 Okay, my aim is to give the shortest introduction at all times for all times. Okay. Welcome everybody You two guys are sitting in the last row. You're not gonna see anything cup forward. I'm serious talking to you two guys Yeah, you too. Okay. All right Please interrupt at any moment and ask questions. Don't wait until the end. Okay, if you don't ask any questions I will turn around and frown at you and Was that Yes, you have to ask questions in fact Otherwise no food Okay, um, you know really it's important. Otherwise, what's the point? You could just as well be on video. Oh Yeah, we have remote participants. So it's important that when you ask a question you remind the speaker to repeat the question Okay for the remote people. I will You know, I will put in the chat all the zoom etiquette that we're following Okay, I hope you have a wonderful time. It's so great to be together in person. Jen. I'm not gonna introduce you Well, thank you for the introduction Good to see all of you great, so I'm talking about Hager flow homology. So I'm a low-dimensional topologist. I study three manifolds one manifold knots and surfaces inside of them and For the next three lectures, I'm going to tell you about my favorite tool for studying these objects Namely Hager flow homology. So this is a package of invariance defined by Ajavathan Zappo about 20 years ago now But and so What is this invariant do so to a three manifold so All of our three manifolds are going to be closed connected oriented So two or three manifold we get an algebraic object So almost to be talking about on the minus flavor. So two or three manifold y We associate HF minus of y. So what kind of algebraic object is this? This is a graded F so Whenever I write Blackboard bold F that'll be the field of two elements So it's a graded F a joint U module. So so graded Module over a polynomial ring in one variable and the degree of U is Minus two so I said you can also use this to study four manifolds. So What does that look like? So suppose you have a Four manifold Corbordism W So it's a Corbordism say from y naught to y1. So what does that mean? It means that the boundary of W is Minus y naught this joint union Y1 and so to a Formatiful Corbordism well, we get a map between the Hager flow homologies of the of the boundaries So this induces a map which I'll denote FW and this goes from HF minus of y naught To HF minus of y1 This is a module homomorphism Maybe if I wanted to be really careful I should say something about spin-C structures on the Corbordism and on the three manifolds but sort of today for sort of the first pass I'll maybe ignore some of those details But but if you if you're worried about that and you know about that then yes I'm sort of Ignoring some of that for the sake of sort of having cleaner statements to give people the the lay of the land This behaves nicely. So for example if W is just a product Corbordism then it induces the identity map and Suppose you have two Corbordisms where they they line up on one end and he stacks them together Well, then the composition of those maps is going to be the same as sort of if you just use that as one big Corbordism it sort of has these nice properties So to a knot K in y well, we also get an invariant associated to that the the version that I'll be talking about this week I'll write it Ah HF KO Right, this works in it. Maybe any three manifold so Maybe let's say null homologous Great. And so what type of algebraic gadget is this? Well, this is a graded module It has more structure than the female three manifold invariant This is a graded module over a polynomial of n and two variables. Well, we will suppose you have a core Yeah, yeah, so Paul asked what's the degree of V? great, so this is actually a bi graded module and so We have two gradings one rating is called the grading the u-grading and The u-grading of u is minus two and the u-grading of v is zero And there's also a v-grading and then you just flip these Hopefully Paul won't be the only person asking questions today If he if he is I'll be very sad and then he'll just have to look at my sad face all week All right Well, you might maybe you'll have a keyboardism of pairs Great, so that's a four manifold w and let's say a surface s From y naught k naught To y1 k1, so this just means well the boundary of w is minus y not just joint union y1 The boundary of s is let's say minus k naught just joint union k1. Well Again, this will induce a Module homomorphism so maybe maybe the cartoon picture Right say this is w. This is why not. This is why one then you have In my one color that looks different than white. I'll draw. Let's say this is Okay, not and then you have some Maybe it has genus and this is k1. So that's a cartoon picture of this so Let me give you some examples great, so What's the simplest three manifold to me? It's s3 and It has sort of the simplest possible. Hey good for homology It's just the polynomial ring and then I'll write my gradings like this So this when I write this zero subscript And if you can't see the subscript, maybe you should move closer because I write my subscript small This tells us that the grading of one is zero And maybe as a remark There's different grading conventions in the literature. I'm gonna use this grading convention that It's that sort of night you want like putting one of them zero is like the simplest thing that you could hope for and you kind of want S3 to have the simplest thing but just so you know in the literature Sometimes this is shifted by two. Sometimes people say this is actually minus two. So just be aware of that That's my warning for all of you if you ever go try to like Actually do with you and with this and you're unhappy about the gradient great Okay, so hopefully this is an interesting invariant. So hopefully some other three manifolds have different here good for homology So hf-minus of the risk or homology sigma 235 so note, this is also Plus one surgery on the right-handed chuck oil It's a very nice three manifold Okay, well, it's a good for homology. It's also just this polynomial ring But but the grading is zero is different on the grading of one is minus two So hey good for homology can tell can tell apart on these two three manifold Let's see. What about what about if I wanted to build a co-orders in between these two three manifold? Wait, so since sigma 235 is plus one surgery in the right-handed truck oil. Well This means that Well plus one surgery on the right-handed truck oil Well, that's the boundary of the one trace of the right-handed truck oil So that's the boundary of what I'll denote X plus one of the right-handed truck oil Right. So what is this? Well, this says take four ball Okay, well, let's take Okay, so the boundary of the four ball is s3. Well, let's take our not an s3 So I'm drawing a cartoon. So my not is going to be an s0 in this cartoon and then let's attach a Plus one frame to handle on the not right. So a to handle is d2 cross d2 and then You attach it along an s1 cross d2 and so the plus one is telling you how to frame that bundle and so I'll draw out like this and then The boundary of this is going to be plus one surgery in the right-handed truck oil. So this is what I'll call x1 Okay okay, and then Then for whatever Orientation convention reasons I'll let W be okay. I want to get a couple orders in between s3 and Plus one surgery in the right-handed truck oil. So I need to take this trace And let's remove a four ball and then because of the direction that I want things to go I'm just going to overall reverse the whole orientation also Okay, so now the way I've set this up the boundary of this is minus Orientation reversal plus one surgery on the right-handed truck oil this joint union s3 So it's a co-border some from here to here And so this four manifold is going to induce a module homomorphism from here to here It's a module homomorphism. So it respects the you action People can see if I write here Yes, wait if you can't then you missed your chance to speak up and so This this map okay, so it's goes from here to here and and This example it turns out this map is great in preserving So it sends the element one here to the element you here. So One Get sent to you another way you might draw that is I like to draw f for join you like this so the vertical lines represent my you action and then Great so and then I guess the way I've drawn it the vertical placement is telling me something about the grading so This is what the map looks like and sort of and then right It's you up a variant because you do you or you go across or you go across and do you yeah I'm just I'm just I've done a calculation and I'm just declaring telling you this is the answer Yeah, yeah, so the point I'm telling I'm giving you some examples just so you can get sort of a flavor of what the invariant looks like and then What's the plan after that the plan is to sketch the definition of the invariant and then The later lectures will talk about some properties of them. Does this map always have to yeah So the question was do these do the cobalt of the maps always preserve grading and the answer is no They don't always preserve grading Right do they they come with some extra structure with these spin C structures and within each spin C structure They're grading homogeneous or possibly with a grading shift. Yeah Great so the question was yeah, so the question is do the maps depend on the spin C structure and the answer is yes I'm sort of But my aim in these lectures is sort of to give people like an overview of Hager for homology So at times I'll maybe suppress them in the details in the ease of sort of like painting a broad picture Do we know different manifolds of the same Hager for homology? Yes So for example, if you take plus one surgery on the chaff oil and You connect summit with its orientation reversal This the Hager for homology of this Looks like the Hager for homology of S3 and there's other examples, but this is maybe the first one that pops into my mind Paul's another question Great, so why it's a math orientation reversal of W. So the way things are set up, right? If I have a co a co-border some from why not to why one the W's to the boundary of W should be minus Why not the string union y1 and then at least my convention for building the trace the boundary of the trace of K is going to be like Plus one surgery on K and I wanted this co-borders and to go from Here to here so that meant that minus plus one minus the surgery had to be one of the boundary components Oh, if you took the reverse The question was what if you take the reverse of W, right? Then you get a co-border them from S3 to plus one surgery and I believe it's the map is zero, but Quickly in my head early in the morning. So I don't know if that's to be trusted But yeah, it's sort of different. It's not it's not just some like dual or simple thing of this More questions if you ask more questions, I'll get used to repeating them. Oh Yeah, I think so the question was can we say anything what the map induced by the other co-border them So reversing the orientation. I think that was Josh's question and It's not it's it's not just like the dual or something nice of us like I mean you you If you gave me a few minutes to sit down and do it I could actually verify if it's zero as I claimed But yeah, I don't know if I understand the question. Oh Yes, does this co-border them up involve the homology orientation of the four manifold More questions. Okay, so I've given you some examples of the three manifold invent I've given you an example of a four manifold co-border them and the map that it induces What about great is there a non trivial co-border them from a manifold to itself that induces the identity probably but Does anyone in the audience want to volunteer one great. Yes, like the ribbon stuff. Yes Yeah, maybe that takes me further a field than I want to discuss right now next example great, so right hander chaff oil and S3 claim that this is well Wait, so to this we're going to associate a chain complex whose chain homotopy type is an invariant of the isotope type of our knot manifold invariant right we got out Finitely generated modules over a PID so those sort of are classified right they sort of look like a Free part plus some torsion part but modules over more general polynomial rings aren't as nice So for this example, I'll describe the chain complex for you So first maybe let me remind you right we have these two variables you and V And we have these two gradings grading view and rating of V And also sometimes we want to consider some linear combination of these ratings, which is called the Alexander rating which is one half the U rating minus the V rating and so Great, and then this is minus one and one. Okay, so the the chain complex that we're going to associate to the right hand and chaff oil well it has three generators a b and c and I've computed the ratings as follows and then maybe a nice exercise It'd be for you to fill these in and then so this is a chain complex So I should tell you what the boundary map is and the boundary of a is zero the boundary of B is U a plus V c and the boundary of C is zero Someone want to tell me what to fill in here? One zero minus one right so I guess maybe maybe I should back up when I talked about these three manifold invariants Right, so we describe our three manifold in some way and out of that data We build a chain complex the chain complex might depend on a lot of choices that we made But the chain homotopy type of that chain complex is an invariant of the three manifold similarly well to an Two or not well, we get a chain complex over the polynomial of being in two variables and the chain homotopy type of that complex is an invariant of the isotope class of our knot Yes Yeah, and today I'll remember well by then to say probably won't remember to review questions, but we can still try Yeah, so the differential has grading minus one minus one. This is a curly D. The mass love rating Yeah, so the thing if the thing that I'm calling the u-grading. That's the mass love rating He has lots of wonderful properties. I'm gonna tell you about two of those wonderful properties So the first wonderful property is that not flow homology So There's not flow homology comes in various flavors. So the flavor called hfk hat This categorifies Alexander polynomial First of all, what is hfk hat in complex over this ring at the join UV and then maybe you're like, oh like That's way too complicated. I don't want to chain complex over polynomial And I just want to chain complex over the field. Well, we can do that Let's just set you and v equal to zero and now we have a chain complex over just f. Okay, so Set you equal to zero and v equal to zero on on the chain level. So On the chain level we'll get out of the chain complex and then so the next step is well Now you have this chain complex over Field and then take the homology Right if I knew still you still have your your grading data So you take homology and then you got the what you get is denoted hfk hat So Will you use this grading and this grading? I mean, it's like the same information is using like these two, but that's just sort of how it works So we'll use the mass law of grading and we'll use the Alexander grading and then It right so this is a so now this is going to be a by graded vector space. So we'll just write it Like this so the subscript will denote the math love grading and then this will denote the Alexander grading So this is a by graded vector space And if you have a by graded vector space you can take the graded or the characteristic So we'll treat M as like our homological grading and then we'll record Alexander grading with the with the variable So what is what does that mean? It means that the Alexander polynomial of k. It's so you sum over the Alexander grading and then you record that with the t to the a and then You take the alternating sum over the mass law of grading and so As an exercise so I'll put the answer on the board and I encourage you to check this in your own time Well, well, I'll just say it out loud But if you say you and V equal to zero the differential becomes trivial So everything is in the kernel and nothing is in the image Yes, yes, it should be the dimension The question was should it have been the dimension of HFK hat and it is and now I've clamped in dimension right here Okay, so you can check that it should look something like this You got copies of F here, right? And then you take the graded or the characteristics. So this is an even mass law of grading and it gets a T to the minus one and this is an odd mass law of grading and then this is an even For those of you familiar with the Alexander polynomial say, ah, yes So that's the first wonderful property of Nautla the question from the chat is does this exist for links as well as knots the answer is yes There's various ways. You can do it. You can maybe get a you and a V through each link component if you want Yes, and it's powerful has lots of wonderful properties, which I won't discuss second wonderful property of Nautful homology is that it detects genus. Okay, so let's take the We'll use the symmetrized Alexander polynomial So sort of the highest positive degree will match the highest negative degree wait, and then you can write your Alexander polynomial and then the genus of K is Bounded below by the largest I such that AI is not zero Wait, so the Alexander polynomial on its own gives us a bound on the genus and then Well, not flow homology improves this to actually detecting the genus This is a result due to our shots and Zabo the genus of K is actually equal to the largest Alexander grating such that The Nautful homology is non-zero and that Alexander grating Oh, I said I was I'm actually going to tell you three wonderful properties about Nautful homology The last one is that Nautful homology detects viburness so the Alexander polynomial obstructs viburness Right, so we'll call that if a knot is fibered Then the Alexander polynomial is Monic and so well when you when you lift this to not flow homology what it tells you is that Right, so this is a theorem of Gagini and me so K is fibered If and only if the not flow homology in the highest Alexander Grating which is necessarily the genus right by the second wonderful property Well, this is just a single copy of it So certainly it's implied that the Alexander polynomial is Monic, but in Nautful homology So that was sort of the overview of the form and properties of these invariants before I sort of dive into the definition Are there any questions? So this is the genus in Oh The question was and I talking with the smooth genus of the topological genus This is the Cypher genus in S3. So this is not a distinction This is the genus the minimal genus of a Cypher surface in S3 But so you might Cypher genus is The question was what do I mean about Cypher genus? What other kinds of genus are there? Right so the Cypher genus is you look for a surface in S3 You could also think about S3 as a boundary before and they look for the minimal genus of a properly embedded Surface in B4 whose boundary is you're not in S3 And then when you look at surfaces in a four manifold like that Then there is a distinction between if your surface is topologically embedded or smoothly embedded So the question was in the theorem about genus when you look at h and k hat of k comma a Do you sum over all mass log weighting and the answer is yes give a sketch of how these invariants are defined it It'll be a sketch So it'll be sort of a jumping off point for you to go and sort of read the definitions deeper I'm not capable of giving you a complete Introduction in three lectures to all pay good for homology. Okay, so I want to talk about Hagar diagrams, right so the The deal is well We need to describe our three manifold somehow and then from that description we feed that into some machine That's fits out of chain complexes chain homotopy type is an invariant So the input into this machine is going to be a Hagar diagram. Okay, so first I want to talk about handle bodies Right. Okay, so let's take a handle body So here this is a three ball and like I want to carry my three ball around with me So like I need some handles on it. So here's Right and maybe maybe I have a friend and my friend also wants to carry it sometimes But so there's two handles on this. So this is a hand this this here is a handle body of Genius to We just care about these things up to homeomorphism. So for example right so here I'm imagining this as Sort of filled in right. This is filled in. This is a handle body of genus three three manifolds Well, these have boundary, but pretty soon they won't So a hey good splitting of a three manifold is a Decomposition of your three manifold into a union of two handle bodies definition Hey good splitting a three manifold. Why is a Decomposition of why into two handle bodies? I Heard from Melissa's Jan that she likes to call well, you have colleagues We have two handle bodies. So it's a handle body and if it's handle buddy they come together Great, so we want we want an invariant that's going to be able to describe any three manifold So maybe the first thing you might wonder is well Does every three manifold admit such a such a splitting? The answer is yes, so theorem Every closed oriented three manifold Admits a hey good splitting great. So what's what's the proof? Well? Okay, well if you believe that every three manifold can be triangulated right So for one of your handle bodies take a neighborhood of the one skeleton to the other handle body Take a neighborhood of the dual one skeleton So that's the one skeleton whose vertices are like centers of your tetrahedron and then whose edges sort of go through each face perpendicularly and Then you can check well, but a neighborhood of a one skeleton Well, that's going to be a handle body and so that gives you your hey good decomposition. So Let's look at some examples No separate numbering my examples and we're on five now, maybe Let's talk about a hay guard splitting of s3 one way to think about s3 is as R3 union a point at infinity handlebodies is going to be a neighborhood of the z-axis together with a point at infinity I'm going to draw it in blue even though So you have a neighborhood of the z-axis to well the z-axis together with a point at infinity is a circle And then you take a neighborhood of that that's going to be a solid torus. So one of so one of our handle bodies is a neighborhood of z-axis Union infinity and then the other this is going to be our second handle body and then the other handle body will be a neighborhood of the Unit circle in the x y plane This will look something like this Okay, and now maybe you stare at this long enough in you believe and this sort of fills in everything and then this is described S3 right here is one of our hand. This is here is one of our genus one handle bodies This is solid torus here and then the other one is sort of everything So the question was if we're triangulating things, that's the continuous case. How do we get there? How do we get to the smooth case? I guess it depends I guess it depends on what what definition of triangulation you're taking but if sort of If if your triangulation is sort of like nice and finite. This is a genus one Hagar splitting of s3 There's also a genus two Hagar splitting of s3. So let me draw that so Here here is your genus two handle body and then sort of outside here I'm claiming if another genus to handle body you could sort of one way to think about is that you can get from here to here just by sort of well tapping on like a Extra handle on to this handle body here and then sort of the compliment will kind of like go through it and give you So we're going to describe our three manifolds with via Hagar splitting But now the question is well, I'm drawing like this Picture for you like how do we how do we feed that picture into some machine? That's going to spit out a chain complex And so the point is that we can describe a Hagar splitting with something called a Hagar diagram a Hagar diagram a triple so The triple denote with the math cal H and it consists of a sort of a closed oriented surface sigma together with a Closed oriented surface of genus G together with a g tuple of alpha curves and a g tuple of beta curves Such that a sigma is a Closed oriented genus G surface. Okay, so this is a collection of G alpha circles So it's a set of pairwise disjoint simple closed curves in sigma such that the compliment is connected and then Same the same for the betas and right, so Just repeat everything here or just replace alpha with beta So that's what a Hagar diagram is and the claim is that a Hagar diagram Describes a Hagar splitting so let me tell you how to from this data how to build a three manifold So go over here and let's so From this data, I'm going to build you a three manifold So I'll give you a description and I'll give you a run-in example the data of a Hagar diagram Here's a closed oriented genus to surface. So my I need a pair of alpha curves Alphas are always red and Betas are always blue You'll have to deal with the blue that doesn't really look blue. These are blue alpha one alpha two Beta one beta two So to build a three manifold Y From this data. Okay, so I'm trying to build a three manifold. Well, this is the surface. So first let's thicken Sigma to sigma across I and so my interval is Zero one and now Right, okay, so now I have two boundary components I have sigma cross zero and I have sigma cross one and so Along on the sigma cross zero side. Let's attach D2 cross interval along each Alpha I cross zero. Okay, so like in this picture Imagine you've thickened the surface I don't know how to draw that in a way that doesn't make the diagram look terrible And then we'll think of the inside of that thickened surface as the sigma cross zero side And so on the sigma on the inside we're along each of these curves. I want to attach a thickened disc. So now Okay, and now because of this condition that the complement of the alpha curves is connected What's gonna happen is that what's left of the boundary on the inside is gonna be? S2 and so there's a unique way to fill that in with a three ball, right? So here if you kind of look on the inside there's like this part here And so that that boundary is an S2 and we'll fill that in with the three ball. So let's attach B3 to the resulting S2 boundary. Okay, and so Basically these alpha curves are telling us how sigma bounds the handle body like on like the inside Okay, and then the beta curves tell us how sigma bounds the handle body on the outside So then the the next step is you just well do do the analogous thing outside So now let's attach D2 cross I to each Beta I announces on the other side and then Okay, so in my picture here Right, so we're attaching a thickened disc like along here and also along here Great, and now again because the complement of the beta curves is connected Well, the result in boundary is going to be an S2, right? It's sort of we sort of like have like filled in the genus on this and now here's an S2 And then we stick in a three ball. His boundary is a B3 Resulting so the definition here the question was the definition here the alphas and betas They don't need to intersect. That's correct. Yeah. Yeah, so I'm principal So for example, here's a perfect the question was so I'm principal the beta circles could be the alpha circles Yes, so for example If you're just talking about a Hagar diagram then for example This is a perfectly good Hagar diagram So there's my alpha. Here's my beta um Later when you get to talking about Associating the chain complex to this it turns out this diagram won't be one will have to sort of wiggle things around gently But if you're just a three-manifold topologist talking about a Hagar diagram This is a perfectly good Hagar diagram and a good exercise to check to see if you are following this to try to Determine what three manifold this describes. Is it true that alpha and beta are a basis for the first homology? And the answer is always Never never great That's right, right, so H1 of a genus G surface See what if you wait, it's neither always nor never if you look at if you look at just the alpha circles That's never gonna be a basis because you only have g of them and you need 2g elements It is true that if the census complement is Connected the alphas are always going to be linearly independent and the betas are always going to be linearly independent And then sort of that that's what you get and then like besides that sort of anything can happen, right? So here they're not a basis and then there they are so sort of Anything can happen which from like basis to not basis how the question is how does the s2 boundary come up? um Yeah, so if it follows from the condition that the complement of the alphas is connected that the boundary is going to be s2 so You can do like an oral characteristic argument or there's probably other ways to see that That follows in this condition. It's a Hager diagram for s3 also a Hager diagram for s3 So we can see just by example that it's possible to have more than one Hager diagram for the same three manifold However, we have the following nice theorem that says that any two Hager diagrams for the same three manifold are related by a sequence of the following moves Isotopy you're allowed to just like wiggle your curves around the first is Isotopy so for example I suppose this is your Hager diagram Well, you can you can wiggle your curves around maybe now my maybe now my alpha curve looks like this and now maybe my beta curve Great, and that's gonna just find the same three manifold All right, that's the first so-called Hager move The second is called a handle slide. So let's suppose We have this diagram Okay, so a handle slide. What's my favorite way to think about a handle slide? Slide this beta curve over this other beta curve Okay, so so that means that this beta curve is the only one that's gonna change So this one stays here. And so how do I think about that? Well, I think about that okay first Let's do like an isotope. So it's really close To the other beta curve and then once you get really close now then just like Tag on to like a parallel copy of the other one. So so the result is gonna look like this Yes, you can I the question is can you isotope each alpha curve and each beta curve separately and yes, you can Okay, so if you notice Neither isotopy nor handle slides change the genus of the surface But well for these two examples of Hager diagrams are actually they had different genera so we better have a move that change that can change the genus and That's stabilization and destabilization so stabilization increases the genus by one and Destabilization decreases the genus by one. So let's say Let's say your Hager diagram looks like this you can think about if you think about stabilization as This part stays the same and then Over on the other side you just get sort of this standard you should a good exercise would be sort of Check that well the three-minute fold your build associated to this is the same as when you get to this but spoiler This is just sort of oh, can you all the question is can you also just apply self-diffure morphisms of the surface? Yeah, I mean yes that won't change the three-manifold But the statement here is that these three moves are sufficient to get between any two Hager diagrams for the same three-manifold. Yeah, but actually going back to the question So yes, you can always apply a self-diffure morphism of the surface So in fact using that fact you can actually deduce that if you want to you can always choose Say your alpha circles to be in standard position So like this is what I call standard position if you do that it might be at the cost of the beta circles become like All interesting and jumbled up But you can always choose either your alpha curves or your beta curves to be in standard position if you so desire The question is how do we get the orientation of the three-minute fold from the diagram? Maybe I didn't say it but you start off with an oriented surface And then if you take an oriented surface across an interval that gets that well The orientation of your surface induces orientation on this product and then that gives you an orientation sort of yeah so maybe a good a good exercise to Understand how you get these females so you should think about well what happens if I what happens if I reversed orientation of sigma? What three-minute fold do I get there relative to the original orientation and also? Say if I swap the rolls of alpha and beta, what does that do to my three-minute fold alpha and beta to be oriented? No change the conditions that the alpha and beta curves are going to satisfy No So that's a good exercise to check that if if the complement of the alpha of the beta curves is connected Well after you do a handle slide the complement of beta curves is still going to be connected more questions No more questions. I'm going to carry on okay So so now you can maybe You can maybe understand at least sort of the big picture of how we can use this to get a three-minute fold invariant Right so to a Hager diagram from a Hager diagram We're going to build a chain complex and then well you just show okay You just show that the chain homotopy type of that chain complex doesn't change under any of these three moves And then then you're happy you've gotten yourself a three-minute fold invariant the plan is that to a Hager diagram we'll associate a chain complex such that the chain homotopy type is Independent of Whatever choices that we made so EG folks in particular will show that it's invariant under Hager Hager moves great For technical reasons we actually need to put a base point on our Hager diagram So it's like literally just like you put a point and give it a label and then and then you Okay, you isotopes are not allowed to cross the base point and your handle slides can't like go over the base point great, so this is for technical reasons We need a base point it's going to live in the complement of the alpha and beta circles and The Hager moves Miss the base point But even if you restrict your Hager moves to ones that miss the base point you can still If two three-minute falls if two Hager diagrams describe the same three-minute fold You can still get to them by a sequence you can get between them by a sequence of Hager moves that misses the base point The question is do we get an interesting theory if we don't require a base point? you Get out something that determines the singular homology is that interesting So now I need to tell you how to build a chain complex from a Hager diagram great, so we're going to build some Complicated manifold then I'm sort of going to do Lagrangian flow homology there and that's how we get a chain complex Okay, so what do we do? We look at the g-fold symmetric product of sigma, so that just means you take Sigma cross itself g times and then you mod out by the symmetric group on g elements Great, so this looks like unordered g tuples of points in your surface It turns out that this is in fact a manifold. This is the surface. That's a nice exercise Okay, and now we're going to be interested in some subspaces of this g-fold symmetric product so the first subspace we're going to be interested in is T alpha so it's just the product of the alpha circles I guess T for torus and Similarly, we'll be interested in T beta So it's product of the beta circles wait, so these are This is Real 2g dimensional these are each g dimensional well one more subspace will be interested in is Vw right our base point. I've named w and then this is the base point Vw is just Right some elements in here are unordered g tuples of points in sigma and this is just going to be At least one of your points is w. Okay, so these are two half dimensional subspaces of simg So generically they'll intersect in points and those intersection points are going to be the generators for a chain complex so Right so to my Hager diagram H. I'm associating a chain complex whose generators are Intersection points between T alpha and T beta It's just generated over this polynomial ring differentials going to count pseudo-holomorphic discs. So so first Wait, so let X and Y be intersection points so maybe Maybe like a maybe I should also draw a cartoon Okay, so I'm drawing on a board which is two-dimensional so if I want to draw two Half-dimensional subspaces. There'll be one dimensional. So My cartoon will look something like this Say this is T alpha This is T beta so this is going to be Homotopy classes of Whitney discs from X to Y So, what do I mean by that? Okay, so let's take a Unit disc in the complex plane and I'm going to decorate my unit disc So that's sort of this right side of it is red and the left side of it is this blue slash white color And then wait, so here I have this disc and I want to map this into simg So that like these decorations match right so that like this right hand bound We should go to the alpha torus this left hand bound We should go to the beta torus and then I guess this should be X and this should be Y All right, and we can We can put some extra structure on things so we can put We can put an almost complex structure on this and then we can look at So let's say we have Fee in here Well, we can look at the Modulize space pseudo holomorphic in The homotopy class of Fee this modular space has some r action Which we can mod out by so I might also talk about m hat of Fee and that's just this modular space mod it up by This our action. Okay, so all the stuff is going to give us Differential and so now while the boundary of X well, we're going to sum over all other generators sum over all Fee in pi 2 of XY Great and then we have some condition on Fee and then we're going to count points in this modular space and then We also count with the u variable how many times we cross the w base point That's the coefficient of Y. Great. What is mu? That's an excellent question. I have not told you so mu of Fee is the expected dimension of The modular space of of Homomorphic representatives of Fee So we want this modular space to be one dimensional and then we're modding up by this our action So when you mod up by this our action you get a zero dimensional space And so those are just points and so this part is just counting points Even better we're working over F2 So our points don't even come with orientation if you really cared they could come with orientation But I don't care and it works over Z With a lot more pain and you still get a lot of mileage out of it just using See my two coefficients. And so I'm happy to use my two questions. Oh, ah, yes What is NW fee that is a great question? I did not tell you so NW a fee This is the okay, so it's like counting It's the algebraic intersection number between Right, so this the sort of this VW and the and the And the image Fee other questions. Yeah, how does this whole thing depend on the almost complex structure that we chose? It doesn't depend on the almost complex structure up to a homotopy Yes, that's another choice that we made. Yes What about the spin C structure that's associated to interest intersection points? Yeah, so this this chain comp like so spin spin C structures on three manifolds are in bijection with h lower one So if you don't know what a spin C structure is well, just they're in one-to-one correspondence with classes of h1 and The this chain complex splits along spin C structures So this chain complex splits as a direct sum over spin C structures and then well the differential then preserves the spin C structure So the question was If We think about these discs and Sigma rather than in SimG. Do we get G discs in Sigma? in some instances Do you might but you could also get more complicated things you can get things that look like Branch covers of discs and In Sigma might look like a branch cover of a disc but in SimG it actually looks like a disc Great. So the question is what are the precise combinatorial conditions? We need to compute the mass love index of a disc in the diagram saying genus 2 Great, so there's like so Robert Lipschitz has a formula gives you sort of An expectation of what the mass love index will be but It's hard in general. It's also hard in general to know if a disc has a homomorphic representative And so if you're like me You very very rarely compute this invariant from the definition other questions Maybe one question you might have while I have this differential Does it squared or zero? It does squared or zero and the reason that it squirts to zero is sort of similar to the It's the same morally the same ideas and more some ology while it's why the differential and more some ology Exes and y's be an index which means degree this is yeah, so this is a graded chain complex You can use sort of the dimension of this modular space to sort of give you the The relative grading so maybe the difference of gratings between two generators and then to pin that down to an absolute grading Is hard you have to look at Corporate some apps all great questions. Let's look at some examples Let's look at some genus one examples why are genus one examples great well Sim one of our surface is just the surface itself. So if we look at genus one examples, they don't have to actually Do anything too horrible. Hey, that's a great three manifold and I need a base point in the complement of the All right, so how many generators does my chain complex here have? Great one. There's one intersection point. I'll call it x. So this chain complex all this h1 Is generated by x. What's the differential in this chain complex going to be right? Yeah, there's there's no discs here I I guess another the differential there's a grading that I haven't really told you much about but the different Differential lowers the grading. So if you only have the thing in one grade and you can't have any differential so the differential is zero Great, so now I take the homology of this thing that gives me what's called hf minus Okay, well the homology of this thing. I have a single generator and no differential So it's just going to be effort join you and then For s3. This is just like right that you have to we have to normalize a gradient somehow So this is just declared that one is in grading zero What if we chose a different Hager diagram for s3? I still want to stick with genus one so we don't have to actually take the symmetric product of anything Take let's call this h2. Where is this chain complex going to have? Three this three intersection points. So let's label them ABC Great, so now we have to compute the differential So we're like looking for discs in a picture that have like these Decorations so on the they go from x to y and as you like leave x to the right You see a red alpha curve and to the left you see a bluish white beta curve Okay So does anyone see such a disc in this picture and if so from where to where does it go? Yeah, B goes to C right so right here right these decorations line up and then for whatever Riemann mapping theorem This doesn't you'd have a homomorphic representative. Okay, so that means that C appears in the boundary of C appears in the boundary of B Does anyone see any other disc B goes to a? right and Do I need? Do I just put a B B goes to a so just put an a here Do I need some decoration on it right across is this one crosses W? Right, and so if you cross W you have to count that with your variable you so it gets it is a you here Great, and then you can check that that's it so these DC and DA or zero and then It's an exercise but you to check that Well the Homology of this thing well it better match up with that and I guess you can check right so a and C are in the kernel and Then well the images C plus UA and then This is isomorphic to effort join you Say generated But you might say what happened to see well It's generated over effort join you by a and when we're C is homologous to you times a oh, how do we normalize the gradient? Yeah, for S3. This is just declared one is declared to be in grading zero That's like the normalization here, but for other for other three-minute folds Right so the way the gradients are defined right for S3 one is declared to be in grading zero and then for these Cabardism maps you can compute the grading shift associated to any cabardism map And then you look at cabardism from S3 to whatever manifold you want And then if you knew then and then that tells you how to grade sort of the other end of that Yeah, so the question was then so now HF plus of S3 That that's sort of the bottom element there has great in to yeah And that was sort of the disclaimer I muttered at the beginning that I'm using these great in conventions But they differ in the literature sometimes. Why am I using these gradient conventions? It makes things like the current formula nicer It's just sort of satisfying for like HF minus of S3 to sort of have this grading plate. And so then The important theorem due to Ashraf and Zabo is that this really is a two-manifold invariant, but H be a Pointed point. It just means you've had that base point. So the HP appointed haga diagram chain homotopy type CF minus of H is an invariant of why so Often I might abuse notation and write this as CF minus of Y. Well, I just mean the Chain homotopy type Orientation preserve and if you're more for some invariant. Yeah more examples so so Here's another wonderful three-manifold haga diagram a good exercise would be to Determine what three-manifold this is? I'll just tell you it's RP3 Check to see if you believe me. Oh, there's two generators X and Y You can check that there's no There's no discs So the boundary of X is zero and the boundary of Y is zero. So this tells us that the Hey, good for homology of RP3. Well, it's Effort join you plus effort join you where you're all intentionally Omit saying what the gradings are. I guess I'll say a little bit about Spin C structures Some remarks HF minus it splits as a direct sum of our spin C structures So if you don't like spin C structures Well, these are in one-to-one correspondence with classes in H lower one of Y and then Maybe the second remark is that if you have a rational homology sphere You get That the Hager flow homology has sort of a Standard form if Y is a rational homology sphere Okay, right the hey good flow homology. Well, it's a Chain complex over effort join you if we want we can invert you ie we can tensor with effort join u u inverse And that gives us this thing called HF infinity great So HF infinity it's we'll just take HF minus and Then this is the effort join you module So tensor with effort join you you inverse So the statement is that if Y is a rational homology sphere this thing always just looks like effort join you You inverse in this RP3 example Right, so there well H lower one of RP3 Is Z mod 2 so right this let's as direct sum of These two pieces I'll just call it s1 plus HF minus of RP3 s2 wait, and then Because of this fact that so the point the point is that those two effort join you some ends They live in different spin C structures. So so sort of for RP3 this is Isomorphic to effort join you and This is isomorphic to effort join you and maybe let's Contrast that with the next example. Let's look at s s2 cross s1 technical reasons we have to we have to have some intersection points in our diagram so Even though the alpha and beta curve so isotope up to one another Well, we'll just like isotope them. So there's some intersections. So how many generators do I have? To wait, so let's put my base point here. So call this x and call this Y x to y can someone want to describe one Okay, it's this I am taking this to mean this there's this that goes around the back and comes over here So that's a disk from x to y. Does anyone see another disk from x to y? Yes, I saw this Come on like this one right here. Yeah. Yeah. Okay. Well, there's two of them We're working Z mod 2 so they cancel out So the boundary of x well, it's y plus y but that's zero mod 2 and then the boundary of y is zero Hf minus That's this example It's also Two copies of Effort join you It turns out that these two generators live in the same spin C structure So really What this is saying is that Hf minus of s2 cross s1 in what all in the spin C structure that I'll label s0 this is two copies of Effort join you and just the the fact is that The grading of one and one of the copies is one half and the grading of one and the other copy is minus one half with People happy I just declare those. Yeah, so there's just two more properties that I want to Tell you about and then I'll wrap up So the first property There's a question Great. Yeah, the point W is outside and what if it was in between that would change. Oh, it's sort of yeah so There was this condition There's something called admissibility This is an admissible diagram admissibility is the condition of what's called periodic domains those are regions in the Hagar diagram whose boundary are full alpha circles and full beta circles and The condition on admissibility would say that if you put the W in there, it wouldn't be admissible anymore first property I want to tell you about is that what happens if you reverse the orientation on why you just take the dual of the chain complex so So if you reverse orientation on why well then That's chain homo to be equivalent to just the dual so you just look at you look a variant homomorphisms from this chain complex to This ring and so maybe I'll maybe I'll sometimes just denote this With a star Does it do well? Okay So so for the three minimum fold invariant you reverse orientation sort of nice controlled things happen And then what's what's another nice operation? You might want to do on three minimum folds might want to take the connected some And so if you take the connected some You also get a nice relationship There's a kind of this type formula which just says that the chain complex associated to the connected some is chain Homotopy equivalent to the tensor product of the respective chain complexes where again the tensor product happens over the 1031 But I think I started at 901 Thanks for attention Right so the question was to try to understand this kind of formula. Well, could you just take the connected some of two diagrams? Yes, so there's a There's a variant HF hat Sort of it's like the simplest form of Hickel homology And in that you just obtain the chain complex for HF hat by setting u equal to zero And so in HF hat then what you said like totally works Right, you just take the connected some near the base point Well, if you say u equal to zero that might meant you were never allowed to cross the base point So taking the connected some well like you're changing some stuff, but you were never allowed to like go there anyway So it's all okay Yeah, so for for the hat flavor paper for homology Yep, that exactly works for the minus flavor. Well now you're allowed to cross the base point So now I don't maybe you like stretch the neck where you do something complicated to me Analogous results for the genus detection result for no homologous knots and other three manifolds Yes, and sort of those generalization in fact Hickel homology to text the Thurston norm and that's sort of if You can you can view sort of the genus detection results is sort of a yes specialization of that But I say one sentence briefly about the one half and the minus one half Well, okay, you know, you Maybe you're not happy to these are rational numbers, but you should be happy that they differ by one because the relative grading of two generators well since there's a And that was one second