 OK. And so if D is an nm-tangle diagram, there's going to be CkH of D is a chain complex over a certain category that I'll call, say, CBN. So this is Barnathan's category. And I'll, again, write the n and the m here. OK. I'm not going to be able to take the homology of this chain complex. OK, this category isn't an abelian category, so there is no homology. But I can still talk about chain homotopy equivalents. And so if D and D prime both represent t, then they're related by right of Meister moves. And CkH of D is chain homotopy equivalent to CkH of D prime. OK, so how are we going to build this chain complex? So to build this chain complex, I have to talk about co-bordisms. So let's suppose I have p0 and p1 are planar tangles. So say they're in p nm. So definition of co-bordism, it's called sigma going from p0 to p1, is an embedded, let's say, sigma inside of r times i. This is the planar tangles space times another i, sort of in the t direction. So I'll call this i coordinate s and this one t, such that the boundary of this sigma, what does it look like? Well, it has a piece that looks like p0 times 0. It has a piece that looks like p1 times 1, but it also has pieces in this boundary coming from the x's. So union, let's say, xn times 0 times i, union xm times 1 times i. And if I have a co-bordism like this, there are actually two kinds of compositions that I can do. Now, I have two ways to compose. So there's what's called horizontal composition. So if I give you, say, if p is in pnm, p0 and p1 are in pnm. And p0 prime and p1 prime are in pnml. And I have co-bordisms sigma going from p0 to p1 and sigma prime going from p0 prime to p1 prime. Then I can stack in the s direction, the direction that I've been drawing when I compose tangles, and get sigma sigma prime. That's a co-bordism from p0, p0 prime to p1, p1 prime. So I also have vertical composition. So if p0, p1, and p2 are all in pnm, and I give you sigma0 from p0 to p1, sigma1 from p1 to p2, then I can stack in the t direction and get sigma1 composed with sigma0. That's a co-bordism from p0 to p2. Here I stack in the t direction. So the fancy way to say this, I have these two different sorts of composition is to say that I have a two-category p where the objects are the xn's. The one morphisms are tangles, and the two morphisms are co-bordisms. And I'm not going to try and write down the axioms for a two-category, but the sort of model thing that you should think of when you think of a two-category is this category of tangles. Yes, these are, sorry, yeah, these are, of course, there's another tang, since this is p. There's another category where I didn't take planar tangles, but let's stick with this one here. OK, and in particular, so for example, the set p and m is itself a category where the morphisms are co-bordisms. All right, so now I'm going to do something totally unmotivated, which turns out to be really handy. And I'm going to enhance this category just a little bit. So I'm going to define a category, let's call it cob, n comma m has objects, again, sort of tnm, just like in this, oops, pnm. And morphisms are pairs, s comma a, where, so if this is in the morphisms from p0 to p1, where what, s is just an ordinary co-bordism from p0 to p1, and a subset of the interior of s is a finite subset. So if d is a tangled diagram, so the cube of resolutions assigns to a vertex v an object, oops, this is a nm tangled diagram. Let's say dv in the objects cob nm, and it sends to an edge e going from v0 to v1, se, that's a morphism, se, the empty set, so no empty finite set going from dv0 to dv1. OK, so I have my cube of resolutions as decorated with objects and morphisms of this category. So the objects are, let's say, in pnm, so the objects are planar tangles. Now, I'd like to build a chain complex out of this cube, and I'd also like to be able to put in those factors of q, I sort of multiplied by q. And I can't do either of these things in this category, so I'm going to have to do a few general category things to make a big enough category that I can do those things in. So let me just say a few general words about category theory. So let's call this a definition. So a category c is pre-graded if there's some function that we'll call q, the grading. What does this do? It takes the set of all morphism spaces in the category to the integers, such that if I take a composition q of alpha composed with beta, that's q of alpha plus q of beta. So as an example, I'll leave this for you to check. Cobb nm is pre-graded, where q of s, a is chi of s minus m plus n over 2 minus 2 times the number of elements in this set a. OK. Now, if c is pre-graded, I can form what can I form? I can form, let's call it CGR, is objects xn. x is an object of the original category, and n is an integer. And I'll say that the set of morphisms from xn to ym, that's the same thing as the set of morphisms from x to y. But the grading is different. So q of alpha in here is q of alpha, where I think of it in here, plus m minus n. So in this category, I can multiply by powers of q. So another way that I could write this thing is I'll call it q to the nx. OK. So now, I need to think about what's a chain complex. So side definition, c is additive if x and y in objects of c. Two things. One is that the set of morphisms, which in an additive category I'll typically write as the set of hams, is a z-module. So it's in a billion group. And two, x direct sum y is an object. So if c is an additive category, I can consider chain complexes that look like, say, ci goes to ci minus 1, goes to ci minus 2, where each of these are objects of the category. And then I have morphisms, di and di minus 1, et cetera, where di minus 1, composed with di, is 0. And that makes sense, because this lives inside an abelian group. We really needed this property. OK. So of course, this category Cobb isn't an additive category either. So hams and seager are the same. That's right. But I have to keep track. The only thing that's different is that I write down the grading changes. This is n and m. Yeah, there are two different letters here, and this is m and this is n. Well, given c, I can form an additive category. Let's call it adc. Objects are formal direct sums, objects of c. And of a direct sum, i equals 1 to n xi direct sum i equals 1 to m, yj is the set of m by n matrices. Let's say fij, where fij is in the z module generated by morphisms from xi to xj. So I just build myself an additive category. All right, and so now we've got two minutes left. So let's make a statement. So if d is a Tangle diagram, I can form c of d. This is a complex over ad of nm gr. At the level of object, this c of d is, again, the direct sum of, let's say, q to the norm of v dv. And let me just say d, as in Kovanov. OK. All right, so at this point, we built a chain complex. But we've sort of done, and I apologize for making you sit through this, right? We've done nothing but formality all lecture, right? And in particular, this chain complex has, the way c of d has no hope of being invariant under the right of Meister moves. So the key step here that we'll do in the next lecture is that we pass to a quotient category, which is this category C, B, N, N, M, OK? And this is not a formal thing. This category C, B, N, N, M is a genuinely interesting category in which you can compute things. And what we'll do is we'll compute and we'll see that in this quotient category, the complex that we get is invariant up to homotopy under the right of Meister moves. OK, so I'll stop here.