Shifted symplectic structures
Tanaka, Hiro Lee
Definitions. Existence on BG, shifted cotangent stacks TX[n] and Parf.
 So thanks to everybody for showing up. Thanks to all of you here for showing up. And thanks to the organizers for letting me give this talk. I think it's a really fun talk to give. I do want to hear what you guys have to say about the frame. So I shouldn't upset it. That's not an action. Okay, so let's get started on something. I don't want to waste your time having fun. So I'm assuming that a lot of people here may not have much experience in synthetic geometries. So I kind of want to start with a motivation for all this, which is actually just by a synthetic geometry for smooth math. Synthetic structures are smooth, actually. So now something that's smooth is playing a totally different role than what happens with the rest of this workshop. By this I mean, you know, some of the metaphors I'm going to get to see in the audience. See infinity and infinity times the correct answer. And you don't want to let it just take your phone out of the speaker. Okay, so what's the motivation? Why should you care about some kind of metaphors? So there's one very good reason, which is as a citizen of the universe. So by that I mean somebody who lives in a physical world. So obviously I can't get into this, but it turns out that almost every single problem you can think of in classical mechanics fits into the framework of the synthetic manifold, for instance the phase space, the configuration space of the system that you're studying automatically has a synthetic structure. And I don't mean things as complicated as a three-body problem. I mean things stupid like a brand, you know, like a brand with a block going down a rank. There's a synthetic manifold that's showing the configurations of the possible states of that system. So maybe a less philosophical reason. So second, a synthetic geometry. I'm not going to talk about any real synthetic geometry, so you're welcome. And there's a reason that this is a field of mathematics. Here's a strong rigidity to a synthetic geometry at the same time of richness that allows you to actually construct synthetic topological invariance. So a synthetic geometry is interesting in its own way. We're going to converge to, this is going to give two more motivations. One is kind of a plug, so if you're into homological mere symmetry. And of course to even state the homological mere symmetry conjecture, you need to know something called the Foucaille Categoryus, and to know something called the Foucaille Categoryus would be better known as a synthetic geometry. But having a final motivation that I think is much closer to the concept of the last couple days of this workshop, I want to motivate by stating a theory of the conceivage. I'm not going to state it with any kind of rigidity, but I'm going to state what most people call the corollary of conceivative disformality. So if you don't ask me what it is, I want to present it. I'll get back to you later when I talk about the red stacks. It's going to be called the conceivative formality. In reality there's a couple of things called the conceivative formality. This is one of them. And so let me just tell you the rough consequence of the first conceivative formality here, which is that synthetic structures of a manifold X are characterized by the definition of the ring of smooth functions on the X, which also goes like on X. So this is the ring of smooth functions on the X as an associative factor. So I'll come back to why this might motivate some of the upcoming talks in the workshop. The point is, it might want to pursue something similar in the case of derived. Derived are the great geometry effects of a synthetic structure. This is some sort of deformation that we can use. So I'll get to that later. Okay, so that's a motivation. Let me now give you a definition. So here's some math. Part two of part one. The definition, that X be a smooth manifold of dimension 2S. And by dimension I mean dimension of the real numbers. You mean a manifold? And that omega be a smooth two form on X. So what does this mean? It's something that eats in, for instance, if you give it two tangent vectors or two tangent vector fields, it'll spit out a function on X and it'll do so anti-symmetrical. So it's just a derivative of four upon X that omega is called symplectic. It's two conditions whole. So first, omega needs to be closed under the derived differential. That's not the real deal, meaning it's whole zero. And second, there's multiple ways of treating this condition. Let's take the usual condition and tell it that it's a cobalt one that we'll meet later on. Which is, well, if you have a two form on a two-dimensional manifold, you can want to answer it to get a top form. And you can ask yourself, is this a volume form? And it has to be a volume form for you to call this a symplectic form. So this top form of omega less than times is number zero. So what does it mean to be number zero? At any point, this actually gives you determinants on the tangent space there. So it gives you a volume form and a manifold. And so here's a proposition relating it to a definition that will move omega up. A proposition. So every symplectic or every two form actually gives you a map. So let omega be a symplectic. And consider the map, now called capital theta, because more or less PTPP, called analogous map and derived shape, which you can see theta, you're following the map from vector field dot x to one form dot x. What does it do? It takes some vector field and gives you the volume form called, plug in E but wait for somebody else to plug in the other vector. And the proposition is theta is an isomorphism. If it only is omega to the n, it's number zero. And the proof of this is totally local. So what does it do? Look, it just gives you, if omega to the n is not a generative, it gives you an identification of a vector space in the universe. So this is somehow the definition of, oh sorry, definition. Let me give you another definition. So this theta depends on omega. If theta is an isomorphism, we call omega not a generative. For the obvious reasons. Okay, perfect. Done with page two. Now let me just give you some dumb examples. You probably think that's funny in the head because you're all smart and useless. Let me give you some examples. Okay, so here's a very dumb example but we'll start with notation. Let's say that your radical X is just number two. whose coordinates are always two times. Okay? So we call it dT, ytq, a decent vector. For the obvious reasons, because the two-manifold, the verb of the verb, has a value of 3.2-manifold and it's clearly defined as a volume formula. Alright, and in this case, the map theta, let's say that some point of X from the tangent space of X at that point, the cotangent space of X at that point, is given by the following. So the tangent space of a point of X is generated by a symmetric vector to the tangent of two, so I'll call it del P and del Q. And this will be searched two months. If you want a basis of a material, it's dq and a basis of a material. So that's a dumb example. There's more to this than one. It's like q gets the bad things. Let's take X to be the cotangent part of this q. And this is actually a construction that we'll be using later on in the right case. So pay attention if you're into it. If you don't pay attention here, I guess you'll have to take the other example later on. Alright, so let lambda be the following one on X. So give it a point, say, now from the cotangent one of the g, to some point of the x, and some vector there. I need to be able to tell you what lambda does in this tangent vector. So lambda at this tangent of the v is going to be the following. Let me draw for you a little picture. So here's the picture of the cotangent model of q. This is the zero section, the map of qc inside, and this is the cotangent interaction. And here's some point out. If we have some sort of tangent vector, the cotangent model, I can take the canonical projection map. So I'll give you a point called pi of alpha. And at this point I'll get some sort of tangent vector called the derivative of the projection map applied to b. It's the tangent vector. But I'll have the one form. It's set above this point so I can evaluate this tangent vector using l. I'll try to evaluate it on that tangent vector concept. Okay. Then the derivative of this one form, which is also omega, is a symplectic form. So every cotangent model comes equipped with a canonical symplectic form. And moreover, this symplectic form arises as the derivative of a one form. The main three models are the two forms. They're easy. And they're proof. It's very similar to some other proof in the derived case. Except that here you can choose local coordinates to be Rn. In the derived case we'll choose local coordinates from a semi-free TGS. So local coordinates. So writing my local coordinates, I choose some open subset of 2 that's in local Rn. And I'll denote the coordinates of Rn that you want to choose. To that, et cetera. Okay. Then let Pi be the dual coordinates for Qn. Okay, so if you like. So the set of numbers Q, how to be, give coordinates on the cotangent model of U. Very standard. And in local coordinates, land can be written as a summation of all the i's of Pi to dQi. So if you like dPi and local coordinates are the height of the cotangent direction, Qi is the point. And if dQi is the same, okay, this differential one form only cares about the horizontal direction of your tangent. So it's precisely the same thing as this. So in our computation for local coordinates, so let's dLand. It says dPi, what's dQi? And it's top four. dLand of tangent power is just dP1, dQ1, where's where's where's where's where's where's where's where's where's, dpn with dqn, which is clearly a top four. OK. It's not a jitter. Well, which is clearly a non-vanishing. OK. And of course, omega is closed because d squared of anything is closed. All right. That's somehow done, isn't it? Let me give you a non-example. A non-example is 1x. Let's say the force is here. OK. So what is that? So, since we have a symmetric form, it has to be closed. So it defines an element in Durand homology. OK. And we also need the condition that the top-wetched power of omega needs to get at the top four. It needs to get volume. In other words, a non-zero element in the top four of all the objects. But of course, we know what the product structure is. It's zero, zero, zero, zero. So it can't define a volume. There's no idea of a symmetric form in the structure of a force here. Even if you're like a pendulum. All right. So that's the end of the review for this smooth piece. Does anybody have any questions? Because I'm going to go up to now Stan and stuff. And I don't think I'm going too fast, which only as I am. I'm going too slow, so I can't help that. So again, just as I did in part two now, the first motivation is shifting simple expressions of the right-flags. So I don't think we're going to get some motivation. So the first motivation is that many marginalized facts that we care about naturally come in close. Well, if they all come in close, we're letting it choose a very natural one. Comparison. Either an obvious or a natural one. Simple expression. And so whenever I say something like this structure of derived stat, it might be shifted. I was playing with this in a second. And then there's a second motivation that relates back to the same actual reality here that I told you about. The second motivation is the quantizing field. So what do we have in the case of the consavency theorem? Consavency told us if we have a sum of quantization of this, we can give rise to definitions of the rate of function of x as an associative element. Under associative, I'm going to have an inference to see if it works E1. You don't know what this means. That's fine. But there's a family of observers called sales, and the rest of them are about to be broken just the same. And I get this as motivation. It's not going to be on the test. Anyway. And so what's the analogous picture that we want here? I can't describe this too. Maybe this is the goal of PCP, but for now it's the same picture. But I think that we have an edge shifting, some type of structure on x of derived stat. And what we'd like to be able to do, which is, I think in Congress, I don't know what the same is, I guess as our mentor. Almost the same. So is there a tree here? No, in the future it's fine. All right. So in case you're not subjected to the future, you're not going to be. This is not going to be the future. OK. What we might want to do is deform not just an algebra but a category. A quasi-colonial sheet of x, and maybe I want to deform a smaller category, but I don't know. Roughly speaking, you want to deform modules over x. So this has a symmetrical model, hence the product, hence the sheet. And you'd like to be able to deform this. So you can't have some of your 10. So I don't know if it's the end or the end minus one. The end minus one. So the end minus one has one. So maybe I should get minus one more. What? Roughly speaking. So let me give you an example because I'm sure that this sounds like all very abstract. But the example might even sound even more abstract. But I figured that I should give the example because we don't have what we're here for. All right. So maybe one comment here. Please make. It's a non-girard case, actually, for a simplified format for even opening to both the form of a mystery. The other is of the case in construction. The end module. The end module. The end minus one. So it's the end. It's the module that's finger-right. The end? Yeah. You know it's almost the end. So it's the end. The thing is that you're tricking us here because the E1 of stairs becomes an easy one downstairs. Because downstairs it becomes the edge of the wall. And downstairs it becomes the module structure that you should put on. You're right. So I'm fooling you. Yeah, that's it. So what this really does is it gives a shift. So if you get an association about the word, you can ask what happens to modules. I think that gives an easier category. People don't know what that happens to you at the time. The easier category of modules. What does that mean? So I'd say just having an associative algebra doesn't look at its category of modules. Well, if it's an intermediate algebra, you might say, oh, we can test together modules. But no, we're an associative algebra if you have one on side. So easier just means there's no multiple structure. So there's no chance for any problems. Just a moment for an equals one, say. Yeah. So as you just said, for any equals zero, we don't deform for X as a district of associative algebra. We deform for any equals one. Then we are going to deform not really the category of modules and modules in the category, but the two categories of categories of modules over a few co-effects as a two-pategority. And this is just keep doing this up to the end. So this is why it's the end only up to some more time. I see. All right. Well, I don't know if this motivation is demo or the other way of thinking. I was thinking all kinds of abstractions about this word, E, N. What does E, N even stand for? It stands for everything. That's true. So an example of this is, for instance, there's a favorite stack called QG. And as we discussed several times in this workshop, there's a rigidification theorem. And so the category of quadric order in chiefs, the complex of the quadric order in chiefs on BG is the same thing as the kind of a representation of G or complex representation. And what we're going to find is that BG has a shifted two shifts. Simplific structure. I think it's a little bit more. Yeah, better than usual though. Okay, good. And so now let's just tell us a bit. Can we deform this category? And this is based on conversations that I've had with Damien. I'm sure that maybe other people are thinking of this as well. I'm sure. So you won't be able to deform this. I'm not quite sure what exactly the right answer is, but it should be something like representation with some quantum. You're asking me to be ready? Yeah, so let's say that it's reductive, maybe even semi-simple. I'm not sure what the... I'm not good enough with representation theory to tell you how important the requirements are. But so what's the category that you might want to deform it to? It might be representations of some quantum group that's in fixed level, a lot of that means. Or it might be the representation of some loop group. I think there's some theorem that says that if you fix... it should be positive energy representation for that. But again, I think there's a theorem that says that these two are quads. So what's the marital structure? So again, here you have this symmetric marital structure called tensioning sheets. And here's the expectation for... actually what we know is that when we have a symmetrical marital structure, we have what's called a graded marital structure. And it's called the fusion product, which I think David just said under his breath. So it's a marital structure called the fusion product. So something that if you wanted to come up for error from the sea after I've done a great geometry, maybe this is some perspective on what's going on outside this sea. Just as a commentary on... So this is somehow a global stat model version of a lot of things that Kevin Costner was trying to do and has done at the formal level. My formal level means something in a formal neighborhood of a point of this stat. You don't know what that means. It's okay because now we're going to get some real numbers. By real, I mean, you know, I don't have to speak inferior to this. Okay, so that's an issue. So let's let it pass. I'm going to give many definitions in this section. It's going to be a derived outness test. And I'm going to assume it's locally a finite presentation. And the only reason that I'm assuming that it's locally a finite presentation is because it's co-tent and complexly perfect. So in fact, I forget whether or not this was stated earlier in the week. I think it was. The co-tent and complex is perfect. Perfect model. Especially over age. So what I mean, I mean, you think of this as a kit to say something like the tangent space that should be finite dimensions. You can do either. And that's how locally finite presentation really means that every single point locally is a finite dimension. So what you can do is... Okay. Since a co-tent and complex model is perfect, you can do it in a finite dimension complex. Now I want to start defining what is the perfect two-forms. So recall from the previous talk today that we're just going to space in two forms of some stacked F of... Yeah, so it's two forms of degree N. So I say the word weight in its broadness, you punch me in the head. So this is a big space, but let's try to pick actually connected components of it. And we saw it here earlier that said that we can actually compute this as a co-amology of some complex. What about co-amology? So let me first present this sheaf co-amology and then I'll try to do this. I think this one won't be possible. So here this is the co-amology of this sheaf called the F of co-tent and complex and the weight in the complex. Of course, this is the same thing as the finite of a home. So I'll go here to the sheaves of structure sheaf to this sheaf. Oh, I get professionally confused by this. This N is outside of the wedge. So what's the point? 22 form. This comes in pretty much all set when you take a vertex in it. 22 form. This is a co-amology and this is a math. The structure sheaf of F is a sheaf where you take the co-tent and complex and you want it twice and then you shift. And so now that this is by the adjunction between the duals, the adjunction between the tending complex, just the fact that the tending complex is the duals of the co-tending complex, we get an apple of capital theta, the exact same as the math of capital theta that I wrote in the smooth case. We get a math. And I'll call theta sub w, terms like the tenets and your choices, so you put in 24, but 24 is nothing like the game. So it's a 24. And the tangent complex as a sheaf, the co-tending complex, shifted by an apple, again as a sheaf. So hopefully this is clear. This is some sort of wedge two of something that eats tangent vectors and spits out some numbers. Or if you like it, it takes in Tf as a Tf and spits out something in structure sheaf. So moving a sheaf over to another side again, if you put a vector, you get a dual to that vector. Not a dual to that vector, but I get something that will wait for another vector before I dial it. Okay, all right. So here's the definition. The definition is number one. So I think that we have a two-fold called omega. On my staff at 3M, so omega is not degenerate. It's theta f. What's that? Theta omega is not up here. It's in a closed. Now again, as I emphasize earlier, this is a map of causing current sheaves. Like it's not a perfect sheave of the same thing. So in what category do you use a q-code as well, if you like, because the full-sub-category is perfect. So now we're very close to our definition of a symplectic form. If you recall in the Smith case, a symplectic form needed to always satisfy two conditions. There had to be not the genital, there had to be closed. It's exactly the same definition we have here. And so now I guess, I'm kind of surprised that this works out. I guess the definition works out pretty nicely. Okay, so I'm just going to give a post-definition. Okay. So here's another definition. We'll see that the space of non-generate two forms is defined to be the union of the connected components inside the space of all two forms of non-generative data. Okay, so being non-generative is a category that's invariant on our whole list. So if you have, you should think of the space of two forms as a bunch of disconnected spaces. You can say some of these spaces have points consisting of non-generate forms. So then in definition two, so you're just thinking in the full-sub-category. Yeah, so if you want to think of this as an infinity group, this is a full-sub-category of those that are in the region. That's exactly right. Okay, and number two, the space of non-generative two forms, let's say the space of n shifting in symmetric forms on your stack F is defined by what will be pulled back. It's kind of silly that people think that way. Okay, so six, F, N, you know what your stack is, so what you're creating. So I told you this is going to be pulled back with something, but what? We can already guess. So we have the space of all two forms, right? And we want to look at those two forms that are non-generative and closed. Well, I forget whether it was the last talk or the talk previous. We saw that there was already a map of the space of all closed two forms, the space of all four forms. Obviously, as we discussed, this is not some ordinary inclusion, it has values given by the space of keys, because being closed is not a property of destruction. We did determine how it's closed. At the same time we have a dumb inclusion here. Here's an actual inclusion of spaces of the non-generative two forms. Of course, if you know your pullbacks, if you want things that are both closed and non-generative, you know you've got to take a pullback. Because of the number of moving pullbacks. Okay. Good. Alright, so hopefully it's kind of clear. So what does it really mean to give a symphonic form? It means that if you just give a two form that happens to be non-generative, then they give some direct closing. Or maybe you could be getting closed forms such that it's, I mean, do you have what the form of the line is? Yeah, it's not going to be closed. Because you mean your top thing that could be an inclusion. Oh, here you go. I don't know, maybe this is all a warning. I don't know what I'm doing. I get the same feeling when I lecture in front of my students, they get no response. And then if I say this is going to be on the exam, they're like, holy shit. So you don't get the idea that if you don't answer these questions, then, so, so, um, because that's terminology, we say that some choice inside of this official set is an initiative to make a similar structure. Okay, so let me just make some remarks so we can get our hands and feet a little bit dirty. What the hell does this mean? For instance, if we have a stack, yes, it's possible that it has many entries and similar structures for different terms. The answer is no. It has an edge of this inflected structure that is unique. So let me get a remark. So a remark. So let's say that, okay, so I'm still on the assumption of before, so I've added some sort of local representative in my stack so that the cotending problem is fair. So if the cotending problem is an amplitude of A, B, and any sympathetic form of some character or space makes sure that the rest of the period is the same. If you like it with a T structure that we have to impose on a quasi-coherent cheese, this means that the co-amology object associated with that T structure is the same degree as A. So if the cotending complex looks inside these degrees, then any sympathetic form has a degree A plus B. Let me just draw a little picture of this to convince you of it. Okay, so here's the cotending complex. It's called homology object that looks inside degrees A to B. So if the cotending complex looks inside degrees A to B, then any sympathetic form of degrees A to B what is its cotending complex like? What's the dual? It's going to look somewhere as a negative A and a negative B. So this drawing is somehow implicitly similar to these things in these parts of the and we'll just find the numbers. And what does that look like? Well, if there's going to be any problems, this negative B needs to find out if something is really A like I said, and this negative A needs to find out if something is really A like I said, okay, just like this. A distance between T and B is not just scale. It's totally arbitrary, and so because of this, for these degrees to match up, I'll play it in my math, right, and ask people because they have negative A and it's like, okay, I understand. Okay, so great. So that's the first remark. So somehow, if somebody tells you that there's an exception to this remark, they start to even fail if the content in confidence is zero. That's a good action. Okay, so I'll come back to that in a second. So let me get some more remarks so that we can understand what's going on. Here's a remark. Okay, so let's say that our aesthetic is actually a spec of some T and B A. When you're redundant and say that's A, then Lf is concentrated in degrees less than or equal to zero. So the corpus of the color of this is that we get a spec A, we get a spec A. It means a simplistic form of degrees zero. Because the pre-strand constraint of this thing F, well, it's a simplistic form of degrees zero. The content in complex, the tending complex must have nothing outside of degrees zero, so it must be simple. So I think it's I'm sure you're thinking you can see the collars turning. So let's get some more attention to this if you can color it. Okay. The shift is always negative on the rarsky. Oh, yeah, that's right. I mean, again, except for this. Except for this. Oh, when you say, what the hell is the content in complex zero in your point? I'm going to say that. Oh, all right. One particular case of the content in complex always being zero, this is a point, and an example of N-shifted symplectic derived stack is a point. And it's N-shifted for a real point. And this is something I'll focus, I don't know, I guess, quality over talking about the derangements in the next slide. It turns out that a derangement stack is N-shifted symplectic. If it only is a Lagrangian instead of a point, but the point with the correct. So that's something crazy. Okay, so here's another reward. So I think I said in the very beginning that symplectic geometry is probably something very rigid. The examples that I'm giving you are supposed to not convey that. Maybe it's not rigid, but certainly it's consistent in the kind of geometry that you want. Here's another remark. So Cf is locally same assumptions. So, locally, try that in presented. So, if F admits that N-shifted symplectic structure and N is odd, then F must have virtual dimension 0. What the hell is the virtual dimension? So it's supposed to be the other characteristic of the tangent complex to the quotient complex. So, like, I mean, virtual dimension, I think is a parameter. And again, this is supposed to be the other characteristic. Is it that? Tangent? No, this is not. Okay. So what are anybody's other characteristics? I mean, this is some sort of, you know, a machine from some crazy derived stack. You know, I didn't get a number out of it. So I think there's some theorems that tell you that a phonetic is a nice enough static and actually you can actually get the other characteristics to get a fiber, because I'm trying to re-enact you to a chain complex. I'm actually not an expert on this. I don't know what that I'm talking about. But you can take the function on the space of, you know, class C points of F, points of where they're due, and that's not the point of the fiber of Tf. The fiber is a context which might be and this function is a very constant. So if the stack is connected, it's not. I didn't follow it up to reproduce that on the board. But there's no Tf that's actually essential. Oh, and so why is this? Is this something that I talked about like 30 seconds ago? Psychically. So, if there's an action to symphonic structure, but that is odd, you know that what happens to the other characteristics if you shift something at times will become this negative. But if there's going to be an identification between the chain complex and this guy, will these have the other characteristics equal to this negative symbol? All right. So now let me give you some examples. I'm going to list the examples and then the rest of my talk will be proving that those examples are the same. So I'm going to list them first. Examples. So is it possible to have a strong indication in the script? I have no idea. Somebody wanted to ask a question. A question on the question. If Bert will mention the interest in the strong indication in the script. No, no, no. You can just add a bit of other chips like minus 1 to come as the other has only been some of the dimensions of the tension complex because I'm in an anti-situation where we have to put a dimension to the other dimension and then minus the dimension of some h1 of the tension complex that is in the same form. This is why it's called the virtual equation. It behaves like something of the dimension zero, but the summation can be very good. So alright I'm going to carry this. I'm going to list them by letters. So that's cool. Alright. So where everything is working on this base K let's take a point. The point has an N-shifted simple structure for all of them. So that's the V. That's right. I mean yes. No, you're right. I have no other say. Hey, see me. So that's a nice beautiful example. Alright, so V. I'm going to define this object in a second. Maybe not a second, but maybe in a minute. So this space is going to be defined by space in this static. So remember earlier we had an example of the co-d tangent bundle of anything smooth that's going to be a simplecic guide. So we can mimic that construction here in fact because when the derivative will be shifted co-d tangent in spaces. I'll tell you what that is in a second. So under some assumptions on X this space has N-shifted simple construction. So what is the assumption? It is. So I'm going to say something pretty strict. I'm going to demand that X be deleted one for it but there's a comment in PTVD that you can probably do this for. And guys as well they just have a bug on that channel. Very good. Right? Am I right? No. If X is a derived deline monfer static and of course everything that we're talking about is locally a finite representation. So that's the tangent complex. It's perfect. So locally the current is presented. The closed form always exists and then we only check that it's not a generic DM stack because we're not in stock. It's suddenly true but you have to go something. It's an exercise there. Did you find it? Did you find it? Don't say it to everyone. The only thing is that we have to check it from this and it should be so often. We didn't do it. Great. I don't think it's a parent or an adult. Should I add that? That's my first email address. There's more examples. This one I think is kind of some of the examples that are most relevant to a lot of mathematics and physics. So let's fix a function. I'll look at the list which we are going to have to manage the symbols of the action physics with no idea. You can consider the first of all two things. Zero session the code taking place exists and you can also consider the graph of the written list if you guys think about school of graph speed or okay, it's not good. Let's say that X is some sort of yeah, that's a really good question. I'll write down the assumptions. Let's say X is some sort of a derived scheme with a zero shift in structure. Does it need to be just derived from this? I'll just start out with X. I'm looking at it because it can be actually the optimum stuff because of the B. You got that? No. Let's just say X is derived from the null fairy-tale. Do you want to go ahead and say that's the assumption and what's the conclusion? So from example D, this has a zero shift in some kind of structure. Conclusion is I can look at the intersection of the zero section with this graph in Dm. So if you do your translation correctly, this just means putting into critical locations of dysfunction. Critical locations of a function where the ds equals zero, and that's where the ds is zero. And of course, because of the run of delts, everything is always repeat. This guy has a shift in some kind of expression. I'll see if we can do this next time. Great. So as a negative point, I'll just take some kind of expression. I could also take a shifted function. It's also positive and then you get higher shifted specifications. I mean, we don't have to write it, just the comments. You can take functions with some degree and you get a change of ideas. And it's an element in hn of flexed description. Oh, and can be... You can take any close for close one form of degree n. And because, well, shifted content in stack classifies those... well, one form of degree n, you have a canonical map from x to t star x shifted by n, then you can go to the graph of this one form and you can go from this. That will be on the exam. Okay. This is actually a very special case of something even more general that we're trying to do on the other set. But in general, this is going to be in class this time. I think. So if I lie, I'm going to be wrong. I couldn't even close the only because we, as a collective whole, maybe the other person doesn't know what that means. There's a lot of notion. And m is n-shift in some way. And where's n-shift? Is n-shift? So, what's the origin of that? What's the opposite of what we're trying to do? Yeah, so the origin is, again, kind of like closed form. It's not some vanishing of something. It's some prescription of how something's n-shift. So let me just make that. Really, because I'm going to give you a little talk about it. Why is it really a good lie talk? Okay, and so now there's a few more advantages. I think that there's so many advantages, I think, of the nicer things about this. And I come away with some sort of nice dictionary. To say that g is an athron smooth smooth smooth smooth, it's just a group. And I'm not quite sure what the actual constraints on g are to be. A lot of authors say reductives. And as a special case of reductives, you might take g to be very simple. And we get bg is a derived orchestra. So we've talked about this, but you made a few appearances. Then, I'll get to this proof later, when it turns out that you can actually find some kind of forms of retitution. And you can actually describe what this set is. It's in bijection. Things that we're going to do with non-degenerative g invariant, I haven't told you what g is, but I've called it a four-time interval, please. Thanks. Do you guys have a word? I'll leave it. There's some kind of interaction here with the effect of geometry and representation here. And the final example, example e, example number e. So we have this stack, a curve, or path. Maybe you can find these examples. I can't remember the problem with the notation on the video, but I proved that this example was a good example. And the point is that this also has retitution. Think about the instruction. Okay. How much time do I have? I don't know. I don't know. 15 minutes? 15 minutes. I'm too boring. All right. We start at almost like 315. 310. Yeah, so you have 15 minutes. Okay. Yeah, no. You know, that's enough. Okay. So now, section of this graph, Lena, part 3, which is the first curve that I want to go through is the dynamic flow. Okay. So I tell you that the quantity of the physics is an angle of 2 to try to have an energy-intensive structure. What the hell does this space have? It's just I think that might be missing in a global section somewhere. But somebody told me again this is the plane to be our spec. Okay. So what is the cotangent level B? It should be something that so x is 1. Oh, so it's a cotangent x. So for now, we'll take it to the right and we'll go to the left. We'll go to the right presentation. It should be spec of something. What are functions of cotangent in the global? They should be like vector elements. Okay. This is going to be this amount of computing I guess taking this into some kind. Because I'm taking a spec of something. The cotangent space will be an alpha in the scheme. This is a problem. So actually it's alpha number of x. That's not how. Well, it's a fun in a generalized sense. It was like in the talk of when you have the spec B of epsilon where it's 1 because the action to bivalvescent can be on the right of the time. It's a fine over x if x is not a fine. Even if x is not a fine on the right scheme, which is our x in this stack. Yeah, no, no, it's an affine stack over x. That's a different one. Yes, affine stack. Affine stack, yes. I was thinking about it. I guess, I apologize because I wrote this down without knowing. But the formula makes perfect sense. It's something that fits over x and you can describe what are the local sections of this community. So if you think of x, the space of lift to start x is just hung from the significant amount of the full back of the tension complex to b. So it just, I mean, this one cannot make a perfect sense to me. It's a relative space. Affine stack over x. All right. So we're going to do the exact same thing that we did in the smooth case. I'm going to give you the last one for you. And the derivative is going to be something like the form, but I actually don't want to spend too much time on this because I can be the part that much works for this. Okay. So, um, I can't figure it out. So, we have a map. Sheens. Sheens. So, going into the one part of the symmetric power, um, for my adhesion, we have a following map. So this is going to be very stupid and easy. And the other thing is that the adhesion is a very, very, very simple word. And this, I'm going to do this in one form eventually. This is the same thing. Same thing. Same thing. Same thing. Same thing. Same thing. Same thing. Okay. So we're going to go ahead and do the projection. Sorry. The only reason I'm doing this is because I want to do the projection. And this map, this is the projection map. This is the projection map. This is the projection. Yeah. We have to hit the coefficient in the middle to X. And so, if you want to understand is that the level of specs, does it have the level of specs? Is it just given by the inclusion of the units? Sometimes everything that I'm doing is almost stupid. And so again by the direction, and of course by the direction, you get a map now of the pullback of the spectrogens, pullback of the co-plating objects, put them like that. And of course, this is your structure sheath, and the co-plating is there. So now we're almost done. So, the co-plating complexes, we've got whites, we've got the octopuses, get our hands, pullback of the structure sheath. If you have a map from B to A, and you have tensor, when you tensor, we think B tensor A or A, we can be, oh okay, this is what you're doing. Okay, so co-plating complexes, we're going to get a friend, and five receipts, okay, and, where's the map from B? So, I'm just not going to go up, there's a lot of co-plating complexes. So this map also has one that I didn't see. What is that? No, you can pull it up, and pull this map. So, we're going to obtain a map of structure sheath. This I'm just talking about by hand sign. This is how the shape of the map ends. So is this, is this part of these factors like these, related to what you told us in the very beginning, the one form that is only out of the parallel vector? No, please, you're all consulting about form on t star x. Yes, but then, these factors will be pulled back of x of n, which means that, yeah, so I didn't say that beginning of comments about the one form to star x being only same of horizontal, because I'm not sure where you can write this, I'll try to make it down to 3, and do some final algebra, but, we can take something like a piece of r as a model to solve, maybe even a bit of a ladder when I do this, and somehow pulling back, I would get exactly how far away it goes, but there's some sort of one form that's not a big part to start, but I'll try to do it in real time, because I don't have much real time. So, actually, we can shoot further as though about half an hour, so that you can have more time, especially if you don't go up early. Great. I apologize that I took so long in the beginning. Alright, so, nobody mind if I now go a little slower. Okay, so the whole point is, what the hell is this? This is exactly one form. The cotangent bottom right shifted. So, one form with an infall shift. Okay, so this defines a number of degree A, you can call this one form, and the next shift by N. So, this is going back to the observation that I made, that A0 of some complex compute pi0 of space of horns and the sum of elements of A0 with the golden section. Okay, so, this is a proposition. The fact that there are months in it, that D to R of lambda, which I'll define as a gear maker, is not a general. I have written out a full proof, but I think that I'm going to write it out as well, at some point, when you get to 15. So, any possible, or non-linear, or how do you do this? I think you want to. You say you turn out only negative, particularly. No, no, no. And the most direct, actually, I think you do. Okay, so here's a sketch of a proof. A sketch, which means I have to skip a page of my notes. So, it affects me. So, again, the claim, but I'm drawing this, it's an exercise for you to try to figure out the case of art, and we're going to stick this to a motor, and that means there's going to be a telecom. Okay. Ui, so in this case, it's an telecom. The program takes a little bit of a long time to finish. The program takes a little bit of an extra. This is what we're going to use for ball theory. That's right. Ball theory, yes. That is true. So, as a result, we can just now review the case. That's the case. And we're making changes in quality. Because these are the high-sod tests in the long term. So, there's something about the line of the production, which is that just construction land that I gave you is supposed to be natural. Obviously, in the form of functionality, you think the grid did it, and it won't. Okay. So, in this case, we're going to go with this case. In this case, the co-edition of the flex between shifting is by definition, which we can expect to this amount, what we're trying to do in the form. It's what you call the Hamilton-Gerfman. And so, one thing to note is, some of these is quality 3 of A, by definition, because it's symmetric and it's a bundle over k. And k is quality 3 over k. So, the B is quality 3 over k. Okay. This is where you skip some of the competition. So, what I'm going to say turns out, so, say, that's what you're going to do with this, in the k-compress, which is to shift it one point of B, because there's a k-compress of B and there's a k-compress of the shift. It's just done with that. I don't know what else to call it. And it just turns out that it's just a matter of shifting back these cases. So, I'm not going to define what these are, for which I apologize. They're all my own words. They're all my key shots. Okay. So, why is this described as being a wide-slimpt on man? So, the point of... Okay. The point is that these are maps of this space. Three models will repeat, with a certain set of generators, but the same type of generator is the same. It turns out that it just sends generators to generators. So, it clearly has more. So, generators, generators. Because you use ideal marionette, you can finish it from the diagonal, and just use the cover. I only use the cover. So, you know those more general things. I mean, there might be some statement about the co-ganded complex. Is that some question? I don't know. The point is, you need to get that problem to deal with some more. Right. But I'm saying, it's like, you can have no sub-questionation from that. I mean, he has that interest, he just sends this problem to us. No motion. There's a BG in under-depart. Does that appear? Do you want to skip the part? I think we want to hear about part. Yeah, we want to hear about both. I think we want to hear about both. Yeah. Actually, BG is almost a special case of part. Which part? BGLN is an open, sort of, attacking part. Yeah. So, many parts are good, no? I don't know. Perfect. I don't know. Do you want me to switch to your side in case you need to do it? No. I'm sorry. But BG is really interesting. What kind of game is this? So, I think, I think the next up, she would need a bit of BG to really quickly... No. No? No? No. Alright. Here we go. The perfect one. Let's remind ourselves of who M.I. is So, let's recall. We define the stacks. We define the stack empty for all BG categories empty. And what's the definition of the stack? So, what the hell is the stack? If you take some sort of CDGA you'll put a space for some visual set. And so, let me just try what it does on objects. It's supposed to take the mapping complex in DG categories of modules. Perfect modules. Well, of DG functions that T got the perfect module. This M.T. was defined. And so, in case T equals 1 for what's 1, there's a category with one object. Some dumb little thing. But then there's a lot of some people that can't hear. But then there's a lot of some people that can't hear. So, this one has a good point. And the one with the data. So, Perf is defined to be n-sub-1. So, obvious computation to not even a computation is a topology. What do you get when you evaluate this on some CDGA? You have maps of DG categories in the unit category. Perfect modules. And this is the space of the problem. It's the largest infinity group void inside the category. Of course, you can do this at the category level. And if you didn't think this as a space or simply so same, you would come to the category in terms of models. I don't think what we need. So, we have a a cool data. What does it say? It tells us how to compute the pullback of the content of 10-day complexes. It's theorem. Well, it's like e to your k-point beside this non-base-stand-collection. According to beta from previous theorem. So, this takes out an object inside a trend inside a DG category T. The pullback is a tangent complex. If we're pulling it back to spec K, so we're just going to get a sheaf on spec K which is just chain complex. There's a chain complex that was given by the endomorphism of the sum of T beside the categories of development. So, if you like stuff about actual probabilities, there's a very nice answer because what should a common state of a tangent complex be an object put back in the space of object should look like a waste of before. And you get that. Come here. Okay, so here's a theorem passed forward. Which is from what's left nicely ascent to the right stack. I'm not quite sure if I need to be hard on somebody who may be telling me I've been given a perfect module over X. What can you say? Pullback to tangent complex on this stack is what's equal to the category of closings in your sheaf. If you like to have a perfect module, it's the same thing. So, this is an object in the category of the sheaf object. Okay. So, now we're going to do this like really annoying categorical now. So, this is a very simple and really annoying categorical nonsense that works out incredibly well. So, by definition, Perth is a stack where if you map into it you get some sort of sheaf on the finger map in front. Well, even any stack there's a canonical map from that stack itself called the identity. So, there's some sort of sheaf over Perth that corresponds to the identity. That's a cool sheaf. Okay, so let's see. Let's take B and have to do the identity. Cool. Now, let's take a gain map from Perth. That's going to give us some sort of complex on it. And I claim that this is going to let us compute. This is going to give us some words to describe the taking complex of Perth. And we have a series of language systems. Tango complexes. Perth Perth and here's the relative. The first one to look at the relative tangent complex of the identity in that one, that thing stands here. What's that? A list sheet. So now we have the tangent complex of Perth as a quasi-convenient sheet called Perth is a complex to write the pullback of T for this. Well, this is a complex but the pullback of the quasi-convenient sheet is. So, like for a thing, the tangent complex of Perth is a complex to our home. We can't have a quality of the cheese on Perth. Which is weird. We can't have a quality of the cheese and then that complex shifted by the. So just abstractly, I have no idea what this sheet looks like. I don't know why. Today's idea is nothing that's perfect. That's nice. We have this first. Okay, there we go. We can read this. The corollary that I stated is this x. Oh, okay. I mean, this is a long, long time to inject some sort of color not to lose anybody. I got to read my sentence. I think your experiment is for k. and mt and now your opinion is T to be Perth and you use the fact that k. and Perth is the same as just the x. and m1. And you have to check that the content of the complex is the corollary of the Udemy and the corollary of the subjunction. If you just think about what the tangent complex will matter to be, you realize that it's not so bad. All right. Now let's make up. Then we have an algebra here. Let's note this algebra probably. Okay. So this is just as simple as what we mean by this. I just want to make note of the fact that the tangent complex is actually similar to this called the Udemy. Remember this is some sort of sheet of an algebra structure but of course in the sheet we remove that algebra structure from the sheet in algebra. But if you remember it as a sheet it's a supporting tangent complex. Okay. Now we leave it around pretty much done through me as the substitution which is such an algebra. Because remember what we need is some sort of biolinder form. What is that biolinder form? So, for first since a we use an anthropomorphism algebra. It has maps called the multiplication. One of the tensor forces them for a x to go up first. And then three since e is perfect it's kind of a weird game of a, that's a problem you check. So I don't know if you can play around much with like dualizability and what e to the t is. You think about it for a second. I mean you can check the cc for instance to find the cc. Okay. And moreover we can actually characterize oh so maybe the word that you prefer is original it's also original. I'll be using that. We can see moreover we can actually characterize the multiplication of a in terms of the unicode in it. Okay. Okay so we have a small multiplication now. Actually has another incarnation. We have tensor a and it's going to rewrite annoyingly. It's a product of four things. And how do I end up with tensor product with two things? Well I think the identity of the outside two and I take the trace map and I know the other one. Identity and the trace. Oh did I call it epsilon or trace? We call it trace. That's not the third thing. So in the topic about rigid objects we had a unit map with a coordinate map. This is a map that takes the middle of the factors and crunches them into a function. Okay. I can see that this is a unit. So if you haven't ever really thought about what multiplication of the matrices looks like this is a funny way to write it. And part D is a similar important part. It's a composite map. The matrix tensor a a okay so we have this map that we tensor a to a but a is the same thing if we check as e the algebra of the reward is a little bit rigid object and we again have the trace map. This is the same as the sheath structure. Okay. So the same as this composite. It's not the gender. It's not the gender. Because e is perfect constraint. There's no difference between the graph to e and the usual e or the whole object is the same. Because of the curves I have in the categorical Why is the utility more than the number of curves? There is a problem. Oh. So it's an iterated application of the triangle identity. For me the most convincing picture is so I apologize if people haven't seen these kinds of pictures before. But the triangle identity most I think it's here in one of the HFCs right? This is the triangle identity. And what this is saying is that what's the multiplication of the HFCs in the graph in the graph? And the whole point is that this this this is moving around like this. There's some non-usual what's the TFT stuff but if you play around the diagram you'll see that this Oh sorry. I forgot the capital. So then you have a trace amount. And this So this is some machine because I'm moving something in a droid but there's a big sphere. Are you with check it? To check that something is not in the head all the moving are perfect you can do it with point files. You take just the 5 or the point and they become complex objects of space, a nice You're not great with math and scissors. It's probably a good one. So what I was just saying is you can now check this point in the graph. I wanted to actually do a reduce the case of computing traces for math to change how much of the data I'll find out in that channel. So I'll just create an agency to check that to get another general matrix. Okay. So the level. Okay. So this is a non-generative form. What do you mean by this? It's a non-generative form. For a non-generative form. Yes. So what is a non-generative form supposed to be? First of all, give us a map. I can see from the tangent complex, tangent twice to the structure shift. Shifted by shift. Okay. We don't want this is. I mean this is by the theorem flash correlator I'm erased. See you want to test it. You want to test it. You want to test it. You want to. And we now have this map to O called multiple pi and then take a trace. It's not the only thing that we need to show that it's closed. It turns out to be not easy so I'm just sketching. To me this is a valid the main reason that am I going to play a wheelchair? I did. I apologize. I really apologize. I can't even look at the show. So in the whole. No, no. They don't make this close. I mean it doesn't kind of seem to be close to the guy. Wait, I thought that this was going to be Oh sorry, I got it. It's not close. It's not really close. Are you happy? No. I want to say that it's not that the lemma doesn't say I can find a leaf. There is one particular leaf and this is the leaf we call the symbol of the Schachter-Montau. Okay. Because this leaf is like a chunk of a leaf and it's not so long. So I don't see it. And that's what you okay. It was the space of the leaves. I have no idea what the space of the leaves is, but I mean we can start to this is getting on the scale. Is that there? Okay. Okay, so I'm just going to sketch this to improve. I'm going to hear the sketch to improve. Thank you to everybody to talk about Schrodinger. Schrodinger is a method. From the stamp of Earth this is an aspect that m controller staples her to get to the surface and so forth. I'm expected that This is actually, this is actually a group of the standard column C. And so this was the one that sends a CDGAA to dope card. so high notes. The class is all cleared in This is, of course, for the first thing I'm going to say is, to do with the chromology of what I gave, I'm going to apply the function of Nc to the static 30-monomes. Taking projections going to the piece, I'm going to reduce that. This is something I'll call here. I'm going to write a chord and first do the same thing. Of course, this is something I'll call point of last time. This is equivalent to finding a finite component. I'm in a space of close to 4 to the degree. Then you check that the zeolute is actually equal. One can check that, oh, I think I defined the non-degenerate two form on the algebra between multiplication and multiplication. One part of the formula is the z-dove. One part of the formula is z-dove. I love the things we're talking about. Do you want me to do this here? No, we can stop. Thank you. Questions? Yes? I think PGLM can be seen as this. I think you can find a 3-chloroposanus, a 3-chloroposanus, so it's a method of GN. Then it sends BGLM to path by sending a vector boundary so that the path back complexity is zero. You get a map from BG to path, if you put it back to form, it's one way to construct. It's also closed. Let me know that it's closed in the BG case is way easier. In fact, you already see your favorite non-generative gene variant in the form here. Multiplications matrices are going to apply to trace. In the classic way, you want to define some gene variant quadratic metric on the line. This is multiplication of matrices in state of choice. It goes back to the standard variable example section. I say this one, you're going to add two more. I'm going to get it. All the equations are on this. All right. Thanks again.
 So thanks to everybody for showing up.  Thanks to all of you here for showing up.  And thanks to the organizers for letting me give this talk.  I think it's a really fun talk to give.  I do want to hear what you guys have to say about the frame.  So I shouldn't upset it.  That's not an action.  Okay, so let's get started on something.  I don't want to waste your time having fun.  So I'm assuming that a lot of people here may not have much experience in synthetic geometries.  So I kind of want to start with a motivation for all this,  which is actually just by a synthetic geometry for smooth math.  Synthetic structures are smooth, actually.  So now something that's smooth is playing a totally different role than what happens with the rest of this workshop.  By this I mean, you know, some of the metaphors I'm going to get to see in the audience.  See infinity and infinity times the correct answer.  And you don't want to let it just take your phone out of the speaker.  Okay, so what's the motivation?  Why should you care about some kind of metaphors?  So there's one very good reason, which is as a citizen of the universe.  So by that I mean somebody who lives in a physical world.  So obviously I can't get into this, but it turns out that almost every single problem you can think of in classical mechanics  fits into the framework of the synthetic manifold, for instance the phase space, the configuration space  of the system that you're studying automatically has a synthetic structure.  And I don't mean things as complicated as a three-body problem.  I mean things stupid like a brand, you know, like a brand with a block going down a rank.  There's a synthetic manifold that's showing the configurations of the possible states of that system.  So maybe a less philosophical reason.  So second, a synthetic geometry.  I'm not going to talk about any real synthetic geometry, so you're welcome.  And there's a reason that this is a field of mathematics.  Here's a strong rigidity to a synthetic geometry at the same time of richness that allows you to actually construct  synthetic topological invariance.  So a synthetic geometry is interesting in its own way.  We're going to converge to, this is going to give two more motivations.  One is kind of a plug, so if you're into homological mere symmetry.  And of course to even state the homological mere symmetry conjecture,  you need to know something called the Foucaille Categoryus,  and to know something called the Foucaille Categoryus would be better known as a synthetic geometry.  But having a final motivation that I think is much closer to the concept of the last couple days of this workshop,  I want to motivate by stating a theory of the conceivage.  I'm not going to state it with any kind of rigidity, but I'm going to state what most people call the corollary of  conceivative disformality.  So if you don't ask me what it is, I want to present it.  I'll get back to you later when I talk about the red stacks.  It's going to be called the conceivative formality.  In reality there's a couple of things called the conceivative formality.  This is one of them.  And so let me just tell you the rough consequence of the first conceivative formality here,  which is that synthetic structures of a manifold X  are characterized by the definition of the ring of smooth functions on the X,  which also goes like on X.  So this is the ring of smooth functions on the X as an associative factor.  So I'll come back to why this might motivate some of the upcoming talks in the workshop.  The point is, it might want to pursue something similar in the case of derived.  Derived are the great geometry effects of a synthetic structure.  This is some sort of deformation that we can use.  So I'll get to that later.  Okay, so that's a motivation. Let me now give you a definition.  So here's some math.  Part two of part one.  The definition, that X be a smooth manifold of dimension 2S.  And by dimension I mean dimension of the real numbers.  You mean a manifold?  And that omega be a smooth two form on X.  So what does this mean? It's something that eats in, for instance,  if you give it two tangent vectors or two tangent vector fields,  it'll spit out a function on X and it'll do so anti-symmetrical.  So it's just a derivative of four upon X that omega is called symplectic.  It's two conditions whole.  So first, omega needs to be closed under the derived differential.  That's not the real deal, meaning it's whole zero.  And second, there's multiple ways of treating this condition.  Let's take the usual condition and tell it that it's a cobalt one that we'll meet later on.  Which is, well, if you have a two form on a two-dimensional manifold,  you can want to answer it to get a top form.  And you can ask yourself, is this a volume form?  And it has to be a volume form for you to call this a symplectic form.  So this top form of omega less than times is number zero.  So what does it mean to be number zero?  At any point, this actually gives you determinants on the tangent space there.  So it gives you a volume form and a manifold.  And so here's a proposition relating it to a definition that will move omega up.  A proposition.  So every symplectic or every two form actually gives you a map.  So let omega be a symplectic.  And consider the map, now called capital theta,  because more or less PTPP, called analogous map and derived shape, which you can see theta,  you're following the map from vector field dot x to one form dot x.  What does it do?  It takes some vector field and gives you the volume form called,  plug in E but wait for somebody else to plug in the other vector.  And the proposition is theta is an isomorphism.  If it only is omega to the n, it's number zero.  And the proof of this is totally local.  So what does it do?  Look, it just gives you, if omega to the n is not a generative,  it gives you an identification of a vector space in the universe.  So this is somehow the definition of, oh sorry, definition.  Let me give you another definition.  So this theta depends on omega.  If theta is an isomorphism, we call omega not a generative.  For the obvious reasons.  Okay, perfect.  Done with page two.  Now let me just give you some dumb examples.  You probably think that's funny in the head because you're all smart and useless.  Let me give you some examples.  Okay, so here's a very dumb example but we'll start with notation.  Let's say that your radical X is just number two.  whose coordinates are always two times.  Okay?  So we call it dT, ytq, a decent vector.  For the obvious reasons, because the two-manifold, the verb of the verb,  has a value of 3.2-manifold and it's clearly defined as a volume formula.  Alright, and in this case, the map theta, let's say that some point of X  from the tangent space of X at that point, the cotangent space of X at that point,  is given by the following.  So the tangent space of a point of X is generated by a symmetric vector  to the tangent of two, so I'll call it del P and del Q.  And this will be searched two months.  If you want a basis of a material, it's dq and a basis of a material.  So that's a dumb example.  There's more to this than one.  It's like q gets the bad things.  Let's take X to be the cotangent part of this q.  And this is actually a construction that we'll be using later on in the right case.  So pay attention if you're into it.  If you don't pay attention here, I guess you'll have to take the other example later on.  Alright, so let lambda be the following one on X.  So give it a point, say, now from the cotangent one of the g,  to some point of the x, and some vector there.  I need to be able to tell you what lambda does in this tangent vector.  So lambda at this tangent of the v is going to be the following.  Let me draw for you a little picture.  So here's the picture of the cotangent model of q.  This is the zero section, the map of qc inside, and this is the cotangent interaction.  And here's some point out.  If we have some sort of tangent vector, the cotangent model,  I can take the canonical projection map.  So I'll give you a point called pi of alpha.  And at this point I'll get some sort of tangent vector  called the derivative of the projection map applied to b.  It's the tangent vector.  But I'll have the one form.  It's set above this point so I can evaluate this tangent vector using l.  I'll try to evaluate it on that tangent vector concept.  Okay.  Then the derivative of this one form, which is also omega, is a symplectic form.  So every cotangent model comes equipped with a canonical symplectic form.  And moreover, this symplectic form arises as the derivative of a one form.  The main three models are the two forms.  They're easy.  And they're proof.  It's very similar to some other proof in the derived case.  Except that here you can choose local coordinates to be Rn.  In the derived case we'll choose local coordinates from a semi-free TGS.  So local coordinates.  So writing my local coordinates, I choose some open subset of 2 that's in local Rn.  And I'll denote the coordinates of Rn that you want to choose.  To that, et cetera.  Okay.  Then let Pi be the dual coordinates for Qn.  Okay, so if you like.  So the set of numbers Q, how to be, give coordinates on the cotangent model of U.  Very standard.  And in local coordinates, land can be written as a summation of all the i's of Pi to dQi.  So if you like dPi and local coordinates are the height of the cotangent direction,  Qi is the point.  And if dQi is the same, okay, this differential one form only cares about the horizontal direction of your tangent.  So it's precisely the same thing as this.  So in our computation for local coordinates, so let's dLand.  It says dPi, what's dQi?  And it's top four.  dLand of tangent power is just dP1, dQ1, where's where's where's where's where's where's where's where's where's,  dpn with dqn, which is clearly a top four.  OK.  It's not a jitter.  Well, which is clearly a non-vanishing.  OK.  And of course,  omega is closed because d squared of anything is closed.  All right.  That's somehow done, isn't it?  Let me give you a non-example.  A non-example is 1x.  Let's say the force is here.  OK.  So what is that?  So, since we have a symmetric form,  it has to be closed.  So it defines an element in Durand homology.  OK.  And we also need the condition that the top-wetched power of omega  needs to get at the top four.  It needs to get volume.  In other words, a non-zero element  in the top four of all the objects.  But of course, we know what the product structure is.  It's zero, zero, zero, zero.  So it can't define a volume.  There's no idea of a symmetric form in the structure of a force here.  Even if you're like a pendulum.  All right.  So that's the end of the review for this smooth piece.  Does anybody have any questions?  Because I'm going to go up to now Stan and stuff.  And I don't think I'm going too fast, which only as I am.  I'm going too slow, so I can't help that.  So again, just as I did in part two now,  the first motivation is shifting simple expressions of the right-flags.  So I don't think we're going to get some motivation.  So the first motivation is that many marginalized facts that we care about naturally come in close.  Well, if they all come in close, we're letting it choose a very natural one.  Comparison.  Either an obvious or a natural one.  Simple expression.  And so whenever I say something like this structure of derived stat, it might be shifted.  I was playing with this in a second.  And then there's a second motivation that relates back to the same actual reality here that I told you about.  The second motivation is the quantizing field.  So what do we have in the case of the consavency theorem?  Consavency told us if we have a sum of quantization of this,  we can give rise to definitions of the rate of function of x as an associative element.  Under associative, I'm going to have an inference to see if it works E1.  You don't know what this means. That's fine.  But there's a family of observers called sales,  and the rest of them are about to be broken just the same.  And I get this as motivation.  It's not going to be on the test.  Anyway.  And so what's the analogous picture that we want here?  I can't describe this too.  Maybe this is the goal of PCP, but for now it's the same picture.  But I think that we have an edge shifting, some type of structure on x of derived stat.  And what we'd like to be able to do, which is, I think in Congress,  I don't know what the same is, I guess as our mentor.  Almost the same.  So is there a tree here?  No, in the future it's fine.  All right.  So in case you're not subjected to the future, you're not going to be.  This is not going to be the future.  OK. What we might want to do is deform not just an algebra but a category.  A quasi-colonial sheet of x, and maybe I want to deform a smaller category, but I don't know.  Roughly speaking, you want to deform modules over x.  So this has a symmetrical model, hence the product, hence the sheet.  And you'd like to be able to deform this.  So you can't have some of your 10. So I don't know if it's the end or the end minus one.  The end minus one.  So the end minus one has one.  So maybe I should get minus one more.  What?  Roughly speaking.  So let me give you an example because I'm sure that this sounds like all very abstract.  But the example might even sound even more abstract.  But I figured that I should give the example because we don't have what we're here for.  All right.  So maybe one comment here.  Please make.  It's a non-girard case, actually, for a simplified format for even opening to both the form of a mystery.  The other is of the case in construction.  The end module.  The end module.  The end minus one.  So it's the end.  It's the module that's finger-right.  The end?  Yeah.  You know it's almost the end.  So it's the end.  The thing is that you're tricking us here because the E1 of stairs becomes an easy one downstairs.  Because downstairs it becomes the edge of the wall.  And downstairs it becomes the module structure that you should put on.  You're right.  So I'm fooling you.  Yeah, that's it.  So what this really does is it gives a shift.  So if you get an association about the word, you can ask what happens to modules.  I think that gives an easier category.  People don't know what that happens to you at the time.  The easier category of modules.  What does that mean?  So I'd say just having an associative algebra doesn't look at its category of modules.  Well, if it's an intermediate algebra, you might say, oh, we can test together modules.  But no, we're an associative algebra if you have one on side.  So easier just means there's no multiple structure.  So there's no chance for any problems.  Just a moment for an equals one, say.  Yeah.  So as you just said, for any equals zero, we don't deform for X as a district of associative algebra.  We deform for any equals one.  Then we are going to deform not really the category of modules and modules in the category,  but the two categories of categories of modules over a few co-effects as a two-pategority.  And this is just keep doing this up to the end.  So this is why it's the end only up to some more time.  I see.  All right.  Well, I don't know if this motivation is demo or the other way of thinking.  I was thinking all kinds of abstractions about this word, E, N.  What does E, N even stand for?  It stands for everything.  That's true.  So an example of this is, for instance, there's a favorite stack called QG.  And as we discussed several times in this workshop, there's a rigidification theorem.  And so the category of quadric order in chiefs, the complex of the quadric order in chiefs on BG  is the same thing as the kind of a representation of G or complex representation.  And what we're going to find is that BG has a shifted two shifts.  Simplific structure.  I think it's a little bit more.  Yeah, better than usual though.  Okay, good.  And so now let's just tell us a bit.  Can we deform this category?  And this is based on conversations that I've had with Damien.  I'm sure that maybe other people are thinking of this as well.  I'm sure.  So you won't be able to deform this.  I'm not quite sure what exactly the right answer is, but it should be something like representation with some quantum.  You're asking me to be ready?  Yeah, so let's say that it's reductive, maybe even semi-simple.  I'm not sure what the...  I'm not good enough with representation theory to tell you how important the requirements are.  But so what's the category that you might want to deform it to?  It might be representations of some quantum group that's in fixed level, a lot of that means.  Or it might be the representation of some loop group.  I think there's some theorem that says that if you fix...  it should be positive energy representation for that.  But again, I think there's a theorem that says that these two are quads.  So what's the marital structure?  So again, here you have this symmetric marital structure called tensioning sheets.  And here's the expectation for...  actually what we know is that when we have a symmetrical marital structure,  we have what's called a graded marital structure.  And it's called the fusion product, which I think David just said under his breath.  So it's a marital structure called the fusion product.  So something that if you wanted to come up for error from the sea  after I've done a great geometry, maybe this is some perspective on what's going on outside this sea.  Just as a commentary on...  So this is somehow a global stat model version of a lot of things that Kevin Costner was trying to do  and has done at the formal level.  My formal level means something in a formal neighborhood of a point of this stat.  You don't know what that means. It's okay because now we're going to get some real numbers.  By real, I mean, you know, I don't have to speak inferior to this.  Okay, so that's an issue.  So let's let it pass.  I'm going to give many definitions in this section.  It's going to be a derived outness test.  And I'm going to assume it's locally a finite presentation.  And the only reason that I'm assuming that it's locally a finite presentation  is because it's co-tent and complexly perfect.  So in fact, I forget whether or not this was stated earlier in the week.  I think it was.  The co-tent and complex is perfect.  Perfect model.  Especially over age.  So what I mean, I mean, you think of this as a kit to say something like  the tangent space that should be finite dimensions.  You can do either.  And that's how locally finite presentation really means that every single point locally  is a finite dimension.  So what you can do is...  Okay.  Since a co-tent and complex model is perfect,  you can do it in a finite dimension complex.  Now I want to start defining what is the perfect two-forms.  So recall from the previous talk today that we're just going to space in two forms  of some stacked F of... Yeah, so it's two forms of degree N.  So I say the word weight in its broadness, you punch me in the head.  So this is a big space, but let's try to pick actually connected components of it.  And we saw it here earlier that said that we can actually compute this as  a co-amology of some complex.  What about co-amology?  So let me first present this sheaf co-amology and then I'll try to do this.  I think this one won't be possible.  So here this is the co-amology of this sheaf called the F of co-tent and complex  and the weight in the complex.  Of course, this is the same thing as the finite of a home.  So I'll go here to the sheaves of structure sheaf to this sheaf.  Oh, I get professionally confused by this.  This N is outside of the wedge.  So what's the point? 22 form.  This comes in pretty much all set when you take a vertex in it.  22 form.  This is a co-amology and this is a math.  The structure sheaf of F  is a sheaf where you take the co-tent and complex and you want it twice and then you shift.  And so now that this is by the adjunction between the duals,  the adjunction between the tending complex,  just the fact that the tending complex is the duals of the co-tending complex,  we get an apple of capital theta,  the exact same as the math of capital theta that I wrote in the smooth case.  We get a math.  And I'll call theta sub w,  terms like the tenets and your choices,  so you put in 24, but 24 is nothing like the game.  So it's a 24.  And the tangent complex as a sheaf, the co-tending complex,  shifted by an apple, again as a sheaf.  So hopefully this is clear.  This is some sort of wedge two of something that eats tangent vectors  and spits out some numbers.  Or if you like it, it takes in Tf as a Tf  and spits out something in structure sheaf.  So moving a sheaf over to another side again,  if you put a vector, you get a dual to that vector.  Not a dual to that vector, but I get something that will  wait for another vector before I dial it.  Okay, all right.  So here's the definition.  The definition is number one.  So I think that we have a two-fold called omega.  On my staff at 3M,  so omega is not degenerate.  It's theta f.  What's that?  Theta omega is not up here.  It's in a closed.  Now again, as I emphasize earlier,  this is a map of causing current sheaves.  Like it's not a perfect sheave of the same thing.  So in what category do you use a q-code as well, if you like,  because the full-sub-category is perfect.  So now we're very close to our definition of a symplectic form.  If you recall in the Smith case,  a symplectic form needed to always satisfy two conditions.  There had to be not the genital,  there had to be closed.  It's exactly the same definition we have here.  And so now I guess,  I'm kind of surprised that this works out.  I guess the definition works out pretty nicely.  Okay, so I'm just going to give a post-definition.  Okay.  So here's another definition.  We'll see that the space of non-generate two forms  is defined to be the union  of the connected components  inside the space of all two forms  of non-generative data.  Okay, so being non-generative  is a category that's invariant on our whole list.  So if you have, you should think of the space of two forms  as a bunch of disconnected spaces.  You can say some of these spaces have points  consisting of non-generate forms.  So then in definition two,  so you're just thinking in the full-sub-category.  Yeah, so if you want to think of this as an infinity group,  this is a full-sub-category of those that are in the region.  That's exactly right.  Okay, and number two,  the space of non-generative two forms,  let's say the space of n shifting  in symmetric forms on your stack F  is defined by what will be pulled back.  It's kind of silly that people think that way.  Okay, so six, F, N,  you know what your stack is,  so what you're creating.  So I told you this is going to be pulled back  with something, but what?  We can already guess.  So we have the space of all two forms, right?  And we want to look at those two forms  that are non-generative and closed.  Well, I forget whether it was the last talk  or the talk previous.  We saw that there was already a map  of the space of all closed two forms,  the space of all four forms.  Obviously, as we discussed,  this is not some ordinary inclusion,  it has values given by the space of keys,  because being closed is not a property of destruction.  We did determine how it's closed.  At the same time we have a dumb inclusion here.  Here's an actual inclusion of spaces  of the non-generative two forms.  Of course, if you know your pullbacks,  if you want things that are both closed and non-generative,  you know you've got to take a pullback.  Because of the number of moving pullbacks.  Okay.  Good.  Alright, so hopefully it's kind of clear.  So what does it really mean to give a symphonic form?  It means that if you just give a two form  that happens to be non-generative,  then they give some direct closing.  Or maybe you could be getting closed forms  such that it's, I mean, do you have  what the form of the line is?  Yeah, it's not going to be closed.  Because you mean your top thing  that could be an inclusion.  Oh, here you go.  I don't know, maybe this is all a warning.  I don't know what I'm doing.  I get the same feeling when I lecture  in front of my students, they get no response.  And then if I say this is going to be on the exam,  they're like, holy shit.  So you don't get the idea that  if you don't answer these questions,  then, so, so, um,  because that's terminology,  we say that some choice  inside of this official set  is an initiative  to make a similar structure.  Okay, so let me just make some remarks  so we can get our hands and feet a little bit dirty.  What the hell does this mean?  For instance, if we have a stack,  yes, it's possible that it has many  entries and similar structures  for different terms.  The answer is no.  It has an edge of this inflected structure  that is unique.  So let me get a remark.  So a remark.  So let's say that,  okay, so I'm still on the assumption of before,  so I've added some sort of local representative  in my stack so that the cotending problem is fair.  So if the cotending problem  is an amplitude  of A, B,  and any sympathetic form  of some character or space  makes sure that the rest of the period is the same.  If you like it with a T structure  that we have to impose on a quasi-coherent cheese,  this means that the co-amology object  associated with that T structure  is the same degree as A.  So if the cotending complex  looks inside these degrees,  then any sympathetic form  has a degree A plus B.  Let me just draw a little picture of this  to convince you of it.  Okay, so here's the cotending complex.  It's called homology object  that looks inside degrees A to B.  So if the cotending complex  looks inside degrees A to B,  then any sympathetic form  of degrees A to B  what is its cotending complex like?  What's the dual?  It's going to look somewhere  as a negative A and a negative B.  So this drawing is somehow  implicitly similar to these  things in these parts of the  and we'll just find the numbers.  And what does that look like?  Well, if there's going to be any problems,  this negative B needs to find out  if something is really A like I said,  and this negative A needs to find out  if something is really A like I said,  okay, just like this.  A distance between T and B  is not just scale.  It's totally arbitrary,  and so because of this,  for these degrees to match up,  I'll play it in my math, right,  and ask people  because they have negative A  and it's like, okay, I understand.  Okay, so great.  So that's the first remark.  So somehow, if somebody tells you that  there's an exception to this remark,  they start to even fail  if the content in confidence is zero.  That's a good action.  Okay, so I'll come back  to that in a second.  So let me get some more remarks so that we  can understand what's going on.  Here's a remark.  Okay, so let's say that  our aesthetic is actually  a spec of some T and B A.  When you're redundant and say that's A,  then Lf is concentrated  in degrees less than or equal to zero.  So the corpus of the color of this  is that we get a spec A,  we get a spec A.  It means a simplistic form  of degrees zero.  Because the pre-strand constraint  of this thing F,  well, it's a simplistic form of degrees zero.  The content in complex, the tending complex  must have nothing outside of degrees zero,  so it must be simple.  So I think it's  I'm sure you're thinking  you can see the collars turning.  So let's get some more attention to this  if you can color it.  Okay.  The shift is always negative  on the rarsky.  Oh, yeah, that's right.  I mean, again, except for this.  Except for this.  Oh, when you say,  what the hell is the content in complex zero  in your point?  I'm going to say that.  Oh, all right.  One particular case of the content in complex always being zero,  this is a point,  and an example of N-shifted symplectic  derived stack is a point.  And it's N-shifted for a real point.  And this is something I'll focus, I don't know,  I guess, quality over talking about the derangements  in the next slide.  It turns out that a derangement stack  is N-shifted symplectic.  If it only is a Lagrangian instead of a point,  but the point with the correct.  So that's something crazy.  Okay, so here's another reward.  So I think I said in the very beginning  that symplectic geometry is probably something very rigid.  The examples that I'm giving you  are supposed to not convey that.  Maybe it's not rigid,  but certainly it's consistent  in the kind of geometry that you want.  Here's another remark.  So Cf is locally  same assumptions.  So, locally, try that in presented.  So, if F admits  that N-shifted symplectic structure  and N is odd,  then F must have virtual dimension 0.  What the hell is the virtual dimension?  So it's supposed to be the other characteristic  of the tangent complex  to the quotient complex.  So, like,  I mean, virtual dimension,  I think is a parameter.  And again, this is supposed to be the other characteristic.  Is it that?  Tangent?  No, this is not.  Okay.  So what are anybody's other characteristics?  I mean, this is some sort of, you know,  a machine from some crazy derived stack.  You know, I didn't get a number out of it.  So I think there's some theorems that tell you  that a phonetic is a nice enough static  and actually you can actually get the other characteristics  to get a fiber,  because I'm trying to re-enact you to a chain complex.  I'm actually not an expert on this.  I don't know what that I'm talking about.  But you can take the function  on the space of, you know,  class C points of F,  points of where they're due,  and that's not the point  of the fiber of Tf.  The fiber is a context which might be  and this function is a very constant.  So if the stack is connected,  it's not.  I didn't follow it up to reproduce that on the board.  But there's no Tf  that's actually essential.  Oh, and so why is this?  Is this something that I talked about  like 30 seconds ago?  Psychically.  So, if there's an action to  symphonic structure,  but that is odd,  you know that what happens to the other characteristics  if you shift something at times  will become this negative.  But if there's going to be an identification  between the chain complex and this guy,  will these have the other characteristics  equal to this negative symbol?  All right.  So now let me give you some examples.  I'm going to list the examples  and then the rest of my talk  will be proving that those examples  are the same.  So I'm going to list them first.  Examples.  So is it possible to have a strong indication  in the script?  I have no idea.  Somebody wanted to ask a question.  A question on the question.  If Bert will mention  the interest in the strong indication  in the script.  No, no, no.  You can just add a bit of other chips  like minus 1 to come  as the other has only been  some of the dimensions of the tension complex  because I'm in an anti-situation  where we have to put a dimension to the other dimension  and then minus the dimension of some  h1 of the tension complex  that is in the same form.  This is why it's called the virtual equation.  It behaves  like something of the dimension zero,  but the summation can be  very good.  So  alright  I'm going to carry this.  I'm going to list them by letters.  So that's cool.  Alright.  So where everything is working on this base K  let's take a point.  The point  has an N-shifted  simple structure  for all of them.  So that's the V.  That's right.  I mean yes.  No, you're right.  I have no other say.  Hey, see me.  So that's a nice beautiful example.  Alright, so V.  I'm going to define this object in a second.  Maybe not a second, but maybe  in a minute.  So  this space is going to be defined  by space in this static.  So remember earlier  we had an example of the co-d tangent bundle  of anything smooth that's going to be  a simplecic guide.  So we can mimic that construction here  in fact because when the derivative will be shifted  co-d tangent in spaces.  I'll tell you what that is in a second.  So under some assumptions on X  this space has  N-shifted simple construction.  So what is the assumption?  It is.  So I'm going to say something pretty strict.  I'm going to demand that X be deleted one for it  but there's a comment in PTVD that you can probably do this for.  And guys as well they just have a  bug on that channel.  Very good.  Right?  Am I right?  No.  If X is  a derived  deline monfer  static  and of course  everything that we're talking about is locally  a finite representation.  So that's the tangent complex. It's perfect.  So locally  the current  is presented.  The  closed form always exists  and then we only check that it's not  a generic DM stack  because we're not in stock.  It's  suddenly true but  you have to go something.  It's an exercise there.  Did you find it?  Did you find it?  Don't say it to everyone.  The only thing is that we have to check it from this  and it should be so often.  We didn't do it.  Great.  I don't think it's a parent  or an adult.  Should I add that?  That's my first email address.  There's more examples.  This one I think is kind of  some of the examples that are most  relevant to a lot of mathematics and physics.  So let's fix a function.  I'll look at the list  which we are going to have to manage  the symbols of the action  physics with no idea.  You can consider  the first of all two things.  Zero session  the code taking place  exists  and you can also consider the graph  of the written list  if you guys think about  school of graph speed  or okay, it's not good.  Let's say that X is some sort of  yeah, that's a really good question.  I'll write down the assumptions.  Let's say X is some sort of a  derived scheme with a zero shift  in structure.  Does it need to be  just derived from this?  I'll just start out with X.  I'm looking at it  because it can be actually  the optimum stuff because of the B.  You got that?  No.  Let's just say X is derived from  the null fairy-tale.  Do you want to go ahead and say  that's the assumption  and what's the conclusion?  So from example D, this has a  zero shift in some kind of structure.  Conclusion is I can look at  the intersection of the zero section  with this graph in Dm.  So if you do your translation  correctly, this just means putting into  critical locations of dysfunction.  Critical locations of a function  where the ds equals zero,  and that's where the ds is zero.  And of course, because of the  run of delts, everything is always repeat.  This guy has a shift in some kind of  expression.  I'll see if we can do this next time.  Great.  So as a negative point,  I'll just take  some kind of expression.  I could also take a shifted function.  It's also positive and then you get  higher shifted specifications.  I mean, we don't have to write it, just the comments.  You can take functions with some degree  and you get a change of ideas.  And it's an element in hn  of flexed description.  Oh, and can be...  You can take any close for close one  form of degree n.  And because, well, shifted content  in stack classifies those...  well, one form of degree n, you have  a canonical map from x to t star x  shifted by n, then you can go to  the graph of this one form  and you can go from this.  That will be on the exam.  Okay.  This is actually a very special case  of something even more general  that we're trying to do on the other set.  But in general,  this is going to be in class this time.  I think.  So if I lie,  I'm going to be wrong.  I couldn't even close the only  because we, as a collective whole,  maybe the other person doesn't know what that means.  There's a lot of notion.  And  m is n-shift  in some way.  And where's n-shift?  Is n-shift?  So, what's the origin of that?  What's the opposite of what we're trying to do?  Yeah, so the origin is, again, kind of  like closed form. It's not some vanishing  of something. It's some prescription of how something's n-shift.  So let me just make that.  Really, because  I'm going to give you a little  talk about it.  Why is it really a good lie talk?  Okay, and so now there's a few more  advantages.  I think that there's so many advantages,  I think, of the nicer things about this.  And I come away  with some sort of nice dictionary.  To say that g  is an athron  smooth  smooth  smooth  smooth, it's just a group.  And I'm not quite sure what the actual  constraints on g are to be.  A lot of authors  say reductives.  And as a special  case of reductives,  you might take g to be very simple.  And we get bg  is a derived orchestra.  So we've talked about this,  but you made a few appearances.  Then, I'll get to this proof later,  when it turns out that you can actually  find  some kind of forms of retitution.  And you can actually describe what this  set is.  It's in bijection.  Things that we're  going to do with  non-degenerative  g invariant,  I haven't told you what g is,  but I've called it  a four-time interval, please.  Thanks. Do you guys have a word?  I'll leave it.  There's some  kind of interaction here with the effect of geometry and  representation here.  And the final example,  example e,  example number e.  So we have this stack,  a curve, or path.  Maybe you can find these examples.  I can't remember the problem with the notation  on the video, but I proved that this example  was a good example.  And the point is that this also has  retitution.  Think about the instruction.  Okay. How much time do I have?  I don't know.  I don't know.  15 minutes?  15 minutes.  I'm too boring.  All right.  We start at  almost like 315.  310.  Yeah, so you have 15 minutes.  Okay.  Yeah, no.  You know, that's enough.  Okay.  So now, section  of this graph, Lena,  part 3,  which is  the first curve that I want to go through  is the dynamic flow.  Okay.  So I tell you that the quantity of the  physics is an angle of 2 to try to have an  energy-intensive structure.  What the hell does this space have?  It's just  I think that might be missing  in a global section somewhere.  But somebody told me again  this is the plane to be  our spec.  Okay. So what is the  cotangent level B?  It should be something that  so x is 1.  Oh, so it's a cotangent x.  So for now, we'll take it to the right  and we'll go to the left.  We'll go to the right presentation.  It should be spec of something.  What are functions of cotangent in the global?  They should be like vector elements.  Okay.  This is going to be  this amount of computing I guess  taking this into some kind.  Because I'm taking a spec of something.  The cotangent space will be  an alpha in the scheme.  This is a problem.  So actually it's alpha number of x.  That's not how.  Well, it's a fun in a generalized sense.  It was like  in the talk of  when you have the  spec B of epsilon where it's  1 because the action  to bivalvescent can be on the right of the time.  It's a fine over x if x is not a fine.  Even if x is not a fine  on the right scheme, which is our x in this stack.  Yeah, no, no, it's an affine stack over x.  That's a different one.  Yes, affine stack.  Affine stack, yes.  I was thinking about it.  I guess, I apologize  because I wrote this down without knowing.  But the formula makes perfect sense.  It's something that fits over x  and you can describe what are the local sections  of this community.  So if you think of x,  the space of lift to start x  is just  hung  from the significant amount of the full back  of the tension complex to b.  So it just, I mean, this one  cannot make a perfect sense to me.  It's a relative space.  Affine stack  over x.  All right.  So we're going to do the exact same thing  that we did in the smooth case.  I'm going to give you the last one for you.  And the derivative is going to be  something like the form, but I actually don't want to  spend too much time on this because I can be  the part that much works for this.  Okay.  So, um,  I can't figure it out.  So,  we have a map.  Sheens.  Sheens.  So,  going into the one part of the symmetric power,  um, for my adhesion,  we have a following map.  So this is going to be very stupid and easy.  And the other thing is that  the adhesion is a very, very, very simple word.  And this, I'm going to do this  in one form eventually.  This is the same thing.  Same thing.  Same thing.  Same thing.  Same thing.  Same thing.  Same thing.  Okay.  So we're going to go ahead and do the projection.  Sorry.  The only reason I'm doing this  is because I want to do the projection.  And this map,  this is the projection map.  This is the projection map.  This is the projection.  Yeah.  We have to hit the coefficient in the middle to X.  And so,  if you want to understand is that the level of  specs, does it have the level of specs?  Is it just given by the inclusion of the units?  Sometimes everything that I'm doing is  almost stupid.  And so again by the direction,  and of course by the direction,  you get a map now  of the pullback of the spectrogens,  pullback of the co-plating objects,  put them like that.  And of course,  this is your structure sheath,  and the co-plating is there.  So now we're almost done.  So,  the co-plating complexes,  we've got whites,  we've got the  octopuses,  get our hands,  pullback of the structure sheath.  If you have a map from B to A,  and you have  tensor,  when you tensor,  we think B tensor A or A,  we can be,  oh okay, this is what you're doing.  Okay,  so co-plating complexes,  we're going to get a friend,  and five receipts,  okay,  and,  where's the map from B?  So,  I'm just not going to go up,  there's a lot of co-plating complexes.  So this map also has one that I didn't see.  What is that?  No, you can pull it up,  and pull this map.  So,  we're going to obtain  a map  of  structure sheath.  This I'm just talking about  by hand sign.  This is how the shape of the map ends.  So is this,  is this part of these factors like these,  related to what you told us in the very beginning,  the one form that is only  out of the parallel vector?  No, please, you're all consulting  about form on t star x.  Yes, but then,  these factors will be pulled back  of x of n,  which means that,  yeah, so I didn't say that  beginning of comments about the one form  to star x being only  same of horizontal,  because I'm not sure  where you can write this, I'll try to make it down to 3,  and do some final algebra,  but,  we can take  something like a piece of r as a model to solve,  maybe even a bit of a ladder when I do this,  and somehow pulling back,  I would get exactly how far away it goes,  but there's some sort of one form  that's not a big part to start,  but I'll try to do it in real time,  because I don't have much real time.  So, actually,  we can shoot further  as though about half an hour,  so that you can have more time,  especially if you don't go up early.  Great.  I apologize that I took so long in the beginning.  Alright, so, nobody mind if I now go a little slower.  Okay, so the whole point is,  what the hell is this?  This is exactly one form.  The cotangent bottom right shifted.  So, one form with an infall shift.  Okay, so this defines  a number  of degree A,  you can call this one form,  and the next shift by N.  So, this is going back to the observation  that I made, that A0  of some complex compute pi0  of space of horns  and the sum of elements of A0  with the golden section.  Okay, so,  this is a proposition.  The fact that there are months in it,  that D to R of lambda,  which I'll define as a gear maker,  is not a general.  I have written out a full proof,  but I think that  I'm going to write it out as well,  at some point, when you get to 15.  So, any possible,  or non-linear,  or how do you do this?  I think you want to.  You say you turn out only negative,  particularly.  No, no, no.  And the most direct,  actually, I think you do.  Okay,  so here's a sketch of a proof.  A sketch,  which means I have to skip a page of my notes.  So, it affects me.  So, again, the claim,  but I'm drawing this, it's an exercise for you  to try to figure out the case of art,  and we're going to stick this to a motor,  and that means there's going to be a telecom.  Okay.  Ui,  so in this case, it's an telecom.  The program takes a little bit of a long time to finish.  The program takes a little bit of an extra.  This is what we're going to use for ball theory.  That's right.  Ball theory, yes.  That is true.  So, as a result, we can just now review the case.  That's the case.  And we're making changes in quality.  Because these are the high-sod tests in the long term.  So, there's something about the line of the production,  which is that just construction land that I gave you  is supposed to be natural.  Obviously, in the form of functionality,  you think the grid did it, and it won't.  Okay.  So, in this case,  we're going to go with this case.  In this case,  the co-edition of the flex between shifting  is by definition,  which we can expect to this amount,  what we're trying to do in the form.  It's what you call the Hamilton-Gerfman.  And so, one thing to note is,  some of these  is quality 3  of A,  by definition,  because it's symmetric and it's a bundle over k.  And k  is quality 3  over k.  So,  the B is quality 3 over k.  Okay.  This is where you skip some of the competition.  So, what I'm going to say turns out,  so,  say,  that's what you're going to do with this,  in the k-compress,  which is to shift it  one point of B,  because there's a k-compress of B  and there's a k-compress of the shift.  It's just  done with that.  I don't know what else to call it.  And it just turns out that it's just a matter  of shifting back these cases.  So, I'm not going to define what these are,  for which I apologize.  They're all my own words.  They're all my key shots.  Okay.  So, why is this described as being a wide-slimpt on man?  So, the point of...  Okay.  The point is that these are maps of this space.  Three models will repeat,  with a certain set of generators,  but the same type of generator is the same.  It turns out that it just sends generators to generators.  So, it clearly has more.  So,  generators,  generators.  Because you use ideal marionette,  you can finish it from the diagonal,  and just use the cover.  I only use the cover.  So, you know those more general things.  I mean, there might be some statement  about the co-ganded complex.  Is that some question?  I don't know.  The point is,  you need to get that problem to deal with some more.  Right.  But I'm saying, it's like,  you can have no sub-questionation from that.  I mean,  he has that interest,  he just sends this problem to us.  No motion.  There's a BG in under-depart.  Does that appear? Do you want to skip the part?  I think we want to hear about part.  Yeah, we want to hear about both.  I think we want to hear about both.  Yeah.  Actually, BG is almost a special case of part.  Which part?  BGLN is an open, sort of, attacking part.  Yeah.  So, many parts are good, no?  I don't know.  Perfect.  I don't know.  Do you want me to switch to your side in case you need to do it?  No.  I'm sorry.  But BG is really interesting.  What kind of game is this?  So, I think,  I think the next up,  she would need a bit of BG to really quickly...  No.  No?  No?  No.  Alright.  Here we go.  The perfect one.  Let's remind ourselves of who M.I. is  So, let's recall.  We define the stacks.  We define the stack  empty  for all BG categories  empty.  And what's the definition of the stack?  So, what the hell is the stack?  If you take some sort of CDGA  you'll put a space  for some visual set.  And so, let me just try what it does on objects.  It's supposed to take  the mapping complex  in DG categories  of modules.  Perfect modules.  Well,  of DG functions that T got  the perfect module.  This M.T. was defined.  And so,  in case T equals 1  for what's 1,  there's a category  with one object.  Some dumb little thing.  But then there's a lot of some people that can't hear.  But then there's a lot of some people that can't hear.  So, this one has a good point.  And the one with the data.  So, Perf is defined to be n-sub-1.  So, obvious computation  to not even a computation is a topology.  What do you get when you evaluate  this on some CDGA?  You have maps of DG categories  in the unit category.  Perfect modules.  And this is  the space of the problem.  It's the largest infinity group void  inside the category.  Of course, you can do this  at the category level.  And if you didn't think this  as a space or simply so same,  you would come to the category  in terms of models.  I don't think what we need.  So, we have a  a cool data.  What does it say?  It tells us how to compute the pullback  of the content of 10-day complexes.  It's theorem.  Well, it's like e  to your k-point  beside this non-base-stand-collection.  According to beta from previous theorem.  So, this takes out an object  inside a trend  inside a DG category T.  The pullback  is a tangent complex.  If we're pulling it back to spec K,  so we're just going to get a sheaf on spec K  which is just chain complex.  There's a chain complex that was given  by the endomorphism of the sum of T  beside the categories  of development.  So, if you like stuff about  actual probabilities, there's a very nice answer  because what should a common state  of a tangent complex  be an object put back in the space of object  should look like a waste of before.  And you get that.  Come here.  Okay, so  here's a theorem passed forward.  Which is  from what's left  nicely ascent  to the right stack.  I'm not quite sure if I need to  be hard on  somebody who may be telling me  I've been given  a perfect module  over X.  What can you say?  Pullback to tangent complex  on this stack  is what's equal  to  the category of closings  in your sheaf.  If you like to have a perfect module,  it's the same thing.  So, this is an object  in the category  of the sheaf object.  Okay.  So, now we're going to do this  like really annoying  categorical now.  So, this is a  very simple  and really annoying  categorical nonsense  that works out incredibly well.  So,  by definition,  Perth is a stack where if you map into it  you get some sort of sheaf on the finger  map in front.  Well, even any stack there's a canonical map  from that stack itself called the identity.  So, there's some sort of sheaf over  Perth that corresponds to the identity.  That's a cool sheaf.  Okay, so let's see.  Let's take  B  and have to do the identity.  Cool.  Now, let's take a gain map from Perth.  That's going to give us some sort of  complex on it.  And I claim that this is going to  let us compute.  This is going to give us some words  to describe the taking complex of Perth.  And we have  a series of  language systems.  Tango complexes.  Perth  Perth  and here's the relative.  The first one to look at the relative  tangent complex of the identity in that one,  that thing stands here.  What's that?  A list sheet.  So now we have the tangent complex of Perth  as a quasi-convenient sheet called Perth  is a complex  to write the pullback of T for this.  Well, this is a complex  but the  pullback of the quasi-convenient sheet is.  So, like for a  thing,  the tangent complex of Perth is  a complex to our home.  We can't have a quality  of the cheese on Perth.  Which is weird.  We can't have a quality of the cheese  and then that complex  shifted by the.  So just abstractly,  I have no idea what this sheet looks like.  I don't know why. Today's idea is nothing that's perfect.  That's nice.  We have this first.  Okay, there we go.  We can read this.  The corollary that I stated is this x.  Oh, okay.  I mean, this is a long, long time  to inject some sort of color  not to lose anybody.  I got to read my sentence.  I think  your experiment  is for k.  and mt  and now your opinion is T to be Perth  and you use the fact that  k. and Perth is the same as  just the x.  and m1.  And you have to  check that the content of the complex  is the corollary of the Udemy and the corollary of the subjunction.  If you just think about  what the tangent complex will matter  to be, you realize that it's not so bad.  All right.  Now let's make up.  Then we have an algebra here.  Let's note this algebra probably.  Okay.  So this is just as simple as  what we mean by this.  I just want to make note of the fact that  the tangent complex  is actually similar to this  called the Udemy.  Remember this is some sort of sheet  of an algebra structure  but of course in the sheet  we remove that algebra structure  from the sheet in algebra.  But if you remember it as a sheet  it's a supporting tangent complex.  Okay.  Now we leave it around  pretty much done through me as the substitution  which is such an algebra.  Because remember what we need is some sort of  biolinder form.  What is that biolinder form?  So,  for first  since a  we use an anthropomorphism  algebra.  It has maps  called the multiplication.  One of the tensor forces them for a x  to go up first.  And then  three  since e is perfect  it's kind of a weird game of  a, that's a problem  you check.  So I don't know if you can play around much  with like dualizability and what e to the t  is.  You think about it for a second.  I mean you can check the cc for instance to find  the cc.  Okay.  And moreover we can actually characterize  oh so  maybe the word that you prefer is original  it's also original.  I'll be using that.  We can see moreover we can actually characterize  the multiplication of a in terms of the  unicode in it.  Okay.  Okay so  we have a small multiplication now.  Actually has another incarnation.  We have tensor a and it's going to rewrite  annoyingly.  It's a product of four things.  And how do I end up  with tensor product with two things?  Well I think the identity of the outside two  and I take the trace map  and I know  the other one.  Identity  and the trace.  Oh did I call it epsilon or  trace?  We call it trace.  That's not the third thing.  So in the topic about rigid objects  we had a unit map with a coordinate map.  This is a map that takes the middle  of the factors and crunches them  into a function.  Okay.  I can see that this is a unit.  So if you haven't ever really thought about what  multiplication of the matrices looks like  this is a funny way to write it.  And part D is a similar important part.  It's a composite map.  The matrix tensor a  a  okay so  we have this map that we tensor a to a  but a is the same thing if we check  as e  the algebra of the reward is a little bit rigid object  and we again have the trace map.  This is the same as the  sheath  structure.  Okay.  So the same as this composite.  It's not the gender.  It's not the gender.  Because e is  perfect  constraint.  There's no difference between  the graph to e and the usual e  or  the whole object is the same.  Because of the curves I have in the categorical  Why is the utility  more than the number of curves?  There is a  problem.  Oh.  So it's an iterated application of the triangle  identity.  For me the most convincing picture is  so I apologize if people haven't seen these  kinds of pictures before.  But the triangle identity  most  I think it's here in one of the HFCs  right?  This is the triangle identity.  And what this is saying is that  what's the multiplication of  the HFCs  in the graph  in the graph?  And the whole point is that  this  this  this is  moving around like this.  There's some non-usual  what's the TFT stuff but  if you play around the diagram you'll see that this  Oh sorry.  I forgot the capital.  So then you have a trace amount.  And this  So this is some machine because  I'm moving something in a droid  but there's a big sphere.  Are you with check it?  To check that something is not  in the head all the  moving are perfect you can do it  with point files.  You take just the 5 or the  point and they become  complex objects of space, a  nice  You're not great with math and scissors.  It's probably a good one.  So what I was just saying is you can now check this point  in the graph.  I wanted to actually do a reduce the case of  computing traces for math to change how  much of the data I'll find out in that channel.  So I'll just create an agency to check that  to get another general matrix.  Okay.  So the level.  Okay.  So this is  a non-generative form.  What do you mean by this?  It's a non-generative form.  For a non-generative form.  Yes.  So what is a non-generative form  supposed to be?  First of all, give us a map.  I can see from the  tangent complex, tangent twice  to the structure shift.  Shifted by shift.  Okay.  We don't want this is.  I mean this is  by the theorem flash correlator  I'm erased.  See you want to test it.  You want to test it.  You want to test it.  You want to.  And we now have this map  to O called multiple pi and then take a trace.  It's not the only thing that we need to show that it's closed.  It turns out to be not easy  so I'm just sketching.  To me this is a valid  the main reason that  am I going to play a wheelchair?  I did.  I apologize.  I really apologize.  I can't even look at the show.  So in the whole.  No, no.  They don't make this close.  I mean it doesn't kind of seem to be  close to the guy.  Wait, I thought that this was going to be  Oh sorry, I got it.  It's not close.  It's not really close.  Are you happy?  No.  I want to say that  it's not that the lemma doesn't say  I can find a leaf.  There is one particular leaf  and this is the leaf we call the symbol of the Schachter-Montau.  Okay.  Because this leaf is like a chunk  of a leaf and it's not so long.  So I don't see it.  And that's what you  okay.  It was the space of the leaves.  I have no idea what the space of the leaves  is, but I mean  we can start to  this is getting on the scale.  Is that there?  Okay.  Okay, so I'm just going to sketch this  to improve.  I'm going to hear the sketch to improve.  Thank you to everybody  to talk about Schrodinger.  Schrodinger is a method.  From the stamp  of Earth  this is an aspect  that m controller  staples her to  get to the  surface  and so forth.  I'm expected that  This is actually, this is actually a group of the standard column C.  And so this was the one that sends a CDGAA to dope card.  so high notes.  The class is all cleared in  This is, of course, for the first thing I'm going to say is,  to do with the chromology of what I gave,  I'm going to apply the function of Nc to the static 30-monomes.  Taking projections going to the piece,  I'm going to reduce that.  This is something I'll call here.  I'm going to write a chord and first do the same thing.  Of course, this is something I'll call point of last time.  This is equivalent to finding a finite component.  I'm in a space of close to 4 to the degree.  Then you check that the zeolute is actually equal.  One can check that, oh, I think I defined the non-degenerate two form  on the algebra between multiplication and multiplication.  One part of the formula is the z-dove.  One part of the formula is z-dove.  I love the things we're talking about.  Do you want me to do this here?  No, we can stop.  Thank you.  Questions?  Yes?  I think PGLM can be seen as this.  I think you can find a 3-chloroposanus,  a 3-chloroposanus, so it's a method of GN.  Then it sends BGLM to path by sending a vector boundary  so that the path back complexity is zero.  You get a map from BG to path,  if you put it back to form, it's one way to construct.  It's also closed.  Let me know that it's closed in the BG case is way easier.  In fact, you already see your favorite non-generative  gene variant in the form here.  Multiplications matrices are going to apply to trace.  In the classic way, you want to define  some gene variant quadratic metric on the line.  This is multiplication of matrices in state of choice.  It goes back to the standard variable example section.  I say this one, you're going to add two more.  I'm going to get it.  All the equations are on this.  All right.  Thanks again.
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So thanks to everybody for showing up. Thanks to all of you here for showing up. And thanks to the organizers for letting me give this talk. I think it's a really fun talk to give. I do want to hear what you guys have to say about the frame. So I shouldn't upset it. That's not an action. Okay, so let's get started on something. I don't want to waste your time having fun. So I'm assuming that a lot of people here may not have much experience in synthetic geometries. So I kind of want to start with a motivation for all this, which is actually just by a synthetic geometry for smooth math. Synthetic structures are smooth, actually. So now something that's smooth is playing a totally different role than what happens with the rest of this workshop. By this I mean, you know, some of the metaphors I'm going to get to see in the audience. See infinity and infinity times the correct answer. And you don't want to let it just take your phone out of the speaker. Okay, so what's the motivation? Why should you care about some kind of metaphors? So there's one very good reason, which is as a citizen of the universe. So by that I mean somebody who lives in a physical world. So obviously I can't get into this, but it turns out that almost every single problem you can think of in classical mechanics fits into the framework of the synthetic manifold, for instance the phase space, the configuration space of the system that you're studying automatically has a synthetic structure. And I don't mean things as complicated as a three-body problem. I mean things stupid like a brand, you know, like a brand with a block going down a rank. There's a synthetic manifold that's showing the configurations of the possible states of that system. So maybe a less philosophical reason. So second, a synthetic geometry. I'm not going to talk about any real synthetic geometry, so you're welcome. And there's a reason that this is a field of mathematics. Here's a strong rigidity to a synthetic geometry at the same time of richness that allows you to actually construct synthetic topological invariance. So a synthetic geometry is interesting in its own way. We're going to converge to, this is going to give two more motivations. One is kind of a plug, so if you're into homological mere symmetry. And of course to even state the homological mere symmetry conjecture, you need to know something called the Foucaille Categoryus, and to know something called the Foucaille Categoryus would be better known as a synthetic geometry. But having a final motivation that I think is much closer to the concept of the last couple days of this workshop, I want to motivate by stating a theory of the conceivage. I'm not going to state it with any kind of rigidity, but I'm going to state what most people call the corollary of conceivative disformality. So if you don't ask me what it is, I want to present it. I'll get back to you later when I talk about the red stacks. It's going to be called the conceivative formality. In reality there's a couple of things called the conceivative formality. This is one of them. And so let me just tell you the rough consequence of the first conceivative formality here, which is that synthetic structures of a manifold X are characterized by the definition of the ring of smooth functions on the X, which also goes like on X. So this is the ring of smooth functions on the X as an associative factor. So I'll come back to why this might motivate some of the upcoming talks in the workshop. The point is, it might want to pursue something similar in the case of derived. Derived are the great geometry effects of a synthetic structure. This is some sort of deformation that we can use. So I'll get to that later. Okay, so that's a motivation. Let me now give you a definition. So here's some math. Part two of part one. The definition, that X be a smooth manifold of dimension 2S. And by dimension I mean dimension of the real numbers. You mean a manifold? And that omega be a smooth two form on X. So what does this mean? It's something that eats in, for instance, if you give it two tangent vectors or two tangent vector fields, it'll spit out a function on X and it'll do so anti-symmetrical. So it's just a derivative of four upon X that omega is called symplectic. It's two conditions whole. So first, omega needs to be closed under the derived differential. That's not the real deal, meaning it's whole zero. And second, there's multiple ways of treating this condition. Let's take the usual condition and tell it that it's a cobalt one that we'll meet later on. Which is, well, if you have a two form on a two-dimensional manifold, you can want to answer it to get a top form. And you can ask yourself, is this a volume form? And it has to be a volume form for you to call this a symplectic form. So this top form of omega less than times is number zero. So what does it mean to be number zero? At any point, this actually gives you determinants on the tangent space there. So it gives you a volume form and a manifold. And so here's a proposition relating it to a definition that will move omega up. A proposition. So every symplectic or every two form actually gives you a map. So let omega be a symplectic. And consider the map, now called capital theta, because more or less PTPP, called analogous map and derived shape, which you can see theta, you're following the map from vector field dot x to one form dot x. What does it do? It takes some vector field and gives you the volume form called, plug in E but wait for somebody else to plug in the other vector. And the proposition is theta is an isomorphism. If it only is omega to the n, it's number zero. And the proof of this is totally local. So what does it do? Look, it just gives you, if omega to the n is not a generative, it gives you an identification of a vector space in the universe. So this is somehow the definition of, oh sorry, definition. Let me give you another definition. So this theta depends on omega. If theta is an isomorphism, we call omega not a generative. For the obvious reasons. Okay, perfect. Done with page two. Now let me just give you some dumb examples. You probably think that's funny in the head because you're all smart and useless. Let me give you some examples. Okay, so here's a very dumb example but we'll start with notation. Let's say that your radical X is just number two. whose coordinates are always two times. Okay? So we call it dT, ytq, a decent vector. For the obvious reasons, because the two-manifold, the verb of the verb, has a value of 3.2-manifold and it's clearly defined as a volume formula. Alright, and in this case, the map theta, let's say that some point of X from the tangent space of X at that point, the cotangent space of X at that point, is given by the following. So the tangent space of a point of X is generated by a symmetric vector to the tangent of two, so I'll call it del P and del Q. And this will be searched two months. If you want a basis of a material, it's dq and a basis of a material. So that's a dumb example. There's more to this than one. It's like q gets the bad things. Let's take X to be the cotangent part of this q. And this is actually a construction that we'll be using later on in the right case. So pay attention if you're into it. If you don't pay attention here, I guess you'll have to take the other example later on. Alright, so let lambda be the following one on X. So give it a point, say, now from the cotangent one of the g, to some point of the x, and some vector there. I need to be able to tell you what lambda does in this tangent vector. So lambda at this tangent of the v is going to be the following. Let me draw for you a little picture. So here's the picture of the cotangent model of q. This is the zero section, the map of qc inside, and this is the cotangent interaction. And here's some point out. If we have some sort of tangent vector, the cotangent model, I can take the canonical projection map. So I'll give you a point called pi of alpha. And at this point I'll get some sort of tangent vector called the derivative of the projection map applied to b. It's the tangent vector. But I'll have the one form. It's set above this point so I can evaluate this tangent vector using l. I'll try to evaluate it on that tangent vector concept. Okay. Then the derivative of this one form, which is also omega, is a symplectic form. So every cotangent model comes equipped with a canonical symplectic form. And moreover, this symplectic form arises as the derivative of a one form. The main three models are the two forms. They're easy. And they're proof. It's very similar to some other proof in the derived case. Except that here you can choose local coordinates to be Rn. In the derived case we'll choose local coordinates from a semi-free TGS. So local coordinates. So writing my local coordinates, I choose some open subset of 2 that's in local Rn. And I'll denote the coordinates of Rn that you want to choose. To that, et cetera. Okay. Then let Pi be the dual coordinates for Qn. Okay, so if you like. So the set of numbers Q, how to be, give coordinates on the cotangent model of U. Very standard. And in local coordinates, land can be written as a summation of all the i's of Pi to dQi. So if you like dPi and local coordinates are the height of the cotangent direction, Qi is the point. And if dQi is the same, okay, this differential one form only cares about the horizontal direction of your tangent. So it's precisely the same thing as this. So in our computation for local coordinates, so let's dLand. It says dPi, what's dQi? And it's top four. dLand of tangent power is just dP1, dQ1, where's where's where's where's where's where's where's where's where's, dpn with dqn, which is clearly a top four. OK. It's not a jitter. Well, which is clearly a non-vanishing. OK. And of course, omega is closed because d squared of anything is closed. All right. That's somehow done, isn't it? Let me give you a non-example. A non-example is 1x. Let's say the force is here. OK. So what is that? So, since we have a symmetric form, it has to be closed. So it defines an element in Durand homology. OK. And we also need the condition that the top-wetched power of omega needs to get at the top four. It needs to get volume. In other words, a non-zero element in the top four of all the objects. But of course, we know what the product structure is. It's zero, zero, zero, zero. So it can't define a volume. There's no idea of a symmetric form in the structure of a force here. Even if you're like a pendulum. All right. So that's the end of the review for this smooth piece. Does anybody have any questions? Because I'm going to go up to now Stan and stuff. And I don't think I'm going too fast, which only as I am. I'm going too slow, so I can't help that. So again, just as I did in part two now, the first motivation is shifting simple expressions of the right-flags. So I don't think we're going to get some motivation. So the first motivation is that many marginalized facts that we care about naturally come in close. Well, if they all come in close, we're letting it choose a very natural one. Comparison. Either an obvious or a natural one. Simple expression. And so whenever I say something like this structure of derived stat, it might be shifted. I was playing with this in a second. And then there's a second motivation that relates back to the same actual reality here that I told you about. The second motivation is the quantizing field. So what do we have in the case of the consavency theorem? Consavency told us if we have a sum of quantization of this, we can give rise to definitions of the rate of function of x as an associative element. Under associative, I'm going to have an inference to see if it works E1. You don't know what this means. That's fine. But there's a family of observers called sales, and the rest of them are about to be broken just the same. And I get this as motivation. It's not going to be on the test. Anyway. And so what's the analogous picture that we want here? I can't describe this too. Maybe this is the goal of PCP, but for now it's the same picture. But I think that we have an edge shifting, some type of structure on x of derived stat. And what we'd like to be able to do, which is, I think in Congress, I don't know what the same is, I guess as our mentor. Almost the same. So is there a tree here? No, in the future it's fine. All right. So in case you're not subjected to the future, you're not going to be. This is not going to be the future. OK. What we might want to do is deform not just an algebra but a category. A quasi-colonial sheet of x, and maybe I want to deform a smaller category, but I don't know. Roughly speaking, you want to deform modules over x. So this has a symmetrical model, hence the product, hence the sheet. And you'd like to be able to deform this. So you can't have some of your 10. So I don't know if it's the end or the end minus one. The end minus one. So the end minus one has one. So maybe I should get minus one more. What? Roughly speaking. So let me give you an example because I'm sure that this sounds like all very abstract. But the example might even sound even more abstract. But I figured that I should give the example because we don't have what we're here for. All right. So maybe one comment here. Please make. It's a non-girard case, actually, for a simplified format for even opening to both the form of a mystery. The other is of the case in construction. The end module. The end module. The end minus one. So it's the end. It's the module that's finger-right. The end? Yeah. You know it's almost the end. So it's the end. The thing is that you're tricking us here because the E1 of stairs becomes an easy one downstairs. Because downstairs it becomes the edge of the wall. And downstairs it becomes the module structure that you should put on. You're right. So I'm fooling you. Yeah, that's it. So what this really does is it gives a shift. So if you get an association about the word, you can ask what happens to modules. I think that gives an easier category. People don't know what that happens to you at the time. The easier category of modules. What does that mean? So I'd say just having an associative algebra doesn't look at its category of modules. Well, if it's an intermediate algebra, you might say, oh, we can test together modules. But no, we're an associative algebra if you have one on side. So easier just means there's no multiple structure. So there's no chance for any problems. Just a moment for an equals one, say. Yeah. So as you just said, for any equals zero, we don't deform for X as a district of associative algebra. We deform for any equals one. Then we are going to deform not really the category of modules and modules in the category, but the two categories of categories of modules over a few co-effects as a two-pategority. And this is just keep doing this up to the end. So this is why it's the end only up to some more time. I see. All right. Well, I don't know if this motivation is demo or the other way of thinking. I was thinking all kinds of abstractions about this word, E, N. What does E, N even stand for? It stands for everything. That's true. So an example of this is, for instance, there's a favorite stack called QG. And as we discussed several times in this workshop, there's a rigidification theorem. And so the category of quadric order in chiefs, the complex of the quadric order in chiefs on BG is the same thing as the kind of a representation of G or complex representation. And what we're going to find is that BG has a shifted two shifts. Simplific structure. I think it's a little bit more. Yeah, better than usual though. Okay, good. And so now let's just tell us a bit. Can we deform this category? And this is based on conversations that I've had with Damien. I'm sure that maybe other people are thinking of this as well. I'm sure. So you won't be able to deform this. I'm not quite sure what exactly the right answer is, but it should be something like representation with some quantum. You're asking me to be ready? Yeah, so let's say that it's reductive, maybe even semi-simple. I'm not sure what the... I'm not good enough with representation theory to tell you how important the requirements are. But so what's the category that you might want to deform it to? It might be representations of some quantum group that's in fixed level, a lot of that means. Or it might be the representation of some loop group. I think there's some theorem that says that if you fix... it should be positive energy representation for that. But again, I think there's a theorem that says that these two are quads. So what's the marital structure? So again, here you have this symmetric marital structure called tensioning sheets. And here's the expectation for... actually what we know is that when we have a symmetrical marital structure, we have what's called a graded marital structure. And it's called the fusion product, which I think David just said under his breath. So it's a marital structure called the fusion product. So something that if you wanted to come up for error from the sea after I've done a great geometry, maybe this is some perspective on what's going on outside this sea. Just as a commentary on... So this is somehow a global stat model version of a lot of things that Kevin Costner was trying to do and has done at the formal level. My formal level means something in a formal neighborhood of a point of this stat. You don't know what that means. It's okay because now we're going to get some real numbers. By real, I mean, you know, I don't have to speak inferior to this. Okay, so that's an issue. So let's let it pass. I'm going to give many definitions in this section. It's going to be a derived outness test. And I'm going to assume it's locally a finite presentation. And the only reason that I'm assuming that it's locally a finite presentation is because it's co-tent and complexly perfect. So in fact, I forget whether or not this was stated earlier in the week. I think it was. The co-tent and complex is perfect. Perfect model. Especially over age. So what I mean, I mean, you think of this as a kit to say something like the tangent space that should be finite dimensions. You can do either. And that's how locally finite presentation really means that every single point locally is a finite dimension. So what you can do is... Okay. Since a co-tent and complex model is perfect, you can do it in a finite dimension complex. Now I want to start defining what is the perfect two-forms. So recall from the previous talk today that we're just going to space in two forms of some stacked F of... Yeah, so it's two forms of degree N. So I say the word weight in its broadness, you punch me in the head. So this is a big space, but let's try to pick actually connected components of it. And we saw it here earlier that said that we can actually compute this as a co-amology of some complex. What about co-amology? So let me first present this sheaf co-amology and then I'll try to do this. I think this one won't be possible. So here this is the co-amology of this sheaf called the F of co-tent and complex and the weight in the complex. Of course, this is the same thing as the finite of a home. So I'll go here to the sheaves of structure sheaf to this sheaf. Oh, I get professionally confused by this. This N is outside of the wedge. So what's the point? 22 form. This comes in pretty much all set when you take a vertex in it. 22 form. This is a co-amology and this is a math. The structure sheaf of F is a sheaf where you take the co-tent and complex and you want it twice and then you shift. And so now that this is by the adjunction between the duals, the adjunction between the tending complex, just the fact that the tending complex is the duals of the co-tending complex, we get an apple of capital theta, the exact same as the math of capital theta that I wrote in the smooth case. We get a math. And I'll call theta sub w, terms like the tenets and your choices, so you put in 24, but 24 is nothing like the game. So it's a 24. And the tangent complex as a sheaf, the co-tending complex, shifted by an apple, again as a sheaf. So hopefully this is clear. This is some sort of wedge two of something that eats tangent vectors and spits out some numbers. Or if you like it, it takes in Tf as a Tf and spits out something in structure sheaf. So moving a sheaf over to another side again, if you put a vector, you get a dual to that vector. Not a dual to that vector, but I get something that will wait for another vector before I dial it. Okay, all right. So here's the definition. The definition is number one. So I think that we have a two-fold called omega. On my staff at 3M, so omega is not degenerate. It's theta f. What's that? Theta omega is not up here. It's in a closed. Now again, as I emphasize earlier, this is a map of causing current sheaves. Like it's not a perfect sheave of the same thing. So in what category do you use a q-code as well, if you like, because the full-sub-category is perfect. So now we're very close to our definition of a symplectic form. If you recall in the Smith case, a symplectic form needed to always satisfy two conditions. There had to be not the genital, there had to be closed. It's exactly the same definition we have here. And so now I guess, I'm kind of surprised that this works out. I guess the definition works out pretty nicely. Okay, so I'm just going to give a post-definition. Okay. So here's another definition. We'll see that the space of non-generate two forms is defined to be the union of the connected components inside the space of all two forms of non-generative data. Okay, so being non-generative is a category that's invariant on our whole list. So if you have, you should think of the space of two forms as a bunch of disconnected spaces. You can say some of these spaces have points consisting of non-generate forms. So then in definition two, so you're just thinking in the full-sub-category. Yeah, so if you want to think of this as an infinity group, this is a full-sub-category of those that are in the region. That's exactly right. Okay, and number two, the space of non-generative two forms, let's say the space of n shifting in symmetric forms on your stack F is defined by what will be pulled back. It's kind of silly that people think that way. Okay, so six, F, N, you know what your stack is, so what you're creating. So I told you this is going to be pulled back with something, but what? We can already guess. So we have the space of all two forms, right? And we want to look at those two forms that are non-generative and closed. Well, I forget whether it was the last talk or the talk previous. We saw that there was already a map of the space of all closed two forms, the space of all four forms. Obviously, as we discussed, this is not some ordinary inclusion, it has values given by the space of keys, because being closed is not a property of destruction. We did determine how it's closed. At the same time we have a dumb inclusion here. Here's an actual inclusion of spaces of the non-generative two forms. Of course, if you know your pullbacks, if you want things that are both closed and non-generative, you know you've got to take a pullback. Because of the number of moving pullbacks. Okay. Good. Alright, so hopefully it's kind of clear. So what does it really mean to give a symphonic form? It means that if you just give a two form that happens to be non-generative, then they give some direct closing. Or maybe you could be getting closed forms such that it's, I mean, do you have what the form of the line is? Yeah, it's not going to be closed. Because you mean your top thing that could be an inclusion. Oh, here you go. I don't know, maybe this is all a warning. I don't know what I'm doing. I get the same feeling when I lecture in front of my students, they get no response. And then if I say this is going to be on the exam, they're like, holy shit. So you don't get the idea that if you don't answer these questions, then, so, so, um, because that's terminology, we say that some choice inside of this official set is an initiative to make a similar structure. Okay, so let me just make some remarks so we can get our hands and feet a little bit dirty. What the hell does this mean? For instance, if we have a stack, yes, it's possible that it has many entries and similar structures for different terms. The answer is no. It has an edge of this inflected structure that is unique. So let me get a remark. So a remark. So let's say that, okay, so I'm still on the assumption of before, so I've added some sort of local representative in my stack so that the cotending problem is fair. So if the cotending problem is an amplitude of A, B, and any sympathetic form of some character or space makes sure that the rest of the period is the same. If you like it with a T structure that we have to impose on a quasi-coherent cheese, this means that the co-amology object associated with that T structure is the same degree as A. So if the cotending complex looks inside these degrees, then any sympathetic form has a degree A plus B. Let me just draw a little picture of this to convince you of it. Okay, so here's the cotending complex. It's called homology object that looks inside degrees A to B. So if the cotending complex looks inside degrees A to B, then any sympathetic form of degrees A to B what is its cotending complex like? What's the dual? It's going to look somewhere as a negative A and a negative B. So this drawing is somehow implicitly similar to these things in these parts of the and we'll just find the numbers. And what does that look like? Well, if there's going to be any problems, this negative B needs to find out if something is really A like I said, and this negative A needs to find out if something is really A like I said, okay, just like this. A distance between T and B is not just scale. It's totally arbitrary, and so because of this, for these degrees to match up, I'll play it in my math, right, and ask people because they have negative A and it's like, okay, I understand. Okay, so great. So that's the first remark. So somehow, if somebody tells you that there's an exception to this remark, they start to even fail if the content in confidence is zero. That's a good action. Okay, so I'll come back to that in a second. So let me get some more remarks so that we can understand what's going on. Here's a remark. Okay, so let's say that our aesthetic is actually a spec of some T and B A. When you're redundant and say that's A, then Lf is concentrated in degrees less than or equal to zero. So the corpus of the color of this is that we get a spec A, we get a spec A. It means a simplistic form of degrees zero. Because the pre-strand constraint of this thing F, well, it's a simplistic form of degrees zero. The content in complex, the tending complex must have nothing outside of degrees zero, so it must be simple. So I think it's I'm sure you're thinking you can see the collars turning. So let's get some more attention to this if you can color it. Okay. The shift is always negative on the rarsky. Oh, yeah, that's right. I mean, again, except for this. Except for this. Oh, when you say, what the hell is the content in complex zero in your point? I'm going to say that. Oh, all right. One particular case of the content in complex always being zero, this is a point, and an example of N-shifted symplectic derived stack is a point. And it's N-shifted for a real point. And this is something I'll focus, I don't know, I guess, quality over talking about the derangements in the next slide. It turns out that a derangement stack is N-shifted symplectic. If it only is a Lagrangian instead of a point, but the point with the correct. So that's something crazy. Okay, so here's another reward. So I think I said in the very beginning that symplectic geometry is probably something very rigid. The examples that I'm giving you are supposed to not convey that. Maybe it's not rigid, but certainly it's consistent in the kind of geometry that you want. Here's another remark. So Cf is locally same assumptions. So, locally, try that in presented. So, if F admits that N-shifted symplectic structure and N is odd, then F must have virtual dimension 0. What the hell is the virtual dimension? So it's supposed to be the other characteristic of the tangent complex to the quotient complex. So, like, I mean, virtual dimension, I think is a parameter. And again, this is supposed to be the other characteristic. Is it that? Tangent? No, this is not. Okay. So what are anybody's other characteristics? I mean, this is some sort of, you know, a machine from some crazy derived stack. You know, I didn't get a number out of it. So I think there's some theorems that tell you that a phonetic is a nice enough static and actually you can actually get the other characteristics to get a fiber, because I'm trying to re-enact you to a chain complex. I'm actually not an expert on this. I don't know what that I'm talking about. But you can take the function on the space of, you know, class C points of F, points of where they're due, and that's not the point of the fiber of Tf. The fiber is a context which might be and this function is a very constant. So if the stack is connected, it's not. I didn't follow it up to reproduce that on the board. But there's no Tf that's actually essential. Oh, and so why is this? Is this something that I talked about like 30 seconds ago? Psychically. So, if there's an action to symphonic structure, but that is odd, you know that what happens to the other characteristics if you shift something at times will become this negative. But if there's going to be an identification between the chain complex and this guy, will these have the other characteristics equal to this negative symbol? All right. So now let me give you some examples. I'm going to list the examples and then the rest of my talk will be proving that those examples are the same. So I'm going to list them first. Examples. So is it possible to have a strong indication in the script? I have no idea. Somebody wanted to ask a question. A question on the question. If Bert will mention the interest in the strong indication in the script. No, no, no. You can just add a bit of other chips like minus 1 to come as the other has only been some of the dimensions of the tension complex because I'm in an anti-situation where we have to put a dimension to the other dimension and then minus the dimension of some h1 of the tension complex that is in the same form. This is why it's called the virtual equation. It behaves like something of the dimension zero, but the summation can be very good. So alright I'm going to carry this. I'm going to list them by letters. So that's cool. Alright. So where everything is working on this base K let's take a point. The point has an N-shifted simple structure for all of them. So that's the V. That's right. I mean yes. No, you're right. I have no other say. Hey, see me. So that's a nice beautiful example. Alright, so V. I'm going to define this object in a second. Maybe not a second, but maybe in a minute. So this space is going to be defined by space in this static. So remember earlier we had an example of the co-d tangent bundle of anything smooth that's going to be a simplecic guide. So we can mimic that construction here in fact because when the derivative will be shifted co-d tangent in spaces. I'll tell you what that is in a second. So under some assumptions on X this space has N-shifted simple construction. So what is the assumption? It is. So I'm going to say something pretty strict. I'm going to demand that X be deleted one for it but there's a comment in PTVD that you can probably do this for. And guys as well they just have a bug on that channel. Very good. Right? Am I right? No. If X is a derived deline monfer static and of course everything that we're talking about is locally a finite representation. So that's the tangent complex. It's perfect. So locally the current is presented. The closed form always exists and then we only check that it's not a generic DM stack because we're not in stock. It's suddenly true but you have to go something. It's an exercise there. Did you find it? Did you find it? Don't say it to everyone. The only thing is that we have to check it from this and it should be so often. We didn't do it. Great. I don't think it's a parent or an adult. Should I add that? That's my first email address. There's more examples. This one I think is kind of some of the examples that are most relevant to a lot of mathematics and physics. So let's fix a function. I'll look at the list which we are going to have to manage the symbols of the action physics with no idea. You can consider the first of all two things. Zero session the code taking place exists and you can also consider the graph of the written list if you guys think about school of graph speed or okay, it's not good. Let's say that X is some sort of yeah, that's a really good question. I'll write down the assumptions. Let's say X is some sort of a derived scheme with a zero shift in structure. Does it need to be just derived from this? I'll just start out with X. I'm looking at it because it can be actually the optimum stuff because of the B. You got that? No. Let's just say X is derived from the null fairy-tale. Do you want to go ahead and say that's the assumption and what's the conclusion? So from example D, this has a zero shift in some kind of structure. Conclusion is I can look at the intersection of the zero section with this graph in Dm. So if you do your translation correctly, this just means putting into critical locations of dysfunction. Critical locations of a function where the ds equals zero, and that's where the ds is zero. And of course, because of the run of delts, everything is always repeat. This guy has a shift in some kind of expression. I'll see if we can do this next time. Great. So as a negative point, I'll just take some kind of expression. I could also take a shifted function. It's also positive and then you get higher shifted specifications. I mean, we don't have to write it, just the comments. You can take functions with some degree and you get a change of ideas. And it's an element in hn of flexed description. Oh, and can be... You can take any close for close one form of degree n. And because, well, shifted content in stack classifies those... well, one form of degree n, you have a canonical map from x to t star x shifted by n, then you can go to the graph of this one form and you can go from this. That will be on the exam. Okay. This is actually a very special case of something even more general that we're trying to do on the other set. But in general, this is going to be in class this time. I think. So if I lie, I'm going to be wrong. I couldn't even close the only because we, as a collective whole, maybe the other person doesn't know what that means. There's a lot of notion. And m is n-shift in some way. And where's n-shift? Is n-shift? So, what's the origin of that? What's the opposite of what we're trying to do? Yeah, so the origin is, again, kind of like closed form. It's not some vanishing of something. It's some prescription of how something's n-shift. So let me just make that. Really, because I'm going to give you a little talk about it. Why is it really a good lie talk? Okay, and so now there's a few more advantages. I think that there's so many advantages, I think, of the nicer things about this. And I come away with some sort of nice dictionary. To say that g is an athron smooth smooth smooth smooth, it's just a group. And I'm not quite sure what the actual constraints on g are to be. A lot of authors say reductives. And as a special case of reductives, you might take g to be very simple. And we get bg is a derived orchestra. So we've talked about this, but you made a few appearances. Then, I'll get to this proof later, when it turns out that you can actually find some kind of forms of retitution. And you can actually describe what this set is. It's in bijection. Things that we're going to do with non-degenerative g invariant, I haven't told you what g is, but I've called it a four-time interval, please. Thanks. Do you guys have a word? I'll leave it. There's some kind of interaction here with the effect of geometry and representation here. And the final example, example e, example number e. So we have this stack, a curve, or path. Maybe you can find these examples. I can't remember the problem with the notation on the video, but I proved that this example was a good example. And the point is that this also has retitution. Think about the instruction. Okay. How much time do I have? I don't know. I don't know. 15 minutes? 15 minutes. I'm too boring. All right. We start at almost like 315. 310. Yeah, so you have 15 minutes. Okay. Yeah, no. You know, that's enough. Okay. So now, section of this graph, Lena, part 3, which is the first curve that I want to go through is the dynamic flow. Okay. So I tell you that the quantity of the physics is an angle of 2 to try to have an energy-intensive structure. What the hell does this space have? It's just I think that might be missing in a global section somewhere. But somebody told me again this is the plane to be our spec. Okay. So what is the cotangent level B? It should be something that so x is 1. Oh, so it's a cotangent x. So for now, we'll take it to the right and we'll go to the left. We'll go to the right presentation. It should be spec of something. What are functions of cotangent in the global? They should be like vector elements. Okay. This is going to be this amount of computing I guess taking this into some kind. Because I'm taking a spec of something. The cotangent space will be an alpha in the scheme. This is a problem. So actually it's alpha number of x. That's not how. Well, it's a fun in a generalized sense. It was like in the talk of when you have the spec B of epsilon where it's 1 because the action to bivalvescent can be on the right of the time. It's a fine over x if x is not a fine. Even if x is not a fine on the right scheme, which is our x in this stack. Yeah, no, no, it's an affine stack over x. That's a different one. Yes, affine stack. Affine stack, yes. I was thinking about it. I guess, I apologize because I wrote this down without knowing. But the formula makes perfect sense. It's something that fits over x and you can describe what are the local sections of this community. So if you think of x, the space of lift to start x is just hung from the significant amount of the full back of the tension complex to b. So it just, I mean, this one cannot make a perfect sense to me. It's a relative space. Affine stack over x. All right. So we're going to do the exact same thing that we did in the smooth case. I'm going to give you the last one for you. And the derivative is going to be something like the form, but I actually don't want to spend too much time on this because I can be the part that much works for this. Okay. So, um, I can't figure it out. So, we have a map. Sheens. Sheens. So, going into the one part of the symmetric power, um, for my adhesion, we have a following map. So this is going to be very stupid and easy. And the other thing is that the adhesion is a very, very, very simple word. And this, I'm going to do this in one form eventually. This is the same thing. Same thing. Same thing. Same thing. Same thing. Same thing. Same thing. Okay. So we're going to go ahead and do the projection. Sorry. The only reason I'm doing this is because I want to do the projection. And this map, this is the projection map. This is the projection map. This is the projection. Yeah. We have to hit the coefficient in the middle to X. And so, if you want to understand is that the level of specs, does it have the level of specs? Is it just given by the inclusion of the units? Sometimes everything that I'm doing is almost stupid. And so again by the direction, and of course by the direction, you get a map now of the pullback of the spectrogens, pullback of the co-plating objects, put them like that. And of course, this is your structure sheath, and the co-plating is there. So now we're almost done. So, the co-plating complexes, we've got whites, we've got the octopuses, get our hands, pullback of the structure sheath. If you have a map from B to A, and you have tensor, when you tensor, we think B tensor A or A, we can be, oh okay, this is what you're doing. Okay, so co-plating complexes, we're going to get a friend, and five receipts, okay, and, where's the map from B? So, I'm just not going to go up, there's a lot of co-plating complexes. So this map also has one that I didn't see. What is that? No, you can pull it up, and pull this map. So, we're going to obtain a map of structure sheath. This I'm just talking about by hand sign. This is how the shape of the map ends. So is this, is this part of these factors like these, related to what you told us in the very beginning, the one form that is only out of the parallel vector? No, please, you're all consulting about form on t star x. Yes, but then, these factors will be pulled back of x of n, which means that, yeah, so I didn't say that beginning of comments about the one form to star x being only same of horizontal, because I'm not sure where you can write this, I'll try to make it down to 3, and do some final algebra, but, we can take something like a piece of r as a model to solve, maybe even a bit of a ladder when I do this, and somehow pulling back, I would get exactly how far away it goes, but there's some sort of one form that's not a big part to start, but I'll try to do it in real time, because I don't have much real time. So, actually, we can shoot further as though about half an hour, so that you can have more time, especially if you don't go up early. Great. I apologize that I took so long in the beginning. Alright, so, nobody mind if I now go a little slower. Okay, so the whole point is, what the hell is this? This is exactly one form. The cotangent bottom right shifted. So, one form with an infall shift. Okay, so this defines a number of degree A, you can call this one form, and the next shift by N. So, this is going back to the observation that I made, that A0 of some complex compute pi0 of space of horns and the sum of elements of A0 with the golden section. Okay, so, this is a proposition. The fact that there are months in it, that D to R of lambda, which I'll define as a gear maker, is not a general. I have written out a full proof, but I think that I'm going to write it out as well, at some point, when you get to 15. So, any possible, or non-linear, or how do you do this? I think you want to. You say you turn out only negative, particularly. No, no, no. And the most direct, actually, I think you do. Okay, so here's a sketch of a proof. A sketch, which means I have to skip a page of my notes. So, it affects me. So, again, the claim, but I'm drawing this, it's an exercise for you to try to figure out the case of art, and we're going to stick this to a motor, and that means there's going to be a telecom. Okay. Ui, so in this case, it's an telecom. The program takes a little bit of a long time to finish. The program takes a little bit of an extra. This is what we're going to use for ball theory. That's right. Ball theory, yes. That is true. So, as a result, we can just now review the case. That's the case. And we're making changes in quality. Because these are the high-sod tests in the long term. So, there's something about the line of the production, which is that just construction land that I gave you is supposed to be natural. Obviously, in the form of functionality, you think the grid did it, and it won't. Okay. So, in this case, we're going to go with this case. In this case, the co-edition of the flex between shifting is by definition, which we can expect to this amount, what we're trying to do in the form. It's what you call the Hamilton-Gerfman. And so, one thing to note is, some of these is quality 3 of A, by definition, because it's symmetric and it's a bundle over k. And k is quality 3 over k. So, the B is quality 3 over k. Okay. This is where you skip some of the competition. So, what I'm going to say turns out, so, say, that's what you're going to do with this, in the k-compress, which is to shift it one point of B, because there's a k-compress of B and there's a k-compress of the shift. It's just done with that. I don't know what else to call it. And it just turns out that it's just a matter of shifting back these cases. So, I'm not going to define what these are, for which I apologize. They're all my own words. They're all my key shots. Okay. So, why is this described as being a wide-slimpt on man? So, the point of... Okay. The point is that these are maps of this space. Three models will repeat, with a certain set of generators, but the same type of generator is the same. It turns out that it just sends generators to generators. So, it clearly has more. So, generators, generators. Because you use ideal marionette, you can finish it from the diagonal, and just use the cover. I only use the cover. So, you know those more general things. I mean, there might be some statement about the co-ganded complex. Is that some question? I don't know. The point is, you need to get that problem to deal with some more. Right. But I'm saying, it's like, you can have no sub-questionation from that. I mean, he has that interest, he just sends this problem to us. No motion. There's a BG in under-depart. Does that appear? Do you want to skip the part? I think we want to hear about part. Yeah, we want to hear about both. I think we want to hear about both. Yeah. Actually, BG is almost a special case of part. Which part? BGLN is an open, sort of, attacking part. Yeah. So, many parts are good, no? I don't know. Perfect. I don't know. Do you want me to switch to your side in case you need to do it? No. I'm sorry. But BG is really interesting. What kind of game is this? So, I think, I think the next up, she would need a bit of BG to really quickly... No. No? No? No. Alright. Here we go. The perfect one. Let's remind ourselves of who M.I. is So, let's recall. We define the stacks. We define the stack empty for all BG categories empty. And what's the definition of the stack? So, what the hell is the stack? If you take some sort of CDGA you'll put a space for some visual set. And so, let me just try what it does on objects. It's supposed to take the mapping complex in DG categories of modules. Perfect modules. Well, of DG functions that T got the perfect module. This M.T. was defined. And so, in case T equals 1 for what's 1, there's a category with one object. Some dumb little thing. But then there's a lot of some people that can't hear. But then there's a lot of some people that can't hear. So, this one has a good point. And the one with the data. So, Perf is defined to be n-sub-1. So, obvious computation to not even a computation is a topology. What do you get when you evaluate this on some CDGA? You have maps of DG categories in the unit category. Perfect modules. And this is the space of the problem. It's the largest infinity group void inside the category. Of course, you can do this at the category level. And if you didn't think this as a space or simply so same, you would come to the category in terms of models. I don't think what we need. So, we have a a cool data. What does it say? It tells us how to compute the pullback of the content of 10-day complexes. It's theorem. Well, it's like e to your k-point beside this non-base-stand-collection. According to beta from previous theorem. So, this takes out an object inside a trend inside a DG category T. The pullback is a tangent complex. If we're pulling it back to spec K, so we're just going to get a sheaf on spec K which is just chain complex. There's a chain complex that was given by the endomorphism of the sum of T beside the categories of development. So, if you like stuff about actual probabilities, there's a very nice answer because what should a common state of a tangent complex be an object put back in the space of object should look like a waste of before. And you get that. Come here. Okay, so here's a theorem passed forward. Which is from what's left nicely ascent to the right stack. I'm not quite sure if I need to be hard on somebody who may be telling me I've been given a perfect module over X. What can you say? Pullback to tangent complex on this stack is what's equal to the category of closings in your sheaf. If you like to have a perfect module, it's the same thing. So, this is an object in the category of the sheaf object. Okay. So, now we're going to do this like really annoying categorical now. So, this is a very simple and really annoying categorical nonsense that works out incredibly well. So, by definition, Perth is a stack where if you map into it you get some sort of sheaf on the finger map in front. Well, even any stack there's a canonical map from that stack itself called the identity. So, there's some sort of sheaf over Perth that corresponds to the identity. That's a cool sheaf. Okay, so let's see. Let's take B and have to do the identity. Cool. Now, let's take a gain map from Perth. That's going to give us some sort of complex on it. And I claim that this is going to let us compute. This is going to give us some words to describe the taking complex of Perth. And we have a series of language systems. Tango complexes. Perth Perth and here's the relative. The first one to look at the relative tangent complex of the identity in that one, that thing stands here. What's that? A list sheet. So now we have the tangent complex of Perth as a quasi-convenient sheet called Perth is a complex to write the pullback of T for this. Well, this is a complex but the pullback of the quasi-convenient sheet is. So, like for a thing, the tangent complex of Perth is a complex to our home. We can't have a quality of the cheese on Perth. Which is weird. We can't have a quality of the cheese and then that complex shifted by the. So just abstractly, I have no idea what this sheet looks like. I don't know why. Today's idea is nothing that's perfect. That's nice. We have this first. Okay, there we go. We can read this. The corollary that I stated is this x. Oh, okay. I mean, this is a long, long time to inject some sort of color not to lose anybody. I got to read my sentence. I think your experiment is for k. and mt and now your opinion is T to be Perth and you use the fact that k. and Perth is the same as just the x. and m1. And you have to check that the content of the complex is the corollary of the Udemy and the corollary of the subjunction. If you just think about what the tangent complex will matter to be, you realize that it's not so bad. All right. Now let's make up. Then we have an algebra here. Let's note this algebra probably. Okay. So this is just as simple as what we mean by this. I just want to make note of the fact that the tangent complex is actually similar to this called the Udemy. Remember this is some sort of sheet of an algebra structure but of course in the sheet we remove that algebra structure from the sheet in algebra. But if you remember it as a sheet it's a supporting tangent complex. Okay. Now we leave it around pretty much done through me as the substitution which is such an algebra. Because remember what we need is some sort of biolinder form. What is that biolinder form? So, for first since a we use an anthropomorphism algebra. It has maps called the multiplication. One of the tensor forces them for a x to go up first. And then three since e is perfect it's kind of a weird game of a, that's a problem you check. So I don't know if you can play around much with like dualizability and what e to the t is. You think about it for a second. I mean you can check the cc for instance to find the cc. Okay. And moreover we can actually characterize oh so maybe the word that you prefer is original it's also original. I'll be using that. We can see moreover we can actually characterize the multiplication of a in terms of the unicode in it. Okay. Okay so we have a small multiplication now. Actually has another incarnation. We have tensor a and it's going to rewrite annoyingly. It's a product of four things. And how do I end up with tensor product with two things? Well I think the identity of the outside two and I take the trace map and I know the other one. Identity and the trace. Oh did I call it epsilon or trace? We call it trace. That's not the third thing. So in the topic about rigid objects we had a unit map with a coordinate map. This is a map that takes the middle of the factors and crunches them into a function. Okay. I can see that this is a unit. So if you haven't ever really thought about what multiplication of the matrices looks like this is a funny way to write it. And part D is a similar important part. It's a composite map. The matrix tensor a a okay so we have this map that we tensor a to a but a is the same thing if we check as e the algebra of the reward is a little bit rigid object and we again have the trace map. This is the same as the sheath structure. Okay. So the same as this composite. It's not the gender. It's not the gender. Because e is perfect constraint. There's no difference between the graph to e and the usual e or the whole object is the same. Because of the curves I have in the categorical Why is the utility more than the number of curves? There is a problem. Oh. So it's an iterated application of the triangle identity. For me the most convincing picture is so I apologize if people haven't seen these kinds of pictures before. But the triangle identity most I think it's here in one of the HFCs right? This is the triangle identity. And what this is saying is that what's the multiplication of the HFCs in the graph in the graph? And the whole point is that this this this is moving around like this. There's some non-usual what's the TFT stuff but if you play around the diagram you'll see that this Oh sorry. I forgot the capital. So then you have a trace amount. And this So this is some machine because I'm moving something in a droid but there's a big sphere. Are you with check it? To check that something is not in the head all the moving are perfect you can do it with point files. You take just the 5 or the point and they become complex objects of space, a nice You're not great with math and scissors. It's probably a good one. So what I was just saying is you can now check this point in the graph. I wanted to actually do a reduce the case of computing traces for math to change how much of the data I'll find out in that channel. So I'll just create an agency to check that to get another general matrix. Okay. So the level. Okay. So this is a non-generative form. What do you mean by this? It's a non-generative form. For a non-generative form. Yes. So what is a non-generative form supposed to be? First of all, give us a map. I can see from the tangent complex, tangent twice to the structure shift. Shifted by shift. Okay. We don't want this is. I mean this is by the theorem flash correlator I'm erased. See you want to test it. You want to test it. You want to test it. You want to. And we now have this map to O called multiple pi and then take a trace. It's not the only thing that we need to show that it's closed. It turns out to be not easy so I'm just sketching. To me this is a valid the main reason that am I going to play a wheelchair? I did. I apologize. I really apologize. I can't even look at the show. So in the whole. No, no. They don't make this close. I mean it doesn't kind of seem to be close to the guy. Wait, I thought that this was going to be Oh sorry, I got it. It's not close. It's not really close. Are you happy? No. I want to say that it's not that the lemma doesn't say I can find a leaf. There is one particular leaf and this is the leaf we call the symbol of the Schachter-Montau. Okay. Because this leaf is like a chunk of a leaf and it's not so long. So I don't see it. And that's what you okay. It was the space of the leaves. I have no idea what the space of the leaves is, but I mean we can start to this is getting on the scale. Is that there? Okay. Okay, so I'm just going to sketch this to improve. I'm going to hear the sketch to improve. Thank you to everybody to talk about Schrodinger. Schrodinger is a method. From the stamp of Earth this is an aspect that m controller staples her to get to the surface and so forth. I'm expected that This is actually, this is actually a group of the standard column C. And so this was the one that sends a CDGAA to dope card. so high notes. The class is all cleared in This is, of course, for the first thing I'm going to say is, to do with the chromology of what I gave, I'm going to apply the function of Nc to the static 30-monomes. Taking projections going to the piece, I'm going to reduce that. This is something I'll call here. I'm going to write a chord and first do the same thing. Of course, this is something I'll call point of last time. This is equivalent to finding a finite component. I'm in a space of close to 4 to the degree. Then you check that the zeolute is actually equal. One can check that, oh, I think I defined the non-degenerate two form on the algebra between multiplication and multiplication. One part of the formula is the z-dove. One part of the formula is z-dove. I love the things we're talking about. Do you want me to do this here? No, we can stop. Thank you. Questions? Yes? I think PGLM can be seen as this. I think you can find a 3-chloroposanus, a 3-chloroposanus, so it's a method of GN. Then it sends BGLM to path by sending a vector boundary so that the path back complexity is zero. You get a map from BG to path, if you put it back to form, it's one way to construct. It's also closed. Let me know that it's closed in the BG case is way easier. In fact, you already see your favorite non-generative gene variant in the form here. Multiplications matrices are going to apply to trace. In the classic way, you want to define some gene variant quadratic metric on the line. This is multiplication of matrices in state of choice. It goes back to the standard variable example section. I say this one, you're going to add two more. I'm going to get it. All the equations are on this. All right. Thanks again.
complex of the quadric order in chiefs on BG is and then you get higher shifted specifications. I