. Okay? So, in this situation, this is an atlas of this stack is nr plus ui of the range. Okay? Subtract each composite, so ui go to this. And then on with alpha and x is n, p. Yeah, in n, how do you think, yeah, it's a morph in between representable stack. So, the last point may seem a little bit bold, but the point is that we want a day in nt to be a local stack. So, I will develop some of the basic properties from the consequences of this. So, all these things implicitly have them for all x, y, g. Yeah, yeah. For every x and for every map. It's the same of the previous situation. I said that for every x and for every map. For example. I thought I said y equals q. So, for map. Not after this. Okay, so, first of all, being n minus 1 representable, or respectively being in n minus 1 p implies that it is n representable. And respectively n p. Okay, for n non-physical. Second of all, n representable and n morphism, okay, Okay, so,reb petty and unrep wealthy and unp росgy and unprom p Kathy involved on the And these two properties, we can already deduce that n-stats is stable under pullbacks, homotopy pullbacks. So n-gimatic, flat, stable under homotopy pullback. And finally, fold property, this is quite useful. So this is the analog of the well-known affinity here of the classical piece. So the factor is a morph between stats, and g is n-gimatic. And moreover, we can find an aphelus of g so that if the state is an n-gimatic, then f is an n-gimatic. And aphelus of g such that if they change, so f, among the people of the circle is n-gimatic, then f is n-gimatic. If you watch about the look of this pointer, this is done as editing by induction. In this case, the hard point is to do the representable case, and it's followed by the general function on the topology. But yeah, it's really more involved the proof with the representable stats than to extend this to the n-gimatic stats. So this is just from being a subcanon. What? I'm saying is it a consequence of the psyche of a subcanon? Yeah, it's essentially that. In fact, with the assumption on topology, subcanonical is not assumed as adoptive topology. And we show subcanonical by this value. And finally, I would like to give you an explanation of stacks as portions. So 2.3 stacks as portions. So this is the construction, let's say, in more or less in one step. What can we say? So in the classical situation, for example, to construct a general space, we consider another equivalence relation. Here we are in an infinity one context, so we should replace equivalence relations with a gupaider or limits of gupaider, a geometric representation of gupaide objects. So what we are going to use is the notion of CW gupaider. So let me call that for you. A simple object in a monogamy model category is a single gupaide. If the following is satisfied, so if 34, the natural map from Xn to X0, X1, this is n times, is a weak equivalence. And second, the natural map from X to X1. This is again a weak equivalence. So here we have considered these five lines. And if X0 star is a single gupaide, we will say in an internal eye context that X star is an NP-CW gupaide. So if it is a single gupaide and moreover, one or two, thank you. And we can write it in the monotonic order. No, no, no, we can write the product of other series or monotonic order. So X0 and X1 are angular. So in this case, C is the capital of X. And for the map, so the face map from X1 to X0 are in X and in X. Okay, so with this notation, we can write the first arrow, so that in your first point you have an arrow and something is written below. This is a weak equivalence. That is correct. Yeah, we can do that. So I can say it in a theorem. So f is static, the following are equivalent. So one, f is n geometric. And second, there exists a single n minus one p gupaide. So n minus one p single gupaide, such that f is weak equivalent to the geneticization of X star. So this by definition is the monotonic limit of Xn. Okay, so you have to do it not hard. On one direction, if f is n geometric, we take an atlas and we form the check nerve of this atlas. So we have the signal coil and we show that the assumption is required by the method. On the other side, we take an n minus one atlas of X0 and we show using product that is also in the top. So the georegulation, but this n minus one atlas becomes an n atlas for us. Maron, should your groupage be in a stack, not in your site? Yeah, I said that from here on this is stack. Okay. So I was saying that we can put this stuff in order to be able to extend geometry on the base site. So extending the geometry. So essentially two different ways to do this. So in one direction, we can look for an enhancement theory. Okay, I won't discuss this. There will be talks about the global transient complex. And on the other side, we can formally extend the notion of certain classes, of being certain classes of both. So for example, in the both, the derived and the right case, we can take flat maps, smooth maps, ethyl maps with respect to the ethyl side with smooth morphisms and extend the notion of being flat for a map between general and flat. So this is really quite formal. We just need to take you, a class of morphisms such that, is stable under conservation, vegidate and equivalences. And the main assumption is that being in Q is local, so local, both on the domain and on the domain. Then the definition of the, what is that essentially. We will say that the morphism between sides is in Q, in NQ. It is unrepresentable and they say that any representable factor is in N-reformity. So F in G is in NQ if it is, so this is the definition. N is representable and for every x is representable, for every morphism. So x is in G, the base say in Q, so x is for G, x is for N-1, 2. So in exercise NQ, the class of morphisms is stable. So this is my equivalence on what we will go back in competition. And second, NQ is local, local on the domain. So this is what I wanted to say about design and theory. So I like to sketch a few problems of geometric sketching that I can use quickly because they don't discuss it essentially. So three, two, one. Okay, so I won't say that the smooth morphism I have talked about is satisfied with the general sense. This is clear in a sense, otherwise we won't be doing the diagonality. So let me recall the contractions of the transition factor even by Peter yesterday. So here we have, I will work with CTGA, so K is the theta characteristic. So we have inclusion of affine schemes into derived affine schemes of K. Here we have T0, calculation. We can extend this to the category of classical stats of K, I of x. And for example, from here, stale and inclusion. And we can extend T0 by the left hand extension. This means completely that we, for the other fine stats, we define explicitly as, and then we require that T0, commuting coordinates. So this gives a functor here, T0, below the began T0. So the capital of the sponsor, now we can extend that nearly. So this is a one with both the number, Alec, and marginal, te также. And that we call, now we will go full table table. I'm not saying that the additive is not paid out. So second, this here stands for angiometric steps to angiometric steps. Third, in visual, together, I'm saying angiometric steps to angiometric steps. I think more is true for those three. I reflect with, if a stack becomes angiometric by... Okay, so another thing I would like to state is a kind of complexity bound. So assume that F is the right stack, which is angiometric. And assume also that A is a CDGA, okay, homologically bounded. So I denote this to this notation. So A is a homology band which is less than M, because of A. Then F of A is N plus M plus 1, okay. The idea to do this is quite... I like it. So we decompose A using the classical tower, and we observe that the more steps. So we first decompose A with the classical tower. Now I will tell you that the more steps appearing in the classical tower are all square-thin extensions. So since it can be shown that the angiometric step has a good intersection theory, a good infinitesimal hypothesis, we can use the intersection theory of A, which will be explained in later talks, I suppose. So use the intersection theory to control the tidal of F at A to... So the M-manifolding theory. And then go by induction, essentially. So the point is that we can show that this fiber is concretely using the intersection theory. So the main idea is that every angle as a positive of the hour, with more fitments, are square-thin extensions, and moreover the section theory allows to control what is the geometric step that allows on the square-thin extensions. So finally we have two four and the intersection. So the last of the three, so the position, always in the right section, so F is the left section, which is angiometric step, then the cohesion of the junction, so this is the presentable. So when I say the presentable, I mean n-presentable for some n, in fact here. So this is first and second for a v-a, which is angle c-v-a, and for every flat morphism from the spectrum, flat them. So we are now going to put that back. And this is essentially saying by definition that, if we are going to use the n-presentable for some n-presentable for some n-presentable for some n-presentable, then we are going to use the n-presentable for some n-presentable. So this is one. So by definition, I've done something like this, instead of pi-0.8, you can do any g-agegraph B, so that the bar from pi-0.8 pi-0.p is subject to. So first of all, let me remark that the n-geometric, and the n-geometric, in the other case, n-geometric, and by n-presentable. So the idea of this proof, the proof is straightforward. The key point is, in fact, the representable facts are zero from n, which is forced into the other case. So this allows to use n-gometric in a function to also, so I can also give this a few. So the base stat is okay, and for n-presentable equal to n-zero, then we fix up and find the representable stat. Can you explain to me the difference between this level and that proposition? What? Sorry? Can you tell me about different contexts for n-zero from the proposition that you stated about? You want to know the definition of this? No, he said that now it's possible to case the proposition. Ah, yeah, yeah, okay. It's a way to put people together. Yeah, yeah, yeah, we're using the input. Yeah, exactly. Yeah, okay. In this case, the proof is much easier. Just use the root space to use n-active flow. So you don't need positive n-optic composition, you don't need the abstraction. You start from the left, huh? And you use the picture theory to see how the proposition is going to be. Okay, there's a lot of further problems. Yeah. No, but okay, I'm discussing on the idea because I want to do the example again. Otherwise, now you've got the position of the other way. Okay, so I will just skip the room for a minute. So the first example I want to discuss is island attack in authors. So we work, for example, in this context with the category of fine schemes, fine schemes with the SVPF probability. Okay, so we fix a sheath of a median use, okay, which goes from commutative calculus to set, sorry, a sheath. So we can define a three sheaths, k with a n, which goes from this time, add to the solution set, setting an island of commutative calculus of r, k of a r and work. This means the island and the space associated with this group, a region group. So a is a... Okay, so yeah, okay, I could... Okay, so this is not the set, but we can specify that. And we can also describe the part of this and the distribution. So I will discuss simply the case of any part one, but it is possible to tonight this construction, just a bit more after. Okay, so using the way to specify definition, k, a, n is the associated stack to k, d, n. So to... So the fact is, basically, there's an associated stack. We use the notion of towards a sheath of a median groups, of groups in fact. So definition, f stack, f is a toaster. Okay, if it has a left a action, okay, m over and class. So for every affine action, I mean action. Yeah, I think it's a median. Okay, because in the case of any part one, we can also pick a monogamy on a sheath. So what I was going to ask is, is it an action in a strict sense or is it a mean sense? I'm the... No, I just want a... So the same thing you mean by action? Yeah, I want a more fit like that. Okay, so for every representable stack, f is a mean action. For every representable stack, f... There exists a cover. Such that f of ui is different from the end set. And second, the natural map... This should be a... Yeah, okay. So with this, we can form a respect in a sheath. So tors A, which goes from commutative algebra to... And here I'm sending r to... So, okay, we have an ocean of tors over r, so we take... This is like a chore that's over the thermal object. Yeah, I'm okay. Sorry. Because you're introducing a... I don't want to... This map, though, should be nothing more than a sheath. Yeah, yeah, thank you. And moreover, this should be a... In my reaction. Okay, also write down the absolute motion and then apply it to the... And then basically... So this will be the final object. So I'm ready to come back to this straight of reactions. This is also an aspect of case of the simplification theorem we've seen. So the group, say... This is a sheath, but it's the same. Take a group, take the... You have a point and a pg, a point and g, so it's another point. Like, we can get the controls from these two central sets the same as pg, from each central set. So this is what we're going to do. So simplification, you need to have time to do your work. But that's the... I mean, the definition of this is the definition. So really this is some logic of reaction. The real thing is the reaction is just some way of accepting it. They've opened up the regions. You write down the regions, you get the question, which is the right one. Okay, so... Yeah, we have the... Do you want to say... Of course, you can say we're going to say that I'm right. But water? Does it matter water? If you place iso-pigopics into... Yeah. I don't know. You want to say it's water. We'll take a look at it. Okay. So... Okay. What can be shown is that this simplification of the sheath is a stack. And it comes in with a more similar of simplification of the sheath. One... What's the reaching uses of the... You're just looking at the nerve. I mean, this is a good point. Obviously, you're really just trying to go around the group of users. But then you get the nerve to get a stack of just... Oh, because the thing is... I don't know if you're really getting the group of users. I mean, this is the group of users. No, what I'm saying is that you're not going to get a higher rate. So this is true. This works, I guess, when A is just a really good group of users. You're going to get a higher rate. Yeah. Yeah, if our presence if A is a sheath of essential group, you need to take the nerve of the equal relationship between the drivers. That would be... And is X a fixed, say, stack? Do you want torsions of A's depend on X? Because you're looking at A torsions over X. A torsions over X means A torsions in the slice......of X. So torsions depend on a transverse. A1? A6. Yeah. X1. Yeah, X is all sides. Yeah, yeah, yeah. That makes me a better... Okay, and... If you can work with the great publishers to do the whole installation, America's back to normal to the top. You're getting a longer break. Yes. I guess you said that we need to do the rest. But do you think this is the existence towards A as a sheath? Yeah, I was thinking about it today. We have an internal map, and this actually tors A as the significant amount of the distribution. So we can identify this with K, A, and Y. Okay. And......Tm......Tm... So if......A is an algebraic space, in fact, is representable by an algebraic space......which is flat......over basically the spectrary......Lan......K, A, and N is NR, in angiomagnetic. Well, you need to look at your flight, I suppose. Look at your flight condition, I suppose. You need to have a great space to look at your flight condition. Who was thinking? Unless it's not you. Because if you take an affine, a very big flat flight, which is a lot of flight types, we may have no more, that K, A, and N, if you start floating on a plane, it won't be a PPL, but it will be flat, so... So you really need to be flat, and look at your flight condition. Okay. I guess that this can be global by induction, so the K is N equal to zero, I think it can be far more than N. N equals one is a......1, and this is a term of 14, already. And then you notice that K1 is a group of objects in all these types. And it's smooth for step K. Okay. So then, for a smooth group of objects, it's pretty clear that the classified space can be too big because there is another one at last. So the only significant case is... And all over. It is......more. Okay. Thank you. I hope for N, sit here. A is for smooth stuff. That's a good thing. The classified stuff for flat groups K of flight orientation is a smooth stuff, for the greatest extent. Thank you. So do I have time on time left or......adjusted? And now......just got a time on this list. So? It was five minutes ago. Okay. Let me try this next couple of statements. So your claim is that this second to five-quad force A isn't at K1. Yes. Okay. And the second is, if A is an algebraic space, it's a group object. Or it'll be in the slide. So......if you want to look at some of the security......I'm not an expert. I won't speak about minimal models. Let me just......explore you that a simply connected space is said to be rational. If it's a amount of news by N, K, with respect to the average point, are......a number of points. So......if you want to look at the number of points that are few vector space for every......another......I'm saying that K is significant. So the parameters......so we'll then......show that the simulation sets can be now in a model structure so that rational spaces are a final object. And we then note this model structure by a point in the rational spaces in this way. And also, Freelance shows that there is an equivalence......freelance equivalence between......simply connected to the other. So......this is a model structure. This is a model structure. This is a equivalence. This is realized by four or five states between an......I don't remember four or five. So what I would like to quickly explain is that in that this thing, Luri, proposes a simplification of this proof using what we call rational types. So... Let me give you the conditions. Sorry, go ahead. To be a two vector space, it's not a property or an interstructure. It's not a... So, a priori, Pi i is a revealing group. Maybe abstractly, it doesn't work in some two vector space, but......is that a property or a structure? It's a......I don't know. I've seen that in the Algos. I don't know. Okay. The point of what we're going to say is exactly to say what's happening over a logical ring. There's an absolute statement because the rational truth is one. Unique, it exists. At least we can find it. But you can replace Q by any ring. Are you going to talk about that one? No. Okay. I would like to talk about it to... explain the......Kanavan and Ian. So, cdh......the visual factor. This is a rational factor. Okay. So, this may be......since the Mr. Algos. We have a client at the axis to create it. Okay. We'll be clear later why we are considering in this exacting, so from the other exacting that we are just restricting ourselves as we look at the object. I mean, extra is equivalent to the IZ or the Z-axis. I've connected. No, okay. And they are. Then we require that for any......community value, not differential, any rhythm, so we'll simply be using zero over zero. X, R is simply an act of that. And... Okay. You're saying that it's really just a......not a random size? And saying that this would be defined over a computer debug. So, Louis said that it's equivalent. He said that this is the kind extension of the easy offense. But it will be clear. I hope why we need to put our stuff in this dynamic. Okay. So, we're going to do a dynamic, but as the dynamic bigger than rule, they......is your K cubed, by the way? What is your K cubed? What is my K? K cubed. Yeah. Thank you. Yeah, so, I'm missing a part. If it is......on the......and this board. So, there's a vector cube, vector face. V, such that R going into pi and X, R is R sent into 3, and so Q, R. So, this by definition, rational type. This is not too far from being a stack on the rules. For example, if all the......a lot of use of X are of an entire......then it can be shown that it is a graphite stack. So, something that could have been presented by a commutative algebra. Commutative address, and data are concentrated in both the two. So, that statement is pi and X are for any case point. What is that? Like, X of R is a point that you're saying with M face point? I can't respond to that. Oh, I didn't see that. Yeah, I had to... But the same is that the point is unique, because of the conditions of the rules. It's simply connected and this pi and X faces. So, the option to have a point is like coherent chronology of spec Q is question directed space. Which is 0. So, what do you think they connected? No, what I mean is that the point is really unique, in the sense that I think the space of... I don't know how to say that. Ah, we mean it's... Okay, you don't need to do a point. A point always exists. You don't need to do a point. If you just mean that the shift pilot is a point, and together with the other conditions, and I feel you can even have I1 as soon as it is like... There are no options, but we have a rule of... Okay. So, we really don't have to have a place one to make. Okay. And... So, what I would like to state, to confuse, is that whenever we have given an irrational type of this kind, so we have a point, we can take the... So, we have Q, going into... I've just seen it as a representable type. Okay. So, we can take the form of condition of X at the point Q. This is... The X is nowhere. What's the X is nowhere? It's an irrational type. What do you mean by... You're for sure F. No, what I mean is what you mean by rational condition of a stack. It's essentially the same thing you do when you associate to a stack. It's form of stack at the point. So, now, okay, I think it is... Let me call this F, Fs. There is a... R is a... F, R over FQ, the point S. This right here. It can be shown that rational type, this has good properties, which makes this into something that looks like a deformation factor. So, in the language of the way this is a formal model, I told them. And we got the... So, this is the result of proving duckpan. And it states essentially that if we start with the formal model, I told them, and we take the tangent spectrum, F, then we realize an equivalence of capitalism. What is proven in duckpan is that this cylinder of functions realizes the equivalent between rational spaces and the general answer. Okay, so this is what I want to state. To design this condition, you probably need an R to give a mathematical answer. I guess it's a little bit of a question. The completion... The question is, does the map from R to FQ, well, do you have a map from R to Q, right? I have... Yeah, because Q is the final object. I can see that. So, sorry, it's the initial object in the category of... I mean, you just need an argumentation so you can talk about the point of R rather than taking the tangent spectrum. Yeah. Yeah, so R is an argument. Sorry. R is an argument to shoot down to R. You're telling us what F hat evaluates on R, and I think R needs to be augmented. I'm going to just give the answer back to R. I think you can respect R to Q, and I think you can respect R to Q, and I'll just respond to what R is. I'll wait for you to do the Q. Wait, I'll do it. I'll do it. You know what point is a subline, or things I'll measure over Q. I mean, I mean, I mean, sorry. Yeah. Yeah, yeah. The key thing is that we're like problems. Rt final, Rt new, CgA, which are kind of augmented. Yeah. Why is the cut cannot be augmented? So, maybe... No, how's it going to... Is it going to look like an extension? Yeah. So, I think space is conflactable, it is. Well... So, in any case, a cut leaves a dust-beam right moment, so it's defined on when it's augmented Q. Yeah. Not anything easy. Yeah, and taking them on. So, if we are working on the capital... on the capital stacks, we have to say they don't combine in parameters, and we get a more efficient problem at R to Fq. No, let's go back to Q. Just because you have such a... you have a more efficient problem. Yeah. All of a sudden. Okay, okay, okay. So, find it should be here. So, we find it here, and in CgA, augmented or augmented. Right? Yes. So, there is a... I mean, there are global versions of this thing. At least there are some people like this. You know, you can always make a... take... you can make a positive image, or no negative image. And there is a spec filter, so, since one of these things would be, say, to map from B dash, you have to have, say, an anti-vape stack, which is your control on rings. Okay? You can show that this control has to be taken from a positive-grade digital response, or K, K is a ring of 30, 0, 2, and an abstract image. Okay. You can compose that to the inclusion, you know. Okay. And you can characterize the image. And isn't the co-development of the stack here on A length? On A length, where A now is inclusive. Yes. And you can show that when K is 2, you can show that this image, I mean, if you are interested to connect it, and simply connect it to G, you can do that with the GD expression on the P-dice. And it's kind of a dysfunction. But in general, you get much more interest. You can get a portion of that find when you include the group schemes. Okay. This is again in the image. So that's why you get, you know, all the fine schemes, like A2 minus 0 is co-affining this sense. Okay. So, I mean, what I mean is, I just don't need it to go to the form of 30, when that first expression of most B30, that's in two steps. Global steps. Okay. Thank you. No, personal comment? So, with the definitions of energy metric, so, I mean, it's inductive. It seems that if you start with the energy metric stack, that, I mean, using the induction and all the addresses you have, you could build a hypercover by appliance. Is the co-skill the functionality of the cybercover related to the end energy metric? So, if you can hand me what you are saying here. Well, I'm saying if I start with, like, we have an encoder of an engine and a stack. Right. To pull a component check, you know, right? Yeah. It starts off at the time that it stops being an appliance, then where it stops, I can then find another cover by appliance and then I can do this and build a hypercover. Yeah. And then you can ask, you know, like, how, how, how, how, how, how can we be truncated if this is cybercover and if it's sort of like, asking, like, at what points can I then restrict to the else we'll see built up and then extend the same thing? And is that, is that, that point will be some positive number? And that's end or close way to end? I, I know I have to think about that. Okay. It sounds very close to me. It can be as close to, as far as, as possible, perhaps, but it sounds very close to me. It sounds very close to me. Yeah. What you can do in cases like, if you have a, if you have an answer, you can always present it as a central scheme. So deciding this post-credits condition plus some, you know, smoothness of the whole extension, where the end is actually this end. So the, the close-coded obvious is again, yeah, but it's a central scheme, so I can find a scheme which is a central scheme. And that's so good, that's really, that's really what I was trying to do. Yeah, I'm going to say that. Okay. This is, this is pretty simple. This is pretty simple. I feel like the central presentation starts. Yeah. And more versions. Are you done? That's it.
under pullbacks, homotopy pullbacks. So say about design and theory. So I like to sketch related to the end energy metric? So, if you can