 you also get an invariant. So the point is there's lots of different algebraic modifications you can do, and as long as you're doing something sensible, well, you're going to get out a non-variant. So now let's talk about the Cronin's formula. So the Cronin's formula is much like the Cronin's formula for three manifolds. So it says that the knot flow complex for the connected sum of two knots is chain homotopic to, well, just take the tensor product over the ground ring of the knot flow complexes for each of your knots. But if you want to work with any of the simpler flavors that are over some simpler ring, you just take the tensor product of whatever ring you're working with. You can just check the algebraically that follows through. All right, so what are some other things we'd like to do with knots? Maybe you take the mirror of your knot. So say if this is k, take the mirror of k. So that's just reversing the orientation of S3. So in terms of a diagram, that's just changing the signs of all the crossings. The knot flow complex behaves nicely under a mirroring in the sense that the knot flow complex of the mirror of k is homotopy equivalent to the dual of the knot flow complex of k. So by dual, I mean F join UV homomorphisms from this complex to the ground ring. And then the gradings do what the gradings do. So let's sketch the proof of this. So let's suppose that h is a Hager diagram for k. So how can we obtain a Hager diagram for the mirror of k? Well, the mirror of k is obtained by reversing the orientation of the ambient three manifold. And then, so if h is the Hager diagram for k, then, well, if we just reverse the orientation of the surface, I guess that was one of your exercises from Monday, which just showed that this reverses the orientation of the three manifold. So this is a Hager diagram for minus k, because reversing the orientation here reverses the orientation of S3. All right, so now we have two Hager diagrams that are very closely related. So in particular, there's a canonical identification between the generators of the chain complex associated to this diagram and the one associated to this one. So we have a canonical identification between the generators, Cfk of h and Cfk of h prime. But now notice we changed the orientation of sigma. So that's going to change the orientation of sim g of sigma. And so now, if you think about what happens, right? So remember, we're counting these disks from x to y. And we've required our disks to sort of c alpha on the right and beta on the left, something like this. So if we saw a disk from x to y in Cfk of h, well, if you reverse the orientation of sigma, now in Cfk of h prime, this is going to be a disk going the other direction from y to x. And so that's going to happen to every disk that we counted, and that's exactly going to give us the dual. So the point is that in pi 2 from x to y, so phi in pi 2 from x to y in Cfk of h corresponds to phi prime going the other way in Cfk of h prime. And sort of going the other way is exactly taking the dual. So that's what happens when we take the mirror. What's another thing that people like to do to not? Well, you can take the reverse of k, which is to reverse the string orientation. So if this is k, then we can take the reverse. So you keep the crossings the same, and then you just change the string orientation. I guess in the case of the trust wall, these are actually the same oriented knot. But in general, reversing the string orientation can change your oriented knot. It did not mean to change the crossings. Thanks. No, the crossings are the same. Thank you. OK, so now what does this do to the not flow complex? And in fact, it actually does nothing. So not flow homology does not see the string orientation. Let's sketch the proof. So suppose we have a Hager diagram for k. So now let's see. If we want to get the reverse of k in the following way. I'm going to reverse the orientation of sigma. I'm going to switch alpha and beta. I'm going to keep my base points in the same order. So let's convince ourselves that this really is the reverse of k. So just in terms of the three manifold, reversing orientation of sigma reverses orientation of s3. But we also switched alpha and beta. And if you switch alpha and beta, that also changes the orientation of s3. So we basically reversed orientation twice. So we kept the orientation the same. But now remember how we formed our knot. We connect w to z in the alpha handle body and z to w in the beta handle body. But now we've sort of switched, which is the alpha handle body and which is the beta handle body. So now, well, here w went to z in the alpha handle body. But in the alpha handle body, now it's going in the opposite direction. So this really is the reverse. As before, in the case of mirroring, there's a canonical identification between the generators. In fact, that identification doesn't just identify the generators. It actually also identifies the differential. Because, well, we reversed the orientation of sigma. But we also switched the alphas and betas. So if you have something, so the point is, if you see, if you saw a disk like this from x to y in simg of sigma, well, now if you switch the alphas and betas, they're right in the blue switch. But if you also switch the orientation of sigma, so now we're going to look at something like this, simg of minus sigma. And so those two switchings cancel out. So we get a canonical identification between the two chain complexes. My choice of Hager diagram to describe the reverse of k, if you think about it, I could have done something else to this Hager diagram, to obtain a w-pointed Hager diagram for the reverse of k. I could have just swapped z and w. If you think about the definition of the oriented not associated to a w-pointed Hager diagram, if you swap z and w to keep everything else fixed, that will reverse the orientation of the not. So this is a Hager diagram of k. Well, if I just swap z and w, this is a Hager diagram for the reverse. So we know from over here that the chain complex associated to the reverse is homotopy equivalent to the chain complex for k. So in particular, well, w tells us what to do with the variable u, and z tells us what to do with the variable v. So what does this observation tell us? This tells us that the chain complex is symmetric under switching u and v. So it's symmetric up to chain homotopy equivalence under switching u and v. So symmetric in u and v, sort of, up to homotopy. So if you look back to our computation for the trap oil, you can see that that indeed is true. If you reverse the vols of u and v, then you also need to sort of reverse the u and v gradings appropriately. But you'll get back the same chain complex. So maybe one question you might have is how do you compute this? So I guess we did one example, the trap oil. And in fact, if you're not has a genus 1 w pointed Hager diagram, then you can just compute it from the definition. So if you use the V-mount mapping theorem I figure out sort of which disks account. And you can just do it. So if k admits a genus 1 w pointed Hager diagram, you can compute this directly. So maybe you're wondering, well, how many knots admit a genus 1 w pointed Hager diagrams? Lots of them do. So for example, all torus knots admit genus 1 Hager diagrams, but they're like a proper subset of all the knots that admit genus 1 Hager diagrams. Well, there's also lots of knots that don't admit genus 1 Hager diagrams. How might you hope to compute the knot flow homology for those knots? So one way is via something called a grid diagram. So these are defined by Manolescu-Ajvat-Sarkar and also Manolescu-Ajvat-Zabointersten. Great. So what is a grid diagram? A grid diagram is sort of a generalization of the Hager diagrams that describe for you on Monday. It's some sort of multi-pointed Hager diagram. So you have a genus 1 surface, and instead of having a single alpha circle and a single beta circle, you have lots of alpha circles, lots of beta circles, and then lots of extra base points, and then you deal with everything and approach the way. And great. So this is some sort of multi-pointed Hager diagram. So the good diagrams is that things become entirely combinatorial. So it's combinatorial, which is great. You can tell a computer to do it. The downside is that the chain complexes that you get are very large. So you have large chain complexes. So you're not drawn in some sort of n by n grid, and the number of generators you get is n factorial. So you have n factorial generators for an n by n grid. And so for example, for the trough oil, you need a 5 by 5 grid. So that's already 120 generators. And then for the figure 8, you actually need a 6 by 6 grid. And then it just gets worse from there. What are the ways that they're to compute not flow homology? So there's a fast algorithm for not quite the full, not flow complex, but the uv equals 0 complex. So this is by Ajrat and Zabo. This is using something called, using bordered algebras. Maybe the caveat, one caveat is I guess I better put quotes around this. Because what's currently available is they define some other not invariant that they conjecture is equivalent to this. And they will prove that it's equivalent. So right now what's written down is that it's conjecturally equivalent to this. And that invariant that's conjecturally equivalent to this, they can compute pretty quickly. And in fact, there's a program on Zoltan's website to compute this. And this is significantly faster than grid diagrams. The complex is that you get a much smaller. OK, so those are the algorithms for computing this. But for other families of knots, you can actually deduce what the not flow complex is from simpler invariants. So for example, the not flow complex of k for k alternating clearly determined by two classical invariants. So by the Alexander polynomial and the signature. For HSK hat, the statement is quite simple. So remember HSK hat is a bi-graded vector space whose graded order characteristic is the Alexander polynomial. And it turns out that for k alternating, HSK hat is supported on the diagonal. So I'm going to use the Alexander and U-grading. So it's supported on the diagonal. m equals s plus sigma of k over 2. So this is with respect to the U-grading and Alexander-grading. And so if you know that it's supported on a single diagonal and you know it's graded order characteristic, that actually entirely determines the graded vector space. So that's one family of knots for which the not flow complex is determined entirely by classical invariants. And then there's another family for which the same statement is true. So if k admits a surgery to an L space, so remember an L space is a Hager flow homology lens space. So a rational homology sphere with the same Hager flow homology as a lens space. Then it turns out that the not flow complex is entirely determined by the Alexander polynomial. So for example, all torus knots admit lens space surgeries. So for all torus knots, you can compute the not flow complex from the Alexander polynomial. And so it turns out that these facts combined with the Cohn's formula actually gives you a large family of examples of sort of interesting not flow complexes. And so that's all I want to tell you today. And so tomorrow, we'll talk about the relationship between this not flow complex and the Hager flow homology of a surgery on your knot. OK, so there's something called bordered flow homology, which is an invariant of three manifolds with boundary. And it has the property, right? It's an invariant of three manifolds with a symmetrized boundary. And then it has the nice feature that if you want to glue together two three manifolds with boundary to get a closed three manifold, you take some appropriate tensor product of the bordered invariant, and that recovers the Hager flow homology of the closed three manifold. And then what AstraZeneca would do is some sort of relative version of that. So they put their knot in bridge position. And then if you put your knot in bridge position, then you can chop it up into pieces that are all either caps or cups or crossings. And then to each of these different pieces that you have, these sort of simple moves, they associate invariant. And then if you want to get your knot invariant, well, you're just gluing these things up. So you're taking some appropriate tensor product of the pieces. And then that gives you back the knot invariant. That's right. So the question is how do these knot flow complexes I described relate to the ambient three manifold? Because you can do the construction. Here we're focusing on knots in S3, but this works more generally for a knot in any three manifold. And then there's a relationship between these invariants and the Hager flow homology of the ambient three manifold. So I guess we saw that if you set V equal to 1, you got back h of minus of S3. Tomorrow we'll actually look, we'll see what happens if we re-invert V so we can tensor with effort during V, V inverse. And then that also, in a suitable way, recovers the Hager flow homology of the ambient three manifold. So basically, you do algebraic modifications with the variables. You think about what that meant in terms of what disks you were counting in the chain complex. And then if you did it sort of in a suitable way, then you'll see that you're actually just counting the things you would have been counting for the three manifold invariant. Yes. So at the end of Monday, I described an algorithm given a knot diagram to produce a Hager diagram for that knot. We started with a projection. And then we built this diagram. And then we put in the right alpha and beta curves. There's a definition of the Alexander polynomial in terms of Kauffman states. So you sort of, it's in terms of a projection. And then you sort of, you decorate your crossings in an appropriate way. And then sort of you, these come with some gradings. And then sort of you sum over states and you get the Alexander polynomial. And it turns out that there is a bijection between these Kauffman states and the generators that come from this Hager diagram. And then it turns out if you also sort of understand the gradings of these generators, they actually sort of exactly line up with sort of the gradings that you compute for the Kauffman states. And right, so if you want to know the graded Euler characteristic of the homology of a chain complex, you actually don't need to know the differential. You just need to know the chain. If you know that it's a graded chain complex and the differential lowers grading by one, it's sufficient to know the underlying vector space together with the gradings. And it turns out that this bijection sort of gives you one proof that the Nauphal homology categorifies to Alexander polynomial. Other questions? Ah, yes. Do all concordance invariants come in from Nauphal homology has the same information as signature? The answer is sort of emphatically no. So there's sort of the first concordance invariant that comes from Nauphal homology is the tau invariant defined by Peter and Zoltan. And the tau invariant, they have a new proof of Milner's conjecture about the unnotting number of tourist knots. In particular, the signature doesn't tell you the unnotting number of tourist knots, but tau does. And then, yeah, sort of I spent a lot of my mathematical life deriving concordance invariants from Nauphal homology and sort of that they can tell you things that signature cannot. For example, they can tell you things about topologically sliced knots, which the signature cannot.