 So I'm going to talk about an invariant of oriented Lantz and links called Havana homology. OK, so what does this invariant do? Maybe before I start, let me mention. So for the first 40 minutes or so, at least, probably more like the first 50, this is going to seem like it has very little to do with the things I talked about in the first lecture. But we'll make some contact with ideas at the end of the first lecture at the end of this one. So Havana homology is an invariant that assigns to a oriented link L inside of S3, a bi-graded homology group, k-h-i-j of L. OK, and if you've read the sort of prelude notes, then I can say that this is related to the Jones polynomial, so I can take the sort of graded Euler characteristic of k-h of L. So that's a sum where I build a generating function in one of these two indices. So I take q to the j and then take the Euler characteristic and the other. And then I take, say, the dimension of k-h-i-j of L. This is the unnormalized Jones polynomial of L. So if you've read the bit about the Jones polynomial already, then you should be thinking, as I define this invariant, how it's related to what you've seen before. If you haven't read it, that's fine. You can take this as the definition of the Jones polynomial, which it is. OK, so how is this invariant defined? So what we do is we start with a planar diagram of our link. And Kovanov tells us how to build a chain complex. Let's call it ck-h of L. Oops, I shouldn't call it ck-h of L, because it depends on D. Different diagrams of the same link will give me different chain complexes. But when I pass to the homology, the answer that I get only depends on L and not the particular diagram that I picked. OK, so I have to tell you how to build this chain complex, ck-h of D. And I'll start with a slogan. So the slogan says that ck-h of D is obtained by applying a certain 1 plus 1 dimensional tqft, let's call it script EA, to the cube of resolutions of D. So this is just a slogan. I have to explain what the words mean, what the cube of resolutions is, what this tqft is, and how I get a chain complex out of it. But the slogan is good to remember, because it encapsulates everything. So let's start with a cube of resolutions. So if I give you a crossing, let's say it this way, there are two ways to resolve a crossing. So every crossing in my diagram D, I can either turn the piece of paper the diagrams on, or if it's on the blackboard, I can turn my head so that it looks like this. And then I can resolve it this way, which I'll call the 0 resolution, or this way, which I'll call the 1 resolution. So if D has n crossings, there are 2 to the n ways to resolve all n. And these are very naturally in bijection with the vertices of the cube, 0, 1, to the n. So maybe let's just do an example. So say my diagram D is this diagram of a hopflink. So it looks like this. So there are two crossings. There's a two-dimensional cube. Even I can draw a two-dimensional cube. It looks like this. Here are the four vertices. So here, I give the cube both crossings a 0 resolution. And I get a picture that looks like this. Over here, I give, say, this one the 1 resolution and the other one the 0. And I get a picture that looks like this. Here, I do it the other way. And I get a picture that looks like this. And up here, I get this. So you'll notice that after I've resolved all the crossings, I have a diagram with no crossings at all, which is just a bunch of circles sitting in the plane. And I'll forget, actually, how those circles are embedded, and just think about them as a one-manifold. So if I have this cube resolutions, so if I have a vertex v of the cube, then to that I see a one-manifold that I'll call vv. OK, so now let's talk about the edges of the cube. So I'll orient the edges so they always point. One coordinate always changes from 0 to 1. I'll draw arrows pointing from the 0 to the 1 like this. And I'll think about an edge, I'll write it this way, e going from v0 to v1. OK, so now what I can do is I can decorate these edges with a co-bordism. Let's call that se from dv0 to dv1. And what is this co-bordism? Well, dv0 and dv1 look exactly the same away from a single crossing. So for example, here as I go along this edge, I see that these two diagrams are the same away from the orange circle. So how do I draw a surface interpolating between these? So let's call this se. So I add a one handle as shown. So I just take a strip of paper, and I paste it along onto these two bits of arcs like this. And when I do that, the upper boundary of the surface that I get is this here. So that's a co-bordism between these two diagrams. OK, now the thing that I want to observe is that if I have a two-dimensional face of the cube like this, so it doesn't matter which order I add the one handles in. I still get the same co-bordism because I'm adding them in different places. So the two-dimensional faces commute. In other words, if I add a one handle here and then add a one handle here, I get the same manifold as if I did it in the other order. So going this way. OK, all right, so now we have the cube of resolutions. It's this cube. All the vertices are decorated by one manifolds, and all the edges are decorated by co-bordisms between them. So now what we can do is we can apply a TQFT. What does that mean? Sort of step two. So a TQFT, I'm going to say a 1 plus 1 dimensional TQFT script DA is a functor from the category whose objects are, let's say, oriented one manifolds and oriented co-bordisms between them. OK, so a picture here, co-bordism S, maybe is the surface here. So this is a co-bordism from the disjoint union of two circles to one circle. And here's a co-bordism from one circle to two circles. I compose morphisms in this category just by gluing the co-bordisms together along their boundary. So here I have a functor from this category to the category of Z modules and Z linear maps. That's what it means to be a TQFT. And let me stick in the word linoidal here, which I'll explain in a few minutes. If you don't know what that means, you can ignore it for now. OK, so what does this do? All right, so now I look at my cube and I hit everything in it with this TQFT. So now over here, I get a group A of dv. And here I get a map. Let's call it A of SE from A of dv0 to A of dv1. And since this is a functor, it's still true that two-dimensional faces commute. OK, so that's good, all right? Because chain complexes are built out of groups and linear maps. So let's build a chain complex. So we can do that down here. So we'll call this step three. So we'll make a chain complex. So the level of groups, I have to tell you what the group CKH of d is. So what I'll do, I'll just take the direct sum of every group that I can see. So I take direct sum over all vertices v, A of dv. And now to tell you what the differential on the chain complex is, it's enough to tell you so what it does on each of these sum ends. So if x is in A of dv, what is dx? It's going to be a sine. So it's the sum over all edges e going from v to v prime. Let's say minus 1 to the sine of e. We'll talk about this sine in a second. And then I just take A of Se and apply it to x. So what does that mean? It means that if I have something in A of this diagram, d of it is something in here plus something in here. If I have something in here, d of it is just something in here. And if I have something in A of this vertex, d of it is 0 for free because there are no edges pointing out. Good question, yes. So you'll notice that there's a, where can I erase? Guess I can, let me keep this for a second. So there's a, let's say, homological grating. Grating of, let's say, x in A of dv is the l1 norm of v. So I just add up all the ones and zeros and see what number I get. So here, the grating is 0. At these two vertices, the grating is 1. And at this vertex, the grating is 2. And so let me pause for a moment. So I remember sitting in this very room 22 years ago, roughly, watching Cliff Taubes give a lecture. And he stopped in the middle of the lecture and said that every talk, you should stop and tell a story in the middle. So let me stop and tell a story for a minute. So when I was, I just finished my PhD. I was giving a talk in Banff. And Yasha Eli Ashborg was in the audience. And I was talking about Kovanov homology. And at some point halfway through the talk, Yasha interrupted me. And maybe there aren't too many of you who know me here. But those of you who know me will know I'm an equal opportunity mispronouncer of people's names. And so Yasha corrected me. And he said that I was saying the name all wrong. And it's not Kovanov homology. It's Kovanov homology. There we go. I still say it wrong. So the point to observe here is that there's a conservation of coves going on. And this invariant isn't Kovanov homology. It's really Kovanov co-homology. Since this is a co-homological grading, right? Gets raised by one by the differential. OK. All right. So we have this grading. And all right. So I guess, yeah, I wanted to talk about this sign. I've told you that we have a chain complex. But in a chain complex, I need to know that d squared equals 0. So why is d squared equal to 0? So say I take d of something in here, or d squared of something in here. What is that? Well, it's what I get by going this way and then this way plus what I get by going this way and then this way. So you see, if I hadn't inserted that sign at all, well, two-dimensional faces commute. So what I get is really just twice the sum of what I get going this way. Now, I suppose I could solve that by working with Z mod 2 coefficients everywhere. But there's a better way to do it, which is to sprinkle some signs, some minus signs, over the edges of this cube in such a way that every two-dimensional face gets an odd number of minus signs. And then d squared equals 0. And so it's kind of a fun exercise, which is on the exercise sheet for this lecture, to figure out why you can always do this and how the sprinkling of signs doesn't really matter. OK. So now I guess what I need to tell you is what is this TQFTA? That's really the only remaining thing that I need to say. So what is A? So again, I need to tell you what A does on objects. So on objects, A of S1 is the free Z module generated by two elements that I'll call 1 and x. So this is a free Z module, but usually I'll just pretend it's a vector space and say things like dimension. And now that word, monoidal, that I had right over there, that's a fancy word, but all that it means is that if I take a disjoint union of n copies of S1, A of this should be the nth tensor product. So let's call this vector space v. It's the nth tensor product of v with itself. OK. And fortunately, that's all the one-manifolds there are, at least all the compact one-manifolds, which is all we get. So I've just told you what A does on objects. So how about co-bordisms? So I can write every co-bordism is a composition of elementary co-bordisms coming from handle additions. And the ingredients that I need are, well, that's adding a two-handle. And here's a morphism, so that's a morphism from S1 to the empty set, v of S1 to v of the empty set. Adding a zero handle goes the other way. So this gives me a morphism from v of the empty set to, oops, A of the empty set. Sorry about that, A of S1. And then I get some more interesting co-bordisms by adding one handles. So I get a co-bordism like this that I'll call m. This goes from, so I guess this is really v to z. And this is z to v. This gives me a map m from v tensor v to v, which is a multiplication that's associative and commutative as it turns out. And also I get a map going the other way, coming from this co-bordism. So we'll call this delta. That takes me from v to v tensor v. And so I need to tell you what these maps are. Well, they're easy to describe, but I should have left myself more space. So I apologize for this. I'm going to allocate another board for this picture. So here, this map here sends 1 to 0 and x. This is a 1 in the vector space v. x goes to the number 1. And this map here sends the number 1 to the vector 1. Now the interesting maps are this multiplication and co-multiplication. So this is v. So m going from v tensor v to v. That sends 1 tensor 1. 1 is the identity element. So 1 sends 1 tensor x and x tensor 1 to x. And it sends x tensor x to 0. So if I look at the multiplicative ring associated with this tqft, it's just z adjoin x mod x squared. Then I also have this co-ordism. That gives me delta from v to v tensor v. And it sends 1 to 1 tensor x plus x tensor 1. And x goes to x tensor x. OK. So at this point, question? Let's see. So here? Yeah, so this is saying there are two objects that I could think about here, 1 tensor x and x tensor 1. Oh, yes. Absolutely. Thank you. Much better. Thanks. OK. All right, so the question? Excellent question. Yeah, that's right. Didn't look like there were any orientations. So conveniently, there's another exercise that explains exactly where the orientations come from. So let's think. You're talking about the orientations on the link or the orientations, the fact that this tqft was, yeah. OK, so let me say the following. So they don't come from the orientations on the link. Instead, they come from the fact that the circles are embedded inside of R3. No, sorry. Let's say that again. Not inside of R3, but inside of R2. These are a bunch of circles in the plane. And there's a natural way that I can orient a bunch of circles in the plane, which is the right way to orient them. OK, all right. So at this point, I've completely defined the homology. You know everything you would need to know to go off and write a computer program to compute kh of your favorite link. Maybe actually, let's make this a little bit more explicit on this diagram over here. So what are the generators that I get? So for example, here, again, I have a pair of circles. So a applied to this has four generators that look like 1, 10, or 1, 1, 10, or x, x, 10, or 1, and x, 10, or x. Here I get two generators, which are 1 and x. Here, again, I get two generators, 1 and x. And here I have 1, 10, or 1, 1, 10, or x, x, 10, or 1, and x, 10, or x. OK, so we have a total of 12 generators, I guess. So now, I said that kh was a bi-graded homology. Where does the other grading come from? Well, so I can define, let's call this maybe five, grading. So I can define a grading on this vector space v by q of 1 is 1, and q of x is minus 1. And then I can extend it to tensor products just additively. Let's call this q twiddle, actually. So q twiddle of a tensor b is q twiddle of a plus q twiddle of b. So for example, over here, 1 in here has grading minus 1. And it's sent to something, oops, 1 has grading 1. It's sent to 0, which I'll say has, 0 is a weird element of a vector space. It has any grading at once. I'll call that 2. x has grading minus 1. It's sent to 1. This is in z. That's grading 0. Similarly, here, this has grading 0. It goes to 1, which has grading 1. So you'll see in these maps, all the gradings went up by 1. But over here, 1 tensor 1 has grading 2. 1 has grading 1. These both have grading 0. x has grading minus 1. This has grading minus 2. Again, 0 has any grading that I want. I'll call it minus 3. Here, this has grading 1. These both have grading 0. This has grading minus 1. This has grading minus 2. What's the difference between these co-bordisms over here and these co-bordisms over here and these co-bordisms here? Well, these have Euler characteristic 1. These have Euler characteristic minus 1. So what's really going on is that the grading Q twiddle of A of SE applied to some element. A is Q twiddle of A plus the Euler characteristic of, let's just call this S. So now I can define a shifted grading on CkH of D. Well, if x is in A of DV, we'll define Q of x to be this Q twiddle of x plus the L1 norm of v. Now as I go along the cube of resolutions, this co-wordism always has Euler characteristic minus 1. And this number always goes up by 1. So this is conserved. So in other words, I see that. So Q twiddle of DX is the same as Q twiddle of x. So that means that I can split as a chain complex, CkH of D. This is a direct sum of chain complexes, say CkH J of D, where this is generated by everything with Q twiddle equal to J. And this is really an honest chain complex here, because D perverse the Q grading. Let's see, yes. So this should, ah, thank you. All right. No wonder everyone's looking confused. All right, should we stop? Are there any more questions? Well, OK. Good. OK, so in particular, I can take the Euler characteristic that I looked at at the beginning. So chi of kh of D, that's the sum of Q to the J chi of khj of D. And that's the sum of Q to the J chi of CkH J of D. And on the other hand, this is the CkH, if you think about it, is really just recording the Kaufman state model for the Jones polynomial. So that's how we get the Jones polynomial on this expression. And again, if you haven't seen the definition of the Jones polynomial, you could just take it to be this Euler characteristic. OK. So now I guess we should ask the question, why do we care? Why is this better than the Jones polynomial, for example? I mean, I guess you can use it to distinguish more knots. But that's almost never a good reason to think that a knot invariant is interesting, because it tells a difference between different knots. So here are two good reasons to care. So one has to do with co-bordisms. So let's suppose L0 and L1 are oriented links in.