 and that phi of e beta lands in t beta. Great. So that's keeping with the convention from yesterday that alphas are red and betas are blue. Well, that's why I labeled e alpha bed and e beta blue. So maybe a cartoon picture of what's going on is that you have sum g and then you have sitting there, you have t alpha and you have t beta. Then you have some intersection point, say x and some intersection point y, and then maybe your disc lands like this. So that's what a Whitney disc is. So we'll let pi 2 from x to y denote the set of homotopy classes of Whitney discs from x to y. So far, I haven't used anywhere that I'm thinking of this disc as sitting in the complex plane, but that will show up. Now, a choice of complex structure sigma induces one on sum g. We can look at the marginalized space, holomorphic representatives of phi, and generically, this is going to be a smooth manifold. I don't really want to get into what I mean by generically. Great. So that's going to be a smooth manifold, so we can talk about its dimension. So we'll write mu of phi to denote the expected dimension of m of phi. Then I will also, w of phi, I'm going to write this to denote the algebraic intersection number between the image of phi and the subspace vw. This is a co-dimension to subspace, so this is going to be a set of points. What else? So there's an r-action on this marginalized space, consisting of complex automorphisms, x plus or minus i. Great. So if this expected dimension is one, we'll then have this r-action, so we can quotient by this r-action. So let's let m hat denote this quotient. Great. Well, if this dimension is one, we'll quotient out by this r-action, or this quotient is going to be zero-dimensional. And this here is a compact zero-dimensional manifold. So in particular, this is a set of points, finite set of points. Now that I've defined all of those things, well, I'm ready to tell you what this chain complex is. So remember, c of hat, this is generated over f by t alpha into t beta. This is just a set of points. And so now I want to tell you what the boundary map is. So the boundary of x. So we'll sum over all the other generators, and then we'll take the sum over phi in pi 2 from x to y. We want mu of phi to be 1, and we want n w of phi to be 0. And then we'll take, right? So this m hat is just a collection of points so we can count them. I guess we're working over the field of two elements, and we'll count them mod 2. And then that's going to be the coefficient of y. So that's what the boundary map is. So in a few minutes, we'll look at an example. But that's what the boundary is. Yes, I'm working on two to avoid signs. You can do this with signs. Yeah. Yes, great. OK, so so far I've defined some map for you. And so we have the following theorem due to Ashraf and Zabo. So if I've been claiming this is a differential, so it better squared to be 0. Great, so d squared is 0. The proof of that is that these points are the ends of a compact one-dimensional manifold. And so if you have a compact one-dimensional manifold, the boundary of that's going to be an even number of points. Or if you orient everything, the sign is going to be 0 points. So that's the proof that d squared is 0. And then the second part is that, well, this actually gives an invariant of the three manifolds. So if h is a Hager diagram for y, then the homology of this chain complex I've just defined is an invariant of y. So in particular, it doesn't depend on any choices that we made. What choices did we make? We chose a Hager diagram. And then we also made this choice of complex structure. But it turns out it's independent of all of those choices. Yes, can I define a z-grading? Yes, that's the next thing I'm going to say. It's almost the next thing I'm going to say. The first thing I'm saying, I already said before, it's that there's a splitting. This chain complex splits along spin-c structures. So there's a splitting on the chain level. And in the problem session, one of the problems tells you how to partition the generators into different equivalence classes. And then you can see that the differential sort of keeps you within your equivalence class. And the second is that there is, so first, let's define a relative z-grading. So relative z-grading means just means, oh, we can talk about the difference, the grading difference between two elements, but maybe we can't pin it down absolutely yet. So if I have two generators. An element phi in pi 2 from x to y. Well, we can define the grading difference between x and y. So this is just going to be mu of phi minus nw of phi. And you can check from this definition of this relative z-grading straightforward to verify that the differential lowers the grading by 1. I haven't yet. That's right. Yes. That is correct. Are there questions? Great. So we have this relative z-grading. And in fact, you can lift it to an absolute grading. So the relative z-grading can be lifted to an absolute. Turns out that it's an absolute Q-grading. So how is that done? Well, I told you that a co-border is in between two, three manifold induces a map between the Hager flow homologies. And so, and in particular, you can sort of understand the grading shift associated to a co-border is a map. And so by looking at co-border is a map, together with a normalization, you can define this absolute Q-grading. So this is from looking at co-border is a map plus normalization. So what do I mean by normalization? I just mean that, well, we define the Hager flow homology of S3 to be in a certain grading. And then the co-border is a map sort of tells us the gradings for any other three manifold. Why is it rational? That's a good question. If you only care about integer homology spheres, it's always going to be integral. Probably. There's probably someone in the audience that can give a better answer to that than I can. Any takers? No takers. OK, well, sorry. So let's look at an example. So let's look at the following paper diagram. Alpha's are red, betas are blue. So we have a, all right, so there's a base point somewhere. All right, so I want to look at sim g. Well, here, g is 1. So sim g just actually, sim g of the surface just is the surface itself. T alpha, well, just is alpha itself. T beta just is beta itself. So the Hager flow chain complex CF hat, well, it's generated by intersection points between T alpha and T beta. There's a single intersection point between them. So this is just a single copy of f. The differential is 0. I guess maybe the easiest way to see that is, well, the differential lowers grading by 1. We only have a single generator. It lives in some grading, so it can't appear in its own differential since the differential lowers grading by 1. Great, so differential is 0. So the homology is just isomorphic to, great. And then this normalization that I mentioned, we normalize this so that this is in grading 0. So that's the Hager flow homology of s3. Another Hager diagram for s3. We'll have, again, this alpha circle. And now this will be our beta circle. Let's put our base point W right here. Great, OK. So now we want to, again, look at intersection points between T alpha and T beta. T alpha just is this alpha circle. T beta just is this beta circle. So now we have three intersection points. So the chain complex CF hat, well, it's f cubed, right? It's freely generated over f by these intersection points. And now I want to compute the boundary. Great, so we want to look for appropriate Whitney disks, say, from C. So if I want to compute the boundary of C, well, I want a Whitney disk from C to some other generator. And in fact, there's one right here. And in fact, a mu of this is 1. So this is going to contribute to the boundary. So the boundary of C is b. And in fact, you can convince yourself that there's no other Whitney disks from C. So in fact, the boundary of C is exactly b. Let's think about the boundary of a. So there's this Whitney disk from a to b, right? Sort of the orientations of the boundary matter, sort of as you leave a, you should see alpha to your right and beta to your left. And you do. What's the problem with this? So remember, in a definition of the boundary, we wanted the intersection between the image of the disk and the subspace vw to be 0. So here, since this is genus 1, the subspace vw just is the point w. And there's, well, and the image of this disk has non-zero intersection number with the base point w. So this doesn't count. So this is 0, right? You might think that the boundary of a should be b. But because this base point sat inside of the image of that disk, it's 0. So this is since nw of the phi i true is non-zero. And then you can also check that the boundary of b is 0. Great. So that's the boundary. And then, well, if you compute the homology of this, well, OK, you better agree with the calculation over there. And indeed, it does, right? Explicitly, well, the kernel is generated by a and b. But b is in the image. All right, so that's cf hat. And cf minus is defined very similarly. So cf minus is really generated over the v in f of join u again by points in t alpha intersect t beta. And so now, well, let me tell you what the boundary math is. OK, so it's going to look very familiar. It's almost the same as before. So again, sum over phi in pi 2 from x to y, again, you want u of phi to be 1. Again, you look at the number of points in m hat of phi. But now, it is something different with the base point. And that's where we use this formal variable u. Now we put u to the nw of phi. So whereas in the definition for the boundary on cf hat, we require the nw of phi be 0. Well, now we allow this to be non-zero, but then we just set that as the exponent of u. Great, and then we again have the theorem due to Ajvath and Zabar that d squared is 0 and that the chain homotopy type of this chain complex is independent of the choices that you made. So independent of your choice of Hager diagram and choice of complex structure. Great, so let's end by computing hf minus of s3. And let's do it. Great, so if you do it from this diagram, well, cf minus is just effigyon u and the boundary is 0. So hf minus is effigyon u, where we normalize a grating, so 1 is in grating 0. Or to compute that from this, Hager diagram. So cf minus, well, we have three generators. This is going to be effigyon u plus effigyon u plus effigyon u. Great, so now it's OK for our disks to cross the base point, but we just record that with the power of u. So over here, where the boundary of a was 0, well, now it's going to be u times b, exactly because of this disk right here that crossed the base point. And boundary of b is still 0 and boundary of c is still b. So if you were to compute the homology of this, well, this is effigyon u if you want to explicitly. So what's in the kernel? b is in the kernel. And then also, if you take a plus ub, that's in the kernel. And then I guess when I'm writing this, I'm writing this as being generated over effigyon u. And so then the image is b. And so, well, then the homology is isomorphic to effigyon u, thought of as being generated by a plus ub. I'll stop there.