The homotopy theory of commutative dg-algebras II
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Oberdieck, Georg
Cotangent complexes, derivations and obstruction theory. Application to mapping spaces of commutative dg-algebras.
 So, maybe I can come to the point. In some ways, it's just a coincidence, so, yeah. This is what I was trying to talk about, and I hope to be able to explain it. So, it's just basically, in the beginning I will shortly record K-Demonstration for the G-Ultras, because I mean, it works the same as for future rings, but just to see. And the set up is the following. We take a pair of G-R-G-M-Mets and G-J-S-Mets and G-J-S-Mets. And that's what we have. We have F-M-M-Mets and what we do. And then there's this thing where it's called the derivation. So, derivation. So, this is A, and G derivation. And then we have the M, and the E-Demonstration. So, my differences in the E-J-M would be, for example, A-D, A-D, A-D, A-D, A-D, A-D, A-D, A-D. So, we can make sure we do. And yes, so building here means, I mean, what are, what are, I mean, so we take one times one for this equation. And we get the E times, E times the capital one. We can see that we take a one zero and from the building here, I see, you see the E, the zero, the odd, the E. The derivation doesn't see functions coming from here. Do you know the sign? This is derivation of degrees zero, I guess. Yeah. I mean, otherwise you can shoot them. I see. Okay. Yeah, so this is degree two, we're going to do this. I mean, it's all you need to find. I mean, I'm just going to do three, I'm just going to hand-ship the problem. Yeah, so it's not going to be more than a second. Okay, so, what is next? So, if you resist and if you're trying to call it derivations, derivations. So, like, like in this definition, and this is in A. And so this gives a sum term, I think, the n modules to n modules, just by, like in the second-hand sections. Do it for me, when you take the version of AM, does it take all derivations of all degrees? Yes. No, just degree zero. I'm just, oh, now I'm just taking the degree zero to be the first one, you know. How is it going to hold this for me? Maybe you can take the whole thing over. Yeah, okay, we can also make it this way. No, it's fine. So the proposition is this one term is called invented by a module of the H. So, yeah, how is it, from the A to N, same module, is the same as derivations. And yeah, so this module is uniquely, determined uniquely, actually, uniquely, as in my, and so I record quickly the proof, so, I mean, it's basic, so, you can find this free module, and then you take the direct sum of all x in A, which are the units, and you take A with x, so, that x is the form of symbol, symbol, or if she's reading, x, degree, and the degree of x. So I think you can find this idea here, I, so, something, generative value, and there is a user, and so, this is the standard construction that we find in house shunt, we define the sub-module, and if you were to, in the standard case, it would be not a standard question. So, since we have a GG word, we do it under the standard direction, and to define this, you can look at this diagram, so you take an n-line shunt, and so you have the direction of the data, so data, they find this one, and you have, and here we have the differential, and to define this differential, if I go into the, yeah, so, yeah, to say this, to the differential, the data acts as the sadness, the edge of the differential, and then you extend this by the average, okay, and then with this differential, you need to check that I is preserved, so check that I is preserved, and then you can design under A of D, by L of N, right, and then, yeah, take the differential, so then you can show this as a problem. And some of this is in nice construction, because, yeah, there's nice universal property that it defines, but somehow to make sense of this in the right word, which we write, this is writing out, so the problem is A1, A1 to a squeaky balloon, and it does not follow the data. So, yeah, so if you want to have an example, you can just take k, so you take D, so can we do the writing, you set check A as what? Check I is preserved by D. So I define this relation to the problem, which is almost similar to the check S, and you define this differential, differential to write down the equation, you want to have, yeah, one hand is the height of the preserved. Okay, so you can have two, oh, there's some of the PGR plus, whose differences are different, and so you can take A1 is to be k, and then two, okay, so, yeah, if we do the construction, we can get the term in minus one, and then, yeah, you just get the term in minus one, so, yeah, so we need to derive, we need to derive, derive, and how to do this is, we consider this, we don't consider the category of DG algebra, but we consider the category of DG algebra over A. So, the category we're working on is DG algebra over A, which are, so objects are the objects, are DG algebra X, which are given map from A and which are DG algebra, okay, so we have DG, DG algebra, okay, and the morphisms are, we have X, so objects, that's what we have to do, okay, so, and if you want to have a co-fragment replacement in this, in this model category, so, since this is over A, you just inherit a natural model structure from DGAB, and the co-fragment replacement in this model structure is actually the semi-period revolution, so, the co-fragment replacement. Okay, so it should be semi-free over B, right, so, since it's the B, DG, algebra, so, the underlying graded algebra should be a DG algebra, okay, and, so, to see what we do is we have to form a definition, so, so that B to DGA, and of course, if we have a single-query resolution of some X, we have automatically enough for A, and so, this is how it works, so, let B and A for A be a single-query resolution, so, we have a definition, and we define, let me just sort of, an eco-fragment resolution has to be semi-free, I mean, what would be the question? What would, like, what would be eco-fragment and what would be semi-free? No, I say, oh yeah, maybe you could think of a eco-fragment solution. So, it's the other way around, it's semi-free, semi-free is a co-fragment of the real presence. So, yeah, but, yeah, so I written some notes, so, we don't want anything, we want to write a little bit, so, I mean, how do we define the co-fragment? We take a semi-free resolution, we apply omega to this semi-free resolution, and enter it back to A, to A, and this is the classical solution, how do we write the conquer, right? So, you place A with something, with a co-fragment resolution, you apply omega to it, and then to get, again, the modules and name of the modules, and then you write the test. So, this is the final, what we can do here. And what you need to check is, of course, the right-hand column, so, when you write this, do you still mean omega p would be something concentrated in R, right? What do you mean? So, the thing on the left of this is a product, you can see it's scratched out to statement that it would be a p module, because you only took degrees of navigation. Are you still taking the degrees of navigation? I mean, I define the relation just in this set, right? But this omega is an A module, I define omega mp as a module. Oh, that's very simple. This one is free, free, okay, great. So, this was the actual one. So, the data we show how this is actually done for construction, we did that category, but for now, I will just use this and do lots of examples so we can download it. It's not too scary at all. So, the proposition we need to prove is that this is quite simple. So, we check two guys, then we know just a co-founder replacement, and then we need to check, so they have got g, and two, the homotopy between two men, and they want to do that. So, we have got g, and g, and g, and g, and g, and g, and g, and g, and g, and the original set, and g. So, the last object is logical and in silicon, Okay, you take some of the freedom from the human being, and you go back into the idea. Yeah, I'm talking here. Well, I mean, yeah, I mean, somehow, the first example I've reviewed was personally for the revolution, this is the only one reason I'm kind of confused. I mean, it's just a different thing. So the first one is sort of anything. Really, it's a good one. And so I record this, so this means that A is the... I don't know which S I should even describe it. So A is A, B is A, and B is A. So A is A, B is A, B is A. Then what you can check is that the only way A, B can just write down to present the U.K. So I mean, in this game, the potential complex is just only the A, B value. And what you can check with this is the other line, the greater algebra, the potential complex is just A, and so it will be plus the differential. So let me write this like this. Plus the differential D. And this is S in the construction. So... You can answer it with D. This is trans-arrollable. So to show you what is the map, so if you start with A, you want to have university integration to this guy. So you want to have university integration to D. And yeah, so you take X. Elements in the second factor. So elements in the second factor are a little bit right as delta D. So then you kind of see what you get. So if you start with the element in the vector space, you send this to one in the w to d, delta D. And if you start with an element in the element here, so for example, in terms of two vectors, then yeah, so you want to be in the form of trans-arrollable. So this is... If you switch those two, then this is just a usual... Yeah, a few different... So somehow... Yeah. Somehow in this case, the total differential is a little bit tricky. So you just start this. And this construction, precisely given with this differential, you take X to the end, and you get the sum. So the next example I want to do is do this conclusion resolution we had last time. So yeah, the second factor is that inside the D, you understand, you need to have a function, which is a regular gap. There's a regular sequence. And then we have seen that there's a total resolution. So in a couple minutes, you take this guy, this three module, and you take the matrix, and we have a natural gap, X to the end, and you send the final line to the gap gap. Yeah, I get the feeling that you gave the example for three problem-rover. So it seems that in the differential you define, the differential of A does not appear anywhere. Yeah, because I simply omitted the differential on this one. So this is this guy, and if you want to define the differential, then you take the same as the differential of A, and cancel the delta thing, plus matrix on to the end, A is also differential. And this guy, you define by the commutation relation. So here also, here is also differential of A, and here is also differential of A. So just the underlying, you know somehow the underlying graded algebra determines the underlying, so the underlying graded algebra of A is the underlying structure of the omega A and D in the differential you define by hand. So we have this map, you get this in a little bit of a position, so you can see the A, the shift of the line, and you still have the final line, and the case is the final variation, and the differential here was defined by the number of degrees, shift A to D, or the slope of D. So if you think D, you can say what, what about D? This is the, what about D? What about D? What about D? This is the, what about D? So this is not a 3, this is a sine wave, because it's been here for years. But then, so this gives you a disguise, and for disguise you know where you are in the case. So the propositional, similar states of this, this is actually a symmetry of motion, so to calculate the contrast and complex of A you just need to, write the, write the machine here, so this is, you get a definition of E correct, or P, E, because, there are many P I think it is. So there's no Y high, there's no direct sum of these, there's no Y high, there's no direct sum of these. That's why I, I have general, I just want to say good morning here. And this is a guy of this, and the contrast and complex of this, between one and the other. So by this guy, so the underlying, the underlying guy is, P and E, and in this way, P is just a, generated by the elements, and we try to calculate it. So this is just a, and yeah, we wrote the delta here for, and then, this is an initial, and so I have, I think same as in between, these just a, and so it's a classical thing, but this is in the, and so this is all a classical example. So what you get is K elements in the beginning of the course. So, so, the potential complex is quite, quite nice. Okay, and then, of course, we should, if we think of this in a geometric perspective, we can see that, and see what it's interacting with. So the difference in the product is about the chain and the second is the, so this one is over B, and this one is over B. So okay, this is a, I mean we can say B, here you take the B, this is the B vector space, and then you put your general size. So, what this says, so what is actually compute, so B, this was AM, and this was V of this other, this is the solution. So what this calculates, is this relative potential complex of this, so, okay, let's V of I or should we question? V of I is the points So this is v of i, i is my idea, v of i is... So to get more information we should consider the potential complex of okay. So you can ask what is a of okay. And yeah, our b was right, it's chosen free, so it's free of okay. And so this p is actually free of okay. So p is free of okay. Mainly p is more than 2 s of this... It's not seen on the way forward. We take this vector space, greater vector space in... Yeah, that's when you write this thing. We take the greater vector space which this is in the preview and this is in the view of the shot. And we take that as well. So doing the same thing again, what you find is... This is then omega p over k, and then d over y over a. So what you find is you have a... So in degree minus one you have k generators in percent of the... k generators on the other side. So it's the difference of v of delta yj equal to delta of dyj. And by this definition I wrote delta of fj. And fj is not mentioned in this. And this is the universal... This is just delta fj over delta xk. This is just the usual differential. So what does it mean? To get a better definition again, this is the contingent complex. By the contingent complex, so the tangent complex, it's just the space there, so it's the rule of the contingent complex. And we can consider this at the point. Take a map from A to k. You can answer the contingent complex with k, I think, to do. And if you can find... And what you get is... These objects are just... We hear... And the unit is just a big differential. And f is just the usual Jacobic rule. So somehow we recover just the usual complex, which is the... As the tangent complex at the point. In classical geometry we just get the tangent space at this, but we get a two-ton complex. So fj0 of this kind is the kernel of this kind. This is the tangent space. And h1 is somehow the excess... And this space. So this is somehow the hook method, which measures somehow how much the equation of degenerate is. This equation of degenerate is given 30 times u bar. And... Okay, so this is enough for the long-term to do about this example. In the smarter way, it is used to use... So the correct way to do it is to use the sequence of graphs. And if you have tools that you know that are associated to this, there's an exact triangle. And from this triangle you can actually do sure this one. So those two guys in normal, from this to the tangent complex of this one. So the perfect example of what we do is the following. So we take three affine schemes. We take a1 to the d-edge point. And I want to consider the direct intersection of those two guys. I don't consider a thing so... So if you think about a1, again, there's a sub-skeam. So you take two sub-skeams, take a a1, take a a2, and you want to define the direct intersection. And to take some idea how to do this, maybe this is the boring you did with... So you use the following graph. And if you have a push-out diagram, you have a... And you have a new section to get introduced to the diagram for the quick end. We have a new section. And you can say it with... And this is the game. So for example, if a1 is a point, then you get an exact time. So you can calculate the potential complex on this guy using those two. So in general, you get a different exact time. So mb can be c or a1. So let me not write over here. So let's look at it with the c's. And so we all consider this always c. So this is the general guess, and this is an exact time. So somehow knowing the potential complex of all the three guys involved in the concept gives you the potential complex on the fourth part. Thank you. Since I don't know much, if I don't, I can take a classical example. So let's take b, just move. Hang on. Close the... Here we are okay. And you take those, the three motions. Then we can see what you get from this sequence as a point. So let's take a point. So let's take a ui. And since we did it better than the tangible, let's dualize the sequence. So we take a point, and then from this we can take the longest exact sequence in chronology. And so if you restrict this, the point, you get the page. So if you write down the sequence, what do you get? So if you take h0 of this guy, you see this is just the usual potential complex. Tension space, if you want to. And here you get the usual coming from the place of v, again. And then we can... Some values, again some concepts. So I didn't say this, but if you have a smooth stream, a smooth dg algebra, that's not on the screen. So if you have a smooth ring, let's say a smooth ring, then the potential complex is reported in dv0. So somehow locally you can see that if you take a position of the local sum, then this is just the usual... Why do you consider the potential complex? Why do you consider that you can't walk with the potential complex? I don't know if I do, but I don't think that's a good question. But of course you can write this differently and just reverse the arrows. Yeah, p1 and p2, I don't know. Yeah, should we say p1? Yeah, h1, so that's what it says. So what do we get? So we have the potential complex is that a1 and a2, and this maps to the potential complex of the swan dg. So somehow if this map is subjected, then the intersection is transverse, and in this case this differential is zero. In this case... So if a1 and a2 is smooth, these two guys are also zero, so if you have a transverse intersection of a smooth arrow, self-parallel, then h1 is zero. So this, then you again, it is smooth. So somehow this sequence has a classic factor, but transverse intersection happens a lot. And in general, this guy measures some construction, which I hope is one of the areas where I'm sure. So maybe I say three, I want to say three most things, but I don't know how much time I have. So let me say the first thing is different derivation. The first one, d of a1, is the same as in the map, same as between a5 and a2. So this is the transverse intersection, it is a transverse intersection. So from this you can define this working space of derivation. So again, how this is a1, that's the mapping space in this category. And again you can show that the guy that you find is actually a core percentage of this kind. I mean I couldn't start over this but it's kind of hard to actually write down what I'm going to do. But at the end of the story, I'm going to start writing down the question. So it's a bit more complex and it's a counter-as in the text of the whole thing that we're talking about. Here you are writing a function of all the categories but more is true now. The function of a module is set as an object in a score of two points. Continue. But this should be a problem. I mean if you talk about it you can have a stronger answer. The second thing I want to say is this is a dilemma. So about what is HR, if you look at the condition of the stage R, it's 8 to 1, a point of 10. And I'm going to now, I'm going to always be over K, with some K, in hours of writing a minute something. This is a problem. What you can do is the quality demo. So the demo is, X1, and M1 means, A, K, A or K, and you can insert it up, and you can hit, and take M and R module, and you will get a small order. So put another one, which will take us on M, and this is just, yeah, in the order. We didn't define it above, I mean we don't know what this means. I think it's more clear to write it this way, I think to write it in the opposite category. So this is that, in the order. How much of a category of F1 seems to be over respecting some MFX, respecting them MFX, from the style of M1 to, so the interpretation of this is you can take A, and M, and the square is the triple extension case, and what you get here is just, you get the X1, and you get the style of A and Y, and this is from X, so this is also the style of the kind of MFX, and this is the Y we had before, and this one is just a more useful spot that we need, and then we can do that. So this is the case X, and classically if I is here, then these are just the function vectors, right? So you have, so I can do, you just get the function vectors, and for higher I's, you get somehow some derived, so you have the same thing as, so let's get to something, let's say this object, and if you look at maps from this small disk, and you precisely get this HIO, and this looks fancy, but it's just a form of consequence of this, so the output is to this, this statement, and this is just, and no shout out anymore. And then I want to come to the, I mean I could look this way, and see the light, at least I want to come here with that point, and since this is not shout out, you can look in one of those extra objects, okay so the third problem is at least I will go this way, and the rest of the structure is, so the structure, and the structure is, so again, this is the extension, extension, and one of the things, you can consider a lifting problem, so I think this, again just in the opposite category, can consider a, given specific extension, and then, okay, and you can think of this as some of the same, this same thing about every while, so we don't have that problem, so the question is, how can we find out if there's a fixed lift from, from spec A to some specific extension, so here's the students inside, and the answer is, that this is precisely controlled by the potential, so this is the task, depending on the piece, and the problem, if the complex of Y over K, and the values on the end of the day, and the value of N, so N, if I add with the kernel, kernel of a plan, as in, as in P, no, it's an A, I mean, yeah, so, no, you can show that the kernel is some A multiple, so it makes sense to take, to consider this in the, and so there's a class in there, and the class is, and the class is, and in the second statement, is more often, the space of lifts, the space of the space of lift, and here again, the same object, so, yeah, so the space of lifts, is just the mapping space of spec A, and so, first of all, we can interpret this problem group as a shorted space for liftings, and as a second space of all lifts, as even the mapping space, instead of a problem, but instead of shift by one, and so there's a difference. And since I have a problem, I, I, it's on the complex, you see right at the section, it sounds like we have, I mean, maybe we can see what this means there. I hope, I don't want to, I have seen a lot of S-type fancy things, and I'm not saying it's okay. So, so we have this, where as we make it, You know, I have a couple of questions. You read the space of lips, but the set of maps and the whole of anything is a set. Yeah, so the space of lips, so I mean there is just high note of the space of lips, is it separate? No, no, no, no, no, I just understood this was an additional note. The sum of the space of lips is the map of these guys. Because this was an additional note, which respect is not given to them, so over here. So here you can write the sum of the letters x of the affine, the affine drafts. So I mean for me, I'm just confused, is there any sum of the letters in the category? I think he writes home from the base of the space and map from the space of it. So here I have home and here I mean actually the pi zero of the mapping space between those two guys. And down here I really mean the mapping space between two guys. I don't know if you can tell me if you're right or not. Great, thanks. That makes me feel more normal. This is a suspicious set, and this is Stephen just saying, suspicious set. So we can consider first the density problem, so the problem here. What you have seen here is precisely h1, dc, this is precisely the x1, x1, x1, x1 of this guy without this and then you should. So this gives you precisely obstructions. So h1, dc, x, gives obstruction. So the first example you can look at is x, x not good point, and you can have n, n could be the trivial. The trivial is n, so what you ask for is that you can. So in this case what you ask for is your living property, you are living at point, and can you find the tangent vector that is point. And this is, the structure of this reason is space, but of course there's always a real tangent. So you can ask a harder question, for example, we're giving tangent vector, and you just take this case. And you look at this extension, a prime, which is just, let's say that you ask, can you extend this tangent vector to second part of the graph. And there's really a good obstruction to this problem, to know that we can extend the tangent vector at the singular point of this intersection. So this obstruction could be same here as in this case. So to give an interpretation of the sequences, you take this obstruction and map it forward to those two values. So if the obstruction, the mapping of the obstruction manages into those two values, there exists the lift of the tangent vector to a1, d2. So suppose this happens, if you can find lifts to a1 and d2, you can extend this tangent vector somewhere. So in this case, the only obstruction that lifts is in the corner, which is in the tangent space, d2. It is going to somehow measure the difference between the two lifts, which you can choose, if any one, and these two. So if you know that there's difference of the lifts, lifts and can be written as a sum of tangent vectors, a1 and d2, we know we can kind of just this lift to actually see in the intersection. So somehow it's you who has processed the illusion of the first line. And then of course you can consider higher components. You can consider n by the 1 and you can consider hn. So again, with this theory, you find the ganglion of this obstruction lifts in h1. Each one of these ganglions goes on the obstruction of the positive phase. And in the derived case, to get higher obstructions, you need to shift the things. So if you pass the n, it's in the square, you can ask for something like... And you can add an element in the line of the line. So it's in the line of the line. And we'll talk to this lifting property, which is in the line of the line. It's measured by x2. So it's measured by the h2 here. So we have hdk and it's measured with g. So the higher hi's we touch. So in the classical case, and it's here, we don't know the interpretation for this higher hi. But here we know precisely how we go into the... And so what else? Let the one to say it. So somehow the collection of this just enjoys the obstruction of things. And you see some power. If you start with something smooth, so a1 and a2 is known, or a1 and a2 are better than the section. Then you know that the higher hi is 0. So a1 and a2 are completely intersection, those guys are not. And c, the co-tension complex of c is just considered in degree 0 and 1. So the intersection of the co-tension complex, the intersection of the two parts, the intersection of the smooth thing is again in degree 0. This is completely different to the classical case. In the classical case, you would understand those two skills just by taking... Just adding up the ideas and looking them up. In this case, you might get a really similar skill with the co-tension complex, as you co-tension approach infinitely in degree 0. So somehow the derived setting kind of lets you control what you say. You kind of get in fun, out of the direction. So to start with something nice, you need to do something more. So this is somehow the model of this conflict. In this sense, nice because it gives you complete control of the co-tension complex in the surrounding areas and the degree of concentration. Questions? I can add a comment. It's about Professor Evian. So we got the diagram there, the extension, but these are the predictions. So it brought up interesting this reason for why he is a modernized program. They wanted to be modernized spaces, so they wanted to be a modernized spaces. So it was this thing. And I think often, if there is a modernized scheme, and they wanted to conduct a modernized scheme, and you can expect a sense of it, it is through the reason of friction, and it is through the reason of the coefficient, and it is in parallel. But the funny thing is that in most cases, this extension complex of the modernized scheme is just completely out of friction. I mean, no one knows what to do with it. I mean, as you said, first it has an infinity many non-zero-quadron group, but I think there are no formula for it. And it's only because you replace Y by a derived version, so it's a function of the degree of concentration, on classical planes, paper-brown, e-organic, I would say, that this extension complex becomes easy to compute. You have a formula, you can write down the formula. So that's not entirely... In fact, friction really exists for schemes, but it's completely useless. You need 40-notch schemes. Most modernized spaces are just not schemes. So try to use these schemes, you know. Get it to the host. Also, maybe you said this today during the capture. This related to this point is, if you're just looking at classical schemes, right, there are these two possibilities. So you have something that's LCI, or something that's worse than LCI, and impossible to go down to parallel. And now we have everything in between, and the soup of this mess. Yeah, so we had to complete the sections, and we were like, a complete... I think sometimes you realize, because I think I wrote a literature of different types of papers where the poem was short theory, with a different complex, like, say, let's take this DG VEA, and do the formation theory for this DG VEA, and Brown, we were covering the formal scheme, the formal structure of why. And this DG VEA has nothing to do with the intention complex of why. There's something completely out of the blue. And the biggest, it's the tension to the underground, of the high-regionalized spaces, which is a problem. So that, and because of what you said, that this DG VEA was, I mean, the content of the complex is impacted so, so much in the boundaries, so there will never be the content of the complex of any scheme of parallel times. They're just, they're just possible. Yeah, I want to just start with you. I have one comment. I think I should ask more questions. That's a good, if you're not behind. Yeah. If you ask questions. Yeah, I do not know. So what kind of problem is, like, this obstruction theory that's solved for... I mean, somehow, you can try to solve this, but it's not really smooth, is what I'm talking about. And how does this thing, can you get the time to make it the higher ones, and then if you can look for all you say, it's infinite, it's infinite, it's infinite. You can prove that the amount of persons that exist, I don't know much about, infinite is not smooth, it's actually smooth. So complexion of the, of the towers, of the mass, by the sky, right? So you take a, you have a, you have a mass mass, and you take a P to I, right? And the potential complex of this guy, and you can characterize the time or smoothness of this map by this potential complex. And somehow, the potential complex, the ex-cruel of the potential complex, precisely gives you a way to get the map. So somehow, classically, the map is the same as the smoothness, to certain degree. This is the way how you, what makes sense is that you have the same goal. So, somehow, obstructions really give you the criteria of the smoothness of the time. So, I don't know, but I think it's under really good care. I was actually thinking about what concrete is happening, is there some logic-like problem which classically is uncomputable, and like this, there is a phase, at least on the example, which is taking the space of representations of the final variable of the, you know, smooth projective complex variety. And there is this term that, if you look at a simple orientation, then you look at the formal information of the space at that point. Then formally, it's given by equality equation. It has only equal to the same parameters. So how do you prove this? Where you compute the potential complex of the jacquard complex space, you find a very specific dg in the n-jama, you use hoshtere to prove its formal, and then you get to, you know, it's just a formal consequence that you have the last two similarities. And the dg in the n-jama as you write down is not the potential complex of the n-jama complex space. You mean the jacquard complex space to this state. So I think this is the frame set of the work. Is it Goldman or Simpson? No, it's not. And then there'll be something similar for, say, follows on the path of the al-Qibla, or, yeah, the representations of a real p-na-ful and the cody-ajar complex will essentially allow you to say something about the lovelace tracker. So we won't be fundraising, but it's a little bit like a curtain locus of a function. So things like this. So I'm going to use the example that you read about, that kind of goes fine, right? Sorry? Are you just criticizing the result because you can always add your simple representations and not a lovelace? Okay. So you're saying more about the conditions. Yeah, I mean, maybe it's... That's a nice thing to say. I don't think it was a criticism, it's one that said the problems you use is the direct information theory. Oh, I see. It did use. I mean, they... They used a different d-shilla, wasn't there your company? No, the thing is that, I mean, you have this complex space, so now find the right one. At this point, there is a potential complex at this point. This is not the one they use. But there is one. I think you can pull the very potential complex as an example, too, that you can take over the formal representation just because of this TGVF. This is not the one they use, they use another one. And they just prove that this other one also controls the same formal ring. I know. But there is no explanation. It's just that, you know, people are used to do this. This TGVF, you know, it controls this formal ring. The explanation is that it's the potential complex at the Ziharki-Uni-Rassauce. I see. Because if you look at the formal ring of the other Harki-Uni-Rassauce, you know, you can use any... So what's my question? What's an example of success? It was an example of success before the Harki-Uni-Rassauce. But it was only the ideas from the TGVF. I think what Chris was saying about the TGVF, you know, there's much more application of the TGVF as it is today. You need some global... It is not just really global.
 So, maybe I can come to the point.  In some ways, it's just a coincidence, so, yeah.  This is what I was trying to talk about, and I hope to be able to explain it.  So, it's just basically, in the beginning I will shortly record K-Demonstration for the G-Ultras,  because I mean, it works the same as for future rings, but just to see.  And the set up is the following. We take a pair of G-R-G-M-Mets and G-J-S-Mets and G-J-S-Mets.  And that's what we have. We have F-M-M-Mets and what we do.  And then there's this thing where it's called the derivation. So, derivation.  So, this is A, and G derivation.  And then we have the M, and the E-Demonstration.  So, my differences in the E-J-M would be, for example, A-D, A-D, A-D, A-D, A-D, A-D, A-D, A-D.  So, we can make sure we do.  And yes, so building here means, I mean, what are, what are, I mean, so we take one times one for this equation.  And we get the E times, E times the capital one.  We can see that we take a one zero and from the building here, I see, you see the E, the zero, the odd, the E.  The derivation doesn't see functions coming from here.  Do you know the sign?  This is derivation of degrees zero, I guess.  Yeah.  I mean, otherwise you can shoot them.  I see.  Okay.  Yeah, so this is degree two, we're going to do this.  I mean, it's all you need to find.  I mean, I'm just going to do three, I'm just going to hand-ship the problem.  Yeah, so it's not going to be more than a second.  Okay, so, what is next?  So, if you resist and if you're trying to call it derivations, derivations.  So, like, like in this definition, and this is in A.  And so this gives a sum term, I think, the n modules to n modules, just by, like in the second-hand sections.  Do it for me, when you take the version of AM, does it take all derivations of all degrees?  Yes.  No, just degree zero.  I'm just, oh, now I'm just taking the degree zero to be the first one, you know.  How is it going to hold this for me?  Maybe you can take the whole thing over.  Yeah, okay, we can also make it this way.  No, it's fine.  So the proposition is this one term is called invented by a module of the H.  So, yeah, how is it, from the A to N, same module, is the same as derivations.  And yeah, so this module is uniquely, determined uniquely, actually, uniquely, as in my,  and so I record quickly the proof, so, I mean, it's basic, so, you can find this free module,  and then you take the direct sum of all x in A, which are the units,  and you take A with x, so, that x is the form of symbol, symbol, or if she's reading, x,  degree, and the degree of x.  So I think you can find this idea here, I, so, something, generative value, and there is a user,  and so, this is the standard construction that we find in house shunt,  we define the sub-module, and if you were to, in the standard case, it would be not a standard question.  So, since we have a GG word, we do it under the standard direction,  and to define this, you can look at this diagram, so you take an n-line shunt, and so you have the direction of the data,  so data, they find this one, and you have, and here we have the differential,  and to define this differential, if I go into the, yeah, so, yeah, to say this,  to the differential, the data acts as the sadness, the edge of the differential,  and then you extend this by the average, okay,  and then with this differential, you need to check that I is preserved, so check that I is preserved,  and then you can design under A of D, by L of N, right,  and then, yeah, take the differential, so then you can show this as a problem.  And some of this is in nice construction, because, yeah, there's nice universal property that it defines,  but somehow to make sense of this in the right word, which we write,  this is writing out, so the problem is A1, A1 to a squeaky balloon, and it does not follow the data.  So, yeah, so if you want to have an example, you can just take k, so you take D,  so can we do the writing, you set check A as what? Check I is preserved by D.  So I define this relation to the problem, which is almost similar to the check S,  and you define this differential, differential to write down the equation,  you want to have, yeah, one hand is the height of the preserved.  Okay, so you can have two, oh, there's some of the PGR plus, whose differences are different,  and so you can take A1 is to be k, and then two, okay, so, yeah, if we do the construction,  we can get the term in minus one, and then, yeah, you just get the term in minus one,  so, yeah, so we need to derive, we need to derive, derive, and how to do this is,  we consider this, we don't consider the category of DG algebra, but we consider the  category of DG algebra over A. So, the category we're working on is DG algebra  over A, which are, so objects are the objects, are DG algebra X, which are given map from A  and which are DG algebra, okay, so we have DG, DG algebra, okay, and the morphisms are,  we have X, so objects, that's what we have to do, okay, so, and if you want to have  a co-fragment replacement in this, in this model category, so, since this is over A,  you just inherit a natural model structure from DGAB, and the co-fragment replacement  in this model structure is actually the semi-period revolution, so, the co-fragment replacement.  Okay, so it should be semi-free over B, right, so, since it's the B, DG, algebra, so,  the underlying graded algebra should be a DG algebra, okay, and, so, to see what we do  is we have to form a definition, so, so that B to DGA, and of course, if we have a  single-query resolution of some X, we have automatically enough for A, and so, this is  how it works, so, let B and A for A be a single-query resolution, so, we have a definition, and we define,  let me just sort of, an eco-fragment resolution has to be semi-free, I mean, what would be  the question? What would, like, what would be eco-fragment and what would be semi-free?  No, I say, oh yeah, maybe you could think of a eco-fragment solution.  So, it's the other way around, it's semi-free, semi-free is a co-fragment of the real presence.  So, yeah, but, yeah, so I written some notes, so, we don't want anything, we want  to write a little bit, so, I mean, how do we define the co-fragment? We take a semi-free  resolution, we apply omega to this semi-free resolution, and enter it back to A, to A,  and this is the classical solution, how do we write the conquer, right? So, you place  A with something, with a co-fragment resolution, you apply omega to it, and then to get, again,  the modules and name of the modules, and then you write the test. So, this is the  final, what we can do here. And what you need to check is, of course, the right-hand  column, so, when you write this, do you still mean omega p would be something concentrated  in R, right? What do you mean? So, the thing on the left of this is a product, you  can see it's scratched out to statement that it would be a p module, because you only  took degrees of navigation. Are you still taking the degrees of navigation?  I mean, I define the relation just in this set, right? But this omega is an A module,  I define omega mp as a module. Oh, that's very simple.  This one is free, free, okay, great. So, this was the actual one.  So, the data we show how this is actually done for construction, we did that category,  but for now, I will just use this and do lots of examples so we can download it.  It's not too scary at all. So, the proposition we need to prove is that this is quite  simple. So, we check two guys, then we know just a co-founder replacement, and then we  need to check, so they have got g, and two, the homotopy between two men, and they want  to do that. So, we have got g, and g, and g, and g, and g, and g, and g, and g, and g,  and the original set, and g. So, the last object is logical and in silicon,  Okay, you take some of the freedom from the human being,  and you go back into the idea.  Yeah, I'm talking here.  Well, I mean, yeah, I mean, somehow, the first example I've reviewed was personally  for the revolution, this is the only one reason I'm kind of confused.  I mean, it's just a different thing.  So the first one is sort of anything.  Really, it's a good one.  And so I record this, so this means that A is the...  I don't know which S I should even describe it.  So A is A, B is A, and B is A.  So A is A, B is A, B is A.  Then what you can check is that the only way A, B can just write down to present the U.K.  So I mean, in this game, the potential complex is just only the A, B value.  And what you can check with this is the other line,  the greater algebra, the potential complex is just A,  and so it will be plus the differential.  So let me write this like this.  Plus the differential D.  And this is S in the construction.  So...  You can answer it with D.  This is trans-arrollable.  So to show you what is the map, so if you start with A,  you want to have university integration to this guy.  So you want to have university integration to D.  And yeah, so you take X.  Elements in the second factor.  So elements in the second factor are a little bit right as delta D.  So then you kind of see what you get.  So if you start with the element in the vector space,  you send this to one in the w to d, delta D.  And if you start with an element in the element here, so for example,  in terms of two vectors, then yeah,  so you want to be in the form of trans-arrollable.  So this is...  If you switch those two, then this is just a usual...  Yeah, a few different...  So somehow...  Yeah.  Somehow in this case, the total differential is a little bit tricky.  So you just start this.  And this construction, precisely given with this differential,  you take X to the end,  and you get the sum.  So the next example I want to do is do this  conclusion resolution we had last time.  So yeah, the second factor is that  inside the D,  you understand,  you need to have a function,  which is a regular gap.  There's a regular sequence.  And then we have seen that there's a total resolution.  So in a couple minutes,  you take this guy,  this three module,  and you take the matrix,  and we have a natural gap,  X to the end,  and you send the final line to the gap gap.  Yeah, I get the feeling that you gave the example  for three problem-rover.  So it seems that in the differential you define,  the differential of A does not appear anywhere.  Yeah, because I simply omitted the differential on this one.  So this is this guy,  and if you want to define the differential,  then you take the same as the differential of A,  and cancel the delta thing,  plus matrix on to the end,  A is also differential.  And this guy,  you define by the commutation relation.  So here also,  here is also differential of A,  and here is also differential of A.  So just the underlying,  you know somehow the underlying graded algebra determines the underlying,  so the underlying graded algebra of A  is the underlying structure of the omega A and D  in the differential you define by hand.  So we have this map,  you get this in a little bit of a position,  so you can see the A,  the shift of the line,  and you still have the final line,  and the case is the final variation,  and the differential here was defined  by the number of degrees,  shift A to D,  or the slope of D.  So if you think D,  you can say what,  what about D?  This is the,  what about D?  What about D?  What about D?  This is the,  what about D?  So this is not a 3,  this is a sine wave,  because it's been here for years.  But then,  so this gives you a disguise,  and for disguise you know where you are in the case.  So the propositional,  similar states of this,  this is actually a symmetry of motion,  so to calculate the contrast and complex of A  you just need to,  write the,  write the machine here,  so this is,  you get a definition of E correct,  or P, E,  because,  there are many P I think it is.  So there's no Y high,  there's no direct sum of these,  there's no Y high,  there's no direct sum of these.  That's why I,  I have general,  I just want to say good morning here.  And this is a guy of this,  and the contrast and complex of this,  between one and the other.  So by this guy,  so the underlying,  the underlying guy is,  P and E,  and in this way, P is just a,  generated by the elements,  and we try to calculate it.  So this is just a,  and yeah,  we wrote the delta here for,  and then,  this is an initial,  and so I have,  I think same as in between,  these just a,  and so it's a classical thing,  but this is in the,  and so this is all a classical example.  So what you get is K elements in the beginning of the course.  So,  so,  the potential complex is quite, quite nice.  Okay, and then,  of course,  we should,  if we think of this in a geometric perspective,  we can see that,  and see what it's interacting with.  So the difference in the product is about the chain  and the second is the,  so this one is over B,  and this one is over B.  So okay, this is a,  I mean we can say B,  here you take the B,  this is the B vector space,  and then you put your general size.  So, what this says,  so what is actually compute,  so B,  this was AM,  and this was V of this other,  this is the solution.  So what this calculates,  is this relative potential complex of this,  so,  okay, let's V of I  or should we question?  V of I is the points  So this is v of i, i is my idea, v of i is...  So to get more information we should consider the potential complex of okay.  So you can ask what is a of okay.  And yeah, our b was right, it's chosen free, so it's free of okay.  And so this p is actually free of okay.  So p is free of okay.  Mainly p is more than 2 s of this...  It's not seen on the way forward.  We take this vector space, greater vector space in...  Yeah, that's when you write this thing.  We take the greater vector space which this is in the preview and this is in the view of the shot.  And we take that as well.  So doing the same thing again, what you find is...  This is then omega p over k, and then d over y over a.  So what you find is you have a...  So in degree minus one you have k generators in percent of the...  k generators on the other side.  So it's the difference of v of delta yj equal to delta of dyj.  And by this definition I wrote delta of fj.  And fj is not mentioned in this.  And this is the universal...  This is just delta fj over delta xk.  This is just the usual differential.  So what does it mean?  To get a better definition again, this is the contingent complex.  By the contingent complex, so the tangent complex,  it's just the space there, so it's the rule of the contingent complex.  And we can consider this at the point.  Take a map from A to k.  You can answer the contingent complex with k, I think, to do.  And if you can find...  And what you get is...  These objects are just...  We hear...  And the unit is just a big differential.  And f is just the usual Jacobic rule.  So somehow we recover just the usual complex, which is the...  As the tangent complex at the point.  In classical geometry we just get the tangent space at this,  but we get a two-ton complex.  So fj0 of this kind is the kernel of this kind.  This is the tangent space.  And h1 is somehow the excess...  And this space.  So this is somehow the hook method, which measures somehow  how much the equation of degenerate is.  This equation of degenerate is given 30 times u bar.  And...  Okay, so this is enough for the long-term to do about this example.  In the smarter way, it is used to use...  So the correct way to do it is to use the sequence of graphs.  And if you have tools that you know that are associated to this,  there's an exact triangle.  And from this triangle you can actually do sure this one.  So those two guys in normal, from this to the tangent complex of this one.  So the perfect example of what we do is the following.  So we take three affine schemes.  We take a1 to the d-edge point.  And I want to consider the direct intersection of those two guys.  I don't consider a thing so...  So if you think about a1, again, there's a sub-skeam.  So you take two sub-skeams, take a a1, take a a2, and you want to define the direct intersection.  And to take some idea how to do this, maybe this is the boring you did with...  So you use the following graph.  And if you have a push-out diagram, you have a...  And you have a new section to get introduced to the diagram for the quick end.  We have a new section.  And you can say it with...  And this is the game.  So for example, if a1 is a point, then you get an exact time.  So you can calculate the potential complex on this guy using those two.  So in general, you get a different exact time.  So mb can be c or a1.  So let me not write over here.  So let's look at it with the c's.  And so we all consider this always c.  So this is the general guess, and this is an exact time.  So somehow knowing the potential complex of all the three guys involved in the concept  gives you the potential complex on the fourth part.  Thank you.  Since I don't know much, if I don't, I can take a classical example.  So let's take b, just move.  Hang on.  Close the...  Here we are okay.  And you take those, the three motions.  Then we can see what you get from this sequence as a point.  So let's take a point.  So let's take a ui.  And since we did it better than the tangible, let's dualize the sequence.  So we take a point, and then from this we can take the longest exact sequence in chronology.  And so if you restrict this, the point, you get the page.  So if you write down the sequence, what do you get?  So if you take h0 of this guy, you see this is just the usual potential complex.  Tension space, if you want to.  And here you get the usual coming from the place of v, again.  And then we can...  Some values, again some concepts.  So I didn't say this, but if you have a smooth stream, a smooth dg algebra, that's not on the screen.  So if you have a smooth ring, let's say a smooth ring, then the potential complex is reported in dv0.  So somehow locally you can see that if you take a position of the local sum, then this is just the usual...  Why do you consider the potential complex?  Why do you consider that you can't walk with the potential complex?  I don't know if I do, but I don't think that's a good question.  But of course you can write this differently and just reverse the arrows.  Yeah, p1 and p2, I don't know.  Yeah, should we say p1?  Yeah, h1, so that's what it says.  So what do we get?  So we have the potential complex is that a1 and a2, and this maps to the potential complex of the swan dg.  So somehow if this map is subjected, then the intersection is transverse, and in this case this differential is zero.  In this case...  So if a1 and a2 is smooth, these two guys are also zero, so if you have a transverse intersection of a smooth arrow,  self-parallel, then h1 is zero.  So this, then you again, it is smooth.  So somehow this sequence has a classic factor, but transverse intersection happens a lot.  And in general, this guy measures some construction, which I hope is one of the areas where I'm sure.  So maybe I say three, I want to say three most things, but I don't know how much time I have.  So let me say the first thing is different derivation.  The first one, d of a1, is the same as in the map, same as between a5 and a2.  So this is the transverse intersection, it is a transverse intersection.  So from this you can define this working space of derivation.  So again, how this is a1, that's the mapping space in this category.  And again you can show that the guy that you find is actually a core percentage of this kind.  I mean I couldn't start over this but it's kind of hard to actually write down what I'm going to do.  But at the end of the story, I'm going to start writing down the question.  So it's a bit more complex and it's a counter-as in the text of the whole thing that we're talking about.  Here you are writing a function of all the categories but more is true now.  The function of a module is set as an object in a score of two points.  Continue.  But this should be a problem.  I mean if you talk about it you can have a stronger answer.  The second thing I want to say is this is a dilemma.  So about what is HR, if you look at the condition of the stage R, it's 8 to 1, a point of 10.  And I'm going to now, I'm going to always be over K, with some K, in hours of writing a minute something.  This is a problem.  What you can do is the quality demo.  So the demo is, X1, and M1 means, A, K, A or K, and you can insert it up, and you can hit,  and take M and R module, and you will get a small order.  So put another one, which will take us on M, and this is just, yeah, in the order.  We didn't define it above, I mean we don't know what this means.  I think it's more clear to write it this way, I think to write it in the opposite category.  So this is that, in the order.  How much of a category of F1 seems to be over respecting some MFX, respecting them MFX,  from the style of M1 to,  so the interpretation of this is you can take A, and M, and the square is the triple extension case,  and what you get here is just, you get the X1,  and you get the style of A and Y, and this is from X, so this is also the style of the kind of MFX,  and this is the Y we had before, and this one is just a more useful spot that we need,  and then we can do that.  So this is the case X, and classically if I is here, then these are just the function vectors, right?  So you have,  so I can do, you just get the function vectors,  and for higher I's, you get somehow some derived,  so you have the same thing as,  so let's get to something, let's say this object, and if you look at maps from this small disk,  and you precisely get this HIO,  and this looks fancy, but it's just a form of consequence of this,  so the output is to this, this statement, and this is just,  and no shout out anymore.  And then I want to come to the, I mean I could look this way,  and see the light, at least I want to come here with that point,  and since this is not shout out, you can look in one of those extra objects,  okay so the third problem is at least I will go this way,  and the rest of the structure is,  so the structure, and the structure is,  so again, this is the extension,  extension, and one of the things,  you can consider a lifting problem,  so I think this, again just in the opposite category,  can consider a,  given specific extension,  and then, okay,  and you can think of this as some of the same,  this same thing about every while,  so we don't have that problem,  so the question is, how can we find out  if there's a fixed lift from,  from spec A to some specific extension,  so here's the students inside,  and the answer is,  that this is precisely controlled by the potential,  so this is the task,  depending on the piece,  and the problem,  if the complex of Y over K,  and the values on the end of the day,  and the value of N,  so N,  if I add with the kernel,  kernel of a plan,  as in, as in P,  no, it's an A,  I mean, yeah, so, no,  you can show that the kernel is some A multiple,  so it makes sense to take,  to consider this in the,  and so there's a class in there,  and the class is,  and the class is,  and in the second statement,  is more often,  the space of lifts,  the space of the space of lift,  and here again,  the same object,  so, yeah, so the space of lifts,  is just the mapping space of spec A,  and so, first of all,  we can interpret this problem group  as a shorted space for liftings,  and as a second space of all lifts,  as even the mapping space,  instead of a problem,  but instead of shift by one,  and so there's a difference.  And since I have a problem,  I, I, it's on the complex,  you see right at the section,  it sounds like we have,  I mean, maybe we can see what this means there.  I hope, I don't want to,  I have seen a lot of S-type fancy things,  and I'm not saying it's okay.  So, so we have this,  where as we make it,  You know, I have a couple of questions.  You read the space of lips, but the set of maps and the whole of anything is a set.  Yeah, so the space of lips, so I mean there is just high note of the space of lips, is it separate?  No, no, no, no, no, I just understood this was an additional note.  The sum of the space of lips is the map of these guys.  Because this was an additional note, which respect is not given to them, so over here.  So here you can write the sum of the letters x of the affine, the affine drafts.  So I mean for me, I'm just confused, is there any sum of the letters in the category?  I think he writes home from the base of the space and map from the space of it.  So here I have home and here I mean actually the pi zero of the mapping space between those two guys.  And down here I really mean the mapping space between two guys.  I don't know if you can tell me if you're right or not.  Great, thanks.  That makes me feel more normal.  This is a suspicious set, and this is Stephen just saying, suspicious set.  So we can consider first the density problem, so the problem here.  What you have seen here is precisely h1, dc, this is precisely the x1, x1, x1, x1 of this guy without this and then you should.  So this gives you precisely obstructions.  So h1, dc, x, gives obstruction.  So the first example you can look at is x, x not good point, and you can have n, n could be the trivial.  The trivial is n, so what you ask for is that you can.  So in this case what you ask for is your living property, you are living at point, and can you find the tangent vector that is point.  And this is, the structure of this reason is space, but of course there's always a real tangent.  So you can ask a harder question, for example, we're giving tangent vector, and you just take this case.  And you look at this extension, a prime, which is just, let's say that you ask, can you extend this tangent vector to second part of the graph.  And there's really a good obstruction to this problem, to know that we can extend the tangent vector at the singular point of this intersection.  So this obstruction could be same here as in this case.  So to give an interpretation of the sequences, you take this obstruction and map it forward to those two values.  So if the obstruction, the mapping of the obstruction manages into those two values, there exists the lift of the tangent vector to a1, d2.  So suppose this happens, if you can find lifts to a1 and d2, you can extend this tangent vector somewhere.  So in this case, the only obstruction that lifts is in the corner, which is in the tangent space, d2.  It is going to somehow measure the difference between the two lifts, which you can choose, if any one, and these two.  So if you know that there's difference of the lifts, lifts and can be written as a sum of tangent vectors, a1 and d2, we know we can kind of just this lift to actually see in the intersection.  So somehow it's you who has processed the illusion of the first line. And then of course you can consider higher components.  You can consider n by the 1 and you can consider hn.  So again, with this theory, you find the ganglion of this obstruction lifts in h1.  Each one of these ganglions goes on the obstruction of the positive phase.  And in the derived case, to get higher obstructions, you need to shift the things.  So if you pass the n, it's in the square, you can ask for something like...  And you can add an element in the line of the line.  So it's in the line of the line.  And we'll talk to this lifting property, which is in the line of the line.  It's measured by x2. So it's measured by the h2 here. So we have hdk and it's measured with g.  So the higher hi's we touch.  So in the classical case, and it's here, we don't know the interpretation for this higher hi.  But here we know precisely how we go into the...  And so what else?  Let the one to say it.  So somehow the collection of this just enjoys the obstruction of things.  And you see some power. If you start with something smooth, so a1 and a2 is known, or a1 and a2 are better than the section.  Then you know that the higher hi is 0.  So a1 and a2 are completely intersection, those guys are not.  And c, the co-tension complex of c is just considered in degree 0 and 1.  So the intersection of the co-tension complex, the intersection of the two parts, the intersection of the smooth thing is again in degree 0.  This is completely different to the classical case.  In the classical case, you would understand those two skills just by taking...  Just adding up the ideas and looking them up.  In this case, you might get a really similar skill with the co-tension complex,  as you co-tension approach infinitely in degree 0.  So somehow the derived setting kind of lets you control what you say.  You kind of get in fun, out of the direction.  So to start with something nice, you need to do something more.  So this is somehow the model of this conflict.  In this sense, nice because it gives you complete control of the co-tension complex in the surrounding areas and the degree of concentration.  Questions?  I can add a comment.  It's about Professor Evian.  So we got the diagram there, the extension, but these are the predictions.  So it brought up interesting this reason for why he is a modernized program.  They wanted to be modernized spaces, so they wanted to be a modernized spaces.  So it was this thing.  And I think often, if there is a modernized scheme,  and they wanted to conduct a modernized scheme,  and you can expect a sense of it, it is through the reason of friction,  and it is through the reason of the coefficient, and it is in parallel.  But the funny thing is that in most cases, this extension complex of the modernized scheme is just completely out of friction.  I mean, no one knows what to do with it.  I mean, as you said, first it has an infinity many non-zero-quadron group,  but I think there are no formula for it.  And it's only because you replace Y by a derived version,  so it's a function of the degree of concentration,  on classical planes, paper-brown, e-organic, I would say,  that this extension complex becomes easy to compute.  You have a formula, you can write down the formula.  So that's not entirely...  In fact, friction really exists for schemes, but it's completely useless.  You need 40-notch schemes.  Most modernized spaces are just not schemes.  So try to use these schemes, you know.  Get it to the host.  Also, maybe you said this today during the capture.  This related to this point is,  if you're just looking at classical schemes, right, there are these two possibilities.  So you have something that's LCI, or something that's worse than LCI,  and impossible to go down to parallel.  And now we have everything in between, and the soup of this mess.  Yeah, so we had to complete the sections,  and we were like, a complete...  I think sometimes you realize,  because I think I wrote a literature of different types of papers  where the poem was short theory, with a different complex,  like, say, let's take this DG VEA,  and do the formation theory for this DG VEA,  and Brown, we were covering the formal scheme,  the formal structure of why.  And this DG VEA has nothing to do with the intention complex of why.  There's something completely out of the blue.  And the biggest, it's the tension to the underground,  of the high-regionalized spaces,  which is a problem.  So that, and because of what you said,  that this DG VEA was, I mean,  the content of the complex is impacted so,  so much in the boundaries,  so there will never be the content of the complex of any scheme of parallel times.  They're just, they're just possible.  Yeah, I want to just start with you.  I have one comment.  I think I should ask more questions.  That's a good, if you're not behind.  Yeah.  If you ask questions.  Yeah, I do not know.  So what kind of problem is, like,  this obstruction theory that's solved for...  I mean, somehow, you can try to solve this,  but it's not really smooth, is what I'm talking about.  And how does this thing,  can you get the time to make it the higher ones,  and then if you can look for all you say,  it's infinite, it's infinite, it's infinite.  You can prove that the amount of persons that exist,  I don't know much about,  infinite is not smooth, it's actually smooth.  So complexion of the,  of the towers,  of the mass,  by the sky, right?  So you take a, you have a,  you have a mass mass,  and you take a P to I, right?  And the potential complex of this guy,  and you can characterize the time  or smoothness of this map by this potential complex.  And somehow, the potential complex,  the ex-cruel of the potential complex,  precisely gives you a way to get the map.  So somehow, classically,  the map is the same as the smoothness,  to certain degree.  This is the way how you,  what makes sense is that you have the same goal.  So, somehow, obstructions really give you the criteria  of the smoothness of the time.  So, I don't know, but I think it's under really good care.  I was actually thinking about  what concrete is happening,  is there some logic-like problem  which classically is uncomputable,  and like this,  there is a phase, at least on the example,  which is taking the space of representations  of the final variable of the, you know,  smooth projective complex variety.  And there is this term that,  if you look at a simple orientation,  then you look at the formal information  of the space at that point.  Then formally, it's given by equality equation.  It has only equal to the same parameters.  So how do you prove this?  Where you compute the potential complex  of the jacquard complex space,  you find a very specific dg in the n-jama,  you use hoshtere to prove its formal,  and then you get to, you know,  it's just a formal consequence  that you have the last two similarities.  And the dg in the n-jama as you write down  is not the potential complex of the n-jama complex space.  You mean the jacquard complex space to this state.  So I think this is the frame set of the work.  Is it Goldman or Simpson?  No, it's not.  And then there'll be something similar for, say,  follows on the path of the al-Qibla,  or, yeah, the representations of a real p-na-ful  and the cody-ajar complex  will essentially allow you to say something  about the lovelace tracker.  So we won't be fundraising,  but it's a little bit like a curtain locus of a function.  So things like this.  So I'm going to use the example  that you read about, that kind of goes fine, right?  Sorry?  Are you just criticizing the result  because you can always add your simple representations  and not a lovelace?  Okay.  So you're saying more about the conditions.  Yeah, I mean, maybe it's...  That's a nice thing to say.  I don't think it was a criticism,  it's one that said the problems you use  is the direct information theory.  Oh, I see.  It did use.  I mean, they...  They used a different d-shilla,  wasn't there your company?  No, the thing is that, I mean,  you have this complex space,  so now find the right one.  At this point, there is a potential complex  at this point.  This is not the one they use.  But there is one.  I think you can pull the very potential complex  as an example, too,  that you can take over the formal representation  just because of this TGVF.  This is not the one they use,  they use another one.  And they just prove that this other one  also controls the same formal ring.  I know.  But there is no explanation.  It's just that, you know,  people are used to do this.  This TGVF, you know,  it controls this formal ring.  The explanation is that it's the potential complex  at the Ziharki-Uni-Rassauce.  I see.  Because if you look at the formal ring  of the other Harki-Uni-Rassauce,  you know, you can use any...  So what's my question?  What's an example of success?  It was an example of success  before the Harki-Uni-Rassauce.  But it was only the ideas  from the TGVF.  I think what Chris was saying  about the TGVF,  you know,  there's much more application  of the TGVF  as it is today.  You need some global...  It is not just really global.
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