Examples of derived schemes and derived Artin 1-stacks
Safronov, Pavel
Derived loop spaces, derived Hilb and Quot, derived mapping scheme, derived stack of stable maps, derived stack of vector bundles and local systems.
 Why D is the constant standard associated with Y. So S1 is a nice space, it's a CW complex, and you can write it as S1. So you can write S0, S1. So this is a homotovaculite. And since you're mapping out the opus, you get a homotov limit of mass into X. So the loop space will be a homotov limit. So it's a pionic homotov limit of opus-pax. So in particular we'll see later that this implies that if X is just this metric, then the favor space is also an metric. So another thing that we'll see later is what happens when X is a derived scheme. So example X derived scheme. In fact there's a different description of the loop space. So the loop space will be a so-called linear stack. So the loop space will be a s-morphic to the addition to the tangent model of X. So this is a linear stack. And let me just remind you how you construct total spaces of vector models. So reminder. When you say linear stack, you're referring to T-shift denies itself? Yeah, okay. The whole thing is a linear stack. Yes, so I'm going to remind you that E of f is a perfect complex. So let's say we have a linear stack on X. Then I define the total space to be the stack of some algebra. So it's going to be a symmetric algebra for OX and A-pool. So this is the total space of f. In this case I'm taking the total space of the shift of tangent model, which is a perfect complex. Or I assume it's a perfect complex in this case. So in this case it's easier to understand what all those spaces are. I'm confused about what you mean. So if you say it's a perfect complex, then you mean it's a perfect complex of sheaves on this, right? Yeah. So when you write sin of this, sin of this will be again some sheave of algebra. Right, so this is a relative stack. So given any commutative decay on X, you get a scheme which is going to be affine, we'll get a stack which is going to be affine on the base, on the base stack, X. So this is the map to X and this map is at Y. So yeah, it's true that X is the value of Y times the mean of X. Yeah, that's right. So I wrote the diagram that I haven't used yet. So if you use the diagram to obtain the diagram was this one. So if you apply the map and stack to that, what you get is a limit, matched from a snot on the x, then you have matched from the actual x, this is x, x here, and here we get the wood space. And this is the full back. So the wood space, oh yeah, snot is just a point. So this is x plus x. So to apply some wood space is just a lot of x plus x over x plus x. So for example, if you have some arganium on stack, then the wood space is going to be an arganium on stack. So I know that any topological space at this constant pre-sheaf where you assign to every dg algebra on that space, is there a way to see what the shape of the case of that looks like? I think if your VDA is connected, then it's just that space, and otherwise it's probably like changing the number of rank and points or something like that. If the only way to understand it is by writing your space as some sequence of pushouts and understanding the whole thing. So you're asking about what this means? No, no, no, I guess I want to know, I'm just curious about why sub-V is for arbitrary space. Is really the only way to understand it is the way of writing and saying why is there a CW complex with a sequence of pushouts, and understanding those pull-back to every empty dg. Now if you write down formula for a certain form of formula, you can take the essential size to create a constant neutral machine, and if you simplify and you get the formula that maps space to a type of topology, which I'm not constantly saying. It can get a complicated effect. Okay, if k is another networking space, it's to use a type of topology to function in a constant neutral machine. So for example if k is not an electric field, then the center from our spaces to our stacks is not very critical. Okay. Yeah? So the next examples are some examples from the last public-private join trade, which are called user schemes. I'm sorry, it's for purpose only. Okay. Let's say x is a single predictive. So let's say x is predictive. So let's say we have a particular scheme. Half a particular sheet. So there is a special equilibrium of any sheet. So it's time to be a little characteristic of this sheet, which is the binomial. So since you're a projector, you have some kind of line-lumbel upon an x-ray of that. So this is something binomial. Okay. And now let me define the code scheme with six-kilometer formula. So I'm going to define some modular problem. So on some affine scheme s, this is ordinary affine scheme. So the group of particular sheets, g, to get to the map from map. So particular sheet g on an x-ray is s. So I'm going to define y over s and the map from map to g. So now f is a sheet from x. So f has a y as a sheet from x plus s. So both of these are sheets from x plus s, which is subjective. And the regular polynomial of g is the total polynomial h. So this is the code scheme and something similar is the Hewler scheme. Again, I'm going to solve the classical version. So this is going to prioritize soft schemes of classical styles such that the total polynomial of z is your fixed-kilomial h. So the code scheme is a concept from affine schemes to group weights. So what's the way you work with this? So morphisms from g are preserving this map. Yeah, I was trying to do that. Yeah, thank you, Rufloid. I think we're going to do a set. Yeah, so let me first state that it's in the Rufloid, so now the main theorem would be that it's actually a scheme. And don't z is flat or is it? Yeah, I'm trying. Yeah, so this is a relative to kilo root of 1, relative to s. So z is flat. S. So the theorem is that both kilo and Hewler's schemes are actual schemes. So that wasn't no Hewler's schemes. So although after you already did this, you're going to have lands in group weights. As Bertrand said, it actually lands in sets. So it's a, let's say, a word model of proof. This will be used in the instruction derived. So the main idea of the Klozscham is that there's this theorem of, of Sarah, that the category, it's a part state of theorem. Let me introduce two rings to the model. So that a be the address of all the sessions of n or s. So this is going to be some gradient of theorem. And this is going to be defined to generate it. And then for any sheaf F, I'll define the associated module or this algorithm, which is just the double sections of the two pieces of f. So this is a gradient model. And so the theorem of Sarah is that the category of different sheaves on x is equivalent to the category of rated n modules. So instead of starting studying the career sheaves on x, we can study modules over this single rated algebra. Which means that you can realize the Klozscham side of some response. So this implies that, excuse me, maybe I'm just wondering, shouldn't you mod that portion to the graded, the ones that are already defined as this portion should you mod the other one? If this includes with the modules. Yeah, most of us don't. Okay, okay, that's what you need with that statement. So this is why I think I realized the Klozscham has a sub-schemes on Breslinan. So this is the Breslinan of A modules inside of the monocles of the map. So instead of studying quotients or career sheaves, you can study quotients of M-M or sub-schemes of M-M. And there's a similar Breslinan embedding for the Klozscham. So it's a, once you go to the Breslinan, it's a scheme, the Klozscham will be some close-up scheme for the Breslinan. So it will also be a resinged Breslinan and the same for the Huber scheme. So let me know in the comments, there's a map from the Huber scheme to the Klozscham of just coax. So given sub-schemes C, let me say the embedding is alia, I send it to just the push-forward of the Sartre scheme. It does obviously coax the more I coax. So this is putting in an esomorphism. So in fact the Huber scheme is a case of the Klozscham. It was a flat-shell and a derived level. And let me say a few words about the instruction of derived Huber and Klozscham. Before I do that, so the language of derived Klozscham was, kind of all of this, there was no derived like John Trey at the time. It was late 90s. So they used the language of the DG schemes. So let me end up with the Sartre. So a DG scheme, X and a DORPX is a scheme of S with a sheave of forces of the graded. Cj's such that, so let me say the HNOT of our X dot is X. I'm going to dig a scheme of the Klozscham. So this is not quite the definition actually. The definition is that the sheave is a sheave of OXCGAs. Yeah, that's right. This is a very big difference between DG schemes and DRC's. In the sense that this should be added to the scheme X because that was a closed Sartre-RC. Yeah, so let me now say how to compare DG schemes to derived schemes. Sorry, sorry, sorry. Don't you want to have a matter of more HCOs in the subject, a nice one? Yeah, that should be added to the S-Prix DG scheme. Okay, yeah, I'm not sure. Let me just... These heave and court to have AIDS there... No. So what is this? That? That's not it. OK. So this is the natural map. It has the best electric. OK, so compared to some two right schemes, four to the right stacks. So let me just say that there is a cover. So this is an offline cover. So here is offline. This is offline. And this is a smooth cover. So four to the right stacks. With the offline. So for example, if the backstakes was impacted and separated, then you have such a cover. That way, what you need to do is you can construct certain solution of the game. So you adopt. So this is the check mirror of you. This is the best. So it's sort of called, for example, you, now it's just you, you want you to use x. And so you can construct. So you can construct o x dot of e dot. This is a, a Cosimblissal DGA. So I'm taking my structure sheet, which is now a, which is a CGGA, and taking it on these local sets. This is going to be Cosimblissal DGA because I reverse the arrows. Now I take stack. This is going to be Cosimblissal stack. And I take general improvisation. So what do you take? Yeah, covenant for the, and this is what they'll call theta x. So this is going to be some puncture from DG schemes to draft stacks. So theta is a puncture from DG schemes to draft stacks. And in fact, I can show you the glass in one direction. Okay. So, so what's the plan and, and so you can, and contain the, they construct DG schemes. It's containing. What's containing? Sorry. Containing. Oh, sorry. I guess. Yeah, okay. Yes. So they construct DG schemes. Let me just say derived code scheme as F, and derived code scheme as X, which are derived from the hands-mits. So this theta is a problem to be fully figured out. So that's the problem is that it's not normally to be. So this is a common problem. Which is expected of B, say. I don't think anyone has 40 stack. So which are other derived from hands-mits? In the sense that the truncations of these derived schemes or DG schemes are ordinary code schemes. Okay. So. What do you mean by stack only? So, what does it mean? So, I X dot, so this dot refers to CGA. This dot refers to the simplification of the fusion. So this is going to be a cost-simplification of the object when the stack is going to be simplitical as that. So it's just the stack of this CD chain. So the simplitical stack. The first thing was the simplitical scheme that you've been doing that much. Right. Yes, the suspect is just a Canadian man. So, yes, let me just write down the tangent complexes. So the tangent complex to the Hilbert scheme, the derived Hilbert scheme at some point is Z. So Z is also a radio mass. This is, so it's a collage of the tangent complex. This will be the sheet homology of Z, which was used in the mobile model. And then the tangent complex of the cyber-henshin G in the right-to-boots scheme. So Z is the X from K to the G and K is the kernel of this map. So these tangent complexes coincide into V0. The crucial fact of the fact that the cold scheme, the sugar scheme, the partner schemes are as a partner but not in general. So if you take, for example, a smooth point in X, this tangent complex is going to be constricting to degrees 0 while this tangent complex, you can do the Z resolution, so it's going to be constricting to degrees. So these are not as far as the various things. So this is an example of same ordinary molecular problems with the different derived molecular problems. So let me say a few words. Well, on the occasions of this cold scheme, the clever schemes. Do they have more than one solution? No. So I can write some guess for the cold scheme. I don't know the guess for the clever scheme, but it's not very valid. So what's the question? The more generalization of these schemes. I don't know. Someone wrote to me that he wrote initially PhD, so the text is lost somehow. And the result was that what you can expect is true. Yeah, you were right. So G is strong over F, and then G is subjective. So this map is subjective by now. The tangent complex is going to be Aga. Exactly. So before I do the applications of these schemes, let me do something easier, which is the more questions of both systems. So let me define all of the genes of sigma. So sigma is something that's defined as a doubly complex. To be the mapping stack from sigma B into BB. So that's very close to the point. So T is what kind of... So the ordinary stack of all the systems is the same equation essentially. So this is all this graph in Hassel. Let's try to compute some examples. So you make two examples. So sigma is S1, then you go to the local systems minus 1. So that's the freedom space of EGA. So it's the modular space of bundles together about mortises. So you can show that this is just that. G by G. And that's the same as ordering the second local systems. So this is always the right kind of function applied to the second local systems. And this is an example of these two. So that's an example that people would get the same... Let me work this out. I'll do parentheses the other way. Oh, no, it's that mapping stack. Where? Sorry. I thought you were going to write a kind of local G, apply it to a small bit of same local G of this one, it's a mapping stack. Yeah, exactly. So I'm going to make it into a driver stack. Okay. So I want to compute local systems minus 2. So it's before the write S2 as a pushout. So if you apply a mapping stack into EGA, what you get is a lot of G by S2 is 0.1G plus 0.1G or G by G. So to compute that, we can rewrite this as 0.1G. So again, I'm doing the driver part of the point. So 0.1 plus 0.1G, that's something I would call the base of those states of G. And you can compute this by this little resolution. And this is going to be the V-algebra in degree minus 1. So that's the drive scheme, mon gene. Okay. So that's the drive stack. And the borders, how do you do the other part? So this is the basic solution. So I think Paul will talk about the field of space, the h-carousel, and the field of space. I will talk about that first. Okay, yes, I will try. Okay, so it's in this way. Yeah. You also have the section of pointing the schools right. Yes, it's a top-famous area of a school. Yeah, that's just the field of space. The seam of the empty space by one. That's actually the shift T tangent shift, the last shift, but not the tangent of BG, right? Sorry. Yeah, that's the formula you wrote at the beginning, the field of space is the Tx shift by minus 1, this is the condition of the formula. Yeah, that's the main thing. That's the final step. Okay, final step. Now, there's two simple methods, so the ordinary stack is just BG. So you see that the drive enhancement of local systems is not... so the obvious drive enhancement is not our model. So our model is much better. We'll see that it's actually a symplectic while the drive enhancement is not the original. Okay, so you said last time you said... So, let me see that this is symplectic. Okay, so let's follow the same old system. Now... Well, that's not the case with the data drive. Yeah, I'm not sure. Okay. Well, I think the G is also symplectic, but those are the same systems. Yeah, I don't see anything like that. Yeah, the order is the same. Okay. So, let me now talk about vector models. Okay. Okay, so now, there's much like problem from vector n of x. So this is going to be the mapping stack from x into vector n. So, like Dan and Sam are sort of really fine. And now, here I assume you're actually a similar progressive. Okay. So the way you prove that... Actually, let me put an R in here. The right mapping stack. So the way you prove that... So the proof that it's actually a zero factor and R from one stack is... Yeah, x is something to not write. Okay. I'll write them all in the same as the code scheme. And the risk scheme, they're also something to not write. Yeah, so to prove that it's zero geometric, let me do the following theorem. So I'm going to state a theorem, but the terms will not be defined in this column. So I'll be able to find a way to talk. Okay, so the right Sam asks his n-geometry. Even though the following theorem is satisfied. So first, the truncation of x is... is marking n plus 1 stack. Then x has an abstraction theory, so this term is not defined in this column. So the thing is that x has a global connection complex and it reserves certain diagrams. So it's going to be cartesian, and then x is convergent. So I'm going to write a cdga, the natural map from x. x is a little bit lighter, but I'll keep using x. So x obeyed to the point of the limit of x. So the limitations of that deviating to this of the positive powers is nice and more visible. Okay, so if this term can be satisfied, you can include the x as n-geometry. So in our case, 2 and 3 are some formal checks. So to check that the monominal system of abstraction theory and check that it's convergent, it doesn't involve any geometry while the mon requires some geometry. So here we have to prove... So you don't have to prove that the monominal system is zero geometry. I want to prove that the monominal system is one stack. So we need to produce some atlas for that. So to produce an atlas of retrobondals, x observed that for any vector bundle 1s, its twist is globally generated, generated by double sections. So we'll be reminding you what the journey of the double sections means. So this means that the formal form of the form of creation map from global sections to global sections is super. So along with studying quotients of some machine, so this is all asked to some power, so can you use the order in? Yeah, thanks. So you can use the order in a quotas scheme to produce an atlas. So in this way you produce atlas, and let me just make a remark that if you have a direct quotas scheme, which actually presents some nice monolid problem, you can avoid using that sort of stability theory, and just use the direct quotas. So in this way you can include that, that and x, these zero-gen are actually a couple of them. So the order in a stack is just assigning the category of retrobondals. So this is the one we want. So this is the one. It's the real dimension, and it's your dimension, and it's not the non-standard. And so the last example I want to give you is the one where I space stable maps. So this is essentially what the quotas scheme is. So you can picture all the different directions that this is just OIS and some power. So this map is equivalent to OIS and some power to supply minus m goes to e. So once you fix the trivialization of all the sections, so this kind of model problem is represented by a quotas scheme, then you forget trivialization, and that's your atlas. In this case it's not like the employment type, so you have to take all possible maps. This is going to be something of a model of application. Okay, and so the last example that I want to give you is more less space stable maps. So maybe people that just, I think we see a proof of the possibility of all this. Maybe we can construct a proof of that. Where people say we need a proof of the possibility of perfect complex. So I'm going to write about the proof of this nucleating. Yes, I'm not immediately because it's a problem. So for instance, text doesn't have to be projected before this. It's most proper. It's going to be a little bit different anyway. Okay, ready? I'm going to write a letter for the vector to put it in the case of the vector. I just think it's in the shape. Okay, so let's define what stable maps and pre-stable maps are. So let energy free the ordinary stack. So when I say ordinary it means underlined by the order of pre-stable curves. So it was a pre-stable curve. So it's a proper curve with no singularities. Okay, and now there's a whole sub-stat called mg bar of green. So mg bar is the modularized space of the stack. So I guess it's the ordinary stack. So it's a pre-stable curve. This one is an ordinary stack. This thing is an ordinary stack. So I'm going to say why it's a pre-stable curve. So it's going to be an ordinary stack of stable curves, or just g. So for example, Cp1 is a pre-stable curve, but it's not in the stable. So it has more than one basis. So m0p is just the sum group, so it's not in the stable number. So this is the regular stable means. So this is going to be a pre-stable curve. Okay, so now we have pre-stable and stable curves. And so mg bar, Cs is the lean-on for stack. So this is the pre-stable curve. So let me define the derived modular problems for stable maps and pre-stable maps. So first, since I have an ordinary stack, I can consider this as a derived stack. So I'm going to just apply this function i to that. And looking to find what the pre-stable maps are, I'm going to say there's a universal pattern. So this is the Cp1, Cp2, Cp3. So now let me define the modular stable maps. So the derived space derived in a less than or pre-stable maps equal x. So what we'll just see with x is smitherectate. Okay, so the derived stack will be the mapping stack. So with mapping stack, we have derived stacks for mg3. So it's a map from this universal family into the triple family. So this means that the refibrated stacks are going to be curved. So this rotation is in our stack and you create x. Okay, so now we can consider a certain size factor of this. So for that we reminded a theorem that I think we've seen today. So the pathologies on this stack and on this stack are the same. So let me remind you of a theorem that Zariski opened in the work of a unit, I think at all. Let me just say Zariski opened the sub-stacks of a stack F, some derived stack, are in digestion with Zariski opened sub-stacks of this truncation. So once we have opened the sub-stack of the truncation, in this case mg3, we'll get some open sub-stack of the derived on this case. So what I'm hearing to the modular space is the table maps. So this is the modular space of table maps. Again, these maps can have some of our morphisms, or not as free of morphisms, while table maps are defined to be more by the equation that the other morphisms are equanimated. So we get an open sub-stack of the derived sub-stack of table maps into X. So this one will be the morphoid model. And similarly, we have certain sub-stacks of all mg3x beta inside of mg3x. So if you look at the great element in the psychological group of X, you can look at the maps, such as the fundamental plus the curve maps to beta. So this is going to be an open sub-stack. And again we get mg4x beta and mg3, the derived one with stats, and mg3x beta. And so that's basically the space that we need to collect. Questions? Yeah, these would be also the indexed, right? No, I think you can do a little longer. I have a double question here, and you should have been following that. What's the proof that these are actually stats and something that they satisfy just set? Is that way too complicated to get into? No, I think that's the way I define them. Well, just there is, because they're all stats. Yes, I think the way they're defined. So stats are defined to be 500 objects in some whole category of tree sheaves, or yes, it's in some category of tree sheaves, and just the fact that this is defined to be the final replacement of that. Yeah, for instance, even when the first example, like the metamene, or the hill over a quote, all that, somehow we do really convince. Well, just for the order of the stack with GLM, you know that you can just literally take a point of positive GLM for every reading. You will not get a sheet. So point one GLM plus pi is trivializable GLM torsors. When you shoot pi that you get more than a trivializable GLM torsors. So you have to shoot pi all the time. So if you do a multiplication, whatever you put in any map is that. So there are a lot of stats that didn't shoot pi. So if I ever want to compute something that actually does not say, you know, even something affine, should I in general use a table to replace that something affine by some resolution by tiny open sense, and then just compute. Well, unless you have the right modular function, so you, like for me, GLM, you know the right moment function. It's not trivializable torsors. It's locally trivial torsors. And that thing will be a sheet. So it's the same thing here. You can try to come in the right modular problem, which is going to be a sheet. So in principle, the way that these things work is like, so I have this pre-sheet. I want a sheet of pi. And it's just a geometric know-how to know that the brain-modulant problem how to actually localize this pre-sheet into natural sheet. Right. You have some tools for this. They are some fundamental, you know, there are two or three fundamental stacks you should know. So you can also pull the stacks out of them. So the first one is the stack of stacks. The next goes to stacks where X is a stack. And that adds to the new moment. So what if you write these nothing-stakes that say, you know, this is the data of a current together with a power company of curves together with a map to X. I mean, a current is a particular kind of stack. So they glue and just have to prove that a stack which locally is a curve, that it is locally is a curve which is more or less true by definition. So you can use this kind of argument. Another one is very quickly-fraud descent. So A goes to this de-modulant, is a stack from the PPF or like I just said, what you're saying. So all these plots, things, and components, I mean, they accept 5% of the total. And then there is also a general firm saying that if you have a group point in stacks and take a portion of this group point, there is an explicit formula of the associated stack by saying they classify total sources of all these group points. So generalization of what you said, there is something very general. If you have a group point in the map stack, take a portion of the stack and it is a formula of the group. So then I have a question. So if two examples are given, like a stack to X, you know, stack to over X, or a stack to A, like DG1, A, those are assigned to the activity categories, not to the group points. And you just take the group points in the stack, or what's the answer? I mean, there are stacks in those sets. There are stacks of categories in the stack, but that is the category. So if you really want them, you can let's take to a DG1, DG1, DG1. But another thing is those stacks also glue as stacks, and they are satisfied to set as stacks. They also have some kind of a different, explicit state of addition. Yeah, because we know it won't be on the side. You could affect in for X stack. You said X is something you did with the R-1. Can you work some of these definitions in the case X is a... Can I give you some presentation of this stack? No, the actual stack. I mean, it's not going to be something nice. It's just going to be something... For example, you take SL2 bundles on Q1. So you know that SL2 bundles are code plus O, O1 plus O1, plus O1 is also a chain, and it's not going to be something we describe easily. But the other thing is you can use the add-was to write down where it also gets the split, and it's going to be something you write down.
 Why D is the constant standard associated with Y.  So S1 is a nice space, it's a CW complex, and you can write it as S1.  So you can write S0, S1.  So this is a homotovaculite.  And since you're mapping out the opus, you get a homotov limit of mass into X.  So the loop space will be a homotov limit.  So it's a pionic homotov limit of opus-pax.  So in particular we'll see later that this implies that if X is just this metric,  then the favor space is also an metric.  So another thing that we'll see later is what happens when X is a derived scheme.  So example X derived scheme.  In fact there's a different description of the loop space.  So the loop space will be a so-called linear stack.  So the loop space will be a s-morphic to the addition to the tangent model of X.  So this is a linear stack.  And let me just remind you how you construct total spaces of vector models.  So reminder.  When you say linear stack, you're referring to T-shift denies itself?  Yeah, okay.  The whole thing is a linear stack.  Yes, so I'm going to remind you that E of f is a perfect complex.  So let's say we have a linear stack on X.  Then I define the total space to be the stack of some algebra.  So it's going to be a symmetric algebra for OX and A-pool.  So this is the total space of f. In this case I'm taking the total space of the shift of tangent model, which is a perfect complex.  Or I assume it's a perfect complex in this case.  So in this case it's easier to understand what all those spaces are.  I'm confused about what you mean.  So if you say it's a perfect complex, then you mean it's a perfect complex of sheaves on this, right?  Yeah.  So when you write sin of this, sin of this will be again some sheave of algebra.  Right, so this is a relative stack.  So given any commutative decay on X, you get a scheme which is going to be affine, we'll get a stack which is going to be affine on the base, on the base stack, X.  So this is the map to X and this map is at Y.  So yeah, it's true that X is the value of Y times the mean of X.  Yeah, that's right.  So I wrote the diagram that I haven't used yet.  So if you use the diagram to obtain the diagram was this one.  So if you apply the map and stack to that, what you get is a limit,  matched from a snot on the x,  then you have matched from the actual x, this is x, x here, and here we get the wood space.  And this is the full back.  So the wood space, oh yeah, snot is just a point.  So this is x plus x.  So to apply some wood space is just a lot of x plus x over x plus x.  So for example, if you have some arganium on stack, then the wood space is going to be an arganium on stack.  So I know that any topological space at this constant pre-sheaf where you assign to every dg algebra on that space,  is there a way to see what the shape of the case of that looks like?  I think if your VDA is connected, then it's just that space,  and otherwise it's probably like changing the number of rank and points or something like that.  If the only way to understand it is by writing your space as some sequence of pushouts and understanding the whole thing.  So you're asking about what this means?  No, no, no, I guess I want to know, I'm just curious about why sub-V is for arbitrary space.  Is really the only way to understand it is the way of writing and saying why is there a CW complex with a sequence of pushouts,  and understanding those pull-back to every empty dg.  Now if you write down formula for a certain form of formula, you can take the essential size to create a constant neutral machine,  and if you simplify and you get the formula that maps space to a type of topology, which I'm not constantly saying.  It can get a complicated effect.  Okay, if k is another networking space, it's to use a type of topology to function in a constant neutral machine.  So for example if k is not an electric field, then the center from our spaces to our stacks is not very critical.  Okay.  Yeah?  So the next examples are some examples from the last public-private join trade, which are called user schemes.  I'm sorry, it's for purpose only.  Okay.  Let's say x is a single predictive.  So let's say x is predictive.  So let's say we have a particular scheme.  Half a particular sheet.  So there is a special equilibrium of any sheet.  So it's time to be a little characteristic of this sheet, which is the binomial.  So since you're a projector, you have some kind of line-lumbel upon an x-ray of that.  So this is something binomial.  Okay.  And now let me define the code scheme with six-kilometer formula.  So I'm going to define some modular problem.  So on some affine scheme s, this is ordinary affine scheme.  So the group of particular sheets, g, to get to the map from map.  So particular sheet g on an x-ray is s.  So I'm going to define y over s and the map from map to g.  So now f is a sheet from x.  So f has a y as a sheet from x plus s.  So both of these are sheets from x plus s, which is subjective.  And the regular polynomial of g is the total polynomial h.  So this is the code scheme and something similar is the Hewler scheme.  Again, I'm going to solve the classical version.  So this is going to prioritize soft schemes of classical styles such that the total polynomial of z is your fixed-kilomial h.  So the code scheme is a concept from affine schemes to group weights.  So what's the way you work with this?  So morphisms from g are preserving this map.  Yeah, I was trying to do that.  Yeah, thank you, Rufloid.  I think we're going to do a set.  Yeah, so let me first state that it's in the Rufloid, so now the main theorem would be that it's actually a scheme.  And don't z is flat or is it?  Yeah, I'm trying.  Yeah, so this is a relative to kilo root of 1, relative to s.  So z is flat.  S.  So the theorem is that both kilo and Hewler's schemes are actual schemes.  So that wasn't no Hewler's schemes.  So although after you already did this, you're going to have lands in group weights.  As Bertrand said, it actually lands in sets.  So it's a, let's say, a word model of proof. This will be used in the instruction derived.  So the main idea of the Klozscham is that there's this theorem of,  of Sarah, that the category, it's a part state of theorem.  Let me introduce two rings to the model.  So that a be the address of all the sessions of n or s.  So this is going to be some gradient of theorem.  And this is going to be defined to generate it.  And then for any sheaf F, I'll define the associated module or this algorithm,  which is just the double sections of the two pieces of f.  So this is a gradient model.  And so the theorem of Sarah is that the category of different sheaves on x  is equivalent to the category of rated n modules.  So instead of starting studying the career sheaves on x,  we can study modules over this single rated algebra.  Which means that you can realize the Klozscham side of some response.  So this implies that, excuse me, maybe I'm just wondering,  shouldn't you mod that portion to the graded,  the ones that are already defined as this portion should you mod the other one?  If this includes with the modules.  Yeah, most of us don't.  Okay, okay, that's what you need with that statement.  So this is why I think I realized the Klozscham has a sub-schemes on Breslinan.  So this is the Breslinan of A modules inside of the monocles of the map.  So instead of studying quotients or career sheaves,  you can study quotients of M-M or sub-schemes of M-M.  And there's a similar Breslinan embedding for the Klozscham.  So it's a, once you go to the Breslinan, it's a scheme, the Klozscham will be some close-up scheme for the Breslinan.  So it will also be a resinged Breslinan and the same for the Huber scheme.  So let me know in the comments, there's a map from the Huber scheme to the Klozscham of just coax.  So given sub-schemes C, let me say the embedding is alia, I send it to just the push-forward of the Sartre scheme.  It does obviously coax the more I coax.  So this is putting in an esomorphism.  So in fact the Huber scheme is a case of the Klozscham.  It was a flat-shell and a derived level.  And let me say a few words about the instruction of derived Huber and Klozscham.  Before I do that, so the language of derived Klozscham was, kind of all of this, there was no derived like John Trey at the time.  It was late 90s.  So they used the language of the DG schemes.  So let me end up with the Sartre.  So a DG scheme, X and a DORPX is a scheme of S with a sheave of forces of the graded.  Cj's such that, so let me say the HNOT of our X dot is X.  I'm going to dig a scheme of the Klozscham.  So this is not quite the definition actually.  The definition is that the sheave is a sheave of OXCGAs.  Yeah, that's right.  This is a very big difference between DG schemes and DRC's.  In the sense that this should be added to the scheme X because that was a closed Sartre-RC.  Yeah, so let me now say how to compare DG schemes to derived schemes.  Sorry, sorry, sorry. Don't you want to have a matter of more HCOs in the subject, a nice one?  Yeah, that should be added to the S-Prix DG scheme.  Okay, yeah, I'm not sure.  Let me just...  These heave and court to have AIDS there...  No.  So what is this?  That?  That's not it.  OK.  So this is the natural map.  It has the best electric.  OK, so compared to some two right schemes,  four to the right stacks.  So let me just say that there is a cover.  So this is an offline cover.  So here is offline.  This is offline.  And this is a smooth cover.  So four to the right stacks.  With the offline.  So for example, if the backstakes was impacted and separated,  then you have such a cover.  That way, what you need to do is you can construct  certain solution of the game.  So you adopt.  So this is the check mirror of you.  This is the best.  So it's sort of called, for example, you,  now it's just you, you want you to use x.  And so you can construct.  So you can construct o x dot of e dot.  This is a, a Cosimblissal DGA.  So I'm taking my structure sheet,  which is now a, which is a CGGA,  and taking it on these local sets.  This is going to be Cosimblissal DGA  because I reverse the arrows.  Now I take stack.  This is going to be Cosimblissal stack.  And I take general improvisation.  So what do you take?  Yeah, covenant for the,  and this is what they'll call theta x.  So this is going to be some puncture from DG schemes to draft stacks.  So theta is a puncture from DG schemes to draft stacks.  And in fact, I can show you the glass in one direction.  Okay.  So, so what's the plan and, and so you can, and contain the,  they construct DG schemes.  It's containing.  What's containing?  Sorry.  Containing.  Oh, sorry.  I guess.  Yeah, okay.  Yes.  So they construct DG schemes.  Let me just say derived code scheme as F,  and derived code scheme as X,  which are derived from the hands-mits.  So this theta is a problem to be fully figured out.  So that's the problem is that it's not normally to be.  So this is a common problem.  Which is expected of B, say.  I don't think anyone has 40 stack.  So which are other derived from hands-mits?  In the sense that the truncations of these derived schemes  or DG schemes are ordinary code schemes.  Okay.  So.  What do you mean by stack only?  So, what does it mean?  So, I X dot, so this dot refers to CGA.  This dot refers to the simplification of the fusion.  So this is going to be a cost-simplification of the object  when the stack is going to be simplitical as that.  So it's just the stack of this CD chain.  So the simplitical stack.  The first thing was the simplitical scheme  that you've been doing that much.  Right.  Yes, the suspect is just a Canadian man.  So, yes, let me just write down the tangent complexes.  So the tangent complex to the Hilbert scheme,  the derived Hilbert scheme at some point is Z.  So Z is also a radio mass.  This is, so it's a collage of the tangent complex.  This will be the sheet homology of Z,  which was used in the mobile model.  And then the tangent complex of the cyber-henshin G  in the right-to-boots scheme.  So Z is the X from K to the G  and K is the kernel of this map.  So these tangent complexes coincide into V0.  The crucial fact of the fact that the cold scheme,  the sugar scheme, the partner schemes are as a partner  but not in general.  So if you take, for example, a smooth point in X,  this tangent complex is going to be constricting to degrees 0  while this tangent complex, you can do the Z resolution,  so it's going to be constricting to degrees.  So these are not as far as the various things.  So this is an example of same ordinary molecular problems  with the different derived molecular problems.  So let me say a few words.  Well, on the occasions of this cold scheme,  the clever schemes.  Do they have more than one solution?  No.  So I can write some guess for the cold scheme.  I don't know the guess for the clever scheme,  but it's not very valid.  So what's the question?  The more generalization of these schemes.  I don't know.  Someone wrote to me that he wrote initially PhD,  so the text is lost somehow.  And the result was that what you can expect is true.  Yeah, you were right.  So G is strong over F, and then G is subjective.  So this map is subjective by now.  The tangent complex is going to be Aga.  Exactly.  So before I do the applications of these schemes,  let me do something easier,  which is the more questions of both systems.  So let me define all of the genes of sigma.  So sigma is something that's defined as a doubly complex.  To be the mapping stack from sigma B into BB.  So that's very close to the point.  So T is what kind of...  So the ordinary stack of all the systems  is the same equation essentially.  So this is all this graph in Hassel.  Let's try to compute some examples.  So you make two examples.  So sigma is S1, then you go to the local systems minus 1.  So that's the freedom space of EGA.  So it's the modular space of bundles together about mortises.  So you can show that this is just that.  G by G.  And that's the same as ordering the second local systems.  So this is always the right kind of function applied to the second local systems.  And this is an example of these two.  So that's an example that people would get the same...  Let me work this out.  I'll do parentheses the other way.  Oh, no, it's that mapping stack.  Where?  Sorry.  I thought you were going to write a kind of local G,  apply it to a small bit of same local G of this one,  it's a mapping stack.  Yeah, exactly.  So I'm going to make it into a driver stack.  Okay.  So I want to compute local systems minus 2.  So it's before the write S2 as a pushout.  So if you apply a mapping stack into EGA,  what you get is a lot of G by S2 is 0.1G plus 0.1G or G by G.  So to compute that, we can rewrite this as 0.1G.  So again, I'm doing the driver part of the point.  So 0.1 plus 0.1G, that's something I would call the base of those states of G.  And you can compute this by this little resolution.  And this is going to be the V-algebra in degree minus 1.  So that's the drive scheme, mon gene.  Okay.  So that's the drive stack.  And the borders, how do you do the other part?  So this is the basic solution.  So I think Paul will talk about the field of space,  the h-carousel, and the field of space.  I will talk about that first.  Okay, yes, I will try.  Okay, so it's in this way.  Yeah.  You also have the section of pointing the schools right.  Yes, it's a top-famous area of a school.  Yeah, that's just the field of space.  The seam of the empty space by one.  That's actually the shift T tangent shift, the last shift,  but not the tangent of BG, right?  Sorry.  Yeah, that's the formula you wrote at the beginning,  the field of space is the Tx shift by minus 1,  this is the condition of the formula.  Yeah, that's the main thing.  That's the final step.  Okay, final step.  Now, there's two simple methods, so the ordinary stack is just BG.  So you see that the drive enhancement of local systems is not...  so the obvious drive enhancement is not our model.  So our model is much better.  We'll see that it's actually a symplectic while the drive enhancement is not the original.  Okay, so you said last time you said...  So, let me see that this is symplectic.  Okay, so let's follow the same old system.  Now...  Well, that's not the case with the data drive.  Yeah, I'm not sure.  Okay.  Well, I think the G is also symplectic, but those are the same systems.  Yeah, I don't see anything like that.  Yeah, the order is the same.  Okay.  So, let me now talk about vector models.  Okay.  Okay, so now, there's much like problem from vector n of x.  So this is going to be the mapping stack from x into vector n.  So, like Dan and Sam are sort of really fine.  And now, here I assume you're actually a similar progressive.  Okay.  So the way you prove that...  Actually, let me put an R in here.  The right mapping stack.  So the way you prove that...  So the proof that it's actually a zero factor  and R from one stack is...  Yeah, x is something to not write.  Okay.  I'll write them all in the same as the code scheme.  And the risk scheme, they're also something to not write.  Yeah, so to prove that it's zero geometric,  let me do the following theorem.  So I'm going to state a theorem,  but the terms will not be defined in this column.  So I'll be able to find a way to talk.  Okay, so the right Sam asks his n-geometry.  Even though the following theorem is satisfied.  So first, the truncation of x is...  is marking n plus 1 stack.  Then x has an abstraction theory,  so this term is not defined in this column.  So the thing is that x has a global connection complex  and it reserves certain diagrams.  So it's going to be cartesian, and then x is convergent.  So I'm going to write a cdga, the natural map from x.  x is a little bit lighter, but I'll keep using x.  So x obeyed to the point of the limit of x.  So the limitations of that deviating to this of the positive powers  is nice and more visible.  Okay, so if this term can be satisfied,  you can include the x as n-geometry.  So in our case, 2 and 3 are some formal checks.  So to check that the monominal system of abstraction theory  and check that it's convergent,  it doesn't involve any geometry while the mon requires some geometry.  So here we have to prove...  So you don't have to prove that the monominal system is zero geometry.  I want to prove that the monominal system is one stack.  So we need to produce some atlas for that.  So to produce an atlas of retrobondals,  x observed that for any vector bundle 1s,  its twist is globally generated, generated by double sections.  So we'll be reminding you what the journey of the double sections means.  So this means that the formal form of the form of creation map  from global sections to global sections is super.  So along with studying quotients of some machine,  so this is all asked to some power,  so can you use the order in?  Yeah, thanks.  So you can use the order in a quotas scheme to produce an atlas.  So in this way you produce atlas,  and let me just make a remark that if you have a direct quotas scheme,  which actually presents some nice monolid problem,  you can avoid using that sort of stability theory,  and just use the direct quotas.  So in this way you can include that, that and x,  these zero-gen are actually a couple of them.  So the order in a stack is just assigning the category of retrobondals.  So this is the one we want.  So this is the one.  It's the real dimension,  and it's your dimension, and it's not the non-standard.  And so the last example I want to give you  is the one where I space stable maps.  So this is essentially what the quotas scheme is.  So you can picture all the different directions  that this is just OIS and some power.  So this map is equivalent to OIS and some power  to supply minus m goes to e.  So once you fix the trivialization of all the sections,  so this kind of model problem is represented by a quotas scheme,  then you forget trivialization, and that's your atlas.  In this case it's not like the employment type,  so you have to take all possible maps.  This is going to be something of a model of application.  Okay, and so the last example  that I want to give you is more less space stable maps.  So maybe people that just,  I think we see a proof of the possibility of all this.  Maybe we can construct a proof of that.  Where people say we need a proof of the possibility  of perfect complex.  So I'm going to write about the proof of this nucleating.  Yes, I'm not immediately because it's a problem.  So for instance, text doesn't have to be projected before this.  It's most proper.  It's going to be a little bit different anyway.  Okay, ready?  I'm going to write a letter for the vector to put it in the case of the vector.  I just think it's in the shape.  Okay, so let's define what stable maps and pre-stable maps are.  So let energy free the ordinary stack.  So when I say ordinary it means underlined by the order of pre-stable curves.  So it was a pre-stable curve.  So it's a proper curve with no singularities.  Okay, and now there's a whole sub-stat called mg bar of green.  So mg bar is the modularized space of the stack.  So I guess it's the ordinary stack.  So it's a pre-stable curve.  This one is an ordinary stack.  This thing is an ordinary stack.  So I'm going to say why it's a pre-stable curve.  So it's going to be an ordinary stack of stable curves, or just g.  So for example, Cp1 is a pre-stable curve, but it's not in the stable.  So it has more than one basis.  So m0p is just the sum group, so it's not in the stable number.  So this is the regular stable means.  So this is going to be a pre-stable curve.  Okay, so now we have pre-stable and stable curves.  And so mg bar, Cs is the lean-on for stack.  So this is the pre-stable curve.  So let me define the derived modular problems for stable maps and pre-stable maps.  So first, since I have an ordinary stack, I can consider this as a derived stack.  So I'm going to just apply this function i to that.  And looking to find what the pre-stable maps are, I'm going to say there's a universal pattern.  So this is the Cp1, Cp2, Cp3.  So now let me define the modular stable maps.  So the derived space derived in a less than or pre-stable maps equal x.  So what we'll just see with x is smitherectate.  Okay, so the derived stack will be the mapping stack.  So with mapping stack, we have derived stacks for mg3.  So it's a map from this universal family into the triple family.  So this means that the refibrated stacks are going to be curved.  So this rotation is in our stack and you create x.  Okay, so now we can consider a certain size factor of this.  So for that we reminded a theorem that I think we've seen today.  So the pathologies on this stack and on this stack are the same.  So let me remind you of a theorem that Zariski opened in the work of a unit, I think at all.  Let me just say Zariski opened the sub-stacks of a stack F, some derived stack,  are in digestion with Zariski opened sub-stacks of this truncation.  So once we have opened the sub-stack of the truncation, in this case mg3,  we'll get some open sub-stack of the derived on this case.  So what I'm hearing to the modular space is the table maps.  So this is the modular space of table maps.  Again, these maps can have some of our morphisms, or not as free of morphisms,  while table maps are defined to be more by the equation that the other morphisms are equanimated.  So we get an open sub-stack of the derived sub-stack of table maps into X.  So this one will be the morphoid model.  And similarly, we have certain sub-stacks of all mg3x beta inside of mg3x.  So if you look at the great element in the psychological group of X,  you can look at the maps, such as the fundamental plus the curve maps to beta.  So this is going to be an open sub-stack.  And again we get mg4x beta and mg3, the derived one with stats, and mg3x beta.  And so that's basically the space that we need to collect.  Questions?  Yeah, these would be also the indexed, right?  No, I think you can do a little longer.  I have a double question here, and you should have been following that.  What's the proof that these are actually stats and something that they satisfy just set?  Is that way too complicated to get into?  No, I think that's the way I define them.  Well, just there is, because they're all stats.  Yes, I think the way they're defined.  So stats are defined to be 500 objects in some whole category of tree sheaves,  or yes, it's in some category of tree sheaves,  and just the fact that this is defined to be the final replacement of that.  Yeah, for instance, even when the first example, like the metamene, or the hill over a quote,  all that, somehow we do really convince.  Well, just for the order of the stack with GLM, you know that you can just literally take a point of positive GLM for every reading.  You will not get a sheet.  So point one GLM plus pi is trivializable GLM torsors.  When you shoot pi that you get more than a trivializable GLM torsors.  So you have to shoot pi all the time.  So if you do a multiplication, whatever you put in any map is that.  So there are a lot of stats that didn't shoot pi.  So if I ever want to compute something that actually does not say, you know, even something affine,  should I in general use a table to replace that something affine by some resolution by tiny open sense,  and then just compute.  Well, unless you have the right modular function,  so you, like for me, GLM, you know the right moment function.  It's not trivializable torsors.  It's locally trivial torsors.  And that thing will be a sheet.  So it's the same thing here.  You can try to come in the right modular problem, which is going to be a sheet.  So in principle, the way that these things work is like, so I have this pre-sheet.  I want a sheet of pi.  And it's just a geometric know-how to know that the brain-modulant problem  how to actually localize this pre-sheet into natural sheet.  Right.  You have some tools for this.  They are some fundamental, you know, there are two or three fundamental stacks you should know.  So you can also pull the stacks out of them.  So the first one is the stack of stacks.  The next goes to stacks where X is a stack.  And that adds to the new moment.  So what if you write these nothing-stakes that say, you know, this is the data of a current  together with a power company of curves together with a map to X.  I mean, a current is a particular kind of stack.  So they glue and just have to prove that a stack which locally is a curve,  that it is locally is a curve which is more or less true by definition.  So you can use this kind of argument.  Another one is very quickly-fraud descent.  So A goes to this de-modulant, is a stack from the PPF or like I just said, what you're saying.  So all these plots, things, and components, I mean, they accept 5% of the total.  And then there is also a general firm saying that if you have a group point in stacks  and take a portion of this group point, there is an explicit formula of the associated stack  by saying they classify total sources of all these group points.  So generalization of what you said, there is something very general.  If you have a group point in the map stack, take a portion of the stack and it is a formula of the group.  So then I have a question.  So if two examples are given, like a stack to X, you know, stack to over X,  or a stack to A, like DG1, A, those are assigned to the activity categories,  not to the group points.  And you just take the group points in the stack, or what's the answer?  I mean, there are stacks in those sets.  There are stacks of categories in the stack, but that is the category.  So if you really want them, you can let's take to a DG1, DG1, DG1.  But another thing is those stacks also glue as stacks,  and they are satisfied to set as stacks.  They also have some kind of a different, explicit state of addition.  Yeah, because we know it won't be on the side.  You could affect in for X stack.  You said X is something you did with the R-1.  Can you work some of these definitions in the case X is a...  Can I give you some presentation of this stack?  No, the actual stack.  I mean, it's not going to be something nice.  It's just going to be something...  For example, you take SL2 bundles on Q1.  So you know that SL2 bundles are code plus O, O1 plus O1,  plus O1 is also a chain,  and it's not going to be something we describe easily.  But the other thing is you can use the add-was  to write down where it also gets the split,  and it's going to be something you write down.
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Why D is the constant standard associated with Y. So S1 is a nice space, it's a CW complex, and you can write it as S1. So you can write S0, S1. So this is a homotovaculite. And since you're mapping out the opus, you get a homotov limit of mass into X. So the loop space will be a homotov limit. So it's a pionic homotov limit of opus-pax. So in particular we'll see later that this implies that if X is just this metric, then the favor space is also an metric. So another thing that we'll see later is what happens when X is a derived scheme. So example X derived scheme. In fact there's a different description of the loop space. So the loop space will be a so-called linear stack. So the loop space will be a s-morphic to the addition to the tangent model of X. So this is a linear stack. And let me just remind you how you construct total spaces of vector models. So reminder. When you say linear stack, you're referring to T-shift denies itself? Yeah, okay. The whole thing is a linear stack. Yes, so I'm going to remind you that E of f is a perfect complex. So let's say we have a linear stack on X. Then I define the total space to be the stack of some algebra. So it's going to be a symmetric algebra for OX and A-pool. So this is the total space of f. In this case I'm taking the total space of the shift of tangent model, which is a perfect complex. Or I assume it's a perfect complex in this case. So in this case it's easier to understand what all those spaces are. I'm confused about what you mean. So if you say it's a perfect complex, then you mean it's a perfect complex of sheaves on this, right? Yeah. So when you write sin of this, sin of this will be again some sheave of algebra. Right, so this is a relative stack. So given any commutative decay on X, you get a scheme which is going to be affine, we'll get a stack which is going to be affine on the base, on the base stack, X. So this is the map to X and this map is at Y. So yeah, it's true that X is the value of Y times the mean of X. Yeah, that's right. So I wrote the diagram that I haven't used yet. So if you use the diagram to obtain the diagram was this one. So if you apply the map and stack to that, what you get is a limit, matched from a snot on the x, then you have matched from the actual x, this is x, x here, and here we get the wood space. And this is the full back. So the wood space, oh yeah, snot is just a point. So this is x plus x. So to apply some wood space is just a lot of x plus x over x plus x. So for example, if you have some arganium on stack, then the wood space is going to be an arganium on stack. So I know that any topological space at this constant pre-sheaf where you assign to every dg algebra on that space, is there a way to see what the shape of the case of that looks like? I think if your VDA is connected, then it's just that space, and otherwise it's probably like changing the number of rank and points or something like that. If the only way to understand it is by writing your space as some sequence of pushouts and understanding the whole thing. So you're asking about what this means? No, no, no, I guess I want to know, I'm just curious about why sub-V is for arbitrary space. Is really the only way to understand it is the way of writing and saying why is there a CW complex with a sequence of pushouts, and understanding those pull-back to every empty dg. Now if you write down formula for a certain form of formula, you can take the essential size to create a constant neutral machine, and if you simplify and you get the formula that maps space to a type of topology, which I'm not constantly saying. It can get a complicated effect. Okay, if k is another networking space, it's to use a type of topology to function in a constant neutral machine. So for example if k is not an electric field, then the center from our spaces to our stacks is not very critical. Okay. Yeah? So the next examples are some examples from the last public-private join trade, which are called user schemes. I'm sorry, it's for purpose only. Okay. Let's say x is a single predictive. So let's say x is predictive. So let's say we have a particular scheme. Half a particular sheet. So there is a special equilibrium of any sheet. So it's time to be a little characteristic of this sheet, which is the binomial. So since you're a projector, you have some kind of line-lumbel upon an x-ray of that. So this is something binomial. Okay. And now let me define the code scheme with six-kilometer formula. So I'm going to define some modular problem. So on some affine scheme s, this is ordinary affine scheme. So the group of particular sheets, g, to get to the map from map. So particular sheet g on an x-ray is s. So I'm going to define y over s and the map from map to g. So now f is a sheet from x. So f has a y as a sheet from x plus s. So both of these are sheets from x plus s, which is subjective. And the regular polynomial of g is the total polynomial h. So this is the code scheme and something similar is the Hewler scheme. Again, I'm going to solve the classical version. So this is going to prioritize soft schemes of classical styles such that the total polynomial of z is your fixed-kilomial h. So the code scheme is a concept from affine schemes to group weights. So what's the way you work with this? So morphisms from g are preserving this map. Yeah, I was trying to do that. Yeah, thank you, Rufloid. I think we're going to do a set. Yeah, so let me first state that it's in the Rufloid, so now the main theorem would be that it's actually a scheme. And don't z is flat or is it? Yeah, I'm trying. Yeah, so this is a relative to kilo root of 1, relative to s. So z is flat. S. So the theorem is that both kilo and Hewler's schemes are actual schemes. So that wasn't no Hewler's schemes. So although after you already did this, you're going to have lands in group weights. As Bertrand said, it actually lands in sets. So it's a, let's say, a word model of proof. This will be used in the instruction derived. So the main idea of the Klozscham is that there's this theorem of, of Sarah, that the category, it's a part state of theorem. Let me introduce two rings to the model. So that a be the address of all the sessions of n or s. So this is going to be some gradient of theorem. And this is going to be defined to generate it. And then for any sheaf F, I'll define the associated module or this algorithm, which is just the double sections of the two pieces of f. So this is a gradient model. And so the theorem of Sarah is that the category of different sheaves on x is equivalent to the category of rated n modules. So instead of starting studying the career sheaves on x, we can study modules over this single rated algebra. Which means that you can realize the Klozscham side of some response. So this implies that, excuse me, maybe I'm just wondering, shouldn't you mod that portion to the graded, the ones that are already defined as this portion should you mod the other one? If this includes with the modules. Yeah, most of us don't. Okay, okay, that's what you need with that statement. So this is why I think I realized the Klozscham has a sub-schemes on Breslinan. So this is the Breslinan of A modules inside of the monocles of the map. So instead of studying quotients or career sheaves, you can study quotients of M-M or sub-schemes of M-M. And there's a similar Breslinan embedding for the Klozscham. So it's a, once you go to the Breslinan, it's a scheme, the Klozscham will be some close-up scheme for the Breslinan. So it will also be a resinged Breslinan and the same for the Huber scheme. So let me know in the comments, there's a map from the Huber scheme to the Klozscham of just coax. So given sub-schemes C, let me say the embedding is alia, I send it to just the push-forward of the Sartre scheme. It does obviously coax the more I coax. So this is putting in an esomorphism. So in fact the Huber scheme is a case of the Klozscham. It was a flat-shell and a derived level. And let me say a few words about the instruction of derived Huber and Klozscham. Before I do that, so the language of derived Klozscham was, kind of all of this, there was no derived like John Trey at the time. It was late 90s. So they used the language of the DG schemes. So let me end up with the Sartre. So a DG scheme, X and a DORPX is a scheme of S with a sheave of forces of the graded. Cj's such that, so let me say the HNOT of our X dot is X. I'm going to dig a scheme of the Klozscham. So this is not quite the definition actually. The definition is that the sheave is a sheave of OXCGAs. Yeah, that's right. This is a very big difference between DG schemes and DRC's. In the sense that this should be added to the scheme X because that was a closed Sartre-RC. Yeah, so let me now say how to compare DG schemes to derived schemes. Sorry, sorry, sorry. Don't you want to have a matter of more HCOs in the subject, a nice one? Yeah, that should be added to the S-Prix DG scheme. Okay, yeah, I'm not sure. Let me just... These heave and court to have AIDS there... No. So what is this? That? That's not it. OK. So this is the natural map. It has the best electric. OK, so compared to some two right schemes, four to the right stacks. So let me just say that there is a cover. So this is an offline cover. So here is offline. This is offline. And this is a smooth cover. So four to the right stacks. With the offline. So for example, if the backstakes was impacted and separated, then you have such a cover. That way, what you need to do is you can construct certain solution of the game. So you adopt. So this is the check mirror of you. This is the best. So it's sort of called, for example, you, now it's just you, you want you to use x. And so you can construct. So you can construct o x dot of e dot. This is a, a Cosimblissal DGA. So I'm taking my structure sheet, which is now a, which is a CGGA, and taking it on these local sets. This is going to be Cosimblissal DGA because I reverse the arrows. Now I take stack. This is going to be Cosimblissal stack. And I take general improvisation. So what do you take? Yeah, covenant for the, and this is what they'll call theta x. So this is going to be some puncture from DG schemes to draft stacks. So theta is a puncture from DG schemes to draft stacks. And in fact, I can show you the glass in one direction. Okay. So, so what's the plan and, and so you can, and contain the, they construct DG schemes. It's containing. What's containing? Sorry. Containing. Oh, sorry. I guess. Yeah, okay. Yes. So they construct DG schemes. Let me just say derived code scheme as F, and derived code scheme as X, which are derived from the hands-mits. So this theta is a problem to be fully figured out. So that's the problem is that it's not normally to be. So this is a common problem. Which is expected of B, say. I don't think anyone has 40 stack. So which are other derived from hands-mits? In the sense that the truncations of these derived schemes or DG schemes are ordinary code schemes. Okay. So. What do you mean by stack only? So, what does it mean? So, I X dot, so this dot refers to CGA. This dot refers to the simplification of the fusion. So this is going to be a cost-simplification of the object when the stack is going to be simplitical as that. So it's just the stack of this CD chain. So the simplitical stack. The first thing was the simplitical scheme that you've been doing that much. Right. Yes, the suspect is just a Canadian man. So, yes, let me just write down the tangent complexes. So the tangent complex to the Hilbert scheme, the derived Hilbert scheme at some point is Z. So Z is also a radio mass. This is, so it's a collage of the tangent complex. This will be the sheet homology of Z, which was used in the mobile model. And then the tangent complex of the cyber-henshin G in the right-to-boots scheme. So Z is the X from K to the G and K is the kernel of this map. So these tangent complexes coincide into V0. The crucial fact of the fact that the cold scheme, the sugar scheme, the partner schemes are as a partner but not in general. So if you take, for example, a smooth point in X, this tangent complex is going to be constricting to degrees 0 while this tangent complex, you can do the Z resolution, so it's going to be constricting to degrees. So these are not as far as the various things. So this is an example of same ordinary molecular problems with the different derived molecular problems. So let me say a few words. Well, on the occasions of this cold scheme, the clever schemes. Do they have more than one solution? No. So I can write some guess for the cold scheme. I don't know the guess for the clever scheme, but it's not very valid. So what's the question? The more generalization of these schemes. I don't know. Someone wrote to me that he wrote initially PhD, so the text is lost somehow. And the result was that what you can expect is true. Yeah, you were right. So G is strong over F, and then G is subjective. So this map is subjective by now. The tangent complex is going to be Aga. Exactly. So before I do the applications of these schemes, let me do something easier, which is the more questions of both systems. So let me define all of the genes of sigma. So sigma is something that's defined as a doubly complex. To be the mapping stack from sigma B into BB. So that's very close to the point. So T is what kind of... So the ordinary stack of all the systems is the same equation essentially. So this is all this graph in Hassel. Let's try to compute some examples. So you make two examples. So sigma is S1, then you go to the local systems minus 1. So that's the freedom space of EGA. So it's the modular space of bundles together about mortises. So you can show that this is just that. G by G. And that's the same as ordering the second local systems. So this is always the right kind of function applied to the second local systems. And this is an example of these two. So that's an example that people would get the same... Let me work this out. I'll do parentheses the other way. Oh, no, it's that mapping stack. Where? Sorry. I thought you were going to write a kind of local G, apply it to a small bit of same local G of this one, it's a mapping stack. Yeah, exactly. So I'm going to make it into a driver stack. Okay. So I want to compute local systems minus 2. So it's before the write S2 as a pushout. So if you apply a mapping stack into EGA, what you get is a lot of G by S2 is 0.1G plus 0.1G or G by G. So to compute that, we can rewrite this as 0.1G. So again, I'm doing the driver part of the point. So 0.1 plus 0.1G, that's something I would call the base of those states of G. And you can compute this by this little resolution. And this is going to be the V-algebra in degree minus 1. So that's the drive scheme, mon gene. Okay. So that's the drive stack. And the borders, how do you do the other part? So this is the basic solution. So I think Paul will talk about the field of space, the h-carousel, and the field of space. I will talk about that first. Okay, yes, I will try. Okay, so it's in this way. Yeah. You also have the section of pointing the schools right. Yes, it's a top-famous area of a school. Yeah, that's just the field of space. The seam of the empty space by one. That's actually the shift T tangent shift, the last shift, but not the tangent of BG, right? Sorry. Yeah, that's the formula you wrote at the beginning, the field of space is the Tx shift by minus 1, this is the condition of the formula. Yeah, that's the main thing. That's the final step. Okay, final step. Now, there's two simple methods, so the ordinary stack is just BG. So you see that the drive enhancement of local systems is not... so the obvious drive enhancement is not our model. So our model is much better. We'll see that it's actually a symplectic while the drive enhancement is not the original. Okay, so you said last time you said... So, let me see that this is symplectic. Okay, so let's follow the same old system. Now... Well, that's not the case with the data drive. Yeah, I'm not sure. Okay. Well, I think the G is also symplectic, but those are the same systems. Yeah, I don't see anything like that. Yeah, the order is the same. Okay. So, let me now talk about vector models. Okay. Okay, so now, there's much like problem from vector n of x. So this is going to be the mapping stack from x into vector n. So, like Dan and Sam are sort of really fine. And now, here I assume you're actually a similar progressive. Okay. So the way you prove that... Actually, let me put an R in here. The right mapping stack. So the way you prove that... So the proof that it's actually a zero factor and R from one stack is... Yeah, x is something to not write. Okay. I'll write them all in the same as the code scheme. And the risk scheme, they're also something to not write. Yeah, so to prove that it's zero geometric, let me do the following theorem. So I'm going to state a theorem, but the terms will not be defined in this column. So I'll be able to find a way to talk. Okay, so the right Sam asks his n-geometry. Even though the following theorem is satisfied. So first, the truncation of x is... is marking n plus 1 stack. Then x has an abstraction theory, so this term is not defined in this column. So the thing is that x has a global connection complex and it reserves certain diagrams. So it's going to be cartesian, and then x is convergent. So I'm going to write a cdga, the natural map from x. x is a little bit lighter, but I'll keep using x. So x obeyed to the point of the limit of x. So the limitations of that deviating to this of the positive powers is nice and more visible. Okay, so if this term can be satisfied, you can include the x as n-geometry. So in our case, 2 and 3 are some formal checks. So to check that the monominal system of abstraction theory and check that it's convergent, it doesn't involve any geometry while the mon requires some geometry. So here we have to prove... So you don't have to prove that the monominal system is zero geometry. I want to prove that the monominal system is one stack. So we need to produce some atlas for that. So to produce an atlas of retrobondals, x observed that for any vector bundle 1s, its twist is globally generated, generated by double sections. So we'll be reminding you what the journey of the double sections means. So this means that the formal form of the form of creation map from global sections to global sections is super. So along with studying quotients of some machine, so this is all asked to some power, so can you use the order in? Yeah, thanks. So you can use the order in a quotas scheme to produce an atlas. So in this way you produce atlas, and let me just make a remark that if you have a direct quotas scheme, which actually presents some nice monolid problem, you can avoid using that sort of stability theory, and just use the direct quotas. So in this way you can include that, that and x, these zero-gen are actually a couple of them. So the order in a stack is just assigning the category of retrobondals. So this is the one we want. So this is the one. It's the real dimension, and it's your dimension, and it's not the non-standard. And so the last example I want to give you is the one where I space stable maps. So this is essentially what the quotas scheme is. So you can picture all the different directions that this is just OIS and some power. So this map is equivalent to OIS and some power to supply minus m goes to e. So once you fix the trivialization of all the sections, so this kind of model problem is represented by a quotas scheme, then you forget trivialization, and that's your atlas. In this case it's not like the employment type, so you have to take all possible maps. This is going to be something of a model of application. Okay, and so the last example that I want to give you is more less space stable maps. So maybe people that just, I think we see a proof of the possibility of all this. Maybe we can construct a proof of that. Where people say we need a proof of the possibility of perfect complex. So I'm going to write about the proof of this nucleating. Yes, I'm not immediately because it's a problem. So for instance, text doesn't have to be projected before this. It's most proper. It's going to be a little bit different anyway. Okay, ready? I'm going to write a letter for the vector to put it in the case of the vector. I just think it's in the shape. Okay, so let's define what stable maps and pre-stable maps are. So let energy free the ordinary stack. So when I say ordinary it means underlined by the order of pre-stable curves. So it was a pre-stable curve. So it's a proper curve with no singularities. Okay, and now there's a whole sub-stat called mg bar of green. So mg bar is the modularized space of the stack. So I guess it's the ordinary stack. So it's a pre-stable curve. This one is an ordinary stack. This thing is an ordinary stack. So I'm going to say why it's a pre-stable curve. So it's going to be an ordinary stack of stable curves, or just g. So for example, Cp1 is a pre-stable curve, but it's not in the stable. So it has more than one basis. So m0p is just the sum group, so it's not in the stable number. So this is the regular stable means. So this is going to be a pre-stable curve. Okay, so now we have pre-stable and stable curves. And so mg bar, Cs is the lean-on for stack. So this is the pre-stable curve. So let me define the derived modular problems for stable maps and pre-stable maps. So first, since I have an ordinary stack, I can consider this as a derived stack. So I'm going to just apply this function i to that. And looking to find what the pre-stable maps are, I'm going to say there's a universal pattern. So this is the Cp1, Cp2, Cp3. So now let me define the modular stable maps. So the derived space derived in a less than or pre-stable maps equal x. So what we'll just see with x is smitherectate. Okay, so the derived stack will be the mapping stack. So with mapping stack, we have derived stacks for mg3. So it's a map from this universal family into the triple family. So this means that the refibrated stacks are going to be curved. So this rotation is in our stack and you create x. Okay, so now we can consider a certain size factor of this. So for that we reminded a theorem that I think we've seen today. So the pathologies on this stack and on this stack are the same. So let me remind you of a theorem that Zariski opened in the work of a unit, I think at all. Let me just say Zariski opened the sub-stacks of a stack F, some derived stack, are in digestion with Zariski opened sub-stacks of this truncation. So once we have opened the sub-stack of the truncation, in this case mg3, we'll get some open sub-stack of the derived on this case. So what I'm hearing to the modular space is the table maps. So this is the modular space of table maps. Again, these maps can have some of our morphisms, or not as free of morphisms, while table maps are defined to be more by the equation that the other morphisms are equanimated. So we get an open sub-stack of the derived sub-stack of table maps into X. So this one will be the morphoid model. And similarly, we have certain sub-stacks of all mg3x beta inside of mg3x. So if you look at the great element in the psychological group of X, you can look at the maps, such as the fundamental plus the curve maps to beta. So this is going to be an open sub-stack. And again we get mg4x beta and mg3, the derived one with stats, and mg3x beta. And so that's basically the space that we need to collect. Questions? Yeah, these would be also the indexed, right? No, I think you can do a little longer. I have a double question here, and you should have been following that. What's the proof that these are actually stats and something that they satisfy just set? Is that way too complicated to get into? No, I think that's the way I define them. Well, just there is, because they're all stats. Yes, I think the way they're defined. So stats are defined to be 500 objects in some whole category of tree sheaves, or yes, it's in some category of tree sheaves, and just the fact that this is defined to be the final replacement of that. Yeah, for instance, even when the first example, like the metamene, or the hill over a quote, all that, somehow we do really convince. Well, just for the order of the stack with GLM, you know that you can just literally take a point of positive GLM for every reading. You will not get a sheet. So point one GLM plus pi is trivializable GLM torsors. When you shoot pi that you get more than a trivializable GLM torsors. So you have to shoot pi all the time. So if you do a multiplication, whatever you put in any map is that. So there are a lot of stats that didn't shoot pi. So if I ever want to compute something that actually does not say, you know, even something affine, should I in general use a table to replace that something affine by some resolution by tiny open sense, and then just compute. Well, unless you have the right modular function, so you, like for me, GLM, you know the right moment function. It's not trivializable torsors. It's locally trivial torsors. And that thing will be a sheet. So it's the same thing here. You can try to come in the right modular problem, which is going to be a sheet. So in principle, the way that these things work is like, so I have this pre-sheet. I want a sheet of pi. And it's just a geometric know-how to know that the brain-modulant problem how to actually localize this pre-sheet into natural sheet. Right. You have some tools for this. They are some fundamental, you know, there are two or three fundamental stacks you should know. So you can also pull the stacks out of them. So the first one is the stack of stacks. The next goes to stacks where X is a stack. And that adds to the new moment. So what if you write these nothing-stakes that say, you know, this is the data of a current together with a power company of curves together with a map to X. I mean, a current is a particular kind of stack. So they glue and just have to prove that a stack which locally is a curve, that it is locally is a curve which is more or less true by definition. So you can use this kind of argument. Another one is very quickly-fraud descent. So A goes to this de-modulant, is a stack from the PPF or like I just said, what you're saying. So all these plots, things, and components, I mean, they accept 5% of the total. And then there is also a general firm saying that if you have a group point in stacks and take a portion of this group point, there is an explicit formula of the associated stack by saying they classify total sources of all these group points. So generalization of what you said, there is something very general. If you have a group point in the map stack, take a portion of the stack and it is a formula of the group. So then I have a question. So if two examples are given, like a stack to X, you know, stack to over X, or a stack to A, like DG1, A, those are assigned to the activity categories, not to the group points. And you just take the group points in the stack, or what's the answer? I mean, there are stacks in those sets. There are stacks of categories in the stack, but that is the category. So if you really want them, you can let's take to a DG1, DG1, DG1. But another thing is those stacks also glue as stacks, and they are satisfied to set as stacks. They also have some kind of a different, explicit state of addition. Yeah, because we know it won't be on the side. You could affect in for X stack. You said X is something you did with the R-1. Can you work some of these definitions in the case X is a... Can I give you some presentation of this stack? No, the actual stack. I mean, it's not going to be something nice. It's just going to be something... For example, you take SL2 bundles on Q1. So you know that SL2 bundles are code plus O, O1 plus O1, plus O1 is also a chain, and it's not going to be something we describe easily. But the other thing is you can use the add-was to write down where it also gets the split, and it's going to be something you write down.
just say derived code scheme as F, and derived same as ordering the second local systems. So let's define what stable maps and pre-stable maps stack of stable curves, or just g. So for