 So for this last lecture, I'd like to talk a little bit about colored polynomials and colored homologies. But before we start doing that, I'd like to start with looking at what we were thinking about in the last lecture with a slightly different perspective. So let's take d is a oriented link diagram and say colored everywhere by the vector representation v of sln. So how do I compute bracket of d? So that'll be a polynomial. And remember what we saw in the last lecture is that we have a relation that looks like the bracket of this is q times the bracket of this minus the bracket of this little diagram with a thick edge, which corresponded to the element say bi in the Hecker algebra. So really what I'm doing here, if I apply this relation, is I'm forming a cube of resolutions. So maybe let's just imagine that our diagram d was a familiar diagram of a hopflink. OK, then we've got two crossings and I'll have a cube of resolutions that looks like this. So here I give the zero resolution to both things. So I guess this is in my notation the zero resolution and this is the one resolution. So here I get a picture that looks like this. Here I get a picture that looks like this. Here I have the same thing but with the thick edge on the downside and here I have two thick edges. OK, and you should notice that this is very similar to the picture that we had back when we were thinking about the Jones polynomial. So I guess really maybe let's make a definition. So in this picture I have these things with these fat edges in them. Really the way you should think about this fat edge is it's sort of a weight two edge as opposed to these thinner edges which are weight one. So let's make a general definition. So a web is a planar oriented weighted trivalent graph with zero flux. I won't explain what that means in a second. The weighted, so the weights should be thought about in the natural numbers. OK, and the condition that there's zero flux just means that if I have something that looks like here I have weights a and b, here I get a weight a plus b. Or here if I had weights a and b this would be a plus b. OK, and in particular all of these pictures here are webs where the thick edges have weight two and all the other edges have weight one. OK, so in the cube of resolutions really what I have is that a vertex v of the cube goes to let me call that wv a web. OK, and then the bracket of d is the sum over all vertices. I guess the way that I'm doing this I should put a q squared here, a minus q here, a minus q here, and a one here. That's just coming from this factor of q here and the minus one here. Let me write this as q to the n, where n is the number of crossings of d. So q to the minus n here I'm going to have minus q inverse to the l1 norm of v. And here I get the bracket of this web wv. OK, that's really what this formula over here is saying. And one thing that I should have pointed out in the last lecture is that there's a really unfortunate convention clash between Kovanov homology and SLN or Humphley homology, which is that the q's go the opposite way. OK, you notice here I have a q inverse. Yeah, that's somehow to do with the fact that in polynomial rings you like to say that x has grading plus two, but the x in Kovanov homology, you want to say has grading minus two because it's like an Euler characteristic. But so you should be aware of this problem. OK, well, but here, now what do I do? You see, before when I was doing the Jones polynomial, it was pretty clear what I did. I just counted the number of circles and took q plus q inverse to that number. And that was how I evaluated these diagrams. But now I have diagrams with these thick edges. It's not quite as clear what's going on. So let's say it this way. To evaluate bracket of WV, we use what are called the MOI. That's for Murakami, Otsuki and Yamada not Morofko, Oshvath and you. Which caused me some cognitive dissonance in my youth relations. OK, and so what are these relations? Well, they say that if I have a circle, I can replace this with quantum n. If I have a way two strand here, let me try and draw this a little bit more clearly. So if I have this, I can get rid of this little bubble over here, but at the cost of introducing a factor of quantum n minus 1. That's that times this. OK. And now actually, where do these relations come from? These relations are exactly the relations for the Oknianu trace that I wrote down in the last lecture. This is the condition that trace of Yoda of sigma was squiggly 0 times trace of sigma. And similarly, there was another condition here that had Yoda of sigma times bn. And that's this. And notice that if I take squiggly 0, that's a to the a minus a inverse over q minus q inverse and set a equal q to the n. And I get quantum n. OK. And so now there's some other relations. So maybe let me write down one more. So if I have a little bubble like this, that's the same thing I can pinch the bubble out at the cost of inserting a quantum 2. That doesn't come from the trace. That's a relation in the Hecker algebra. That's the relation that bi squared is quantum 2 times bi. OK. Now there's one more diagrammatic relation that I won't draw the diagram for. But it just corresponds to the relation that bi bi plus 1 bi minus bi is equal to bi plus 1 bi bi plus 1 minus bi plus 1, which is the other interesting relation in the Hecker algebra. OK. So we can use these relations here to evaluate the bracket of any of these webs. OK. So for example, maybe let's just do the most complicated one. So if I take the bracket of that thing over there, close it up. Well, I can pinch this out at a cost of quantum 2. And then I'm left with the bracket of this kind of diagram here. I can kill this circle at the cost of a quantum n minus 1. And I'm left with just a plain circle. And then that's this, which I think it's really better to write as quantum 2 times squiggly 1 times squiggly 0. OK. All right, and see, actually, we've evaluated this one as well on the way. So we could now work out exactly what these things are. OK. So now we'd like to talk about colored polynomials. And it turns out that the situation is much better for colors that live in colors that are exterior powers of the vector representation. So for now, let's consider coloring my diagram d with, let's say, lambda to the k of v, where again, v is cn, is the vector representation of sln. OK, now we want to ask, how can we compute in this situation? So for example, I just erased this, unfortunately. So what's the analog? So here, let's first remark. So we'll label so strands with this number k, which is a natural number. OK, so for example, the bracket of this, where I label both of these strands by 1, this is the vector representation on both of them. We saw before that that's q times the bracket of this. Both of these are labeled by 1 minus the bracket of this, where this is a 2 and these are 1s. OK, so now I could ask, what's the analogous formula if instead of 1s, I had 2s here? So each of these are now labeled by the second exterior power of the vector representation. Well, maybe let's just think for a moment. I'm going to tell you an answer, but OK, so it's an exercise. To see that if I look at wedge 2v, tensor wedge 2v, this has three irreducible summands. So that means that the vector space that I associate to a tangled diagram like this, where now this is a 2 and this is a 2, this is a 2, this is a 2, this thing should have dimension 3. And in fact, what I can do is I can express what this is as a sum of things in a three sort of think about these two things are basis vectors for a two dimensional space. I'll write down some basis vectors for a three dimensional space, which are like this. So this is q squared times here, here I put 2 and 2, minus q times, you know, all right. I'm going to write this again big in a second, so I apologize if it's hard to read. So here I'm going to have a web that looks like this. So the weights at the bottom and the top have to be 2. I put a 1 on each of these, so here I get a weight of 1 and 3. And then minus, sorry, plus the bracket of now the thing where here I have a 4 and these are each 2. All right, and so in general, this is an incidence of a big cube over, you know, so let's say this. In general, I have the bracket of, say, k, l. And here let's assume that k is bigger than or equal to l. This is given by a formula that looks like the sum from h equals 0 to l of minus q to the l minus h. And then I have a web that looks like this. These are usually called ladder webs. So here I have k, here I have l, here I have k, here I have l. I've drawn, no, I've drawn this wrong at the bottom. This was k over here and this is l over here. Now I carry each units of mass over this way and then carry them back over here. You can figure out what all the other entries are. So I think not standing up here, though. We're just going to take this. So I'm going to explain a little bit about how you could work with this. But I think maybe I'll make a comment in a second or two. All right, so now if you give me my diagram labeled by these exterior colors, I can apply this relation and evaluate its bracket as a sum of a huge number of different webs summed over this. Now it won't be a cube of resolutions, but rather a sort of cuboid of resolutions where the sides have possibly different lengths depending on exactly what colors show up here. So I get some gigantic sum. But now I still need to be able to evaluate what the bracket of a web is. OK, so the bracket of a web is still determined by MOI relations. They're just more of them. I'm not going to write down all of these relations, but maybe just for illustrative purposes, let me write down one. So for example, there's a relation that says that if I have a bubble, kind of like the one that I drew before, so here maybe I had k, l, and then these were both k plus l, then I can get this to be, so k plus l choose k times the bracket of this thing just labeled with k plus l, where bracket k, l is the quantum, right? So I talked about this a little bit at the beginning of the last lecture. So what we have here is all of these, and I've somehow managed to turn my tangles 90 degrees to where they should have been relative to every other picture I've drawn. I'm sorry about that. Yeah, these tangles should be turned 90 degrees. But we said in the last lecture that if I have a bunch of points labeled by representations here, then I get a vector space, and the bracket of my tangles lives in that vector space. That vector space was called, say, v lambda bar mu bar, where the lambda bars and the mu bars were these representations. And the dimension of this is the same as the dimension of harm of the tensor of the lambdas to the tensor of the mu's. OK, now, yes, OK. This picture, yes. It's maybe better to look at this picture over here, which is drawn a little bit bigger. So we have a trivalent graph. So for example, this would have weight L minus H. Yeah, so some of the weight gets carried over to here, and then it comes back. No, this is an H up here. Maybe that's why you don't understand. I need to get K up here. So this is K minus L plus H or something like that. OK, knowing that this weight is H, you can work out all the rest of them. And remember, the point is these are all webs. OK, so for negative crossings, good question. So there's a similar formula. I'm not going to write down all of the formulas, but there's a similar formula. One of the things about this subject is that if you want to show everyone all the relations, you pretty much have to put them on a bunch of computer slides. But I think rather than trying to write down all the relations, I just want to point out some of them. OK, so we were looking at this quantum binomial coefficient. So what does that mean? That means that this is, well, it's like a regular binomial coefficient. It's quantum K factorial over quantum. This was K plus L, huh? Let me just K plus L factorial over K factorial. And quantum L factorial means you take quantum L, multiply it by quantum L minus 1, all the way down to quantum 1. In particular, if you set q equal to 1, you just get the regular binomial coefficient. And similarly, another relation that's kind of nice is that if I have just a K-colored circle, the bracket of this is quantum N choose K. All right, so now I've told you that Murakami Yatsuki and Yamato wrote down a whole bunch of relations that let you evaluate the bracket of these webs. But now there's a real problem if you phrase it that way. Here's the problem. I can write down these relations. But how do I know the relations are consistent? In other words, I can use these relations to simplify my web until eventually I get down to the empty web and I get some polynomial in the q's times 1. But maybe if I applied the relations a different way, I'd get a different polynomial. That could be bad. So how do I know that that's not the case? Well, Murakami Yatsuki and Yamato solved this problem in the following way. What they actually did to find bracket of my web is, I'm going to say it's a sum over A, which is a state of w, q to the, let me say, f of A. And now this expression makes some things very clear. So one thing is that this is going to be a polynomial of positive coefficients, which looks really promising for categorification. Now I should explain a little bit about what the states are. I'm not going to tell you exactly what this function is. You can read their paper, which is very nice. But the states, where a state of w is a function, let's call it A, from the set of edges to the set of subsets of the numbers 1 up through n, such that, 1, the size of this subset is determined by the weight on the edge. So if E has label k, then the number of elements in A is exactly k. And 2, there's a flux condition, which you should think about as generalizing the sum condition that we had before. So for example, here, if I had something like this, here I have A of E1, here I have A of E2. And well, this has size k, this has size l, let's say. And here I'm supposed to get something of size k plus l. Well, there's an obvious thing I can do. This should be A of E1, union A of E2. That works as long as these are disjoint. If these weren't disjoint, then this wasn't a legitimate state. And so now, for example, you can see that a lot of least, I haven't told you what this function f of A is. But if you look over here and set q equal to 1, a lot of these relations look very natural. So for example, this says, on this closed circle, I just need to pick k labels in that set 1 through n. That's that n choose k. Over here, I have a set of k plus l labels. And I need to sort of split them into two parts, one of size k and one of size l. That's that k plus l choose k. So you explicitly, Murakami and Yomata, explicitly write down this function f of A. And then they checked that bracket satisfies the MOI relations. And indeed, one way to think about this is that this gives us another construction of the Humphley polynomial. You define these functions. You check that it satisfies the relations that you need. And then you prove that the Humphley polynomial exists. OK, so that's a formula for closed webs. I should have made that clear. This is for closed webs. What do I do if I have an open web? What's the vector? So how do I figure out what the vector space that my web lives in? OK, so for example, how do I determine maybe v of this thing where, say, here I had two things labeled by 2 and 2 by minus 2 and nothing over here? And if you think about it, that's also naturally identified with the vector space v, where I've got two points labeled by 2 and two points labeled by 2 over here. Well, there's a universal construction. And again, I'm going to say these people's names and write down their initials. Due to Blanchet, Haberger, Mazbaum, and Vogel that operates as follows. So define v of this thing.