 the same picture, really. But here, I know that I can simplify this by getting rid of this circle. And then when I do that, I'll get q squared because I had a factor of q here times this plus one times this. Okay, and now this isomorphism here, we wrote it down before. It's the matrix that looks like a to x, a to, okay. Now this is supposed to be an isomorphism of chain complexes, so it should be a chain map. So this map down here has to make this square commute, but that's easy to do. I just pick a to x, s and a to s. All right, and then I claim this is an isomorphism of chain complexes. Okay, but you know, a to s, what is this? This is a one-handle addition followed by a canceling two-handle. Okay, so I added this one-handle and then I capped that off. Okay, that co-bordism just cancels to give me the identity co-bordism. Okay, so another way that I could write this complex is that it's isomorphic to, I have this one object here. It goes to q squared times this plus this. And here the map is multiplication by x. That's adding a thought and the identity. So now we're almost done, but we need a really important lemma. This lemma is called the cancellation lemma. Sometimes this is also called Gaussian elimination. Okay, it says that if, say, i going from b to b prime is an isomorphism, then, now I'm gonna write down a big chain complex that looks like cn plus one goes to cn plus b goes to cn minus one plus b goes to cn minus two. Okay, now I'm gonna write down the maps in this chain complex. So let's call this dn plus one. This I'm gonna put a question mark because actually I don't care what it is. I'm gonna call this alpha, beta, gamma. And here I'm gonna use the isomorphism Yoda. And here I have dn minus one and, again, something that I don't care what it is. Okay, so here's a chain complex. This chain complex is homotopy equivalent to the chain complex that looks like cn plus one goes to cn, goes to cn minus one to cn minus two. And here I have the obvious things, dn minus one and dn plus one. And in the middle here, I should have written a smaller homotopy equivalent sign. I have alpha minus beta Yoda inverse gamma. Okay, so this is a single most useful fact about homological algebra that you won't find in a standard algebraic topology textbook. It's really handy. If you've never seen it before, you should prove it during the exercise session. Why would I call this cancellation? Well, for example, a common situation which this might apply is if you have a Morse complex. All right, so maybe you know that if I have a Morse complex and I've got two critical points and a unique flow between them, then I can cancel those two critical points to get a Morse complex with two fewer critical points. Okay, that's exactly what's going on here. Okay, this Yoda here is the isomorphism between those two points. And this complex here is what you would get in your new Morse complex after you perform that cancellation. Okay, but back over here, what does that mean? It means that I have a one here. This is an isomorphism. So that means that I can cancel this object with this object. And I'm left with just nothing goes to Q squared times this which is let's say CKH of this tangle diagram up to shifts. And these shifts are the same shifts that you should have gotten when you did the exercise yesterday. Let's try the right of Meister 2 move. Be a little bit quicker with that, but it's worth looking at. So, okay, so now say I look at CKH of this diagram here. This is a right of Meister 2 diagram. Let's just draw the cube of resolutions. So down here, I'm gonna get a diagram. I give this the zero resolution and I have to be very careful here. This crossing does not look like this one. I have to make sure that I turn my head 90 degrees when I give this the zero and the one resolutions. So down here at zero, zero, for example, I get something that looks like this. Over here at zero, one, I get something maybe that looks like this. Up here, I get this. And over here, I get something that looks like this. Okay, and now I'm gonna simplify this object, but before I do, I wanna look for a second at this complex here. You see this orange complex is really nothing other than CKH of this diagram right here, okay, which is also a right of Meister 1 move. I think it's actually the right of Meister 1 move we did right over there. So, wait, no, this is totally wrong. I've circled the wrong thing. I'm sorry. Let's try this again. How's that look? That looks more like this complex. Okay, and this complex over here looks like CKH of this diagram. Okay, so that's good. I guess I only computed, this is kind of the other right of Meister 1 move, but maybe you'll believe me that the differentials look kind of similar. So, in this complex, what do I get? I'm gonna get something, oh, and I didn't write in any Qs, did I? Q, Q squared, Q. So, here I get something that looks like this goes to, let me write this as Q plus Q inverse times this. I was parenthesizing these, wasn't I? This goes to, oops, but there's a factor of Q, so actually this is Q squared plus one. This goes to Q squared times this, and over here I get Q times this. And if you work out these matrices, well, these matrices we worked out before, this is X and one, and over here, I think, and maybe let me mark, let's call this, remember in this picture, we're sort of operating on the right-hand angle, so this is X right, it's multiplication by adding a dot on this component. And similarly here, I think you'll find that this is X left and one, no, one and X left. One and X left, okay? And again, we see that we have these entries that are isomorphisms, so I can cancel this with this and this with this, and I'm just left with this, which is CKH up to a shift of the other right-and-meister diagram, okay? All right, so you can imagine you can do the same thing with a right-and-meister three move. I won't do it here. So proposition, so if D and D prime are related by a right-and-meister move, then CKH or D, let me write a twiddle here that means chain homotopy equivalent possibly with a shift is the same as CKH or D prime, okay? And the proof is basically the same as for bracket, so we check for individual right-and-meister moves, so these very particular diagrams, then we note that if we add some straight edges, it doesn't matter, and then we factor, right? And then we use the fact that sort of C of D D prime is C of D tensor C of D prime. Let's maybe do one more quick example. So we looked at, let's take a interesting angle now, something where I have two crossings going the same way. Okay, well this is going to look, how's it going to look now? I think what I'm going to see is, I'll see this down here going to one of these is long, over here the other one is kind of long, and up here I get this, and I have factors of Q and Q, oops, Q and Q squared, so what is that isomorphic to? So now it looks like Q inverse plus Q times this, this goes up to Q times this, this goes over to also Q times this, and both of these here go over to Q squared times this. These are saddle morphisms, one-handled addition, and these maps here, look, let's see, over here this was the thick side, so this here is probably something like, oops, this is a matrix that goes this way, isn't it? So this is X left in one, and this is also a matrix that goes this way, this is probably X right in one. Okay, and now you see that I can cancel this object, well I could have canceled it with this or with this, I could have done either one, doesn't matter, but I can't cancel this one away. So this complex here is homotopy a complex equivalent to a complex that looks like Q inverse times this goes to Q times this, goes to this, whoops, with a factor of Q squared, this is a saddle morphism. If you work out what this is, it's the difference between XR and XL. Okay, I'll leave you to convince yourselves of that, but let's at least check the D squared equals zero. All right, see here XR and XL are different morphisms because the thought is on either the left-hand side or the right-hand side, but once I compose with this saddle, I get a connected co-bordism from here to here. Okay, and I can move the dots around, so they're the same morphism on here, so I really do get D squared equal to zero. Okay, so in the exercises, you're asked to do the same thing for a similar braid and I'd advise you that rather than writing out a n-dimensional cube, it's better to take this complex and tensor it with one more crossing like this, which you can do by this relation here. Okay, so now I'm gonna make one more statement and then I'm gonna stop and talk about the big picture for a couple minutes and then we'll be done. Okay, so the remark that I wanna make, maybe let's call this a proposition. So if I take CKH of D, this is homotopy equivalent to what's called a minimal complex. So no components of D are the identity. Well, that's easy, right? I just use the cancellation lemma. If I see a component of D that's the identity, it means I can cancel and get something smaller and I just keep on doing that until there's nothing left to cancel. But moreover, any two such minimal complexes, so minimal complexes, homotopy equivalent to CKH of D are actually isomorphic, okay? So I can think about, I don't have to think about sort of abstract homotopy equivalents. If I wanna tell if two tangles, two diagrams give me the same chain complex, I just cancel down to a minimal complex and look and see if they're isomorphic as complexes. So CKH of T1, which is by definition this minimal complex, maybe let's say makes sense. Okay, so the minimal complex that I get doesn't depend on the diagram of the tangle that I picked, okay. So now I seem to have a few minutes left, so let me indulge in a little big picture discussion. So if we go back to the Kaufman bracket, what I see is that if you give me a pair xn and xm, I get a vector space, v and m. This is actually the same as vn plus m. So the picture here is that if I have, for example, a simple planar tangle that looks like this, I just glue these two sides together and say that I could stretch this out, oops, I should have said maybe vn plus m comma zero to look like this. And a tangle t going from xn to xm gives me bracket of t, which up to some shift factors that I've been lazy about is a well-defined element in here, okay. So really what I have is I have sort of a functor from the categories objects or these xn's and these morphisms or tangles over to the category of vector spaces and linear maps. So this is sort of a relative tqft, okay. And so what does Kovanov homology do? Well now, so now Kovanov homology lets me go from xn, the pair xn, xm to a category, say this is really the category of chain complexes of CBN and m, okay. And a tangle t from xn to xm gives me an object of this category, CKH of t in here. But now I can go one step further. So if I have a co-bordism sigma from t0 to t1, okay by the same kind of movie move business that we talked about yesterday, I can use this to get a morphism from chain map from CKH of t0 to CKH of t1, all right. And so if you think about it, what I really have here is sort of a two-funkter between two categories, all right. So for example, the fact that this thing behaves well with respect to horizontal composition as part of the statement that that's a two-funkter, okay. And really what categorification from a topologist's point of view is about is it's taking say a relative two-plot plus one-dimensional tqft and turning it into an extended relative, let's call it two-plus one-plus one-dimensional tqft, okay. That's sort of the whole goal of the categorification game from a topologist's point of view. So it lets you upgrade this kind of tqft to this kind of extended tqft, okay. So there are lots of, yep, yeah, great question, right? So that's the million-dollar question. So the answer is no, certainly not that we know how to do. And so Riemann surfaces times r, even there it's difficult, right. So what you're really talking about is sort of categorifying the skein module of that Riemann surface. And I think there are partial results in that direction, but it's not the case that everything is happy and obvious. But yeah, so maybe since you asked the question, let me remark that fluoride, so this picture here was about the Jones polynomial, right. And so the one case where we do have a story that extends like this is in fluorohomology, okay. And that's in some sense related to the fact that the Alexander polynomial has this really robust definition, right. The Alexander polynomial is an invariant of pi one, okay. So it's easy to define the Alexander polynomial of a nought and a three manifold. I'll just look at pi one of the complement, okay. So similarly here, the one place that we do have this picture extending is with the Alexander polynomial where the extension is given by fluorohomology and not fluorohomology, and by maps induced by not fluorohomology by co-bordisms, which were studied by Mjuhas and Zemke and Marengen, for example. Yeah, but it is, right. So it's a central problem and it's been a problem for a long time as to whether this picture extends into the world of three manifolds. And I think in some sense there are interesting recent ideas in physics, so I think Pavel Putrov will talk next week about physical ideas which are perhaps, you know, offer a way to think about extensions like this, no promises, but you know, yeah. So we do not know how to do this for tangles in an arbitrary three manifold. Okay, and I think it's lunchtime. So I'll postpone the rest of the big picture to the start of the next lecture. So the other part of the big picture is to say a word or two about general quantum invariance, but I think that'll go well at the start of next lecture.