 So I'm going to go ahead and start, and so this lecture I'm going to talk about the Jones polynomial for tangles and then start to talk about the Havana homology for tangles. Now if there's one thing that I'd like you to sort of come out of the course feeling like you understand, it's the Havana homology for tangles. So rather than rush through it, I think I'm going to take this lecture, I feel like the last two lectures have been going kind of fast. So I'm going to take this lecture and the next lecture to talk about Havana homology for tangles. All right, so maybe let's just start with some definitions. So we'll talk about tangles. So for each, say, positive integer, let's fix some set xn, which is a subset of r, with n elements. You could think it's like the first n integer points or something, but it doesn't matter much. So now we'll make a definition. So nnm-tangle is a proper smooth embedding, let's call it t, that takes me from a disjoint union of intervals. Maybe I have r intervals together with some number of copies of s1 into r2 times the interval, I'll think about the coordinate on the interval as being given by s, such that the boundary of t, the image of the boundary of these intervals, is exactly xn times 0 inside of r2 times 0, union xm times 0 inside of r2 times 1. So this is maybe x and y. So if I have a tangle, I can represent it by a diagram that I project onto the xy plane. So here's a tangle diagram. So this is a diagram of a 4, 2 tangle, and maybe I'm going to draw for the moment the sort of edges of this strip like this. And then maybe the tangle diagram is something like this. So that's a 4, 2 tangle. And we'll consider the tangles are equivalent if they're isotopic through tangles. So remember that isotopic means that I can find a map of this space here times the interval into here, such that at each time slice, I get a tangle. And in particular, that means that the ends of t stay fixed through the isotopy. Because my requirement is that the ends of the tangle always lie on this sec. So over here, I had x4 was these four points. And over here, I had x2 is these two points. So the ends never move under the isotopy. And maybe let me just remark, this number r here is necessarily m plus n over 2. OK, so this is a similar notion. Similarly, we have planar tangles. That's a map. And we'll call it p, again, from some number of arcs and some number of circles, but now just into r times i. And the boundary of p is xn times 0 union xm times 1. So this is a planar n comma n tangle. OK, so now if I give you a tangled diagram like this, I can form its cube of resolutions, just like I did when I defined the Kovanov homology. Maybe I'll just draw a picture. So say I started out with this 2, 2 tangle here that looked like this. OK, so there are two crossings. The cube is two-dimensional. And what I see here, I give each crossing the zero resolution, and I get a picture. So at each vertex now, I'm going to get a planar tangle. So I'll get a picture that looks like this. Over here, maybe I'll have a picture that looks like this. Here I'll have this, and here I'll have this, let's say. And so now what I can do, well, if I think about the Jones polynomial, when I define the Jones polynomial, I sort of divide it out by the relation where a little circle gave me a q plus q inverse. So I can still do that here. So let's make a definition. So we'll take r to be the ring z adjoin q plus or minus 1. And we'll define, say, v and m. This is, let me, I missed a piece of notation that I want to add. So let me add just a little bit more notation. So sometimes I want to think about the set of tangles as a set. So I'll call that tau n comma m. This is the set of n comma m tangles. And similarly, I could form p nm. This is the same, but with planar. If I want to think about isotope classes of tangles, then I'll write it as I'll put a sort of absolute value around it. So this is isotope classes nm tangles. So I'll take v and m to be, what do I do? I take the free r module generated by a set of isotope classes of nm tangles. And then divide out by a local relation that anywhere that I see a circle, I can replace it, and I can erase the circle and replace it by q plus q inverse. OK, so then it's not hard to see that v and m, this is, again, a free r module generated by isotope classes of, let's call them, simple nm, simple planar tangles, where simple means that there are no closed components. So for example, the space v22 is spanned by a pair of tangles. This is maybe not big enough to see. Let me try and draw a little bigger. Spanned by a pair of tangles that look like this and like this. And so if p is a planar tangle, we'll write square brackets of p to mean the image of p in v and m. So now I can define the analog of the Jones polynomial. Maybe I'll call this the Kaufman bracket still. Let's call this the Kaufman bracket. So let's say that the bracket of d, if d is a planar tangle diagram, is the sum over all vertices of the cube of resolutions. I'll put in a minus q to the l1 norm of v times the image of this dv. So for example, over here, what would that look like? Well, here this is 0. So here I would multiply by a minus q. Also here I'd multiply by minus q. Here I'd multiply by q squared. So if this is t, then the bracket of t is q plus q inverse times, let's just for short call this x and y, times x minus 2q times x plus q squared times y. So in other words, that's q inverse minus q times x plus q squared times y. Yep. Sorry. Yes, x is definitely the thing without the circle. So now the very first exercise that everyone should do in this subject is to see that if, say, r i and r i prime are the tangles for the i-th Ritemeister move, then the bracket of r i is up to some factor of plus or minus q to the k, the same thing as the bracket of r i prime. So maybe I'll do this for the Ritemeister one move, so you see what it means. And then you should check it for the other two. Ritemeister one move, I have to evaluate the bracket, say, of a tangle that looks like this. Well, what is that? They're just two different resolutions. So one, the zero resolution, I think, is, of course, I've managed to draw out a funny way. I think the zero resolution looks like this, this one-dimensional cube. And the one resolution looks like this. So the bracket here is this minus q. Now I can erase this circle at the cost of a q plus q inverse times this. And we see that this term here gets canceled by this q times q inverse. And I'm just left with minus q squared times this. So a nice thing about tangles is that we can compose them. So for the exercise, the exercise is asking you just to compute for these particular tangle diagrams. So this Ritemeister one diagram is a diagram of a one-one tangle. There's a two-two-tangle diagram, which represents the Ritemeister two move, and a three-three-tangle diagram, which represents the Ritemeister three move. And this bracket gives you a recipe for calculating what the bracket of each of those tangles is. And you should calculate and see that you get the same answer on both sides of the equation. So now we can talk about composition. So there's a map, let's say, from tau nm times tau ml to tau n comma l that sends, say, I have a pair t and t prime. It sends this to a tangle, which I'll write as tt prime. So maybe t looked like this, and t prime looked like this. The number of n's coming out on the right has to match out with the number of n's coming out on the left of t prime. And then I just stick them together to get a picture like this. OK, so maybe as a kind of interesting example, let's notice that this means that t, scripty t, nn, set of isotopy classes is a monoid. And I can compose elements in here as much as I want. And inside, there's a group called the braid group generated by the elements that look like this. So I'll call this sigma i, where this is the first, the second, up through the i-th strand here, down through the n-th. And similarly, if I put the crossing the opposite way, I'll call this i-th strand sigma i inverse. And these really are inverses, because if I stick them together, I can separate them to form the identity, which is just this tangle. So the braid group as a group is generated by these sigma i's. And actually, you can show that it has a presentation called this Brn, that's braids on n strands, has a group presentation that looks like well it's generated by sigma 1, sigma 2, up through sigma n minus 1. And we have relations that say that sigma i, sigma j is the same thing as sigma j, sigma i. If the distance between i and j is bigger than 1, this just means you can slide independent crossings past each other. And then there's an interesting relation that says that sigma i, sigma i plus 1, sigma i is sigma i plus 1, sigma i, sigma i plus 1. This here is the right of Meister 3 move for these diagrams. OK. So similarly, composition makes, let's say it this way, lets us define a multiplication, say, from Vnm times Vml to Vnl. That just sends, say, p times pair of p, p prime to the image of the composition p, p prime inside of here. This makes Vnn into an R algebra. And let's make a definition. This is called the temporally-leave algebra. The temporally-leave algebra, Tln, is Vnn with this multiplication. For example, if n equals 2, Tl2 is generated by two tangles. There's this tangle and this tangle. I'll call this 1, because that's what it is. And I'll call this b. And then I see that b squared. So that's this, which is q plus q inverse b. So in general, Tln is generated by tangles that look like this. So I have a pair of these things. And this, again, is in the i-th position. And we'll call this bi. And it has an algebra presentation. So this isn't a group. It's an algebra. So it's generated as an algebra by b1 up through bn minus 1 with relations that, again, have far commutativity. So bi bj is bj bi. If the distance from i minus j is bigger than 1, there's the relation bi squared that we saw over there. That's equal to q plus q inverse bi. And then there's the relation that says that bi bj bi is equal to bi if the distance from i to j is 1. And as an exercise, you're asked to check that this relation here makes sense. So an important feature of these compositions is that they behave well with respect to the bracket. So let's call this a proposition. So if I take the bracket of the composition of two-tangle diagrams, that's an element in v. This is the same thing as taking the bracket of d and multiplying it by the bracket of d prime. And I won't write out the whole proof. But the point is that the cube of resolutions, the diagram dd prime, is the same thing as the cube of resolutions for d times the cube of resolutions for d prime. And so then I just match up the terms at each of the vertices. And so let's remark. So the prop implies that if d and d prime are related by a righto meister move, then bracket of d is the same thing as bracket of d prime up to sort of plus or minus q to the k. And so the proof proceeds in three steps. So the first step is to observe that it's true if d is ri and d prime is ri prime. That was the exercise. It's a calculation. True. It's true if we add maybe some extra straight strands. So if I have the tangle diagram that looks like, here's the tangle diagram for r2. And I add a couple more strands. Well, the bracket of that is going to be the same as the bracket of r2 prime for obvious reasons. These straight strands aren't doing anything in the picture. And so then the third step is, in general, we can factor. Say d is d0, say ri bar, d1. And d prime is d0 ri prime bar, d1, where ri bar is as in 2. And then we just use this proposition here to get what we want. So now the bracket of d is the bracket of d0 times the bracket of ri bar times the bracket of d1. And that's the same thing up to scale as the bracket of d prime. OK. So this is a sort of uncategorified story for tangles. So now we'd like to categorify it and see what happens if we try to build a Havana of homology. So maybe let me, again, tell you the slogan. And then we'll work through what all the pieces mean. So let's categorify.