 V of, let's say, zero lambda bar, so this would be V of, maybe I should call this empty set, the empty label, the empty label here and two, two, minus two, minus two over here. So what do I do? This is, actually I did this way back when we were talking about this definition for the Jones polynomial. So I just take the free R module, spanned by all webs w, this is really isotope classes, w going from x empty set to x lambda. Okay, so that's an enormous, lots and lots of webs. Okay, and then I divide out by a relation, okay, where I say that w naught is equivalent to w one, if so, the bracket of w naught, w prime is equal to the bracket of w one, w prime, for all w prime going from x lambda bar to x empty set. Okay, so these compositions now are closed webs, and I know how to evaluate them. And what I get is some vector space, which is the space V zero lambda bar that I want. Yep, so will I recover the, no. So from this over here? Okay, so yes, okay, all right. So yes, you'll recover the web relations from this statement, right? Because the web relations hold in any closed diagram. Okay, so once you know that they hold in a closed diagram, then if you make this definition, then they, almost for free, hold in an open diagram as well. Okay, all right, okay. So similarly, if I give you some web going from x lambda to x mu, this induces, let me just call it Vw going from V zero lambda to V zero lambda to V zero mu just by sending a web w in here to the equivalence class, oops, bad choice, w naught, to the equivalence class of w naught, w. So knowing how to evaluate closed webs, just in this universal way, lets us figure out what the vector spaces are and what the maps between them are. Okay, yep, are there, oh yes, that's a, yeah. So there are definitely, yeah, it's no longer, right? So when we thought about the Jones polynomial, there was sort of a canonical basis for this space, V zero lambda, say, that was composed of the simple planar tangles, okay? It's unfortunately not the case. There are more, there's not some set of simple planar webs that forms a natural basis for this, okay? So the, yeah. All right, so now we could try to categorify this picture. Okay, and to do that, we should have, for example, a closed web w should go to some vector space a of w, okay? This is like in Kovanov homology, we had a bunch of circles, and we assigned to it the vector space that was the TQFT applied to those circles. And so what I wanna talk about is how, you know, how do you construct this vector space a of w? And similarly, all right, so then a co-bordism, let me put this in quotes, say f going from w naught to w one, say these are both closed webs, goes to a of f mapping from a of w, a of w one, okay? So we wanna understand how we can construct these things. Okay, and so first we need to say some words about what the analog of a co-bordism is in this world. So analog of co-bordism is a foam definition. A foam is a subset. So like a co-bordism, it lives in r times i times i, okay? Okay, now a co-bordism, I could say that it's locally modeled on something that looks like the Euclidean plane inside of this three-dimensional space, right? That's what, you know, manifold is. So here we're gonna have some more local models just because webs have more local models, right? They have these trivalent vertices. And they also have these colors modeled on. So there are three sort of possible local models. Maybe I'll draw them in orange. So one is that locally near a point in your foam, it just looks like a plane, I didn't wanna dot there, I wanted a number with a number, which is the weight, just like the sort of weight that we had on webs, okay? So another possibility is that I could have a trivalent vertex times i, or maybe times, since zero, one, it's kind of an open model. And again, here I need to have sort of A, B, A plus B, a weight constraint on these edges. And now, since I've gone up a dimension, I have to have one more sort of singularity, which is, the third local model is the cone on the one skeleton of the tetrahedron, okay? And on the tetrahedron, I have to label some of these edges. Here I have, say, A, B, C. This is A plus B, I think. This is maybe B plus C, and this is A plus B plus C. Those are the weights on the cones of those edges, assuming I've gotten this right. Okay, and again, okay. The level of detail we're operating at, that's probably sufficient. Yep, yeah, yeah, that's right. So they're supposed to be orientations on all of these things, and I've been sloppy about, maybe I should say, with orientations. And I'm just gonna be sloppy about describing them. So, then we can form what's called the Ricard complex, okay, which is an analog of that complex that we defined in the category Cobb and N, okay? So it's kind of a formal complex in the category of these foams. And what it looks like is we just take the formula that we had for the crossing and turn it into a complex. So maybe here I had K and L, and K was bigger than or equal to L, and then I think I get something that looks like K, L, gosh, I'm confused, no. You know, let me, here, which I have this weight. Here, maybe I'm gonna put this H here. I'm gonna do sort of H is zero, and then go to H is one. Okay, and so forth and so on. And then down to H is L, okay? So I just take the whips that I saw in that sum, I arrange them in this natural order, and then it turns out that there's a nice foam co-bordism from one to the next. Okay, but you know, this is still like that sort of completely formal complex that we had in Cobb and N for the Kovanov homology. Okay, so to really get an invariant, we're gonna need to pass to a quotient of this foam category in the same way that we passed a quotient to get the Barnathan category. And how are we gonna get that quotient? Well, maybe let me just describe how we get, for example, this vector space here, okay? So how to define W when W is a closed web. Okay, so in fact, it turns out there are lots of different ways to describe this category in terms of representation theory, in terms of geometry. Okay, so these categories have actually been pretty intensively studied, but I think maybe the simplest way to describe it is due to Robert and Wagner, okay? And their approach just mimics the Murakami-Atsuki Yamada method. So, okay, before I say that, let me say one thing. So to get a Nodin variant, we must pass to a quotient foam, let's say, so this is the analog of Cobb, foam lambda bar, mu bar, okay? Satisfying some foam relations, analogous to the relations we wrote down to define Barnathan's category, right? There we have the sphere relations and the neck cutting relations. So here there's going to be some really eye-turningly complicated looking diagrams that these foams have to study, satisfy. I'm not going to try and write them on the board, okay? Okay, so how can you find this quotient and see that it's well-defined? Okay, so, Robert and Wagner's approach is to define an evaluation on closed foams. So let's call it bracket of F, where this is now a closed foam, okay? And this is again, as a state sum, kind of the way MOI defined the evaluation of a web, but now I have to choose, you know, on each face of this surface, I have to choose labels from one to N, okay? In the same way that, you know, here I'd have to choose N elements of the set, one up to big N. Okay, and then, you know, again, I have to write down some power of Q associated to that thing, which I won't do, but Robert and Wagner did. So you define some relation, some evaluation on closed foams, check that it satisfies, satisfies the foam relations you want, and then apply the Blanchet-Havegger-Masbaum-Vogel construction to get A of W, okay? All right, so we've got a few minutes left. I had foolishly thought I would talk about what to do in the case where the colors are not exterior, but I think that with the amount of time remaining, that's a terrible idea. So let me say something else that I wanted to say, which is some, you know, why, right, so at this point, the categorifier's job is done. He's defined, you know, some vector space, you know, which categorifies the lambda K polynomial. But what, you know, why would you care? What should you do with this information? And so one thing that I want to say is that very often these categorifications have a lot more structure than the original polynomial, and that structure lets you understand things about the polynomial that you might not otherwise have understood. Okay, so, so, let's say, categorified polynomials extra structure, which lets you see things that, you know, you might not have guessed about the original polynomial. So for example, here's a conjecture of Gukhov and Stosech. So let's say it this way. If K P over Q is a two-bridge knot, then let's say the dimension of H, the wedge K, homology of K P over Q is just the cave power of the ordinary wedge one homology of K P over Q. Okay, this sort of behavior is totally invisible on the polynomial level. At least, you know, does not seem to have been observed before. And I think, you know, one of the most intriguing of such observations is a conjecture of, let's see, Stosech, maybe I'll just write initials again. So Stosech, Kucharski, Sulkowski, and Reinecke, this is just the order their names occur to me. And it says that actually all wedge K Humphley polynomials are determined, right, so according to the definition I'm giving, these things ought to look really terrible, right? I mean, you have to write down this web and work in this huge cube of resolutions and evaluate the web and see what you get and it looks really complicated. So, but nonetheless, the conjecture says that these are determined by a relatively small amount of data. Let me say it this way. A vector space equipped with a quadratic form and two linear forms. So, it's not quite how they phrase this. They think of it as saying that you have a quiver, but I think this is maybe a little bit easier to say. In particular, it's true. This is a theorem of Marcos Tosic and Paul Lendryk that this is true for two-bridge knots. And moreover, that the dimension of this vector space is just the determinant of the knot. You know what that is. It's the Alexander or the Jones polynomial evaluated at minus one, which is P, okay? And you might look at this and say, well, why should I believe, you know, two-bridge knots are super special? And indeed, it's two-bridge knots are super special. So, it's a really interesting question whether this statement is true. There's an incredibly strong statement about these polynomials. It's an interesting question whether this is true more generally. And one reason that you might think or hope that it's true, which was the original motivation of these guys, is that it implies the LMOV conjecture, which is about sort of the integrality of some rearrangement, you know, some generating series that you build out of these wedge K-homphly polynomials. So, if you know that the precise version of this statement is true, then you also know the LMOV conjecture for that knot. So, you know, if you believe the LMOV conjecture, you might hope that, you know, this is true too, and then it's explaining why. Okay, so I think I'm gonna stop now. Question? You know, I think that's, I mean, I think it's at least as hopeless. I mean, you could ask, I mean, is the answer known for all the colored, you know, if you look at all the colored Homphly polynomials, are those complete knot invariants? I mean, is that true? So, certainly the Homphly polynomial itself is not, right, not a complete knot invariant, but I, you know, I thought that if you looked at, like, say, the Homphly polynomial with all possible colors, then it was, it might be open whether that, you know, if that's not true, you should tell me. Okay, yeah. Okay, so I think I don't know these complexes, so maybe you should, okay, all right, this, yeah, okay. Yes, okay, and so, all right, so the question again is, you know, I think, yeah, I think somehow, you know, on the mathematics side, that project has a lot of analysis that's required to go into it, and I think in some sense people have thought quite a bit about the analysis, and maybe not so much about the geometrical, you know, I certainly, you know, don't know anything, but yeah, okay, any other questions? All right then.