 So today I'd like to talk about the Barnaton category and the Kovanov homology of a tangle. So maybe let's just recall. So say D is a nm-tangle diagram. And if we had this category, which I had this horrible name, add of Cobb nm-gr. That's far too long to write all the time. So let's just abbreviate that by C bar of nm. And now we have the cube of resolutions. So well, in this category, I'm allowed to shift gratings by multiplying objects by powers of q. I'm allowed to take direct sum of objects. And I'm also allowed to add morphisms together. That's why we went to all this categorical work, so I could do those things. So now to a vertex v, I can just assign, let me say, q to the l1 norm of v times dv. This is a perfectly good object of our category C bar nm. And to an edge, e from v0 to v1, I have a morphism. Let's call it Se, let's say Se empty set. Because remember, these morphisms in here came with a set of marked points, finite set of marked points. So here, there are no marked points at all. Going from q to the v dv to q to the, oops, this is v0 and v1, dv0 and dv1. And then two-dimensional faces commute. So now we can form a chain complex in this category. C of d, at the level of objects, it's just the direct sum over all v of q to the l1 norm of v, dv. And the differential d is given by taking the sum over all edges, e of, let's say, minus 1 to the sigma v. And then this morphism Se empty set. So this is an endomorphism C of d. And because two-dimensional faces commute, d squared is 0. So it's a totally formal construction. But one thing that I'd like to point out, which is still formal but useful, is that horizontal composition gives a functor. Let me call it h for horizontal composition from c and m, tensor c bar and l to c bar and l. And the proposition, which categorifies the proposition we had last time about horizontal composition, is that if I take c of dd prime, so if I compose two-tangle diagrams, I get c of d tensor c of d prime. And rather than write down a proof, I'm going to observe a few things. So let's notice that c of, for example, this crossing here, this is a complex that looks like, let me write it this way. The zero resolution goes by the one-handle co-bordism S. That's where I add a one-handle like this to the one resolution. So it's a short complex. Now, say I have two short complexes. Say, c0 goes to c1 by some differential d. And I tensor this complex with c0 prime goes to c1 prime by d prime. Well, what is this? This is just a complex that looks like this. c0 tensor c0 prime goes to, say, c0 tensor c1 prime. Here I have c1 tensor c0 prime. And here I have c1 tensor c1 prime. And for example, this map is one tensor d prime. And this map is d tensor one. You can fill in the others. You'll notice that taking a tensor product of short complexes gives me a cube. And that's really what the cube of resolutions is. It's a tensor product of short complexes associated to these crossings. OK. So now we wanted to pass to the Barnathian category. So maybe let me just say a little categorical nonsense. So definition, so c is an additive category. So an ideal i in c is a collection of things that look like ixy for pairs x, y, which are objects of d, such that 1, ixy is a subgroup of harm x, y, and 2, maybe I'll take script EI to be the union of all of these things. So if beta is in ixi, that means that alpha compose beta and beta compose gamma are in i. Whenever these expressions make sense. So if I have an ideal, I can take a quotient category. How should we call that? Maybe c mod i given by, I have to tell you what the objects are and also what the morphisms are. Well, the objects of c mod i are exactly the same as the objects of c. But harm in c mod i of x, y is, well, it's the only thing it could be. It's harm in c, oops, wrong place, harm in c of x, y modulo this ixy. So the thing to understand, if you've never seen this before, is that the objects are the same. But objects that weren't isomorphic in c may well become isomorphic in the quotient. So a canonical example, maybe let's say c is the category of chain complexes over some ring r. i is the set of chain maps that look like dh plus hd, so the null homotopic maps. And then the quotient, say c mod i, this is the homotopy category of chain complexes. So two chain complexes that are homotopy equivalent are isomorphic in this category. Let's make a definition. Let's call it, say, ibn is the ideal in c bar. I guess if I'm keeping score, I should maybe write some nm's down here, generated by some local relations. And the local relations are as follows. So if some component of my co-bordism is a sphere or a sphere, so now I'm going to draw, remember we have this set of mark points on our co-bordisms. I'm going to draw the mark points as dots. So a dot means a point in A. So we have this co-bordism, s comma a. This is also equal to a sphere with three dots on it. And they're all equal to 0. So any time I have a co-bordism with a component that looks like that, that's the zero-morphism in this category. On the other hand, a sphere with one dot on it, I can erase it and just have the number 1. So I can just erase that sphere at no cost. Those are sort of sphere relations. And then the most interesting relation is usually called neck cutting. And it says that any time I have a piece of my co-bordism that looks like this, that's the same as the sum of these two co-bordisms, plus almost the same thing, that I switch the dot to the other side. And now these relations may look like they come out of nowhere, but actually the place they come from is the TQFTA that we used yesterday to define the Kovanov homology. There's an exercise in which you're asked to sort of relate these relations to the TQFT. And let me just remark. So here in this category, we impose local relations on morphisms. You should compare that to define the Kaufman bracket. We imposed local relations on diagrams. We said that this was this minus Q times this. So when we pass to this categorified version, instead of working with local relations on non-diagrams, we work with local relations on co-bordisms. So now it's obvious what I'm going to do. I should have said it's a terrible form, but let me just do this. Let me remark over here that there's an obvious functor, sort of projection functor, say pi from c to the quotient whenever I have a quotient category. OK. So then we define c b n, n comma m. That's this category c bar of n m. Modulo, this ideal of Barnathan relations. And c k h of d is going to be the image of the complex c of d under this projection to the Barnathan category. So now what we're going to do is we're going to study the structure of this category c b n. Now the structure c bar is really kind of woolly. It's got lots and lots of objects. And instead, what we'll see in this category is basically everything can be expressed in terms of finitely many objects. And the key point here is the following proposition. So let's just write it this way. So if I have a diagram, I'm going to write my planar diagrams with parentheses around them. So if I have a planar diagram just with a circle in it, and if I were working with a Kauffman bracket, I'd say that that was equal to just q plus q inverse, where I erase the circle. OK. Here what I can say is this is isomorphic to q times the object where I erase the circle, plus q inverse times the object where I erase the circle. OK. So how do I prove this? Well, I have to write down some isomorphisms. So I need to write down a morphism, f from this object to this object. And what is it? Well, this will sort of be a matrix with one column and two rows. And what the matrix is is it looks like this. So here I cap off this circle with a two-handle, and I put a dot on it. And down here, I don't put any dot. And I'll write that as sort of eta x and eta where x corresponds to a little cylinder where I've put a dot on it. And eta is the two-handle co-bordism. OK. So now I need a morphism going the other way. Let's call it g from q empty set plus q inverse empty set to the diagram where I have a circle. OK, that's going to be a matrix with one row and two columns. And what is it? Well, it's this matrix. So now I put the dot over here. OK. And I could write that as maybe epsilon and x epsilon, where epsilon is the zero-handle co-bordism. All right. And now I need to check that when I compose these two things, I get the identity map. So it means to be an isomorphism. So for example, if I multiply the matrices fg, that means I'm going to take this matrix and multiply it by this matrix. And that'll give me a 2 by 2 matrix. You can do the multiplication, and you'll see you'll get four pictures each with a sphere. This one, the dot is on top. This one, the dot's on the bottom. Here, I have dots on both the top and the bottom. And here, I have no dots. But that's really just the matrix 1001, which is the identity. So here, I use these sphere relations. And similarly, if I look at g composed with f, that's when I multiply these matrices to get a 1 by 1 matrix. Well, that's the 1 by 1 matrix. It has two different terms in it. One of them looks like this. And the other one looks like this. And conveniently, the other relation was just made to tell us that this is the cylinder here. And what's the cylinder? It's none other than the identity co-bordism for the circle. So these things really are isomorphic. And again, notice that in this category, we don't have to impose the relation that these two objects are isomorphic. We get it from the relations that we imposed on the morphisms. So we calculate that this is true. So a corollary of this is that every object of CBN is isomorphic to a direct sum of things that look like, say, q to the i pi, where pi is a simple planar diagram. So I can express all the objects in my category in terms of this finite number of objects, which are the simple planar diagrams. So now let's think about morphisms. So maybe first dilemma. So if S a has a component with more than one dot, so more than one element of a, then it's equal to 0 as a morphism. So now, from now on, I'm always working in this category, CBN. And the proof, well, so maybe here's my component with two dots on it. It looks like this. So I apply the net cutting relation and cut this blue neck right here. And then I see that that's a sum of two things, each of which involves sort of the rest of the co-bordism in a sphere. One of them has three dots here, but that's 0. And the other one has one dot here and two dots here, but that's still 0. So the whole thing is 0. OK. So now let's make another definition. Right, so the relation that I just used, so to pass from here to here, I use the neck cutting relation. So there's around this blue circle, there's a cylinder. And so I applied the neck cutting relation on that cylinder, and I got these two terms. Is that the question you're asking? OK. Question? Yeah, that's right. So remember, the objects in this category cob were planar-tangle diagrams. Yes. OK, so definition on, let's say, SA is simple if every component of, let's just say, SA is a disk with either 0 or 1 dots. OK. And then the proposition is that HOMP0P1 is spanned, maybe has a basis even, consisting of simple co-bordisms. So this is a basis, I guess, as a Z module. OK, so the proof, well, 1 is if S has a component with mon0 H1, so maybe I have something like this. So I choose an embedded curve representing a non-zero class in H1. So I apply neck cutting, a non-zero class, to reduce H1. Well, so after I've done that, I've got a whole bunch of components with no H1, so there's spheres and disks. So then I kill spheres with sphere relations and the disks with the lemma. So if I have a disk with more than 1 dot on it, then I know it's 0. OK, so here the corollary is, and now we come back to our old friend, that TQFT. So if P0P1 are simple planar diagrams, then HOMP0P1 is actually isomorphic to this TQFTA applied to let me draw this diagram big so you can see it. So let me do it down here. So this is A applied to the following diagram. So I take P0, I stick it together with P1, but I have to kind of turn P1 around like this. OK, can everybody see the operation that I did on P1? OK, and why is that? Well, proof, let me call this P0P1R diagram here. So if I look at co-boardisms from P0 to P1, well, this diagram P0P1R is none other than P0 times 0 union P1 times 1 union xn times 0 times i union xm times 1 times i. OK, in other words, it's the boundary of any co-boardism between P0 and P1. I've just the boundary of this curve here sits on a cylinder and I've taken and squashed this cylinder flat onto the blackboard and that's why I had to turn this P1 around. OK, and now, homophism is supposed to be, well, I need to choose a disk for each of these and then I put an x on it or not. So that's like labeling each circle in this picture with either a 1 or an x and that's what this is. And if I wanted to get the gradings right, I should probably put in a factor of Q to the minus m plus n over 2 here. OK, so now we really know everything about this category. We know what its objects look like. We know what the morphisms look like and we can just go compute. So let's do that. So maybe let's think about proving right-of-meister moving variance. So this is the categorified analog of the exercise that you did yesterday. So say that I look at, let's call this right-of-meister moves. Say I look at CKH of this diagram here, which I know I can untwist. OK, so what is this? This is, well, when I do the zero resolution, I get a curve that looks like this. The one resolution gives me a curve that looks like this. Can people see this curve here? Is this large enough for, OK. And then there's a morphism in between them, right? I have to remember these aren't just objects. There's a morphism in this chain complex, which corresponds to adding a little one handle right here and cutting off this disk. OK, so now this chain complex in the category, I'm gonna write down an isomorphism to another chain complex. Oops, and I forgot I should have put a Q here, shouldn't I? So here I have this object. Well, this is just the identity. This is the.