 So welcome back to day two of Hager-Florhomology. So what did we do last time? Last time we defined Hager diagrams, and we proved that if you have two different Hager diagrams for the same three manifolds, then they're related by a sequence of isotopes, handle slides, and stabilizations. And so while we pointed, we noted that if you wanted to define a three manifold invariant, well, you can just define one and you can define some invariant of a Hager diagram. And then as long as that invariant doesn't change under isotopes, handle slides, and stabilizations, well, then you've got yourself a three manifold invariant. So that's what we're gonna do today. So goal, so from a Hager diagram H, build a chain complex, we'll actually build two different chain complexes. So the first will be called C of hat of H. This is gonna be a finitely generated, graded chain complex over the field of two elements. I'll also build a chain complex CF minus, which is a finitely generated, graded three chain complex over the ring F of join U. Our U is just some formal variable, U has degree minus two. And then while it turns out these chain complexes, the chain homotopy type is gonna be unchanged under Hager moves, so in particular, the homology is gonna be invariant of the three manifold described by H, so, homology. So H of hat, that'll be the homotopy type. So H of hat, that'll be the homotopy type. The homology of the chain complex CF hat, or H of minus, which will be the homology of the chain complex CF minus is an invariant of Y. So here H is a Hager diagram for Y. Oh, and so all of this, basically everything that I'm gonna say today is due to Algevof Sabo. So I'll write that once right here, but then I probably won't say it every time. Basically every statement I say is due to Algevof Sabo today. They owned, right, so there's these two different flavors, and if I'm gonna, sometimes I'll talk about sort of, I'll say some fact that's true for either of them, and so sometimes I'll write this little circle where a circle can mean hat or minus. There's some other flavors plus and infinity, but they can be determined from, actually from H of minus, so sort of knowing H of minus is sort of, I'm gonna tell you everything that you're gonna wanna know. All right, so before I give you the definition of these chain complexes, I wanna tell you some properties and sort of what structure they take. So let's talk about some properties of these invariants. Great, so the first is that we have the following splitting. So they split over spin C structures on Y. So it just means that, well, it's a direct sum of pieces like this. Okay, and if you don't know what a spin C structure is, well, it's good enough for our purposes to think of them, they're in one-to-one correspondence with second cohomology classes, or if you like first homology classes. So C of minus, C of hat is a finitely generated graded chain complex over F, so in particular, it's homology is also going to be finitely generated, graded vector space. Great, and so we'll know what a finitely generated graded, finitely generated vector space looks like. It just looks like F to the N for some N, and so then in each grading, this just looks like F to the N for some N. So that's what that looks like. Say that again. Do I mean, oh, so the homology splits as a direct sum, and in fact, the splitting comes from the splitting on the chain level. Yeah, so it comes from a splitting on the chain level. So I could have made a statement for either HF or CF. That's right, wait, okay, and then HF minus, right? So it's the homology of a finitely generated graded free chain complex over F, R, and U. So that means that, well, the homology is gonna be a finitely generated graded module over F, R, and U. F, R, and U is a PID, so there's a fundamental theorem of finitely generated graded modules over the PID, which is in flavor exactly the same as the fundamental theorem of finitely generated abelian groups, which says that every such group is a direct sum of free parts and torsion parts. And so you have the same statement here that every finitely generated module of a PID is a direct sum of free parts and torsion parts. We're here by torsion, we mean U-torusion. So what am I saying? I'm just saying that any finitely generated graded module over F, R, and U is of the following form. It's a direct sum of some free parts plus some torsion parts. So the torsion parts are all gonna look like F, R, and U, mod some polynomial, and that polynomial has to be homogeneous in degree because this thing is graded, and so since U is degree minus two, every homogeneously graded polynomial is just a monomial. So it's just U to the nj for some j. So just algebraically, this is what hf minus looks like. I guess there are some gradings everywhere. So I'll put some subscripts to denote the gradings. So here, this denotes the grading of element one in this ring, and then U lowers graded by two. Great, okay. So that's sort of the algebraic structure of these invariants. If y is a rational homology sphere, then we can actually say even more. So if y has the same rational homology as S3, then Ajavar Sabo tell us that there's exactly one free summand. So what do I mean by that? I mean that hf minus of y looks like F would join U plus some U torsion pieces. Okay. And so some of you may have heard of something called the Dn variant, coming from Hager-Full homology, and that's exactly this D right here. Maybe you're worried that this splitting is not canonical, so you can also define D more invariantly as the maximum grading of an element X in hf minus such that U to the n times X is non-zero for all positive n, eight. Since U lowers graded by two, you can check that this D was exactly gonna agree with whatever the grading of one is right here. So another term you may have seen in the literature is something called an L space. An L space is something called an L space, which is something called an L space. So this is something called an L space, and L space is a Hager-Full homology lens space. That is a three-manifold that has the same Hager-Full homology as a lens space. But I guess we can define that in terms of whatever in here. So this U torsion part has their name. It's called hf red, wait. And we say that a rational homology sphere Y is an L space if hf red of Y is zero for every spin state structure. You might be wondering if there's some sort of relationship between the hat flavor and the minus flavor. And there is. So it comes from the following relationship that you have on the chain level. So on the chain level is a short exact sequence, like this, cf minus to cf minus to cf hat. For this map here, so remember, these are chain complexes over f of join U. So there's this action by U, multiplication by U, and this map here is multiplication by U. And so you have this short exact sequence on the chain level, and so this induces an exact triangle on the level of homology. So you have the following exact triangle. So exact just means that the kernel of this, the kernel of any map is the image of this map coming before it. Another fact about Hager-Full homology is that if y is a rational homology sphere, the dimension of Hf hat of y is always greater than or equal to the order of H1 of y. And then it turns out that if you have equality, well, that's equivalent to being an L space. And in fact, you can prove that this if and only a statement are using this exact triangle and the definition over there of an L space. What other properties do we have? You might wonder, how does this invariant behave under connected sum? It satisfies a Cronin's type formula. Cronin's formula. So if you wanna compute, if you wanna know the chain complex of the connected sum, oh, great, so since the chain complex associated to a Hager diagram, well, the chain homotopy type of that is an invariant of the three manifolds. I'm often gonna abuse notation and write the chain complex of a three manifold. But really, I mean, pick a Hager diagram for that three manifold, compute that chain complex. Since this pick any chain complex is chain homotopy to that. Great. So this is homotopy equivalent to the tensor product, Cf hat of y1 and Cf hat of y2. And then you get an analogous statement for Cf minus. And in both cases, the tensor product should be taken over the ground ring. So for Cf hat, that's F, and for Cf minus, that's F adjoin U. All right, another nice property of Hager flow homology is that it behaves nicely with respect to cabortisms, which Chipman alluded to yesterday morning. So if you have a four manifold, W, smooth compact, and the boundary of W is the distant union of y0 and y1. Well, oh, and so let's put some extra structure on here. So let's also equip our cabortism with a spin C structure, S. S induces a map from the Hager flow homology of y0 to the Hager flow homology of y1. So this induces a map from the Hager flow homology of y0. So this denotes the, if you have a spin C structure on the four manifold, well, you can restrict it to a spin C structure on either boundary. And then these cabortism maps behave nicely under composition, if you stack two cabortisms next to each other, lots of composition of the two respective cabortism maps. If you have a product cabortism, that's going to induce the identity map. Another property is that there's a surgery exact triangle. So what's the setup here? The setup is that we have a compact three manifold Z with torus boundary. And now we're going to have three different simple closed curves on the boundary of Z. And these simple closed curves are going to satisfy a certain relationship. So we have gamma naught, gamma one, gamma infinity. These are simple closed curves on the boundary of Z, such that if you look at the algebraic intersection numbers, so let's look at the algebraic intersection of gamma naught and gamma one, that's going to be equal to the algebraic intersection of number of gamma one and gamma infinity. I'm just cyclically permuting them. It's going to equal the algebraic intersection number of gamma infinity and gamma naught. These should all be equal to minus one. So in case you're wondering why did I choose these as my subscripts like zero, one, and then I went to infinity, well, let's look at, let me give you an example of such curves. So here is a torus. Let's say you can check that as I call this gamma naught, this gamma one and this gamma infinity, and then maybe you can choose, I guess the oriented simple closed curves. And then you can check here, will these satisfy this condition? And then while the subscript is just denoting the slope. So that's why I named them that way. And so now, right, so Z has torus boundary, so we can fill that in with this, do a day in filling along each of these curves. So what do I mean by that? Let Y i be Z union of solid torus such that gamma i bounds the disc such that gamma i is equal to a point across D2. Ashroth and Sabo tell us that we get an exact triangle of the Hager flow homology. All right, so as you'll see in the problem session, well, this can actually be enough. So sometimes, like, in certain special cases, if you know two of these and you know that they fit into an exact triangle, that's enough to deduce what the third group is. You know where these maps come from, right? And so if you notice in this setup, right? So why not and why one differ by how we glued the solid torus into Z. And in particular, there's a two-handle cubortism from Y naught to Y one. And this map is exactly the cubortism map coming from that two-handle. Oh, yes, thank you. So if you noticed over there, for that exact triangle, I stated it in terms of HF hat. You're probably wondering, well, is there also an exact triangle, an analogous exact triangle for HF minus? The answer is basically yes. For some sort of annoying technical reasons, you actually have to, instead of working over the polynomial of F or join U, you have to work over the power series thing. So we have an analogous exact triangle for HF minus, except you need to work over the power series thing instead of the polynomial. All right, maybe I'll give you a more concrete example of a family of Y naught, Y one, and Y infinity that fit into an exact triangle. So for example, if you take Z to be a knot complement, and then you take, you can take your curves to be a meridian, an n-frame longitude, and an n-plus one-frame longitude, you can check that those fit into those satisfy the conditions. So that means that you have an exact triangle like this, where this denotes n-frame surgery, and then you can take your curves to be a meridian, n-frame surgery, okay, as you heard about this morning. So those are the formal properties that I wanted to tell you about before I move on. Are there any questions? Great, so now we're going to dive in and define the chain complex from a Hager diagram, associated to a Hager diagram. Great, so the setup is that we have a Hager diagram, so that's a closed surface of genus G, G-alpha circles, G-beta circles, and a base point. Wait, let's assume that our alphas and betas all intersect transversely. From this, we're gonna build, we're gonna build a space. What space are we gonna build? We'll build a sim-G, a sigma. So this is the G-fold symmetric product. So what do I mean by that? We'll take sigma cross itself G times, and then quotient out by the symmetric group on G elements. Points in this space consist of unordered G-tuples of points in sigma, right? So this, well, here's a G-tuple of points in sigma, and then we quotient out by the symmetric group for saying, well, now it's an unordered G-tuple. So this is a space consisting of unordered G-tuples of points in sigma. Now that since we're taking the symmetric product of a surface, this is actually a smooth manifold. So that's one of your exercises in the problem session. I'm gonna describe some subspaces of sim-G for you. Great, so the first will be T-alpha. So this is just alpha one cross alpha two cross alpha G, and then similarly, you could define T-beta, which will be beta one cross beta two cross beta G, and you'll notice these are both half dimensional subspaces. So define, we haven't done anything with the base point yet. So we'll also, we'll use that now to define a subspace, which we'll call VW. So this is gonna be, well, it's gonna be points in sim-G where at least one of the points is your base point W. So a slightly abused notation, but I'll write that as W cross sim-G minus one. Great, so this is just saying, well, at least one of your points is W. This is a co-dimension to subspace. Let's also look at the subspace T-alpha intersect T-beta. So we assume that the alphas and betas intersected transversely, and then we'll T-alpha and T-beta are both half dimensional subspaces. So this is gonna be zero dimensional, so it's gonna be a collection of points. Right, so these are points in sim-G, but we can see them in the Hager diagram, right? So, well, what are these, if you wanna sort of see these in the Hager diagram, what are they gonna look like? Okay, well, it's gonna be a G tuple of points that are intersections between the alphas and betas. Moreover, each alphas circle has to be used exactly once and each beta circle has to be used exactly once. So, at the end of yesterday's lecture, we ended with this Junus-4-Hager diagram for the trough oil, but if you forget one of those base points, well, it was just a Hager diagram for S3, and so a good exercise to sort of make sure you can understand these points in T-alpha intersect T-beta is to sort of find all of those points in that Junus-4-Hager diagram, right? So, it's a G tuple of intersection points between the alphas and betas, where each alphas curve is used exactly once and each beta curve is used exactly. I'm making such a big point about that particular subspace. The reason I'm doing that is because the chain complex, CF hat, so it's generated over by T-alpha intersect T-beta. Any questions so far? Let's move along with the definition then. So, let X and Y both be intersection points between alpha and T-beta, and D is gonna be the unit disk in the complex plane. We'll make the following definition. Whitney disk, we'll say from X to Y is a continuous map phi from this disk to SIM-G of sigma, satisfying the following four properties. So, maybe I'll draw a picture of this disk. So, here's minus i, here's i, and then let's also label the boundary of the disk. We'll call the part of the boundary the right part, but we'll call this e-alpha, and we'll call the left side e-beta, and then we'll require that phi sends minus i to the intersection point X, that phi sends i to Y, that phi sends e-alpha, that that lands in T-alpha.