Differential forms
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Ziegler, Paul
The global cotangent complex, differential forms and descent. Relation to Hochschild homology and derived loop stacks.
 So, the final list, an object, which we already received in action. So, in all types, there will be the global co-touch complex for the implementation stacks. And so, throughout, I will read my stack, just take the right stack. Okay, I guess pretty much everything that I'm going to say today, I would actually make a general rack context, but I don't want to do this. Okay, so, we want to define the co-touch complex for the stack. So, I'll just buy out some rest for the stack, yeah? And of course, the idea is, we define this by finding the whole stack to each outline, which maps the rest. Okay, to see what this code actually means, let's look first at the case of usual range. So, let's say, A is just usual range. I guess you can make it into a BG range, if you like. And then, I'll have to write it into A, this is an object A, to the B, okay? And for some A module M, we want to push it to the C, so, we'll just have amorphous over A to the top star of our B. Okay, so, I'm going to start this out, and then we'll get explicit. This is the number of places that we can make from the homogeneity, and so we'll be, okay, if you look at this, you'll see is that, what this is, it's a relation from B to M. If you look at M as an A module by the map, M from B to A. Okay, so, let me start this out, the question is B from B to M, which I wrote in the first one, and, oh no, this is personal. Such that D of M, D, F of M, B of B, plus B of A. Okay, answer that. Now, I want to rewrite this, the other way that you can write the relation is, this is also the same, taking map from B to a 0 extension of A by M. Yes, I mean this is an algebra, I do it in a usual way. And then we'll just ask, is that, take the projection down here to A, and take the projection, and this map is F. Okay? So now, this is the tool that we don't need to do the spectrum of, something, I, yes. So now that we want to do it, so this is, so here the spectrum of A, is now what we want to replace by F. Okay? So this is, so, somehow we want to replace this, this description, we want to replace A by F, and so this is a useful, a problem we mentioned. B by F. Ah, so, yes, I was, so, it's true, but yes, that's why these words stack. Yes, so this is the whole, like, stack A, and B, and B plays through all that. Thank you. So first I want to show you what it'd be like. Okay, so I take the stack, some map from, it's from A, which is from X, so F, and then M, some A right here, so this should be A. A now is a bit of a EG, I guess, so I take some M, some A, and some M positive, and then M. So now I want to define the relations, and follow, so I take F of A, tensor M, yes, so this is maps here from, so it corresponds to this map, and then what you want is that, this M is bound to F of A, here, but this should just be, you know, the section that we started out with, and this is, yes, I should give this a name, yes, this is given by F. So your X, it's okay? So the reason that you put that degree restriction on M is that it squares your extension still in our category of CDGs below zero. Sorry, can you say that again? Yeah, the reason that you can pose a degree restriction on M, the module, is to make sure that that square zero extension is still in our category of CDGs. Okay, now, I mean, just as you, for example, we have started in the event of the code of the collection complex, and I'm sorry, I'm not, it didn't really get in the way of the circumstances. Okay, so the first part is that F of A, and then D of government minister, there exists an F as contra production complex, starting from the first L, so the F, F, F, if there exists a logic, L, and F, then after we can take over this variable, I've noticed that really here I need to come down to the real and this is a general goal, I'll give it some way to the right. That's such that we're over f, we're going to a flatribe we're jumped there until we're off. I guess this was someone who... It's not the status application of PULIDA, because it doesn't belong to the category of which we're supposed to define. That's serious. It can prove that I tune it in fine. Wait, this is in the blogging algorithm? Alright, this is LFN. L might be a mold. So I can't be unbounded. This round, if you put a blank here, this is just a factor in the category of bounding in models. But the status of the method of less is uniquely defined. Oh, so we're talking in a complex, I'm sorry, in fine-division. So, okay. The best point is from the standard teaching modules to the filters on the actually rated teaching modules. That's what I mean. Okay. Okay, that's what I mean. This is the case, and we also have a bunch of other ones. Which is better? What am I learning on theЯ? Okay, so this is the tension complex at the point. Okay, I mean when copetent exists globally, it exists at every point, and the compatibility is going to be the second part, so F as global tension complex of every field as probability F is my K, F has more diagrams, kiss, it's my D, it's my K, F is my F, this is F, I'm going to be at primary, this is my U, so what you always have is F, okay so what you always have is the family's interrelations about that kiss, and this should be the new difference, okay? Okay, so now I'm going to do this, I'm going to do this, I'm going to do this, I'm going to do this, okay so now I'm going to do this, so the fundamental result is now that the drag's heavy, and the energy's back, as a global tension complex, all over the sky, so the energy gives you a balance, okay? Okay, so let me just say, F, F, F, F, F, F, F, F. Okay, so I can't quite read that. Okay, so I'm going to say it exists in the quasi-quadric, yes, yes, I mean after what's going on, I can't quite read that, but I can't quite read that, but I can't quite read that, but I can't quite read that. Okay, so I'm going to say F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, so I can't quite read that, but I can't quite read that, but I can quite read that, So that is the correct case for the volume of MQ. We have the left of the variation of F. I'll put it in the middle. And then the right. And now... So there's always a natural MQ. And furthermore, here in here, in here, I'm sorry. And then there's the denominator of zero variation. And the rule is just... which is that the variation is transferred into the zero model. So this is the variable. So what you do is this, this, and then, OK, the relative variation is adjusted. I'm not going to be liable for this. So this is the actual thing. And then, OK, let me do as before. So you can define the variable here, the current action, and then go to contact, go to contact assessment. Just say that there are two variations, which are represented by the difference. So this is the... the right. The right from G to F, so the potential. And then, I mean, also, the right from G exists. So what you do is put the problem after F to G, small f, then what you do is say you take the callback from G to F, and then ask before to what you're matching back, but here to F to F, and, OK, it just takes the amount to be called back. So it follows the right from the definition of this guy, but with some... if you have global potential complex, it's sort of a stax thing. And then, I think it's a lot of difference. OK. So let me sort of briefly give you an idea of how to prove existence with the potential complex. So, I mean, of course, the argument is an inductive one, and so for any... for any postmaster, I mean, of course, it's just a theory of this. And to go back, then you will see that the... so this is the argument of our theory of the show, the potential complex, the cheerifying effect, exactly as the right argument to do, I think it's the relations as a pressure force. So now for... and then for one person, somehow you will be able to explicitly construct the potential complex. So if you go back to the potential complex, you will see the f line. So for this, I mean, try to give you the t, how this is done. So, you actually write the definition. So let me for a moment just assume that the potential complex is all exist that we want. Then I want to look at the diagonal map from f to f cross f. So let's look at this Bible sequence, which we have here, and so I mean, the point is what you should expect is that if you go back to the potential complex on f cross f or on f cross f, just to say it happens as an divisive interest, and then you can divide it down if you call this a divination. And so what you get is the privacy that is to follow f and f, so I will call it f and f. This is the pullback. Then you go to f and f, and here you will be able to call the potential complex at delta, which I will write as this. So what should this map be? What should this map just be? Addition is unfortunately running under the backward. So let me go right to the shift. It's awesome, yeah? And so now if you try to compute, and you think about what people are wondering about the first-hand shift, I'm sorry. I mean, this is how five receivances were in the Bible and this is how you compute the outcome. So this is a trade-off category, a non-sense or a trade-off category for one. So I guess somebody can say, let me just expand your view on this. So if I were to see people like this, the homophobic curve of this map is going to be down with that furniture. So it's minus one. It's minus one? Yeah. Okay, and so now, so say you want to compute the homophobic curve of this map. So you can quickly have this computer take this, the coparace is just the mapping problem. The coparace is like this. The coparace is always just the homophobic coparace. It's always just the mapping problem. So this is going to be the guy who shifted the parameter to minus one. So if you try to compute this, what you see is that what you actually get is just an LF, and you send the meter to it. And this is just the map which is going to be broken into minus one. Okay? And so why am I telling you this? The problem is now that LF should be the same. It has L delta, and now I'm going to be coming with the shift to minus one. Okay? So this tells us that if the LF really exists, you can compute it by pulling back the relative coparace of this map. Sorry, I was going to use more of a while. Do you want to make that an application? Well, so it seems that if you write on this map, you can compute this. I mean, it's not... I think it's just the fact that if you have a complex, just take the map which goes from the sum of this, twice the complex to itself, while the derivative of the fiber of this is the complex. You have a map from C plus C to C, while the derivative of fiber is C. Okay, so now let me tell you about the complex of the LF. The LF is a group because there are some axes which match the angle. Some LF to LF. Okay, so LF is this edge of magic, and we want to actually construct the full-body coparace. So I look at this diagram. This is the vector map from H plus X plus X here. So I'm going to use some n-pages. So then I also write down the fiber product. So I'm going to use the fiber product for that. This is the one, the fiber product. Okay, so now we have some delta, and actually I'll show you some other things. We have a product, the map page, you just have to... okay, so I'm going to use some of the stuff. This is the map page. So I'm going to use this one here. I'm going to use a lot of the stuff. When I set the form, it's that the delta start and the map delta is LF. Okay, so now it's not typical to show that we've got the potential to come back to the system and then come back with the code back. So if you tell us that the delta, delta, delta is the format of X plus X, you should expect this to be just a product of H. Okay, so now what I want to do is I want to pull back everything so that this is not the key. If I pull this back I'm done. I get that D star, you have to fill this star and then fill X plus X. I expect it to be pulled back to this map, but this is just expected to be an X star. This should just be an X star. Okay, so now the point is that this guy here, we actually have it. Y, so X plus X, so this F is the number, soama adult pad. But that just goes back to the attack. That should be the value of that S factor Y plus S factor X. Okay, so this would be the number of chips. This reduction could be All-Kauf NOnes. Sorry. Sorry. It is... It is quite sometimes you put the... After the object and the rest of the... Oh, sorry. Sorry. Sorry. Sorry. Sorry. Sorry. Sorry. Sorry. Sorry. That should just be the same. Yes, this should... This time it is going to be two times the back of here. This should also... This should also be the same. Just a two-star angle. This is so that I try to control that I am so unusual. This is good in this ball steps, right? because after you can factor it in each of these. Right? Good. Okay? Great. So this is, if you like, yes, this is a two-star which starts out with L. So the point is that this should be just the L, okay? So this should be the star L, L, L. Okay, that's enough. Sorry. Well, why is that? So then this is this is generally a property for the co-financial companies that have such a amount of problems. And the co-financial companies here today are just going to be this rather complicated. You know, if you write the properties of the potential complex, there are amounts from this spec A plus M2 or something. So if it maps to a final product, it becomes a final product. And this is the system of this. If you know it exists, it's a lot of problems. Okay, so the point I'm trying to make is that this guy can actually, I mean, he actually had it. Yes, of course, and X plus Fx on the sum of the financial management. Right, because as soon as the problem is on the reaction to a stack, the dyagonal map is representable by M minus one-shell matrix stacks. So this tells exactly that if you take the FM map like this one here, then this co-back is going to be M minus one-shell matrix. Okay. And so, so what is this given as a function of the dyagonal is a case of what is the stack and then also for the the map and the thing that we have, we have the potential complex. So I can write down the corresponding five seconds. This will be the the table star of the stacks. Then I can write down the sum of the FM and then write that down. Okay. And now we have a an impossible stack. This is the sum of the actually have. So I take the T star and the tilde star and X plus X. T star and X plus X and X and tilde. Okay. And so that is expected this guy here to be to pull back the LF complex. Okay. So I just said these two guys actually have another map. So you can just take the how to be full and how to be co-fibre. Okay. And then you shift by one. So take the how to be co-fibre shifted by one and then you actually have a candidate. Okay. And so then you actually have to go and prove that this guy does it. Which requires a different work. So this is the idea of Can you explain a lot of the new right that you said? Because those are smart parts of it. Okay. Okay. Just a candidate. Yes. So I have this one. I want to be I want to be I want to be co-fibre of co-fibre of co-star co-star of x. x. L x. x. So then you shift by how by one and you can take the time for a change of the time. Okay. So for example this shift by minus one that happens here tells you exactly that you can do this for an engine and you start with the co-tension complex here to get exactly goes up to the end. Because at the beginning of the induction the co-tension complex, the error of the construct of this one just goes up to zero. And as we check that this guy has the correct result of the co-tension, you have to use that F as smooth as possible. Because then you have to prove that this is a good case to be the co-tension complex you have to prove it is with another complex. But we never use that F as smooth as now. Sorry, I have a very bad question. I hope that any question. Why do you think we should respond to the class that's enough? Yes, we we never use a smooth. No, we don't need to do smooth. When do we use that F as smooth as we can? Oh, we get the pop out there as a minus one change of the system. We get the this is M minus one change of the system. We don't actually know that that is a co-tension. No, but that is just a diagonal way of responding. So we never use here that there is a smooth side to that. That's something we are. Well, okay, so I mean it stops using the results of these guy words. So the point is that you need this one assumption. You know, some of them taking their relations is compatible with shifts. So this one assumption is here, which is generally the case from the fact that we have an obstruction theory. And then here for this, okay. No, I was worried that we claim that anything that has a problem to lag not on the computer. So we see not what people do. Okay. You need that to show that the shifts are compatible with what? Sorry? So, yeah, I mean, so now you actually have to go on proof of this. And what you need is a certain statement of the variations are not compatible with the way the shift and if you want I can show you. You can show me how to do it. Yeah, I mean, okay, so there is this one condition which only follows so later I will tell you what is an obstruction theory and so from this fact from this fact, I mean this condition follows. So ultimately what you really do is you probe both this existence of a potential complex and the existence of the obstruction theory which I will tell you later simultaneously by induction. This is something I really like about that. Yeah, so you need something like stress cure condition. You need to prove that F satisfies some kind of stress cure condition and for this the way you prove it is very simple. Okay, so this just have a no problem with the structure. Yes, okay, so this and to this this somehow does get a candidate to take some of the actual show that this is a no problem. Okay, so yes, let me briefly say something on this side. So if you look and this is what that concept before really and the potential complex is going to be a classical theory because you will see you look at the function spec A spec A to X as F to I want to say to F to F was F and this is a certain it's a category of this I mean if you do this before letting A vary if you use it on the K-1 those are the Well, I mean so I guess in one category you should have a feature notice on the base train. This is also to get a few of the classical and the A that's actually the base train. Well, for example yes, I guess. So the point is that so this is a sheet for the etal topology if you look at an etal map the potential complex is going to be an ancient so this test will be full back for a moment, let's just be like that. So this is not a sheet for the last one. So I think we're going to wait for it to be a lot more open to something. So I mean the point is it is a name that we can go and forget about and right now we can use it a little bit. What? It is a label in the sense that Benjamin and others Yes, so what's the question? What we're trying to do is to make this into a sheet on the etal side of the map. No, on the deep side of the find seems correct. Okay, either way. I mean, but right now it's depending on the domain of the map, it's centered to A in modules and then the next thing is into B modules so there's a map that's actually correct. Yeah, so there's a section of the fiber that's going to be there. So the find keeps correct. The fiber at the stake here is just K-Mod and this F star F star F defines a section and the obatubility says that it's a Cartesian section and that's another definition of a positive component. Yes, I guess this is the definition of the type of the It's a section of the fiber. Okay, so there's one proposition which has a stake which will begin later and then the other one is the the one that's going to be there. So, this is the the one that's going to be there. So, this is the which will begin later. So, it tells you that I mean that the global section of this is positive career in sheet and then you can compare it if you look out at the global sections or all the alphanets which map the essence of itself. And that's something here. So, we've got the mapping space in the positive career sheet so we're going to go left and we're going to go to this and this is somehow the global section which is sheet. So, you have this guy from one minute and we're all maps from the correct A to the left and now here I take the mapping space in say K models from K to then so this is in fact the easy to do. If I should say that is check. When using the map like A, I wonder if you have the same length of numbers on it. Well, okay so I mean I didn't write it that way but the point is that I mean, how do you make this into a diagonal? So, I guess what you do is you put it into F and what you can do is you take the push forward of A to that of all these guys and then I can map the two different states and then you take the name of such a state then I don't know where to say but just say the thing on the top map as a tool is also a whole new limit but now the whole new limit is with maps in A1 from A to the full map of LF array and there is a non-sense map of LF array to falls on A which is just pulling back on this and so the two things are so okay so let's move on to the second key player which is the question key so here so the point is and this is again the generalization but here in the last part the last part was hard so the question is the following so I'm going to start with the question so the question is the following A5 sounds clear 0 extension of CDG less than the difference F left left right we have some action here the question is when we have F yes and F is the one last case of something and this is the most basic question of the formation theory where you can be forward to objects for just 0 extension and whether there is some nice bacteria in front of this so what we will see is for geometric stacks in the case of the geomorphology classes which is termed with F and F so that means F is much very good for the geomorphology now I'm going to end up with magic rings it's a good for very good affine schemes I'm sorry and so somehow the basic trick here is to be following so that such as 0 extension to plane then there exists a variation from A to N to 1 by 1 so by variation I think it's this way this is sorry so we dive inside so there is no variation like this such that the square is F like this so this here if you look here it's the 0 variation so it's just the inclusion of the first bacteria there is always a variation which gives such a amount of space so this is this is some of the HCO this is what's right this is what I use all the time so if I have a structure here I only talk about such squares but this fact tells you that the 3-tables space is 0 extension so this situation I also want to be known to stay prime by with a different term and oh sorry I forgot to finish the program you can take N to the Ukraine of A prime this is if you can actually make it to N1 with this and you can take N to the Ukraine to make the actual prime so what is the actual standard there is this variation and this is such that each square is 0 extension you can write the first one and the other one and the end that you use is exactly the same oh that A prime is the same I thought that was A prime so A prime in any square is 0 extension so you can recover the square yes I'm saying that is 0 extension so just to find out the equivalence between models I don't know I don't know why do you just win derivation some variations ok now I'm going to this one is an extension ok so now it's your prime start with 2 so that's the start here so once that is usually I estimate the region if the wall is in the G state so I have a little more space and while there is a speed from the M2 to the wall square just to get by 5F to the diagonal just slowly so this is an F of E this is the F of 0 for the region and this is this is ok so so we want to look at all these so if you estimate the region first here and then 2 I just want to come here F has some instructions here if the decision is passed on to the region and and then So, I'm going to simplify it to you why this is called construction theory. This is called the position. So, I'm going to write the stack and then pay and use as a profit the definition. So, the first step is to use the actual class and the actual class. Okay, this is the corresponding x. OK, so that's the corresponding to x. x and little else in the same? What are you going to do? Yeah, right. Sorry, it's x. No, it's x. So this is the class. We're on the stretch of the edges. I'm going to leave x in the middle of the edge. To add extension. OK, so this is the discussion of the class. So I'm going to show you the class and to add an extension. Just because we're looking at this and we're following this, the same is class 1, OK. So here we get this class as an x that you need to keep at least. So this is part of the text. Sorry, I just said no way. Yeah, the second thing is if you actually have this, you have this to 0. So the set of this is going to be like this. I'm going to take the arm and add some little stacks. It's like I'm going to add some new to that. This is going to be the set of this. So this is the individual full circle under omega plus the maps. You know. And then x, you put some of this on the x and then you add that to the x or the size. So we get this back there. Small access. OK, so this is the discussion of the group. This initial set of the groups is exactly the full set of the groups. So this is an example of the group that you have in the initials of the class. Which tells you where you are at the plates and if it is, you can write it in some description. You can describe the plates. I don't know if you're just interested in the pi zero of this and then this will exactly be x0. So you can see that you get the right place in one isob. Do you think that n has a more extreme zero for this equation? Oh, sorry, yes. Yes, sorry. And then as a group. Sorry? As a group. Oh, as a group means A, D. But I think we're going to spend a lot of time on this. I want to spend a little bit. OK. So this is the hands on the top of the answer. It's really explained to you how the problem for one goes. So if you look at the square, you get that line. So if you look at the square, you get that line. So I mean, look at this square is only a quantity partition. And for the second thing, you can get here and just pretend that this is partition. OK, so then we have an object x in here and the other one in the right side. OK, so because this is partition, what we need is, we find the right sum object down here, which has the same image as x and is really exciting to be able to do it. But some other points that we don't actually have a choice. Because by that definition of derivation, if you look at the projection map from A, you can actually take the 1 to 8 and here, and here I write the identity. OK, so I mean, the only candidate that we can take here for the object, here is again x. And so I mean, sometimes maybe this again d and this to 0. So the point is that this is again, this is just the image of the zero derivation and so on. This is the zero derivation in here. And so what we need for this is that d of x is the zero of x. And OK, this here is zero. This is the zero derivation. So this d of x leads to the class of the object x. I remember that the derivation is the guy in here, which maps the object x from here. This is exactly what we have. There is a d of 0 of x. It shows the same derivation, so we need this guy in here. This guy is very, very nice with the class. And OK, because I'm in that light, this procedure will not be a multi-parillion. So I'm in that light, and I'm not a two-parillion. What sort of you're saying now, this is how you view as a double-tip? I want to view this all the x inside of where you said it is. So we have to use the universal property to find the particular class. Oh, yes, yes. What I'm going to say is, of course, so now I just need to sort of dig into the lesson with the decent part. And this guy here is the same as this one. Now we have a derivation, and the only one is the alpha x, and alpha x, and the error, after x, the m should be 2 by 1. This is by definition the same as max and y. Ah, so because of the condition. OK, that's why. Yes, and because I like and assume that this is condition, I don't get the class error. Yes, sure. What do you mean, like, so this f a or some d or m is the same thing as d? 5 root of 5 is the condition. The relation of the definition of 5 over x is this now. OK, and then I have such a value. So in this case, you'll get that dx is all about the d to 0. So in addition to this, it's not that dx is truly equal to 0, but it's all about the d to 0, and then you still get the element in 5 0. I guess I don't have space for that. OK, so that's the, I mean, the space of 5 to the space of 5. Yes, this is why I'm asking you also. What's the good part behind the zyber problem? Well, I mean, so if you want to, if you want to look at the lift, so essentially what you have to do is you take the 5 pullback. You look at the mark from points to this, you take the 5 pullback. And then if this is 0, you have the star and the star with this guy and this guy, and this will, this will, OK, pullback, sorry, pullback for exactly the space of this. OK, I should write the r problem. So you look at the mark from star to infinity, and you take the pullback of everything on the disk now. Yes, so the pullback up there, as you can see, right in the head, is exactly the space of this. Here, here is the life star, and here you have the pullback of this under the map from star to here, which will just be the derivation. So exactly, and this will basically tell you that the space of lifts is the root group associated with the space of derivation. I don't know if you see this, but the space of this is the space of identification between the x and 0. So that is the space of what is the space of the x and 0. So it's path from x to 0 inside this derivation space. If you have two points, the space of path is also over the space of roots right now. So that's where I'm starting from. APPLAUSE More questions or comments? Thanks again.
 So, the final list, an object, which we already received in action.  So, in all types, there will be the global co-touch complex for the implementation stacks.  And so, throughout, I will read my stack, just take the right stack.  Okay, I guess pretty much everything that I'm going to say today,  I would actually make a general rack context, but I don't want to do this.  Okay, so, we want to define the co-touch complex for the stack.  So, I'll just buy out some rest for the stack, yeah?  And of course, the idea is, we define this by finding the whole stack to each outline, which maps the rest.  Okay, to see what this code actually means, let's look first at the case of usual range.  So, let's say, A is just usual range.  I guess you can make it into a BG range, if you like.  And then, I'll have to write it into A, this is an object A, to the B, okay?  And for some A module M, we want to push it to the C,  so, we'll just have amorphous over A to the top star of our B.  Okay, so, I'm going to start this out, and then we'll get explicit.  This is the number of places that we can make from the homogeneity,  and so we'll be, okay, if you look at this, you'll see is that, what this is,  it's a relation from B to M.  If you look at M as an A module by the map, M from B to A.  Okay, so, let me start this out, the question is B from B to M,  which I wrote in the first one, and, oh no, this is personal.  Such that D of M, D, F of M, B of B, plus B of A.  Okay, answer that.  Now, I want to rewrite this, the other way that you can write the relation is,  this is also the same, taking map from B to a 0 extension of A by M.  Yes, I mean this is an algebra, I do it in a usual way.  And then we'll just ask, is that, take the projection down here to A,  and take the projection, and this map is F. Okay?  So now, this is the tool that we don't need to do the spectrum of,  something, I, yes.  So now that we want to do it, so this is, so here the spectrum of A,  is now what we want to replace by F. Okay?  So this is, so, somehow we want to replace this, this description,  we want to replace A by F, and so this is a useful, a problem we mentioned.  B by F.  Ah, so, yes, I was, so, it's true, but yes, that's why these words stack.  Yes, so this is the whole, like, stack A, and B, and B plays through all that.  Thank you. So first I want to show you what it'd be like.  Okay, so I take the stack, some map from, it's from A, which is from X,  so F, and then M, some A right here, so this should be A.  A now is a bit of a EG, I guess, so I take some M, some A,  and some M positive, and then M.  So now I want to define the relations, and follow, so I take F of A,  tensor M, yes, so this is maps here from, so it corresponds to this map,  and then what you want is that, this M is bound to F of A, here,  but this should just be, you know, the section that we started out with,  and this is, yes, I should give this a name, yes, this is given by F.  So your X, it's okay?  So the reason that you put that degree restriction on M is that it squares  your extension still in our category of CDGs below zero.  Sorry, can you say that again?  Yeah, the reason that you can pose a degree restriction on M, the module,  is to make sure that that square zero extension is still in our category of CDGs.  Okay, now, I mean, just as you, for example,  we have started in the event of the code of the collection complex,  and I'm sorry, I'm not, it didn't really get in the way of the circumstances.  Okay, so the first part is that F of A,  and then D of government minister, there exists an F as contra production  complex, starting from the first L, so the F, F, F,  if there exists a logic, L, and F, then after we can take over  this variable, I've noticed that really here I need to come down to the real  and this is a general goal, I'll give it some way to the right.  That's such that we're over f, we're going to a flatribe  we're jumped there until we're off.  I guess this was someone who...  It's not the status application of PULIDA, because it doesn't belong to the category of which we're supposed to define.  That's serious. It can prove that I tune it in fine.  Wait, this is in the blogging algorithm?  Alright, this is LFN.  L might be a mold.  So I can't be unbounded.  This round, if you put a blank here, this is just a factor in the category of bounding in models.  But the status of the method of less is uniquely defined.  Oh, so we're talking in a complex, I'm sorry, in fine-division.  So, okay.  The best point is from the standard teaching modules to the filters on the actually rated teaching modules.  That's what I mean.  Okay.  Okay, that's what I mean. This is the case, and we also have a bunch of other ones.  Which is better?  What am I learning on theЯ?  Okay, so this is the tension complex at the point.  Okay, I mean when copetent exists globally, it exists at every point, and the compatibility  is going to be the second part, so F as global tension complex of every field as probability  F is my K, F has more diagrams, kiss, it's my D, it's my K, F is my F, this is F,  I'm going to be at primary, this is my U, so what you always have is F,  okay so what you always have is the family's interrelations about that kiss, and this should be the new difference, okay?  Okay, so now I'm going to do this, I'm going to do this, I'm going to do this, I'm going to do this,  okay so now I'm going to do this, so the fundamental result is now that the drag's heavy,  and the energy's back, as a global tension complex, all over the sky,  so the energy gives you a balance, okay?  Okay, so let me just say, F, F, F, F, F, F, F, F.  Okay, so I can't quite read that.  Okay, so I'm going to say it exists in the quasi-quadric, yes, yes, I mean after what's  going on, I can't quite read that, but I can't quite read that, but I can't quite read that,  but I can't quite read that.  Okay, so I'm going to say F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F,  so I can't quite read that, but I can't quite read that, but I can quite read that,  So that is the correct case for the volume of MQ.  We have the left of the variation of F.  I'll put it in the middle.  And then the right.  And now...  So there's always a natural MQ.  And furthermore, here in here, in here, I'm sorry.  And then there's the denominator of zero variation.  And the rule is just...  which is that the variation is transferred into the zero model.  So this is the variable.  So what you do is this, this,  and then, OK, the relative variation is adjusted.  I'm not going to be liable for this.  So this is the actual thing.  And then, OK, let me do as before.  So you can define the variable here, the current action,  and then go to contact, go to contact assessment.  Just say that there are two variations,  which are represented by the difference.  So this is the...  the right.  The right from G to F,  so the potential.  And then, I mean, also,  the right from G exists.  So what you do is put the problem after F to G,  small f, then what you do is say you take the callback  from G to F,  and then ask before to what you're matching back,  but here to F to F,  and, OK, it just takes the amount to be called back.  So it follows the right from the definition  of this guy, but with some...  if you have global potential complex,  it's sort of a stax thing.  And then, I think it's a lot of difference.  OK.  So let me sort of briefly give you an idea  of how to prove existence  with the potential complex.  So, I mean, of course,  the argument is an inductive one,  and so for any...  for any postmaster,  I mean, of course, it's just a theory of this.  And to go back, then you will see that the...  so this is the argument of our theory of the show,  the potential complex, the cheerifying effect,  exactly as the right argument to do,  I think it's the relations as a pressure force.  So now for...  and then for one person,  somehow you will be able to explicitly construct  the potential complex.  So if you go back to the potential complex,  you will see the f line.  So for this, I mean,  try to give you the t, how this is done.  So, you actually write the definition.  So let me for a moment just assume  that the potential complex is all exist that we want.  Then I want to look at the diagonal map  from f to f cross f.  So let's look at this Bible sequence,  which we have here,  and so I mean,  the point is what you should expect is that  if you go back to the potential complex  on f cross f or on f cross f,  just to say it happens as an divisive interest,  and then you can divide it down  if you call this a divination.  And so what you get is the privacy  that is to follow f and f,  so I will call it f and f.  This is the pullback.  Then you go to f and f,  and here you will be able to  call the potential complex at delta,  which I will write as this.  So what should this map be?  What should this map just be?  Addition is unfortunately running under the backward.  So let me go right to the shift.  It's awesome, yeah?  And so now if you try to compute,  and you think about what  people are wondering about the first-hand shift,  I'm sorry.  I mean, this is how five receivances were in the Bible  and this is how you compute the outcome.  So this is a trade-off category,  a non-sense or a trade-off category for one.  So I guess somebody can say,  let me just expand your view on this.  So if I were to see people like this,  the homophobic curve of this map  is going to be down with that furniture.  So it's minus one.  It's minus one?  Yeah.  Okay, and so now,  so say you want to compute  the homophobic curve of this map.  So you can quickly have this computer  take this, the coparace is just the mapping problem.  The coparace is like this.  The coparace is always just the homophobic coparace.  It's always just the mapping problem.  So this is going to be the guy  who shifted the parameter to minus one.  So if you try to compute this,  what you see is that what you actually get  is just an LF, and you send the meter to it.  And this is just the map which is going to be  broken into minus one.  Okay?  And so why am I telling you this?  The problem is now that LF should be the same.  It has L delta, and now I'm going to be coming with the shift to minus one.  Okay?  So this tells us that if the LF really exists,  you can compute it by pulling back  the relative coparace of this map.  Sorry, I was going to use more of a while.  Do you want to make that an application?  Well, so it seems that if you write on this map,  you can compute this.  I mean, it's not...  I think it's just the fact that if you have a complex,  just take the map which goes from the sum of this,  twice the complex to itself,  while the derivative of the fiber of this is the complex.  You have a map from C plus C to C,  while the derivative of fiber is C.  Okay, so now let me tell you about the complex of the LF.  The LF is a group because there are some axes which match the angle.  Some LF to LF.  Okay, so LF is this edge of magic,  and we want to actually construct the full-body coparace.  So I look at this diagram.  This is the vector map from H plus X plus X here.  So I'm going to use some n-pages.  So then I also write down the fiber product.  So I'm going to use the fiber product for that.  This is the one, the fiber product.  Okay, so now we have some delta,  and actually I'll show you some other things.  We have a product, the map page,  you just have to...  okay, so I'm going to use some of the stuff.  This is the map page.  So I'm going to use this one here.  I'm going to use a lot of the stuff.  When I set the form,  it's that the delta start and the map delta is LF.  Okay, so now it's not typical to show that  we've got the potential to come back to the system  and then come back with the code back.  So if you tell us that the delta, delta,  delta is the format of X plus X,  you should expect this to be just a product of H.  Okay, so now what I want to do is  I want to pull back everything so that this is not the key.  If I pull this back I'm done.  I get that D star,  you have to fill this star and then fill X plus X.  I expect it to be pulled back to this map,  but this is just expected to be an X star.  This should just be an X star.  Okay, so now the point is that this guy here,  we actually have it.  Y, so X plus X,  so this F is the number,  soama adult pad.  But that just goes back to the attack.  That should be the value of that S factor Y plus S factor X.  Okay, so this would be the number of chips.  This reduction could be All-Kauf NOnes.  Sorry.  Sorry.  It is...  It is quite sometimes you put the...  After the object and the rest of the...  Oh, sorry.  Sorry.  Sorry.  Sorry.  Sorry.  Sorry.  Sorry.  Sorry.  Sorry.  That should just be the same.  Yes, this should...  This time it is going to be two times the back of here.  This should also...  This should also be the same.  Just a two-star angle.  This is so that I try to control that I am so unusual.  This is good in this ball steps, right?  because after you can factor it in each of these.  Right?  Good.  Okay?  Great.  So this is,  if you like, yes,  this is a two-star  which starts out with L.  So the point is that this should be just  the L, okay?  So this should be  the star L,  L,  L.  Okay, that's enough.  Sorry.  Well, why is that?  So then this is  this is  generally a property for  the co-financial companies  that have such a amount of  problems.  And the  co-financial companies here  today are just going to be this  rather complicated.  You know, if you write  the properties of  the potential complex,  there are amounts from this  spec A plus M2 or something.  So if it maps to a final product,  it becomes a final product.  And this is the system of this.  If you know it exists,  it's a lot of problems.  Okay, so  the point I'm trying to make is that this guy  can actually, I mean, he actually had it.  Yes, of course,  and X plus Fx  on the sum of the  financial management.  Right, because as soon as  the problem is on the reaction to a stack,  the dyagonal map is  representable by M minus  one-shell matrix stacks.  So this tells exactly that if you take the FM map  like this one here,  then this co-back is going to be M minus one-shell matrix.  Okay.  And so, so what is this  given as a function of the dyagonal  is a case of what is the stack  and then also for the  the map and the thing that we have,  we have the potential complex.  So I can write down the corresponding  five seconds.  This will be the  the table star  of the stacks.  Then  I can write down the sum of the  FM and then  write that down.  Okay.  And now we have a  an impossible stack.  This is the sum of the  actually have.  So I take the T star and the  tilde star and X plus X.  T star  and X plus X  and X  and tilde.  Okay.  And so that is  expected this guy here to be  to pull back the LF  complex.  Okay.  So I just said these two guys  actually have another map.  So  you can just take the how to be full  and how to be co-fibre.  Okay.  And then you shift by one.  So take the how to be co-fibre  shifted by one and then you actually  have a candidate.  Okay.  And so then you actually have to go  and prove that this guy does it.  Which requires a different work.  So this is the idea of  Can you explain a lot of  the new right that you said?  Because those are smart parts of it.  Okay.  Okay.  Just a candidate.  Yes.  So I have this one.  I want to be  I want to be  I want to be  co-fibre  of  co-fibre  of  co-star  co-star  of  x.  x.  L  x.  x.  So then you shift by how by one  and you can  take  the time  for  a  change of the  time.  Okay.  So for example  this shift by minus one that happens here  tells you exactly that  you can do this for an engine and you  start with the co-tension complex here to get  exactly goes up to  the end.  Because at the beginning of the induction  the co-tension complex, the error of the construct  of this one just  goes up to zero.  And as we check that this guy  has the correct result of the  co-tension, you have to use that F  as smooth as possible.  Because then you have to prove  that this is  a good case to be the co-tension complex  you have to prove it is with another  complex.  But we never use that F as smooth as  now.  Sorry, I have a very bad  question.  I hope that any  question.  Why do you think we should  respond to the class that's enough?  Yes, we  we never use a smooth.  No, we don't need to do smooth.  When do we use  that F as smooth as we can?  Oh, we get the  pop out there as a minus one change of the  system.  We get the  this is M minus one change of the  system.  We don't actually know that that is a co-tension.  No, but that is  just a diagonal way of  responding.  So we never use here that there is a smooth  side to that.  That's something we are.  Well, okay, so I mean  it stops using  the results of these guy words.  So the point is that  you need this one assumption.  You know, some of them taking their relations  is compatible with shifts.  So this one assumption is here, which  is generally the case from the fact that we have  an obstruction theory.  And then here for this,  okay.  No, I was  worried that  we claim that anything that  has a problem to lag  not on the computer.  So we see not what people do.  Okay.  You need  that to show that  the shifts are compatible with what?  Sorry?  So, yeah, I mean, so  now you actually have to go on proof  of this.  And what you need is a certain statement  of the variations are not compatible  with the way the shift  and if you want  I can show you.  You can show me how to do it.  Yeah, I mean, okay,  so there is this one condition  which only follows  so later I will tell you what is an obstruction theory  and so from this fact  from this fact, I mean this condition follows.  So ultimately what you really do  is you probe both this existence  of a potential complex and the existence  of the obstruction theory which I will tell you later  simultaneously by induction.  This is something I really like about that.  Yeah, so you need something like  stress cure condition.  You need to prove that F satisfies  some kind of stress cure condition  and for this the way you prove it is very  simple.  Okay, so this  just have a  no problem  with the structure.  Yes,  okay, so this  and to this  this somehow does get a candidate  to take some of the  actual show that this  is a  no problem.  Okay, so  yes,  let me briefly say  something on this side.  So  if you look  and this is what that concept  before really  and the potential complex is going to be  a classical theory  because you will see  you look at the  function  spec A  spec A to X  as F  to  I want to say  to F  to F was F  and this is a certain  it's a category of this  I mean  if you do this before  letting A vary  if you use it on the K-1  those are the  Well, I mean  so I guess in one category  you should have a feature  notice on the base train.  This is also to get a few  of the classical and the  A that's actually  the base train.  Well, for example  yes, I guess.  So the point is that  so this is a sheet for the  etal topology  if you look at an etal map  the potential complex is going to be an ancient  so this test will be full back  for a moment, let's just be like that.  So this is not a sheet  for the last one.  So  I think  we're going to wait  for it to be a lot more open to something.  So I mean  the point is  it is a name  that we can go and forget about  and right now we can use it a little bit.  What?  It is a label in the sense that  Benjamin and others  Yes, so what's the question?  What we're trying to do  is to make this into a sheet  on the etal side of the map.  No, on the deep side of the find  seems correct.  Okay, either way.  I mean, but right now  it's depending on the  domain of the map, it's centered to  A in modules and then the next thing is  into B modules so there's a map that's  actually correct.  Yeah, so there's a section of the fiber  that's going to be there.  So the find keeps correct.  The fiber at the stake here is just K-Mod  and this F star  F star F  defines a section  and the obatubility  says that it's a Cartesian section  and that's another definition of  a positive component.  Yes, I guess this is the definition  of the type of the  It's a section of the fiber.  Okay, so there's one proposition  which has a stake  which will begin later  and then the other one  is the  the one that's going to be there.  So, this is the  the one that's going to be there.  So, this is the  which will begin later.  So, it tells you that  I mean  that the global section of this  is positive career in sheet  and then you can compare it if you look out  at the global sections or  all the alphanets which map  the essence of itself.  And that's something here.  So, we've got the mapping space in the positive career  sheet so we're going to go left  and we're going to go to this  and this is somehow the global section  which is sheet.  So, you have this guy  from one minute  and we're all maps from the correct A  to the left  and now here I take  the mapping space  in say K models  from K to  then so this is in fact the easy  to do.  If I should say  that is check.  When using the map  like A, I wonder if you have the same length  of numbers on it.  Well, okay  so I mean  I didn't write it that way  but the point is that  I mean, how do you make this into a diagonal?  So, I guess what you do  is you put it into F  and what you can do  is you take the push forward  of A to that  of all these guys  and then I can map  the two different states  and then you take the name  of such a state  then I don't know where to say  but just say  the thing on the top map  as a tool is also a whole new limit  but now the whole new limit  is with maps  in A1  from A to the full map  of LF array  and there is a non-sense map  of LF array to  falls on A  which is just pulling back on this  and so the two things are  so  okay so  let's move on  to the second key player  which is  the question key  so  here  so the point is  and this is again the generalization  but here in the last part  the last part was hard  so the question is the following  so I'm going to start  with the question  so the question is the following  A5  sounds clear  0  extension  of CDG  less than the difference  F  left  left  right  we have some action here  the question is  when  we have  F  yes and  F is  the one last case of something  and this is the most basic question  of the formation theory  where you can be forward to objects  for just 0 extension  and whether there is some  nice bacteria in front of this  so what we will see is  for geometric stacks  in the case of the geomorphology classes  which is termed  with F and F  so that means F is much very good  for the geomorphology  now I'm going to end up with magic rings  it's a good for very good  affine schemes  I'm sorry  and so  somehow the basic trick here  is to be following so  that  such as 0 extension  to plane  then  there exists a variation  from  A to N  to 1 by 1  so by variation  I think it's this way  this is  sorry  so we dive inside  so there is no  variation like this  such that the square is F like this  so this here  if you look here it's the 0 variation  so it's just the inclusion  of the first bacteria  there is always a variation  which  gives  such a  amount of space  so this is  this is some of the HCO  this is what's right  this is what I use  all the time  so if I have a structure here  I only talk about such squares  but this fact tells you  that the 3-tables space  is 0 extension  so this situation I also want to be known  to stay prime by  with a different term  and oh sorry  I forgot  to finish the program  you can take N to the Ukraine  of A prime  this is  if you can actually make it to N1  with this  and you can take N to the Ukraine  to make the actual prime  so what is the actual standard  there is this variation  and this is such that  each square is 0 extension  you can write  the first one  and the other one  and the end that you use is exactly the same  oh that A prime is the same  I thought that was A prime  so A prime in any square  is 0 extension  so you can recover the square  yes I'm saying that is  0 extension  so just to find out the equivalence  between models  I don't know  I don't know  why do you just win derivation  some variations  ok  now I'm going to  this one is an extension  ok so now it's your prime  start with 2  so that's the start here  so once that is  usually  I estimate  the region  if  the wall  is in the G state  so  I have a little more space  and while  there is a speed from the  M2 to the wall  square  just to get by 5F to the diagonal  just slowly  so this is an F of E  this is  the F of 0 for the region  and this is  this is  ok so  so we want to look at all these  so if you estimate the region  first here  and then 2 I just want to come here  F has some instructions here  if  the decision is passed on to the region  and  and then  So, I'm going to simplify it to you why this is called construction theory.  This is called the position.  So, I'm going to write the stack and then pay and use as a profit the definition.  So, the first step is to use the actual class and the actual class.  Okay, this is the corresponding x.  OK, so that's the corresponding to x.  x and little else in the same?  What are you going to do?  Yeah, right.  Sorry, it's x.  No, it's x.  So this is the class.  We're on the stretch of the edges.  I'm going to leave x in the middle of the edge.  To add extension.  OK, so this is the discussion of the class.  So I'm going to show you the class and to add an extension.  Just because we're looking at this and we're following this,  the same is class 1, OK.  So here we get this class as an x that you need to keep at least.  So this is part of the text.  Sorry, I just said no way.  Yeah, the second thing is if you actually have this,  you have this to 0.  So the set of this is going to be like this.  I'm going to take the arm and add some little stacks.  It's like I'm going to add some new to that.  This is going to be the set of this.  So this is the individual full circle under omega plus the maps.  You know.  And then x, you put some of this on the x and then you add that to the x or the size.  So we get this back there.  Small access.  OK, so this is the discussion of the group.  This initial set of the groups is exactly the full set of the groups.  So this is an example of the group that you have in the initials of the class.  Which tells you where you are at the plates and if it is, you can write it in some description.  You can describe the plates.  I don't know if you're just interested in the pi zero of this and then this will exactly be x0.  So you can see that you get the right place in one isob.  Do you think that n has a more extreme zero for this equation?  Oh, sorry, yes.  Yes, sorry.  And then as a group.  Sorry?  As a group.  Oh, as a group means A, D.  But I think we're going to spend a lot of time on this.  I want to spend a little bit.  OK.  So this is the hands on the top of the answer.  It's really explained to you how the problem for one goes.  So if you look at the square, you get that line.  So if you look at the square, you get that line.  So I mean, look at this square is only a quantity partition.  And for the second thing, you can get here and just pretend that this is partition.  OK, so then we have an object x in here and the other one in the right side.  OK, so because this is partition, what we need is,  we find the right sum object down here,  which has the same image as x and is really exciting to be able to do it.  But some other points that we don't actually have a choice.  Because by that definition of derivation,  if you look at the projection map from A,  you can actually take the 1 to 8 and here,  and here I write the identity.  OK, so I mean, the only candidate that we can take here for the object,  here is again x.  And so I mean, sometimes maybe this again d and this to 0.  So the point is that this is again,  this is just the image of the zero derivation and so on.  This is the zero derivation in here.  And so what we need for this  is that d of x is the zero of x.  And OK, this here is zero.  This is the zero derivation.  So this d of x leads to the class of the object x.  I remember that the derivation is the guy in here,  which maps the object x from here.  This is exactly what we have.  There is a d of 0 of x.  It shows the same derivation, so we need this guy in here.  This guy is very, very nice with the class.  And OK, because I'm in that light,  this procedure will not be a multi-parillion.  So I'm in that light, and I'm not a two-parillion.  What sort of you're saying now,  this is how you view as a double-tip?  I want to view this all the x inside of where you said it is.  So we have to use the universal property  to find the particular class.  Oh, yes, yes.  What I'm going to say is, of course,  so now I just need to sort of dig into the lesson  with the decent part.  And this guy here is the same as this one.  Now we have a derivation,  and the only one is the alpha x,  and alpha x, and the error,  after x, the m should be 2 by 1.  This is by definition the same as max and y.  Ah, so because of the condition.  OK, that's why.  Yes, and because I like and assume that this is condition,  I don't get the class error.  Yes, sure.  What do you mean, like, so this f a or some d or m is the same thing as d?  5 root of 5 is the condition.  The relation of the definition of 5 over x is this now.  OK, and then I have such a value.  So in this case, you'll get that dx is all about the d to 0.  So in addition to this,  it's not that dx is truly equal to 0,  but it's all about the d to 0,  and then you still get the element in 5 0.  I guess I don't have space for that.  OK, so that's the, I mean, the space of 5 to the space of 5.  Yes, this is why I'm asking you also.  What's the good part behind the zyber problem?  Well, I mean, so if you want to,  if you want to look at the lift,  so essentially what you have to do is you take the 5 pullback.  You look at the mark from points to this,  you take the 5 pullback.  And then if this is 0,  you have the star and the star with this guy and this guy,  and this will, this will, OK, pullback,  sorry, pullback for exactly the space of this.  OK, I should write the r problem.  So you look at the mark from star to infinity,  and you take the pullback of everything on the disk now.  Yes, so the pullback up there,  as you can see, right in the head,  is exactly the space of this.  Here, here is the life star,  and here you have the pullback of this under the map  from star to here, which will just be the derivation.  So exactly, and this will basically tell you  that the space of lifts is the root group  associated with the space of derivation.  I don't know if you see this,  but the space of this is the space of identification  between the x and 0.  So that is the space of what is the space of the x and 0.  So it's path from x to 0 inside this derivation space.  If you have two points,  the space of path is also over the space of roots right now.  So that's where I'm starting from.  APPLAUSE  More questions or comments?  Thanks again.
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