 r3, let's make a definition. So a co-bordism, let's call it sigma from l0 to l1, is a smoothly embedded orientable surface sigma inside of r3 times i with the boundary of sigma is l0 times 0 union l1 times 1. So it's hard to draw a really good picture of this, because we don't have four dimensions available. But roughly speaking, you should think that it looks kind of like this, where down here I have l0 and up here I have l1. So if I have a co-bordism like this, then I could look at kt, say that's sigma intersect r3 times t. So usually this will be a knot. There'll be some values of t for which I get a sort of singular thing. But generically, it'll be a knot. And I could take dt as a diagram of kt. So if I want to keep track of what this co-bordism looks like, I could look at this diagram and watch how it changes. And if I picked a generic projection here, and I set up the surface around just a little bit, I'd see that the diagram changed really in one of two different kinds of ways. So let's try and make that precise. So we'll call this, there's a movie. So I take some of these dt's. So my original diagram, d0, is dt0. And then it goes to dt1. And then after a finite number of steps, I get to dtn, where each dt0 to dt1 is either a righto meister move. So let me say what those are. So I could have something that looks like I have an arc. And I add a little kink. Two, I could have two strands. And I could pass one strand over the other, or three. I have, I'm supposed to draw a big. So let's do three over here. Three looks like this goes to this. So what I've done is, let's see if I managed. Yeah, and I've drawn these two wrong, haven't I? Yes, OK, so I should have this and this. OK, so what I've done here is I've taken this crossing here and passed it over this strand, OK? So that's one option. And so theorem of Kovanov says that if d, d prime differ by a righto meister move, there exists a chain homotopy equivalence, say ckh of d, to ckh of d prime, OK? So this is how Kovanov proved that the invariant that you get only depends on the underlying link and not the diagram. OK, so that's one thing that can happen. Let's see, have I managed? Let's see, I've drawn all of these going the wrong way. OK, there we go. Sorry. OK, yeah, OK, good, right. So let me just say here, up to global grading chess, which I'm going to be a bit lazy about. OK, so you get these by looking at the orientation on your link L, but I'm going to be lazy and not say what these are. That's one sort of movie move that I can see. The other sort is that I can have more smooths, where the topology of KT changes. So I could add a 0 handle. I could go from sort of having nothing to having a little circle. I could add a 1 handle. I have a bit of my link that looks like that. I could add a 1 handle along here. And then I got something that looks like this. Or I could add a 2 handle. And so here I have a circle, and I cap it off. Little circle, and I cap it off, and I get nothing. So let's note that if d goes to d prime is a Morse move, then for every vertex in the cube of resolutions, so I get a co-bordism, say dv goes to dv prime. There's a little co-bordism, which is just given by this handle addition. So let's say a of dv goes to a of dv prime by a of s. So this induces a chain map from a ckh of d to ckh of d prime. So both of these types of movie moves give me chain maps between the covena of complexes associated to them. So what does that mean? So it means that we can define, let's say, a map phi sigma. Remember, sigma was my co-bordism from l0 to l1. Find phi sigma going from kh of l0 to kh of l1. So phi sigma is the map induced by the composition of these little individual maps. So let's say phi n minus 1 composed with phi 0 star, where phi i maps ckh of dti to ckh of dti plus 1. Now at this point, there's a question you should really ask yourself, which is, how do I know this is well-defined? So if I give myself a co-bordism, there are going to be lots of different ways to cut it up into a movie. How do I know that I get the same answer? And the answer is that just like there are these Reitemeister moves, which relate different diagrams of the same knot, there's a set of movie moves which relate different movies of the same co-bordism. That's a little bit bigger. I think there are about 23 elements. Kovanov looked at the number 23 and thought that it was big, so he didn't bother to check whether this was well-defined when he wrote it down. But Jacobson did, so there's a theorem of Jacobson that says that phi sigma is well-defined, well, not quite, up to sign. So there's really a genuine sign ambiguity in the way that this is defined. OK, so maybe let's summarize what we have here. Is we could think that we have a category whose objects are oriented links in R3 and co-bordisms in R3 times i. And what Kovanov homology does, well, up to this annoying sign is it provides a functor from this category to the category of Z-modules and Z-linear maps. So really what we've done is we've taken this 2 plus 1 dimensional TQFTA that we started with, and we've upgraded it to a kind of relative TQFT for links in R3 and co-bordisms in between them. So that's kind of interesting. Another thing that's interesting is that we can think about deformations of this chain complex. And let me explain by what I mean by that. So what if we use another TQFT A prime instead of A? Well, the whole complex, the whole construction, with these chain complexes work the same for any TQFT. Really the point where A was needed was to prove this theorem of Kovanov that the answer that I got didn't depend on which diagram that I picked. Now there's another exercise that says that if you want something that's invariant, you better choose dim of A prime of S1 is 2, which means that I should think about deforming the ring, let's say, A, which was Z of x as the multiplicative ring of my original TQFT to another ring, A prime, which is say Z of x mod x squared mod x squared minus Ax minus b. Let me just write down what the multiplication and co-multiplication maps look like for that TQFT. So I have 1 tensor 1 goes to 1, 1 tensor x and x tensor 1 both go to x and x tensor x now goes to Ax plus b. Now remember, this had grating minus 2, this has grating minus 1, and this has grating 1. So whatever I get when I do this is no longer going to have that Q grating on it. It's not going to be a graded TQFT. Similarly, the delta looks like this. 1 goes to 1 tensor x plus x tensor 1 minus v of 1 tensor 1, and x goes to x tensor x plus A of 1 tensor 1. Again, this has grating minus 1, this grating has gone up, this is grating 0, this is grating 2, this is grating 2. So what I see is that on the other hand, minus 1 and 1 aren't equal to minus 2, but they're both bigger than minus 2. And 2 is not equal to minus 1, but it's bigger than minus 1. Actually, what I have here is a filtered TQFT. So in fact, this is essentially the only A prime that I can have given that it's two-dimensional as a vector space. Good, OK, so we're requiring two-dimensional because if you don't, then you immediately see that you don't have an invariant, OK? So that's an exercise that you should do after this lecture. So what should we say here? So A prime is a filtered TQFT with associated graded. So the associated graded TQFT, I get it by throwing out all the terms that raise the Q-grading. OK, so this means that the chain complex that I would build is a filtered chain complex with associated graded chain complex, which is just the original CKH of D, OK? And so that means that I get a spectral sequence from the homology KH of D to this deformed homology, write the spectral sequence this way, KH A prime of D. OK, so now on the other hand, it turns out that this deformed homology is easy to compute. So there's a theorem, which is due originally to leave, but let me also cite Turner that says as follows. So if this polynomial x squared minus ax minus b has distinct roots in some field f, then the deformed homology with coefficients in f always has dimension 2 to the number of components of my link. OK, and the homological gradings are determined by linking numbers of the link l. So in particular, if K is not the dimension of KHA prime of K is 2, and we get a bonus invariant by looking at the filtration gradings of these two surviving generators. So the filtration gradings of the surviving generators in the spectral sequence, well, you can prove actually that they always differ by 2 are, let's say, s of K plus or minus 1. Take that to be the definition of the bonus invariant s of K. So another way that you can think about this, what does this really mean? i.e. this number s of K plus 1 is the maximum value of q of x such that the class of x is non-zero in KHA prime of l. And now, if I have a co-boardism, say, sigma going from K0 to K1, well, this induces in much the same way a map phi sigma from KH of KHA prime of K0 to KHA prime of K1. And this satisfies, so, q of phi sigma of x is bigger than or equal to, since this is a filtered map, q of x plus the Euler characteristic chi of sigma. So now, you can prove that if sigma is connected, phi sigma is an isomorphism in much the same way that you prove this theorem of Lee and Turner. And this means that s of K1 is bigger than or equal to s of K0 plus chi of sigma. So now, if I give you sigma mapping from K0 to the n-lot, it's easy to compute that s of the n-lot is 0. So this equation here says that 0 is bigger than or equal to s of K1 plus chi of sigma. This is minus 2g of sigma. So I get that 2g of sigma is bigger than or equal to s of K1. So you can tell from the noise outsize that I should be done. Let me make a definition state of theorem, and then I'll stop. So the definition says that I could think about what's called the smooth four-dimensional slice genus of K. So this is the minimum of g of sigma such that sigma, let me just, since I'm short of time, let me say it this way, sigma is a co-bordism from K to the n-lot. Then what we've done, plus the fact that s of minus K, so if I take the mirror of K as minus s of K implies that, actually, this invariant gives a bound on this smooth four-genus. So I get that 2g star of K is always bigger than or equal to the absolute value of s of K. So this is an interesting theorem because it tells us that Kh tells us about the topology of smooth surfaces inside of our four. So let's remark. So I could consider g for top. This is the popological four-genus. It's the same thing. But now I just require that sigma is, let's say, locally, flatly embedded. So every point has a neighborhood where the neighborhood is homeomorphic to r4. And the surface intersect the neighborhood looks like r2 inside of r4. This is another reasonable kind of surface that you might think about. And these two numbers are not the same. So I could look at s of that p minus 3, 5, 7 pretzel knot that we had at the end of last lecture. This is 2 and just compute this since Kovanov homology is very combinatorial. But the theorem of Friedman says that g for top of this knot, in fact, of any knot with Alexander polynomial 1 is 0. So this manifold bounds topological disk, but not a smooth one. OK, so I'm going to stop now. Sorry for running over. And yeah, so we'll come back in the afternoon and we'll talk about Kovanov homology for tangles and how that helps us prove Reitemeister invariance. OK.