OK, thank you very much. So yeah, so before I said, so my role in the sort of asteroid belt of minor bodies that is orbiting a conceivage is the following one that I take a small slice of his things, mainly from the late 90s. And I combine them with a small slice of Foucaillat's and a small slice of Donaldson's things. So the rules for this game have been known, especially for a long time. There is a geometric ingredient, which is the classical, which is the geometry of left shits vibrations, which is a very rich thing. And I think we don't really understand how complex it is. And the main goal of the theory is to try to capture some of that complexity as much as it can. So that's Donaldson's contribution. And then there's Foucaillat's contribution, which is the associated theory of pseudoholomorphic curves, which is also very complex. But in some sense, in a way that we understand. So if you have a problem that you want to solve using pseudoholmorphic curve theory, you have to decide which family of remand surfaces you're going to use, which auxiliary geometric data. But it's a finite amount of games. And then there is one of the things that the conceivage has contributed is a sort of particular flavor of homological algebra, which you will use to sort of formulate your results and capture the information that pseudoholmorphic curves give you. So there are these three kinds of building blocks. And then there is homological mirror symmetry, which is like the building plan that will help you tell you where things should go. So it's a little bit like putting together IKEA furniture. And so my success so far is really equal to my success in putting together a IKEA furniture, which means that at each step, I'm convinced that this is exactly the way the things are supposed to go. And then when I look back when I'm finished, it turns out that not only was this kind of clumsy what I did, but it's actually an obstruction to proceeding to the next step. So this has iterated several times. This is my latest iteration. I'm not particularly sanguine about it. But hey, that's what I can do. There won't be a particular amount of theorems in this thing, but I'm trying to make some kind of picture. So the basic geometry here in the case of classical case of, say, a smooth projective variety, we have a line bundle, and we have two sections. And just classical, well, I don't want any of these sections to be 0. I also don't want them to be linearly dependent. And in fact, I don't want the 0 sections to have 0 sets of common components. So let's say if you take both 0 sets, you intersect them, it has at least dimension to it. You can assume that it's a generic pencil if you want. So obviously, there's a hypersurface x0, which is where s0 vanishes. There's a hypersurface x infinity, where s infinity vanishes. And then there's a whole pencil of hypersurfaces xz, which are defined by vanishing of some linear combination. And those hypersurfaces intersect all at the same thing b. So if you don't mind that they intersect, you can think of the whole thing as being given by some not quite well-defined map to the projective line whose fibers are these xz. Or if you mind the fact that the fibers intersect, then one thing you can do is you can remove one fiber. Let's say the one at infinity. And then the quotient of these two things becomes a well-defined function, which lands you in the affine line. The fibers of that function, w, are not the original, not these xz, but the open parts where you remove the base locus. So there are two different ways that I'm interested in to look at these things. One is to look at their topology and look at their symplectic topology. So there we look at the singularities. We look at vanishing cycles and monodromy representations. And since the vanishing cycles are Lagrangian submanifolds, it's natural to use symplectic topology to it. And look at pseudoholomorphic curves, so pseudoholomorphic maps into x or into the fibers with various boundary conditions. I think about this. It's still essentially topological. So you don't really use the algebraic geometry very much. In particular, you don't really care about the single numbers. For instance, where exactly the critical values are located on the complex plane isn't really important. If you deform things slightly, that will be sort of indistinguishable from this topological point of view, which is green. So the other point of view, which is sort of algebraic, which is purple, is to really look at these things here as defining a family of algebraic varieties, xz. And you can look at them from the point of view of deformation theory, from the point of view of cycles, from the point of view of Hodge theory, yes, money in connection. And to a large, I mean, not the Hodge theory, but everything else I said, you could do it over a graph. You don't need the fact that the ground field was the complex numbers. You could do it over a variety of other fields. Thank you. Sonsicutromark, you asked t-shirt of the piece of green size in the green color. That's right. I'm definitely there. Yeah.  My t-shirt is green. So here, actually, the numbers really matter, right? So they will be in your family. There will certain parameter values, z, where special things happen. Maybe a cycle becomes algebraic. And those things are relevant. OK, so it's a well-known thing that sort of these two points of view are exchanged under mirror symmetry. So I will do an example which is very close to the most classical one rather than trying to formulate things sort of abstractly, OK? So the example is this thing here that on the sublactic topology side, I look at a pencil of cubics on CP2. So the line bundle is O of 3, which is the anti-canonical bundle, which is always the case in mirror symmetry. It is not quite a generic pencil. So I could have done it for a generic pencil, but this is slightly simpler situation. So one section, I choose x0, x1, x2 on P2. So the zero set of that is a chain of 3, whoa, of 3 plus 1 curves. I wish I could edit this thing here, but it's too late. So this will stand as a mark of my shame. And S0 now will choose generically. And then it doesn't matter very much which one you choose, OK? So the base locus of these things are nine points. And the generic fiber is an elliptic curve. So since I'm thinking topologically, let me say it's a torus. And there are nine critical points. In fact, if you remove the fiber at infinity, you just get the open torus, C star cross C star, OK? So it's a little bit confusing, just a special feature of this example that the mirror looks vaguely similar, but you shouldn't attribute any importance to these similarities. So the mirror is, roughly speaking, it's an orbifold. So it takes Cp2, divide by Z3, in a way that gives you crepent singularities. A2 singularities, which admit crepent resolutions, red resolve them, or you can just think of working on the orbifold. And you look at essentially the same thing, except the fiber at infinity will now look differently because you did this resolution. So it's a chain of nine rational curves. And because of this Z3 quotient, you now have three base points and you have three critical points. And in fact, if you write down the function w in explicit coordinates, it is this one here, which is well-known player role mirror symmetry for P2. So this is this geometry here. So this, maybe naively speaking, I would say it appears that this x check is slightly bigger than x on account of its total betty number or something. But in the way that the complexities accounted for, they actually become equal. So this is the way that the complexities accounted for. And I have to warn you that now there will be sort of, when this subject started, there was like a symplectic and an algebra geometric side. And each side, there was a single category. And as the subject developed, there's sort of various versions you can do, which give rise to all sorts of categories. So later, when we get to a couple of slides forwards, you will see that there are a lot of boxes. Now I just make a list of stuff, and later we'll try to put the stuff into boxes. So on the algebraic geometry side, maybe the simplest thing you could do is that you could take your variety x check, it's smooth, and you consider its derived category, which by which I mean the bounded derived category of coherent sheeps. So this thing here, it's k theory has rank 9. And so the corresponding thing on the symplectic side is that you take this w, which is this left sheds vibration over the affine line. And there is a focaya category associated to it whose objects are essentially left sheds symbols. Since we have nine critical points, we have a basis of nine left sheds symbols, which generate everything. And so this focaya category will be the mirror of this derived category here. Obviously another thing, which is maybe more obvious on the symplectic geometry side, is instead of considering this vibration by itself, why don't we just consider a single fiber? So the fibers of w are this xz minus the base locus. So in this case here, it's a punctured elliptic curve, t2 minus nine points. And this, roughly speaking, this corresponds to the fiber at infinity of the mirror pencil. Now this x check infinity is singular. So if I write d of x check infinity, it's not particularly well behaved. So I replace it by something which is slightly better behaved, which is the subcategory of perfect complexes. But maybe I shouldn't have made a distinction and I should just call everything d and just said if it's singular, then d is perfect complexes because that's what I like and it's well behaved. It's a question about which finiteness properties you're going to impose. What do you think of something being bounded and compact, in a sense? OK, so these do correspond. And then the next version is something which is where the subject actually, so now we're sort of going away and we're converging, going backwards in time towards the version that the subject actually started. So this xz minus b, this is a punctured torus. So I can put the punctures back. So the way that you do it, put the punctures back, is in the original thing, you considered curves on this punctured torus and you considered polygons as holomorphic maps on the punctured torus. Now let's consider holomorphic maps on the torus and see how often do they actually go over the punctures. If they don't go over the punctures at all, now we're working in the complement of the punctures. If they go over the punctures one, you count them with a formal parameter q. If they go twice, you count them with q squared. So that gives rise to what's called the relative focure category of the torus relative to these punctures. It's a category which has a formal parameter q. And if you set q equals 0, you recover the thing that you had before. And so it's a formal one parameter deformation of the previous thing. And so unsurprisingly, if you have the mirror pencil, you have the fibrate infinity. The fibrate infinity is the fibrate infinity in this family of fibers. So instead of checking just the point at infinity, you can take a formal disk near infinity parameterized by some parameter q. And so that gives rise to a scheme over this spec c double bracket q, which is this family of elliptic curves, which are q equals 0 at singular, and otherwise it's non-singular. So these correspond to here. But I haven't told you, here there's a natural choice of parameter q. I haven't told you how I'm choosing the parameter q on the other side. That is much trickier. And so then the next step is relatively straightforward, which is to have this formal parameter q. Let's invert q. So we introduce q inverse on the algebra geometric side. That means that I pass to the generic fiber, which is now a smooth elliptic curve. So it's over this ground field. So this is where this is conceived its original version of a homological mirror symmetry, which says that in this case here, it just means that we work over the compact torus and we forget the things. Allowing inverses of the parameter q basically allows your objects, which are curved, to sort of pass over the punctures. So it's effectively as if you actually worked on the actual torus, OK? Now, so this is the classical version. Now, another thing which you can do certainly would be on this side here, plop. You could just remove the fiber at infinity, OK? And then so you get an open variety and you consider its derived categories. This is smooth, so I don't have to worry about what I'm doing. And at some point, it was realized that there is a corresponding thing on the symplectic geometry side, which looks quite similar, which is where I take my CP2, I remove this fiber here, and now I do something which is called the script W. So it's another version of the Foucaille category, which I will script F. It's called the Wrapped Foucaille category. So it has non-compact Lagrangian submanifolds, just like the previous left shed symbols, which were also non-compact. But the left shed symbols are only allowed to go to infinity in a particular specified way that's dictated by the left shed vibration. Here, they can go to infinity in a rather freer way. And the price that you pay between them to make sense of this is that the morphism spaces are infinite dimensional, which is pretty good because the morphism space is here, also infinite dimensional. OK, and the last thing I'm missing from my collection is how about mirror symmetry for CP2 itself? Of course, that's something that you might want to do instead of passing to fibers or removing parts of it. So there is a Foucaille category of CP2. There's also actually a one parameter family of these things here that depends on a complex number, Z. So I'm kind of for a bulk parameter. And so what happens here is that there should be a W check here type of. So here you have these dual paths. You remove the fiber at infinity. And then you have this function, W check to A1. And there's a way to associate to it a category, the category Landau-Ginsberg brains. So locally speaking, what's happening here is that we have the total space, which is a smooth algebraic variety. We consider Z2-periodic complexes of vector bundles on it. And with a differential, where the differential does not square to 0, but it squares to W. Or if you can put in a parameter, which is a constant Z, they square to W minus Z. Oh, do you really mean bulk? Do you mean like central charge? I never. Oh, I'm saying, can't you use a bulk is not this? Well, this is bulk, yes. It's a constant bulk. It's a multiple of the identity that I'm using as, no? Maybe it's not. But probably wrong word, OK? It's a constant identity, if that's what you're asking. OK, anyway, so you get this thing here, which is called the Landau-Ginsberg category here. So and this is a mirror to this thing here. Now, when I say mirror, in most cases, there are certain closures and the formal operations, which are supposed to do to make things match up. But I'm just omitting all of them. And there's one more thing, which we have no idea how to make an actual geometric counterpart, which is, this isn't actually you get an actual family of elliptic curves. We have a good mirror to the fibrous infinity, which was right here. We have a mirror to the fiber near infinity, which is right there. So it's a natural question to ask, what are the mirrors to the other fibers, which are actually nice, smooth elliptic curves of a C? And we have no idea. I have a doubt in the previous slide. Yes? Yeah. It says you can go back. When you can see the relative focac, you can count the dissension with this B. You should get kind of a curve, a formal family of Calabi also. But in flat coordinates, it should be not an algebraic curve, not your original family of a.  No, no, but after optimization, it's made so that I don't really believe it, because here you can see that if you count here intersection with B, you get a curve, which is a straight curve in the flat coordinates. And let's suppose you have not fiber something, you curve it a little more, and this straight curve, it's not the germ of algebraic curve in the model Calabi. Because here, on the mirror side, you can see the algebraic curve in the model Calabi. They don't repeat here. I don't, I mean, there's only one parameter anyway, so. Oh, no, no, you have to get one part of the family here to leave the curve when you minus something. But in more complicated examples, the sum of gray curve in the high dimension model space. Yes. And it's not a straight line of flat coordinates near the castle. Well, I'm not claiming that this is a description of what happens in general. OK? This is just this case, because if I do general things, I have to put in caveats and assumptions and everything all over the place. OK? But it will be a description in some cases. And if you don't think that's true, then you should tell me. OK? So I'll remind you of your objection later.  So OK, so right. So OK, so who invented all these things and who came up all these things? And the answer is obviously not it wasn't all me. So the same thing with names. OK, so let's see. So the focaya categories that I'm using, just in their plain flavor, say for a puncture torus, obviously due to focaya. The version for left shits vibrations is with these left shits symbols. This is as far as I know due to conceivage. And I sort of wrote it down, big deal. The relative version I also wrote down, but it's also no big deal. On the algebraic geometry side, well, this is Verdi's thesis. I don't know who invented perfect complexes. But anyway, I don't think anybody's going to ask me to assign precise credit for this thing here. So the version of mirror symmetry, in this case here, well, actually, there's a computation in conceivage's original paper, which essentially proves it. And then there was a lot more computations, which sort of, I don't know, done by Polisiu Ganzaslow. And the limiting version where you have sort of a puncture torus, I mean, nobody really knows who should take credit for it. The one here is in a paper by Oru Katzarkov and Olof. No, but if now that's really a good example, I'm not general writing, it's in view of it's that sort of, it's not for me to define the direct category of lifto-curve. It's not for you to define the derived category of, well, in this case here, it's of the CP2 orbit fold. I mean, I don't know. OK, well, so over here, well, this is now, this is the focaya category of the actual elliptic curve with this parameter here. I don't know. I just wrote focaya order. And obviously, their theory is much more general than this. So in this case here, it was already in conservage and in Polisiu Ganzaslow who also proved this thing here again. Well, these wrapped focaya categories were introduced by a bunch of people, among them Abu Zaid and myself. And actually, this particular special case here you could do by hand. But these kind of arguments that actually compute the wrap category probably also should be credited to the two of us. This thing here, this FZ done by Yong and Oh and focaya. And Cholian Chou and a number of other people to prove this thing here. Essentially, in this particular case here, these categories are rather simple. So you can prove that match objects one to one. The missing thing is to show that you have accounted for all the objects for CP2. That's where this long list of names here comes in. Oh, and so matrix factorizations were invented by Eisenberg. And the categories, the triangulated categories, there's a long paper by Buchweiz, which is somewhat relevant. And then there is Olaf who gave a different definition to develop the theory to a very large extent, especially beyond the case of FI varieties, which was the focus of the original theory. OK, I hope I didn't. Does anybody want to correct me on the names? I tried my best. On both sides, we have all sorts of different categorical structures. How are they all related to each other? OK, fun. Let's look at the algebraic geometry side. So we have the total space of the pencil. And we have one of the hypersurface. It'll say the one at infinity. That's what appeared. Obviously, there's restriction and inclusion factors. So then we have the fiber at infinity and the fiber in a formal disk near infinity. And obviously, there's some kind of deformation theory there. And then when you want to pass to the punctured formal disk, you just invert the parameter. That's just a simple process. You just invert Q. There's nothing going on here. What happens when you try to remove x infinity and pass to the complement? You can actually do it by a sort of, well, this is a localization. You pass to an open subset. There's a correspondence of categorical notion of localization, where you take the image of Perth inside here and you quotient out. And then you get the derived category of the complement. And then the passage from this thing here to the Landau-Ginsberg category, you can think of it in a way as a deformation and then passing to the generic fiber. You have to be somewhat careful, because the deformation parameter has degree 2. So the theory doesn't quite look like ordinary deformation theory. But I'm just going to allow myself to just say this. And exactly the same thing holds on this side here. You have left shift symbols. They have boundaries, which are vanishing cycles in a fiber. So this is some of the restriction functor. There's an adjoint functor. Little bit more complicated to describe. Once you are in a fiber, there's a formal deformation that gives you to the relative category and so on. And maybe most substantially, for the point of view of symplectic geometry, you can obtain the wrapped category as a sort of categorical localization of this category of the vibration. So it would seem that we are all set. There's a vibe of relationships. There's a corresponding vibe of relationships. Now, this is really not good. And why is it not good? Well, for two reasons. One is I can make funny arrows here. But the funny arrows don't mean that things actually determine each other. So this arrow here with the dots, this is like a little reproduction of the picture here on the right. This funny thing here with the dots and this thing here is the dot are deformations. Well, just because I say it's a deformation, it doesn't mean that you know what the deformation actually is going to be. There has been some success in using abstract deformation theory as a shortcut for computations. But still, that just doesn't determine this one here. And also, the question is, where's the center of the picture actually? So if you look back, it seems like with the arrow going, it seems that this pair here is the center of the picture. But really, if you look at it from a point of view of geometry, the center of the picture is given by the pencil itself. And there's nothing here which is just pencil to pencil because the objects that are here in the middle actually depend only on what the topology is after you throw away the fibers at infinities. And I'm talking about the green part, right? So if I focus on this square here, I actually forgot that this thing here extends over CP1. And clearly, I cannot hope to recover all the data if I forgot this crucial geometric fact. So actually, let me skip this thing here in the interest of clarity. I'll show it to all the specialists later, secretly. So then you have to also forget this box here. But let me say it as much as this thing here. So there are cases where on the symplectic geometry side, it looks essentially the same. You have a pencil. But on the algebraic geometry side, you get something which isn't quite classical algebraic geometry. It could be some kind of non-commutative deformations. Or it could be something where instead of the variety, which was supposed to be the x check, is itself not a variety by the Landau-Ginsberg model. And then you have some kind of vibrations of that. And so formally speaking, things sort of look the same. But on the algebraic geometry side, it's not really a pencil in the sense of classical algebraic geometry. And this is actually a good thing about this game here, which is because it gives you a window into what non-commutative algebraic geometry and various exotic versions should actually look like, in corresponding to still symplectic geometry. So what we're going to think about is what is actually a pencil. So let me be slightly less ambitious and think, what is actually a divisor, an effective divisor, a hypersurface? So what is it? So I have a variety. Now the notation is going to switch temporarily because I'm going to write x, but I'm going to do algebraic geometry on it because I'm going to forget about mirror symmetry for a while. So you have a variety. You have your line bundle. You have a non-zero section that gives rise to divisor. But instead of looking at the actual hypersurface that it defines, I prefer to look at a sort of DG resolution of this. So I take the structure sheaf and I take the direct sum with the invertible sheaf L placed in degree minus 1. So I think of this as the exterior algebra on L. So this is a Z graded sheaf over x. I equip it with a differential that's given by S. So this is a sheaf of DG algebras, which I think of as some kind of differential graded scheme. And this is actually quasi-izomorphic to the hypersurface, which means this thing here is acyclic everywhere outside the hypersurface. And on the hypersurface, it's just its homologous O. And it's actually somehow quasi-izomorphic in some DG algebras hence. So the motto is that when I think of this hypersurface here, I don't want to think of the hypersurface. I want to think of this hypersurface here as an extension of x by this line bundle. And the same thing is, if I have a pencil, I want to think about it as a family of extensions of x by line bundle that's parametrized by the projective line. So it's parametrized by two homogeneous coordinates. Happiness? Excellent. Also mean hypersurface being smooth? No, I didn't say that. Did I say that? I didn't say that. OK. I mean, for instance, not all fibers of the pencil can be smooth, right? Sorry. You said family. I'm still processing. Excuse me? So you said family, family parametrized by P1. By P1. So it's parametrized by two. It has two homogeneous variables. Motion of DG scheme over P1. Is that what you're thinking? Yeah, yeah, yeah, pretty much, yeah. I mean, over A2 with some homogeneity property. That's pretty simple. OK. So let's see what this, so now I'm going to take this stuff here. And I'm going to translate it all into abstract non-commutative geometry, OK? So instead of my original variety X, I'm going to have a non-commutative space here, which is, for me, I got addicted to this language of A infinity structures. So I would like to this to be an A infinity algebra. What if you don't like or don't want to like A infinity structures? Well, you can assume that it's a DG algebra, OK? But you'll have to pay for it in the next step. There's no loss of generality. But you'll have to pay for it in the next step. So the next step is to say, OK, this guy here, now I need something which replaces the line bundle itself. So I'm going to replace it by a bi-module, which is invertible with respect to tensor product. So which is another bi-module, which you tensor them together, you get the diagonal bi-module. Or let's say, tensoring with P is an invertible operation. And so what you will have to pay is that in the A infinity world, any A infinity bi-module is homotopically flat and homotopically projective and homotopically anything. Whereas if you work in the DG world, you have to put in a bunch of assumptions on P. So you have to find a suitable resolution. But there is no actual substantial difference, OK? So what do I want to do is now I want to make an extension. So I take A, I add P, shift it down in degrees by 1, and I want to make this into an A infinity algebra, OK? And this I will call a non-community divisor. There's some conditions that I want to have. So if you know an A infinity algebra, it's given by multilinear operations on B, OK? So my first condition is obviously I have to somehow have to have something to do with the original A. So the condition is if you take your multilinear operations and all your inputs are in A, then your output is in A and it recovers the original structure of A, OK? So A sits inside there. The next simplest case would be to say, I take all the inputs in A except one, OK? And oops, I didn't say it. And the one lies in P. And here I didn't write it right. So when you have the one input in P, you could land in A or you could land in P. If you just consider the component that lands in P, it recovers the bimodule structure of P, OK? So you could do the same thing here in a more classical context here. So then it's a kind of extension of this DGA by this DG bimodule. But it's not a square zero extension, OK? So you can have operations which takes several P's and outputs something non-trivial, OK? So this thing here, I decided now to call it a non-commutative divisor, but it's not entirely new. There's a version of this where P is the dual diagonal bimodule, which, for instance, appears in Konsevich Blasophilus. Excuse me? Would the second salmon be an ideal in P or not? Would the second salmon be an ideal? No, absolutely not. In fact, here I've wrote it in a confusing way, but you'll see in a second. Yeah, there you go. So let me consider. So there is, in fact, there is a part where you take all inputs in A except one in P and you land in A, OK? So let's look at this part here. And the way to say that part here is let's consider B itself, not as an algebras, as a bimodule over the subalgebra A, OK? So here's B. So this means that we consider only those operations where all inputs except possibly one lie in A, OK? So then there's the assumptions say that there's A sits inside of it as a bimodule over itself. The quotient bimodule is P, but the extension is not necessarily split, OK? And in fact, the boundary map of this extension is, because of the way that I shifted the grading, is a bimodule homomorphism from P to A. Since A is invertible, it's also a morphism from A to P inverse, which you want to think of as a section of P inverse, OK?  And so the corresponding thing is I actually want to use geometric language. So when I say A, I just call it the ambient space where the divisor lives. P, I call the line-bundle. So this sigma here, I call the section of the inverse. That's what defines the divisor. And B, I just call the divisor by itself. That's what it is. It's this extension. So if we forget about A, just consider B as an infinity algebra. That's the divisor by itself. So there's a slight surprise here that sigma in general is not really everything, right? So there are higher order terms. And those higher order terms, you can analyze them into obstruction theory. There are obstruction groups. The pieces of those obstruction groups are harms between higher tensor powers of P and A. They lie in certain degrees, OK? So you can ask them, why didn't I see those pieces in classical algebraic geometry? In classical algebraic geometry, you give the line-bundle, you get the section, and you're done, right? There's no extra data needed. It's because in the classical algebraic, these all live in negative degrees, the higher order obstruction. The classical algebraic world, there are no negative degree harms, and you will not see these things. These are harms of bimodules. These are harms of bimodules, yes. OK? Well, these are effective divisors, right? This is always effective divisor. When I say divisor. Do you think about metamorphic sections of these guys? Oh, not yet. OK, when I say divisor, I always mean effective divisor. So one thing that has occurred before in the story on the algebraic geometric side is where I removed the divisor and considered it's complement, right? And I said that on the level of categories, this can be interpreted as a categorical localization or quotient construction. And the same thing is true here. You basically take a bimodule, sorry, you take a modules. And b is an a module.  And so this gives you a localization construction in the sense of Keller and Drinfeld. And I just write it geometrically as saying, you take a and we remove b, OK? But you could write, or that's obviously a highly non-traditional way of writing it. OK, so now we know what a non-commutative divisor is. What is a non-commutative pencil? Well, it shouldn't be very hard, right? It's a family of non-commutative divisors that's parametrized by two homogeneous variables. So we just have to get the homogeneity right. So OK, so let me fix a two parameter space. If you want, you can set it equal to c squared. But as usual, maybe you don't want to. There could be a space and a dual space. And so I want to have the same thing. I start with a, which is my space, my line bundle, which is this bimodule p. And I do the direct sum, as usual, which gives this b. Now I want to have operations here. But the operations are parametrized by these two variables. So they land an additional tensor product, which is the symmetric algebra. So you have to take care of the homogeneity, right? So you have two homogeneous parameters. If you set the two homogeneous parameters to zero, you're supposed to recover some of the trivial non-commutative divisor, which is just given by taking a and p and putting them together with nothing. So the way to do it is to say, well, b is a direct sum p. You introduce some kind of weight grading. Well, this thing here is weight 1 and that has weight 0. And then v has to have weight 1. And then you ask for this thing here to preserve weights. And you also ask, before that, there were some conditions that certain restrictions recover a or recover p. And let me just say that if you take any w and you pair with w, so you make this thing here into scalars by evaluating a w, then you recover a non-commutative divisor. But you could also formulate it in a general way. So this is a generalization of an a infinity structure. You can think of it as an a infinity structure over this symmetric algebra, but that doesn't quite capture all the structure because of the homogeneity that's involved, which is non-traditional. So it doesn't quite reduce to something that you've seen before, but it does produce a family. You can specialize to any w and then it produces a non-commutative divisor, which we saw before. And it actually, if you rescale w, then you get quasi-izomorphic specialization. So it's really actually parametrized by the projective line, the non-trivial part of it. And so obviously, this is non-commutative geometry. One test is that the ordinary commutative geometry fits into it. So if you have a smooth algebraic variety and you have a pencil of hypersurfaces as defined before, it actually gives rise to a non-commutative divisor, a non-commutative pencil. Oh my god. So where a corresponds to the derived category somehow and p corresponds to l and everything is like that. And did I have to work hard on this? No, you don't have to work hard because, as I said, there's some obstruction theory. All the higher obstructions vanish. So it's kind of straightforward. This is not a one-to-one correspondence. For instance, when I first defined divisors, I kind of wanted sections to be non-zero and so on. But in principle, you can let the sections be zero or other degeneracy. And you will still land in something that's in non-commutative geometry. I don't know what, I don't think of that as being a problem. OK. So just as before I said, the non-commutative divisor, you can think of it as this bimodule map, which is the first order part and there's obstruction theory. So the same thing is true for the pencils here. So you look at a part of it, so you get a two-parameter family of bimodule maps or if you just choose two basis vector, you just get two different bimodule maps, which are called the first order part. So these are thick of sections of P inverse. So those are your two sections that define them. And then there is a sort of deformation theory treatment. Actually, the way to understand this deformation theory is in a reasonable way is to put it into the framework of the Maura-Cartan formalism. So when you actually, there's a bi-graded DG-Li algebra, such that if you have a non-commutative divisor structure with fixed A and P, that corresponds to a solution of Maura-Cartan in this DG-Li algebra. Or what is the same thing is that a solution of Maura-Cartan is the same thing as an L infinity map from a trivial one-dimensional Li algebra into your target G. And so what we're doing here is just simply instead of having one, we have two. And the reason why you see, from this point of view, why you see symmetric powers here, that's just part of the formalism of L infinity maps. Yes? Sorry, because Li algebra depends on the line bundle, yeah? No, no, no. You fix the, yes, the Li algebra depends on the line bundle, that's right. The bundle in pencil is high. No, no, it's a fixed line bundle. Fixed line bundle, yeah. Okay, so now I have defined a non-commutative pencil. What can I do with it? Okay, well, you know, originally I said, if you have a divisor, you know, you have the space, you have the infinity outward correspond to the divisor, and you have the complement, okay? So now we can have more fun, obviously. We have the fiber, we can take the, you know, there's an infinity algebra which corresponds to the divisor associated to any point in CP1, but you can also do more general points. So for instance, you know, instead of considering a single value of Z, I can consider a formal disk around some value, and then I get an infinity algebra, which is associated to that, which is over a one parameter, you know, it's parameterized by this formal disk, and is a deformation of this BZ, okay? Obviously, what you can also do is, before you can take this quotient construction which formally removes the fiber, you could do it to any fiber, of course, but, you know, the fiber infinity will be more interesting to us, and most importantly, actually, in parallel, so you know, when I first, slide number one, I said a pencil, I said you remove the fiber infinity, and then you have this function W, okay, which is S0 divided by its infinity, and there is actually an analog of this thing here, which I call the non-commutative Landau-Ginsberg model, which is sort of a formal deformation of this guy here, but with a deformation parameter of degree two, I don't understand it very, very well so far, I've wrote down the definition for it, and it seems to make sense in examples, but so this is maybe the most, this is something where you really need, for both this B-hat and this one here, you really need the pencil, except here, you just need the pencil in the formal neighborhood of the point, this really uses the entire pencil, okay? Is there some kind of two-periodic to the two-periodic category of motion? Yes, yes, yes. It's a two-periodic infinity algebra or something, which is, and usually when I say the algebra I secretly mean some kind of category associated to it. Okay, so now we finally get back to my original thing of symplectic geometry, right? How many parameters are in this definition? One. And in all examples of B-1? In all examples of B-1, I mean there is no obstruction to introducing non-commutative higher-dimensional linear systems, they follow the same thing, but I haven't found any fantastic use for them at the moment. Okay, so let's go back to symplectic geometry, okay? And let's do some stuff, okay? So let me look at a left-shift pencil, now from a symplectic geometry point of view. I'm gonna, now, because I wanna state an actual theorem, I will make all sorts of simplifying assumptions. So I'm gonna take a symplectic manifold, and the symplectic class, it's like the symplectic counterpart of Fano, so the symplectic class is equal to the anti-canonical class, and now I wanna take a pencil and now I don't care anymore whether it comes from algebraic geometry or not, as long as it qualifies as a pencil and a symplectic geometry, a pencil associated to this, L to this anti-canonical line bundle, which is ample, and so it has fibers, which are symplectic hypersurfaces, it has a base locus, and but now I'm gonna assume that it's really a left-shift pencil, in particular the fiber at infinity is smooth, and the other fibers have generic singularities. So my assumptions that I've done here, this assumption here and the fact that I'm using the anti-canonical implies that the fibers have a zero first-churn class, so they are symplectic calabiows. It also implies that if I remove this fiber at infinity, the symplectic form becomes exact on the complement, and all of them help me to state stuff in a reasonable way. So as we saw, so the basic category I'm gonna use is the focura category of the left-shifts vibration, which is the one that uses left-shifts symbols, and the one that's supposed to be mirror to the total space if we had mirror symmetry, okay? And the theory is that this comes with a canonical structure of a non-commutative pencil. So how does this pencil relate to the one that comes from Donaldson's theory? No, I mean, look, okay. So the word pencil appears on this slide in two different ways, right? So you start with a pencil. This could be what comes out from Donaldson theory here, right, the green thing. It's exactly, that would qualify, yeah? And except, well, you know, Donaldson theory usually requires you to pass to a power of this thing here, but here for simplicity, I mean, you know, it wouldn't break my heart to pass to a power, but I have to modify the formalism a little bit, okay? And then you get a pencil, but not in the ordinary sense, but in this kind of non-commutative geometry sense, a pencil living on this category here. So how is this pencil made up of, okay? So to make the pencil on this thing here, we need a bi-module. That bi-module is always the same. It's the dual diagonal bi-module, which corresponds to the anti, it's some sort of non-commutative geometry version of the, excuse me, of the canonical line bundle. So this has no additional information. It's just constructed canonically associated to NEA. And then I said that there were two, going to be two bi-module maps here. So there are two bi-module maps from A dual to A, if you wanna think of it in categorical terms, there's two natural transformations from the serfunctor to the identity. And they actually play wildly different roles. So the one that I call sigma infinity, exists actually for any left-shift vibration over C. And it has to do with the, it's hard to say, the serfunctor in this category here has a geometric interpretation which involves rotating your direction in which the left-shift symbols go. And because it's part of a one-parameter family, that family gives rise to this sigma infinity. Now the sigma zero is totally different. It actually only exists for pencils that actually can be extended over infinity. Maybe it can't be extended smoothly over infinity, it may not be necessary. But definitely there's some, it encodes the fact that, if you have this guy here that maps to C, the monodromy around a large circle is essentially the identity, okay? And you will use this to construct this thing here in more, maybe in more intrinsic ways. It sort of counts holomorphic sections of your pencil that go through the fiber at infinity. Okay? Okay, so you said the left-shift pencil, the symplectic left-shift pencil gives rise to a non-committive pencil. What if you just take a symplectic anti-economical device, a single one? Will you get a non-committive device? That's right, but that's not new. I mean, well, I have to be a bit careful, sorry. Principle, yes, as usual, the literature is sort of sketchy. Will you say that the pencil, so if you take the generators of the original pencils, take the two symplectic devices? No, no, no, no, no, no. That's the whole point I'm trying to make here. This is a pencil of a P1, but the P1 is not homogeneous. The point infinity plays a special role in this story here. And so does the point zero, in fact, to a slightly lesser extent. But so this is really a parameterized P1 and the point at infinity is important. Okay, so now I have this symplectic geometry situation and I associate to it a non-commutative pencil. On the other hand, the non-commutative pencil has all sorts of things associated to it. So let's see what they could mean geometrically. So non-commutative pencil, one thing you can do, you can take a fiber, say fiber at infinity, and that should correspond to the focaya category of a fiber of W, so part of the fiber minus the base locus. But on the other hand, we could take the fiber of the formal Disney infinity and that should correspond to the relative focaya category where you put in the base locus. After a suitable change of, I mean, there's a question of what the deformation variable is which is rather tricky. And so this at least is internally a consistent picture because this relative focaya category is a formal deformation of this one here and this one here is certainly a formal deformation of that one here. And it seems to work out well in example. And then obviously, if you take the fiber at a punctured formal disc of the non-commutative pencil then that would correspond to inverting the deformation parameter so you would land here. On the other hand, you can remove the fiber infinity just by this quotient construction that could just correspond to the VRAP category of X minus this divisor and then there's the most sophisticated thing which is this, you have this non-commutative pencil, you can do this non-commutative Landau-Ginsperr model and that should actually correspond to the focaya category of the original thing, X that you started with. Or maybe I should say it should be closely related to and you can also get these versions of the focaya category which have what I maybe erroneously called bulk and there's just same thing that the Landau-Ginsperr model where you take, you subtract the constant which is not a big deal. So the reason why all my statements tend to be a bit sketchy but here I am particularly careful for the following reason that the objects of the focaya category of X as we define it these days are closed Lagrangian submanifolds of X. So if I made a statement that says this is this, this actually amounts to a way of constructing closed Lagrangian submanifolds inside X which actually nobody knows how to do in general. There's no way to predict whether this will be zero or not. Now my point of view is that there's just a definition of F of X isn't right. But you have to be careful what you say here. Okay, so this is my wishful thinking thing and here you can make your objection. So I like this a lot better than 10 slides back. Why? Because we have one object in the middle and all these arrows that I've drawn they are strictly the object in the middle determines these things by strict construction. You have this guy here and you get everything and from a symplective geometry point of view this is interesting because the difficulty of computing the things in the outside boxes actually varies enormously. For this one here, strangely enough for these graph categories and for this one here we have pretty good computational approaches. This one here is very hard. This one here is very hard in general. So, okay, so this was as far, was that? In the constructions of all these arrows. Was it really essential that the original symplectic pencil you took was lectures?  So for instance, if you have a toric funnel and one of the generators of the symplectic pencils is the toric device and they just take any other. Yeah, yeah, yeah. So what I said is the condition that the fiber at infinity is smooth is probably not essential. What happens is I used it improving this theorem here. So I have to think again. But if you think of this, the reason why I formulated it in this way here is to open up the idea that, you know, if I just consider case five at infinity is smooth I would have written here sigma zero depends on the fact that the monodromia at infinity is trivial. But I didn't quite want to say this. So this is for in case you want generalization. Okay, so this is how it looks like in my most optimistic moments where is the things now and where I have the problem. So at least I proved that in some cases this non-committal pencil structure exists. There's this relation here with the fiber with the category of the punctured fiber. I proved this one here to first order with respect to this weight thing. So ignoring the higher order parts of the pencil which is a reasonable indication I think. The relation to the VRAP categories is proved in a large part by Mohammed and myself. But these two things that remain are by far the most interesting things. So this thing here, the relation between the structure of the non-committal pencil and the actual structure of the Foucaille category of XZ or if you want the relative Foucaille category this is completely conjectural. I have some ideas how to do it. That doesn't mean that I know how to do it. And there are a couple of warning signs. So one thing that this, I mean, so this thing here, this is a formal family of categories in this parameter Q. So you can look at the Horschel term and say periodic cyclic homologies and it carries a connection which in many examples is equal to, well, there's a map and if that map is an isomorphism then that connection is equal to the sort of quantum connection or a model connection. And obviously, if this is true then that corresponds, that a model connection is the non-commutative Gauss-Mannien connection that's associated to this family here in a form of Disney infinity which means that if you change variables back to the variables that are rational on the P1 then this connection extends rationally. And so I don't know that this is always the case from a subjective geometry point of view. If it's not always the case, this is clearly wrong. On the other hand, if you know how to prove that this is always the case I would love to hear this because that would help my confidence as far as proving this is concerned. So there's a prediction here which says that the a model connection after a suitable change of variables extends rationally. And in fact, mod sum open work in non-commutative geometry this should extend to a regular connection away from infinity. So, and yeah, this one here is also conjectural but at least here because you have a formal deformation with a parameter of degree two the amount of reparameterization that you could be wrong by is very small, essentially none. So, and so, and in a sense, there's a Denier-Ru's kind of interpretation of mirror construction in terms of discounting going through the fiber at infinity should basically be a model for what I'm supposed to do here. Okay, and now, so I've, now I talked about symplectic geometry for a while. So the, there may be overly naive picture of homological mirror symmetry is that, you know, you have two pencils on both sides and one, both pencil give rise to a non-commutative pencil structure. This one here in a more straightforward way, you take the, you know, the derived category and, you know, you encode the commutative algebraic geometry into a non-commutative one. And this one here in this way that I've described where you pass through Foucaille categories. And so the hope is that these are actually the same, you know, there's an isomorphism of non-commutative pencil structure which then by this big thing here would imply all the other equivalences of categories. So we sort of cleaned up this relationship. And, you know, so on this side here, obviously there are cases where you have a symplectic, especially if you drop this assumption here that it's a pencil, anti-canonical pencil. You get cases where the mirror is in all sorts of weird and wonderful versions of algebraic geometry. But, you know, here you can generalize things easily because, you know, it's just algebra. So there's a hope that they fit all in. But, you know, I'm not claiming this right. Okay, so this is the end. Yeah, but we see that on the A-mortal side, it's, you don't have continuous parameters in the sense, yeah? That's right. Yeah. So it means it should be kind of algebraic pencils, kind of code given defined to re-integers and, you know. Well, it is, I mean, this side here, yes, is defined over the integers or, you know, at least over Q. Yeah, no, because it's kind of in the same direction which you could have, let's say, a planar variety. You can see them in a symplectic sense. You can see the quantum products with this canonical class. You get some integer matrix. Or maybe some integer matrix. Also, you get gradient matrix. You get integer matrix. Then you can build from them some connection on, like we have the gradient matrix, right? Yeah, and it should have rebels and rash group and bells money connection. It should follow from all the speech, but it will be really object over integers which are structured. And these are objects, really canonical pencils to build the right, yeah? Do you have? Yeah. I mean, yes, I, I, that.   Somebody didn't, didn't hear it. Or rather the, the comment was that, you know, this thing here is, has a canonical parameter. So there should be a, you know, it's defined over z. So here there should be, you know, a canonical pencil. And especially if you have multiple, you know, if you look at the modular space of Kalabiyao's, you know, it's not clear that there is, you know, where the line should go through, right? But however, this predicts that there should be some way to, you know, single out a, a canonical line. Yeah, also some remark, we, we, for two years, I'm still preparation, I'm preparing this Tony and Whitmill, maybe you know, about Varno, how to predict hodge numbers. And there are plenty of conjectures about the duration of hodge to the run for, you know, exponential, since almost always I'm pruned by Sabha and Zaito and so on. But something is still not pruned, I can, I can share. It also have tables, very similar tables. Plenty of categories, yeah. Yes. Other questions? If multiple is thankful again.


Going on everyone Sam here with late sports cards doing the rant or ripping 10 bucks Tuesdays 10 bucks select honors first off the line Football mixer number 159 random team style. There's a look again at your random results Zoe's can find these on the website under the results tab as well as the link in chat So good luck everybody. We'll start with the leaf. Let me get 10 bucks Pulled up leaf metal jumbo. We'll start with now. We'll start with hobby. I guess it makes more sense to do that We're now live on tick-tock again. Let's go, baby And any guy not in the NFL will be a random to all at the end. I'll check it out, too Check things out here. All right. We've got a Dwayne McBride to start us off Dwayne McBride Vikings on that one One of five Got Zach Charbonette nice one there go to the Seahawks. I know that one six of seven go to the Seahawks Devon a chain Devon a chain is Devon Devon Devon, where are you? Dolphins Dolphins on that one one of three Christian Gonzalez nice one there. I believe that is a Patriots if I'm not mistaken Yes, go to the Patriots on the redemption And go to the Vikings Jordan Addison four of six rats on that Minnesota four of six Jordan Addison All right, lead metal jumbo now. I don't think we have any Wemby You're ripping product with Wemby in it. I'm Friday Woman's best university Marvin Mims gonna be our first one here that is going to the Broncos Marvin Mims five of five We've got a K-Shawm beauty Go to the Patriots nice one there and I'll see you three of three going to the Pats One of one is going to be Israel Abba Abba Nacanda is somewhere the Israel is on the jets and that's a one of one We've got armed and dangerous Jalen Milrow Jalen Milrow It's a random to all four four God Xavier worthy up next Xavier worthy is No, actually Texas random to all so that'll be the second card or second. Yeah card that gets random Who is it? To type out That's gonna be Chris Rodriguez, I believe he's a commander Don't check that. Oh Chris Rodriguez going to commanders nice one there. That is two of six Xavier Hutchinson that is one of one press proof Xavier Hutchinson Go to the Texans This I think will be a random between these two teams Brian branch is a lion Will Anderson is a Texan so unless it's owned by the same person Lions Benjamin L Texans is Austin B that'll be a random between those two teams two of three Another random to all bow Nicks bow Nicks five of five And then this will be a random as well to all is Gonna be a one of one Quinn Ewers Nice one there again. That'll be random to all at the end Quinn Ewers grads on that one there. Everybody good luck. All right. Oh, what are we gonna do next? 21 mosaic hobby now And we'll do that with those randoms at the end same way we do all of our leaf randoms Is this split? No, I'll catch my homes mosaic red Brady Herbie James Harrison and it's a win it's Greg. 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There's a Brady mosaic red Sir tan there Henry Chris Carson overdrive nice one there Etn rookie debut That's a Patrick Jalen camp Pink David Montgomery nice Monty for the Bears to 49 am I saying your name right to thrill silver Keem Hicks Ferris or Fires It was Mills Chris Evans There's nice Barry Sanders on the busted super short print About one a case on those for the Lions Nice one there on the die cut congrats Detroit Benjamin L and Tony O Gibson montage digs to uh Bateman rookie Brady Marshall Jamar Jefferson 21 chronicles next 21 chronicles next See prism, uh, wouldn't worry about it Yeah, it's up Nick Chubb Appreciate that bought from us before. Thank you. Thank you. Thank you All right, 21 chronicles hobby. Here we go Monro St. Brown silver score our legacy update. There's Tony Trevor Lawrence rookie orange prism black is There's Marshall to 75 Get your hits Mack Jones And Jamar chase nice. 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We got two jersey cards in this sky more Yes, we did to go into the Chiefs and Belish Jones Going to the Bears nice two, uh, jersey cards in that very nice Joe Perot to 199 Mystique to 299 is Danny Gray There's Traylon Burks 299 for the Titans I love when we get In 10 bucks. He's uh, I look forward to it every week Keenan Allen to 499 Funkadelic man Third one DK Metcalf. Let's go Michael Pittman George Pickens Heinz war. Nice. That's actually kind of nasty day job boot 12 of 75 for the, um Uh Steelers and operation detonation in Derek Henry. There's red Michael Pittman jr. 399 and mystique James Cook What do we got next 22 prism now? We'll do two selects after 22 Got a 2020 honors first off the line coming up at the end everybody Last group break box of the night Suva Derwin James purple 2125 Michael Thomas slint man Daniel Bellinger Damian Pierce Suva Treson Abner rookie nice Orange Justin Simmons to Um Is that 249? Yes 249 Brian Robinson jr. And George Carlaftus Suva lockdown buddha baker purple power amari Um, oraware to 49 for the lines Got Danny gray and jelani woods So we got an auto coming up here Suva steve young Auto 34 of 149 is Randall cob nice one there for the Packers Now a jet rogers Back sale eric azucanma Desmond winner emergent rookie green scope a j torrell to 75 for the falcons Marie Saul rookie and kairi New recruits Garrett wilson Red wave jalen hurts to 149 the eagles Sam williams and kenneth walker. I get to reveal the trash list this week Silver miles Garrett There's nine of ten gold alec pierce Very nice congrats on that one.  Yeah, this is hobby Blue wave de andre swift the 199 lions germane johnson and bilish jones silver treyvon walker emergent blue ice trish and worths the 99 Got gerome ford and mackerel Silver davante adams fireworks purple ice j jettas the 225 for the Vikings nice color match there josh pascale and dakoby durant a firework sharpie josh pascale rookie purple ice 225 lions k-dotton and t-random specs tdp uh not usually so like base cards yes like hobby exclusives autos not really uh davante Wyatt 2175 on the hyper like you couldn't pull this card in retail uh bailey zappy and i say likely probably closer to the national would be my guest brian So we'll do the next almanac uh 10 bucks is Or two boxes a select hobby now 22 select hobby It's going to jeffrey s and the niners good one of one brock birdie Holy cow that might be the biggest car we pulled out select this year so far That's insane That's that's insane. We got a card stand that one. 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It's massive gold color wheel chris olave Silver trevon walker alvin kumar. Yes, it is cam. Yes. It is my friend. There's a gold now one of 10 Eden hutchinson gold patch nice one there Cleo Shakir silver trey hen trickson They is a brock bird This is uh, that's a one of one alvin kumar silver And drake london rookie white john mechie 235 silver coby bryant and snoop connor Damian pierce rookie I yeah, and the trey lands crown royale. I forgot about that. It's true Good point thrill. Uh garratt wilson rookie signatures light blue prism going to the jets ice one there The insane two boxes and a party rookie party field level zebra and a party 101 and uh, oh my god, dude Like no words that brian cook to 99 lettered floyd silver cmc and courtland Sutton silver daven cook silver de force buckner just in fields field level field fields field Uh, it's lime green pierce strong jr. 249 silvage kwan brisker and joshua, uh, zulu All right, that was the best two boxes of select. I'm ever going to open Holy dude, holy. All right, we've got, uh 22 honors hobby now I got you. I got you.  I'm kind of jealous of one of these guys Gordon davis rookie to 75 And they read derrick henry, uh, three of 10. That's one there for the titans First auto is aiden hutchinson 25 of 25 going to the lions I just won their lines Yes, it is bill that and uh That and that's all for the same team the gold as well One of one marquise brown on the flawless rookie shadow signatures going to the ravens. That's dope, dude. That's 2019 flawless rookie auto congrats ravens All right, 2020 honors photo everybody 2020 honors first off the line. Good luck Okay, I did not look at the buyback. I'll be honest. I just saw I have the rookie auto here Uh bayan was 309 dollars a spot Justin jepperson rookie one of 15 nice one there Uh for the vikings and karsten wence. That is five of seven first off the line exclusive Yes, bill. Yes, they did to say the least is right Uh trevon digs that is going to be eight of nine rookie auto for the cowboys And then a one of one rob gronkowski Superbowl impeccable auto 2018 impeccable I just went there for the patriots Go into the paths All right, gotta do a quick random here guys got the lions texans brine branch will Anderson and then we have uh Four from leaf to do as well. So we're gonna stop tiktok. See you later um, oh my god, dude that the lord lions Texans Texans everybody for the All right guys way this is gonna work is we're gonna start with the dual that's brine branch and will Anderson between the lions and the texans I think let me let me double check really quick. You said Uh brine branch. Yes, lions. Okay Sorry, shon. Sorry for not believing you Roll the dice to go nine times. That's very valid one two three four five six seven eight nine Going to the texans on that one And then the way this will work is the first card we pulled was the jail in mill row Then the worthy then the nicks and then the ewers. So teams number one two three four will get the cards in that order after four times one Two three and four. So you've got raiders browns getting that one The broncos getting the worthy The packers getting the nicks And then congrats to the saints getting the one of one, uh, when you were All right guys, let's do a quick recap here start with our autos Or start with our jerseys. We had the mill row and the worthy and the mix. We had the gold aiden hutchinson Uh, matt chorale that's cool for the Steelers skymore vealish jones and kaderis tony We had autos Is bryan branch will anderson garret wilson bryan earlocker charles hayley randall cob treylon berks bricehall trey lance alvin harper michael parson's jaymond davis zavier hutchinson chris rodriguez israel, uh, a body canda It's kshawn beauty jordan adison marvin mimms christian gonzalez devin ashain zach charbonate dewayne mcbride george pickens Out onto our mag stuff. Good lord. What a break. What a break first of all for the Steelers. That was incredible Quinn ewers at a kenny pickett rpi to 99 at a gold alec pierce auto for the colts one of one marquise brown that's going to the ravens we had the Aiden hutchinson from honors to 25 as well as the treyvon diggs eight of nine from honors first off the line 2020 We had the grunk one-of-one auto and then how about just color wheel gold nine of ten christian mccaffrey field level zebra brock purdy And one-of-one neon icons brock purdy Unbelievable jeffrey asks congrats, man. Congrats to everybody who hit. Thanks again for uh jumping in next up is personal


Jerry at Faroaks. Do you think it'll be all right? Jerry, no unnecessary talking. You can talk, Jerry. What were you going to say? I don't know now. I forgot. You understand why you can't talk in the library. A lot of fellows are studying there and doing research. Sure, I know. I just forgot. What do you think of our library? Sure is a nice one. Say you've got almost any kind of book you'd want. Yes, it's very complete. Gee, it sure is pretty here at night. All these big oak trees and the moon shining through so quiet. Boy. A lot different than the circus, huh, Jerry? I'll say it is. Hey, what time is it? Just a little after 9.30, I guess. We go to bed at 10. That's right. And sure is early for me to be going to bed. Why, when I was with the circus, sometimes I didn't get to bed until way after midnight.  And here you are getting ready for bed. Yeah. I wonder if I'll be able to sleep tonight. I'm sure you will. Why shouldn't you? Well, tonight's the first time in a long, long time that I haven't gone to bed on a train. Well, I'll fix that for you. What do you mean?  That's an idea. And ring a bell and go to a toot toot. Oh, hey, good, Jerry.  Uh-huh. Well, that's the cadet officer of the day. Cadet Captain Duncan. Now, don't forget to salute him. Oh, I won't. You going to talk to him? Sure. I want you to meet him. Here he comes now. Captain Duncan, I'd like you to meet Cadet Jerry Dugan. Glad to know you, Dugan. You're the new man that just came in today, is that right? Yes, sir. What quarters have you been assigned? He's rooming with me. Oh, fine. Well, I'll see you later. Good night. Good night. He seems like a nice fella. Yes, Duncan's all right. Very strict, though. Nothing keeps him from doing his duty as he sees it. That's the way to be, I guess. A cadet officer is just supposed to be just like any other officer. That's right, Jerry. Hey, look, isn't that Major Davis just coming out of Custis Hall? I can't quite see. Oh, yeah, that's him. I wonder if he wanted to see me again today. When I left his office, he said he'd see me later. Well, did he tell you to report back to him? No. I guess he didn't want to see you. See, everything's done by direct orders here at Faroaks. He's coming this way. If he stops us, be sure to come to attention. Yeah, I know. Well, that's not talking. Good evening. Well, Jerry, I see you were fitted out in your uniform. Yes, sir. Looks very good on you, Jerry. Have you seen all of the campus? Almost all, I think. I tried to show him as much as I could, sir. Did you get down to the lake? No, sir. We didn't have time.  Yes, sir.  And during your study periods tomorrow, look the book over. You'll be able to understand it, all right? But if there's anything that you can't get quite straight, we will help you out of it. Yes, sir. You going to your quarters now? Yes, sir. We just came from the library. Try to get all your things put away tonight before it taps, Jerry. You'll have a very full day tomorrow. You can help him if he needs any help. Yes, sir. I'll see you in the morning, Jerry. Good night. Good night, Major David. Good night, sir. Well, what do you think of the Major Jerry? I think he's one of the nicest men I've ever met. He reminds me a little bit of Mr. Randall. Now, don't forget to get that book of rules and regulations right after breakfast tomorrow. No, I won't. Hey, what's in the book, Lee? All about what to do and what not to do? That's right. And once you get your book, you're supposed to memorize every rule in it. It tells all about how to conduct yourself in the different buildings, about the cadet officers, the bulletin board, the different bugle calls, formations, inspections, and passes, and everything. There must be an awful lot to it. Oh, it's not hard. You sort of learn things by watching the other cadets, too. What do you mean by inspection, Lee? Well, every Saturday morning, our rooms are inspected by cadet officers. Believe me, they have to be just right. And then, too, they're inspections in the line during drill. To see if our uniforms are all right, huh? And your hands and nails and hair, too. Oh, look, Jerry, before we go in, I want you to see Davis Oak by moonlight. Davis Oak? See that big one there by the fountain? You mean the biggest one of all? That's it, the biggest and the oldest. Yes, that's Davis Oak. That tree, Jerry, is the reason for Faroaks being where it is. I don't understand. Well, it's like this. Major Davis' father founded this school many, many years ago. When he was looking for the property to build on, he came out here. But this was all open country then, except for that big tree you're looking at. Weren't all these other beautiful oak trees here then? Yes, but they were just little trees then. Davis Oak was the only full-grown tree. Major Davis' father, so the story goes, pictured that tree right in the center of the quadrangle. So the school was really built around that tree. And isn't it beautiful? Sure is. Well, come on. The other fellows will be coming back from the library in a couple of minutes now. You're OK. But gee, I could stay out here all night. It's so still that, well, you know, Leigh, I kind of miss the circus right now. I mean, now that it's bedtime and... Say, I was lonesome too for the first few days. Come on, Jerry. Well, that was Jim Cooper and Glenn Radford. I guess I was wrong. We're not the first ones back to quarters after all. They're going in the room next door. Yeah, that's their room. Yeah, where's Tubby's room? Just down the hall from ours. Let's see. It's the third one, I think. And Harold Linwell, where's his room? Down at the other end of the hall. Well, here we are. Hey, where'll I put my suitcase while I unpack on the bed? Oh, no. Here on this bench. Don't ever put anything on the bed. Want a hand? No, I can make it. Look, Jerry, this is your drawer here. OK. And this right-hand top drawer is yours too. That's for socks and handkerchiefs. Hey, what's that? Well, that's Ted too. That means get ready for bed. Here, I'll put these things away for you. Well, look at this. What? These pictures of this elephant. Hey, and that shoe up on his head. Sure. That's El Mundo. That's the elephant I did my act with in the circus. All by yourself? I didn't know you did that. Why, sure. I thought maybe you were just part of some act. I didn't think you were an elephant trainer. See this picture here? Or I'm swinging from his trunk. Some trick, all right. Yeah, and look at this one, where El Mundo's standing up and I'm standing right under him. Oh, wow. Did you make him do that yourself? Sure, I did. Besides, I had to give all the commands in Spanish. In Spanish? Before? Because he was a Spanish elephant and didn't understand English. Can you talk Spanish? Oh, only a few words. Just the commands I had to give to El Mundo. I learned how from Carmen Bendini. She was a tightrope walker with a circus. Well, that's sure interesting, Jerry. Did you ever do any horseback riding? Well, there was a little coat by the name of Splendor, and I rode him. Not in the show. I just used to ride him for fun. I broke him in too. And maybe it wasn't hard getting him used to the saddle and bridle. You sure must have had fun with a circus. I did. I even jumped with Splendor. High fences, too. Well, say, you might make one of the polo teams if you can ride that good. Oh. Hello, Morrison. At ease, Phillips. Attention, Duggan. What's all the chatter going on in here? I was just talking to Lee. He was just telling me about some of his experiences with a circus. Oh, so you were with a circus, huh? He did his own act with an elephant. Go on, show Red the pictures, Jerry. No, never mind the pictures. Just tell me about it. Tell me that you were the man on the flying trapeze that flew with the air with the greatest of ease. I didn't do an act on a trapeze. I've got to put this suitcase away now. No, you don't. You put it away for him, Lee. I want to hear all about the circus, Duggan. Go ahead. Go ahead, talk. Oh, let me alone. I've got to get ready and go to bed. Now that's no way to treat company. I want to hear about the circus. Go on. Well, once we had a fire, it was in the horse top. And all the horses were trapped. Didn't look like they could be saved. And then?  Throw it on the fire and put it out. Hooray, Duggan is a hero. Oh, you don't want to hear anything. Sure I do. This is all very interesting. What else? I'm not going to say anymore. Oh, yes you are. Tell me something exciting about the Lions and Tigers. Didn't anything ever happen to take your breath away? I want to hear about a real thrill. I'll tell you something that really happened. But don't kid me about it. It's true and... Go ahead, Duggan. Well, once we had generator trouble, and the light started to flicker. It went out for a second or so, and then went on again. That's a good story. Go ahead. Well, just as Jason, that's the line tamer with the circus, went into the steel. Steel? What's that, Jerry? That's the big enclosed ring where the wild animals work. Okay, go ahead. I know that. You just listened, Phillips. Well, his act had just started, and all the Lions and Tigers were in the steel with him. He was just about to put them through their tricks when the lights went out and didn't come on again. They stayed out. The audience went wild, and even the performers got scared. They figured the Lions would jump on Jason. Okay, that's good. Now I'll finish it. And just then, you ran right out among all the wild animals, and you had a big birthday cake with a lot of candles on it. You lit the candles. The wild animal trainer finished his act, and everyone lived happily ever after. Oh, what's the use of telling you anything? Because I like it. Here at Fair Oaks, we don't get to hear about a circus every day. Go on, Dugan, one more story. Uh-uh, that's all. Oh, so you won't talk, huh? I'll let up, Red. You keep out of this, Phillips, or I'll be asking on you for some favors. Well, Dugan, as long as you won't talk, how about turning on the radio for me? You know we aren't allowed to play radios after tattoo sounds, Red. Yeah, turn on the radio, Dugan. Now you're learning to take orders the way you should. We'll get in trouble for this, Red. That's all? Turn it on louder, Dugan. Louder, I said. There is that louder now? Yeah, that's fine. You want to dance? Turn that on. What are you doing in here, Morrison? Just visiting with a new man, Duncan. Tattoo and cold quarters are both in. Go to your room, Morrison. Yes, sir. Well, good night. Now, who turned that radio on? Well, you see, sir, it was... I did, Captain Duncan. You know better than that, Phillips. You're not setting a very good example for a new man. I'm going to have to give you two demerits for this. Yes, sir. Get ready for bed. Taps will be sounding any minute now. What did you take the blame for, Lee? Oh, I was afraid you were going to tell Duncan the truth. Sure I was, and he ought to know. No, Jerry. That's one thing we don't do here at Fair Oaks. And we settle all our scores among ourselves. But you got two demerits for something you didn't do. That's nothing, Jerry. I didn't work them off with two hours of extra study. And I can use the time anyway. Just forget about it. Oh, I'm getting to bed. I'll switch the lights off. Someday that red morsel... Forget it, Jerry. Well, good night, Jerry. Good night.


For the next three weeks, this campus-like setting will be your home away from home. The Federal Bureau of Prison Staff Training Academy, located on the grounds of the Federal Law Enforcement Training Center in Glencoe, Georgia. If it's vaguely reminiscent of college campuses you may have attended or visited, it would be altogether appropriate. But this 1500-acre center is a scene where more than 20,000 students train, improve existing skills, and develop new ones each year. These students are all part of more than 60 federal law enforcement agencies from throughout the nation. And BOP is especially proud because its academy is the largest at the center and the best. You'll be putting in a lot of hard work while you're here at the academy, but what you learn will stay with you forever. By the way, you'll also find time to enjoy some of the beautiful surroundings of historic and scenic southeastern Georgia. The beaches, the resorts, the golden aisles, all part of your home away from home. Hello, I'm Mary Ann Christensen and welcome to the Bureau of Prison Staff Training Academy. The training and instruction you experience here are designed to give you the knowledge, skills, and techniques necessary to become an effective correctional worker. When you graduate and return to your facility, you will find yourself better prepared to carry out your responsibilities as well as the mission of the Bureau. The mission of the Bureau of Prisons is to maintain safe, secure, and humane correctional institutions while developing and operating correctional programs. This mission is accomplished by correctional workers who are committed to being proud, professional, and proficient. Well, if you feel they're okay, tell the inmates so. There's nothing wrong with these cuffs. What we'd like to do now is take you on a brief tour of the center to give you an idea of what you can expect while you're here. One of the first things you'll notice is that the center is very large and activities seem to be going on all of the time. But first things first, driving in your first stop will be the front gate where you'll register your car. Once you've shown identification, security will give you a permit which will allow you to park in designated areas around the center. Then you'll be on your way to registration where you'll pick up your ID badge and your dormitory assignment. Keep your badge with you at all times and make sure you wear it in the dining halls so you won't be charged for meals. The first thing you'll probably want to do is locate your dormitory. You'll find them centrally located and within easy walking distance to classes and the dining hall. You should find the rooms comfortable and functional. Community areas supply all the extras.  There's also a lounge where you can relax and watch TV. A break area offers a selection of snacks including soda and ice machines. Not only do correctional officers do security checks or inspections. Your academic courses will be conducted in rooms like this one which are modern and spacious. You'll receive the latest technical and field information from Bureau of Prisons instructors. They will utilize both interesting and up-to-date teaching methods. These instructors are some of the best employees in the Bureau of Prisons. They are committed to the mission and core values of the Staff Training Academy. The instruction will involve such areas as security, communications and the inmate classification system. Don't lose your cool. Take the courses in communication skills for example. They'll help teach you to read the communications of others. How to convey an order. Apply proper levels of assertiveness. How to communicate both verbally and non-verbally. And how to prepare reports. The courses in the fundamentals of correctional law will review real-life cases concerning inmates and correctional workers. Your legal rights and duties will be outlined as they apply to existing policies and procedures. Good sorry. Tell me what is the most wrong. Understanding all of the academic material is necessary to ensure your success as a correctional worker. Remember you will be tested on all courses and you will be required to make a passing score in order to graduate. But there's more than just academic classes. You will also be given tests to determine your overall level of physical fitness. These are called PEB or the Physical Efficiency Battery. Instructors will direct you in performing exercises and stretching, agility and endurance. Your level of performance in each one will be measured so that you can assess your current level of fitness. Ready? On guard. Another aspect of your extensive training will be daily classes and self-defense. The principles of Akita will be taught to you so you can protect yourself and others. Through techniques based upon this martial art, you'll learn how to make an opponent's own momentum work against him. The mastering of these self-defense techniques will help build your self-confidence. Like in the academic courses, you will be tested on these important abilities in self-defense. A passing score is required. What do you think you're doing out here? Can you stop right there? Yeah, I'm going to stop right here. I want to know what's going on. One of the favorite activities for all of the students is the practical exercises. These allow you to put into action what you've learned in the classroom. But don't be surprised if a few interesting details pop up, things you may not have thought of before. Do you know where we got it? No, we don't, ma'am. Instructors will place you in a mock field setting where you will be called upon to apply any number of skills. Search procedures, problem resolution, knowledge of agency policy, communication skills, and proper methods of assertiveness will be evaluated. Good job, guys. Let's try another one. Who's next? Here's an example. In this practical exercise, students are searching for contraband in the Glencoe Jail. Through exercises such as this, you will become aware of what to look for when conducting a thorough search of an individual or an area. You'll begin to see the surprising range of possibilities that arise when trying to outthink a prison inmate. Another chapter in your training will be the handling and maintenance of firearms. It's a standardized program where you'll learn to use a variety of weapons, including handguns, rifles, and the shotgun. Indoor and outdoor shooting ranges offer you diverse environments in which to develop your skills. Instructors will take time to help you become more comfortable and confident, assisting you along the way in developing your expertise. As your abilities develop on each weapon, you'll be tested. As always, a passing score is required. We have a well-rounded curriculum here at the Staff's Training Academy. The instructors here at the Academy are concerned for the students and are committed to helping the students develop to their fullest potential. As instructors at the Academy, we have a broad base of experience. Instructors have worked in various occupations in the field and have brought to the Academy a wealth of experience. Instructors are charged with the responsibility of helping the students to develop and not taking this responsibility lightly. We would never tell a student to do something that we haven't done ourselves in the field. The Academy is a place to get a lot of serious work done. So a lot of thinking and planning has gone into making sure that can happen in an atmosphere that's structured yet enjoyable. You've heard the story about all work and no play. Well, we want to make sure you take the opportunity to relax after hours. And yes, enjoy yourself. In the Student Center, you'll find friendly conversation and a host of activities to entertain you. Take advantage of the large indoor and outdoor pools or take a jog around the fitness trail. Enjoy a game of golf on our nine-hole course or reserve time for a game of tennis or racquetball. At the Convenience Center, you'll find almost everything you need. The post office is open during midday hours, making it easy to send or receive mail. The credit union will cash your checks and take care of your banking needs. And there's a convenience store for all kinds of supplies and things you need. Snacks, beverages, specialty items, you name it. One place you'll find yourself a lot is the dining hall. It's open seven days a week for three meals a day. There are a variety of appetizing setups, hot meals, cafeteria style, a grill that offers hamburgers and hot dogs, a sandwich deli where you can create your own, and the popular salad bar with all of the trimmings. As you attend the Staff Training Academy, you'll find that the program is designed to be thorough and well-rounded. The many skills expected of a correctional worker are diverse. You need to be assertive. Know the job by the book. Be adept at handling many demands. Every BOP graduate has worked hard to qualify in all of these important areas, but it's not just the hard work. BOP employees add a special sense of dedication and belief in what they're doing. They're proud, proficient and professional. That's pretty evident. Here's what a few graduates have to say about the training. I work with inmates every day, face to face. Going to the academy helped me build up my confidence and make me a better officer. I got a lot out of Glencoe. Don't get me wrong, you've still got a crack of books, but there's a lot to do there. The facilities are fantastic. I learned quite a bit. The academy at Glencoe helped prepare us to work in a large institution like this with a lot of inmates. I had a lot of practical exercises to deal with emergency situations because in a place like this, anything can happen and will happen. One thing I hadn't anticipated was the tricks and manipulation games that inmates will try to play on. Not too long ago, an inmate here tried to manipulate me and because of my training at Glencoe, I was able to resolve it before it got out of hand. I think one of the most important things I learned at Glencoe was communicating. Glencoe has really helped me to improve my communicating skills. I communicate with the inmates on a daily basis in the factory here. Prior to going to Glencoe, I didn't know what the inmates were authorized to have or not to have, so I think Glencoe has really helped me in that area. You're embarking on a very important phase of your career. As a correctional worker for the Federal Bureau of Prisons, your knowledge and expertise will cover a wide variety of subjects. This important mixture of skills and abilities is vital to your success as you set out to help fulfill the agency's mission. At the end of this three weeks, you will renew the oath of office and become a member of the Bureau of Prisons family. It won't come easy. It will take hard work and dedication, but most assuredly, it will be something of which you can be proud. We hope you've enjoyed this tour of FLETC. We look forward to your visit to the Staff Training Academy and you're becoming a member of the BOP family.


My name is Cecilia Vicuña, in New York there is a sort of boundless energy from the unfulfilled desires of millions of immigrants that have been coming here for perhaps two centuries or more and so this bubbling energy that you feel here can be very oppressive if it becomes pressure but at the same time it can be something that suits you into becoming yourself because how else are you going to be able to say something when there are 20 million people trying to say something? Only by going very deep into your own way of being.