 All right. So now we're going to need to consider some more maps. Okay. So let's let iota sub v denote the inclusion where we invert v. That's a v. We can also do the same thing with u. So that's just the inclusion. Remember, we also have this chain homotopy equivalence between these. That was this v sub s that I wrote right here. So we have this map. So these are homotopy equivalent via v, which will just be the direct sum of these v sub s's. Okay. Great. So we've done a bunch of algebraic things. We have this homotopy equivalence there. And so now I'm going to build a chain map. So let's define a chain map dn. It's going to go from here to here. So it's going to go from the knot flow complex to the one where we've inverted v. Great. So what is dn going to be? Okay. So it's going to be iota sub v. Okay. Maybe you're tempted to just take this composition here. If you want some dependence on n somewhere, so we're actually going to add to that v to the n composed with v, composed with iota u. Great. So here's just some chain map we've concocted. Remind you of some construction from homological algebra. So recall, if you're given a chain map, so remember a chain map, well, it commutes with the differential, you can form the mapping cone of a chain map. So it's a chain complex called cone of f. So the underlying vector space group module, what have you, it's just the direct sum of your original ones. Maybe you want to put a grading shift down here, but I'm going to ignore that. And now, so I've told you what the underlying vector space or module is. So now I need to tell you what the boundary map is. So the boundary of x, y, well, so x lives in x, so you take the x boundary of that, and now this is going to be f of x plus the y boundary of y. And this is, I'm working over characteristic two. So if you're not working over characteristic two, you need some plus or minus signs depending on the gradings of things. But over characteristic two, you don't have to worry about plus or minus signs, so life is great. Well, we have a chain map here, and it turns out that the mapping cone of that chain map is exactly what we want. So theorem. Let's do the Ajvat and Zabo. I guess Ajvat and Zabo stated it for hf plus, and we're doing it for hf minus, so I guess that's stated by Manolescu Ajvat. So this says that, well, let n be any integer, hf minus, so remember I said something about completing coefficients. So we need to work with completed coefficients, so you can just do that as follows. This is isomorphic to the homology of the mapping cone of this map dn. And so as before, right, this is a f of joint w module, and then here we should think of this as being a module of f of joint w instead of u. And in fact, you can actually upgrade this to a statement about graded modules, which we won't do. But remember, you can't upgrade to an isomorphism of absolutely graded modules. And then, well, okay, so we did this for integer surgery, but in fact, there's a very similar formula for rational surgery. So there's a similar formula for rational surgery. Great, and so we've stated this for hf minus. There's also a similar formula for hf hat, right? So sort of the rule for hf hat is, well, just on the chain level, set your variable equal to zero. So we have the following corollary. So n is any integer. hf hat of n surgery on k is isomorphic to the homology of the cone of dn hat. dn hat is just perhaps overall everything from the previous formula. So this is a u. So more explicitly, so basically anywhere where you had your not-floor complex, instead of thinking of it as a module over effort-join uv, think of it as a module over effort-join uv mod uv equals zero. So for example, this map here is a map from to that same complex where we've inverted u, right? So just anywhere where you used to have an ordinary not-floor complex now just portioned by uv equals zero. The proof of this map and cone formula sort of uses the large integer-surgery formula that I stated at the beginning of today combined with the exact triangle, right? So sort of we saw that these maps, iota u and iota v, well, those are really cobaltism maps, right? In fact, they correspond to the cobaltism maps in a certain exact triangle and then there's some homological statement that if you have like an exact triangle, the exact triangle comes from some distinguished triangle and so you sort of have this map and cone statement that goes into some homological algebra. Okay, so great. So we spent this week talking about these invariants coming from, you know, this package of invariants, Hagrid for homology, but you know I'm a low-dimensional topologist and so it'd be nice to know that these invariants actually tell us some new things about maybe some three manifolds. So I want to tell you about some applications that rely on this map and cone formula. So there's something called the sort of cutely named the cosmetic surgery conjecture which says that let K be a non-trivial knot in the three sphere. Our surgery and our prime surgery give you homeomorphic manifolds. This is orientation preserving. The surgery coefficients actually had to be the same. So here these can be rational numbers. And so using the map and cone formula, Ne and Wu were able to prove the following statement. So this is in 2010. So suppose K is a non-trivial knot and suppose that our surgery is orientation preserving homeomorphic to our prime surgery. Well then they proved that R and R prime had to differ by a sign and that they were a rational number of the form P over Q where P and Q were co-prime and Q squared is congruent to negative one mod P. And so a part of their proof was using the map and cone formula to understand what the Hager flow homologies of each of these looked like. So in particular as you vary the surgery coefficient, well you know that these have to be homeomorphic so that their Hager flow groups are the same and sort of studying what that said through the map and cone formula allowed them to get at this result. So they proved that the surgery coefficients have to have opposite signs. Maybe worth remarking that Ne and Wu's result was recently improved within the past month or so by Jonathan Hanselman. So he recently improved that result to actually significantly improved it. So R and R prime differ by a sign and they have to either be plus or minus two or plus or minus one over Q. And so he did this using bordered flow homology which is a generalization of Hager flow homology to three manifolds with boundary. And in particular, Jonathan Hanselman, Liam Watson and Jake Lesmeson have a reformulation of bordered flow homology in terms of immersed curves and that's what he uses to get this improvement. So let me give you one more example of an application of the map and cone. So this is an application to surgery obstructions. So that's a theorem which I understand that the undergraduate summer school is currently proven due to licorice and wallace which says that every closed oriented three manifold can be obtained by surgery on a link in S3. So once you know this fact well the natural question is well can everything be maintained by surgery on a link? What about surgery on a knot? And so in fact, well the answer is no, right? So can every three manifold be obtained by surgery on a knot in S3? And so the answer to this question is no. So note that H1 of P over Q surgery on K is cyclic. So in particular, if you have a three manifold of H1 knot cyclic, so for example RP3 connects some RP3 this cannot be obtained by surgery on a knot because its first homology is not cyclic. Okay, the only look at three manifold in which this obstruction vanishes i.e. what if you only look at homology spheres? Can you obstruct a homology sphere from being surgery on a knot? And in fact one can. So this is first done by Dave Ockley. So he showed that there exists integer homology spheres that are not obtained by surgery on a knot in S3. So he gave a hyperbolic example and a toroidal example and in fact he can generate his technique is infinitely many such three manifolds. But then using the map and cone formula one can give a new proof of this result and in fact show that there are these homology spheres that are not obtained by surgery on a knot. So this is joint work with Charakara Kurt and Ty Lidman. So he proved that the integer homology spheres sigma 2k 4k minus 1 4k plus 1 for k greater than or equal to 4 are not obtained by surgery on a knot in S3. And so he proved this theorem by coming up with a Hager flow obstruction to being surgery on a knot in S3. So using the map and cone formula we show that if you're a three manifold that's obtained by surgery on a knot in S3 there's a particular relationship between your D invariant and your reduced Hager flow homology. So I'll stop there.