Okay, my aim is to give the shortest introduction at all times for all times. Okay.  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?  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?  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.  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?  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


Hello, my name is George Lu and this talk will be about new techniques in replica encodings and client setup So replica encodings are this primitive that were introduced in a paper last year by Damgard Ganesh and Orlandy and the Ideas you have a client with the message and he wants to produce several replicas of this message Where each from each replica independently you should be able to recover the original message But you have say you have an adversary that takes a look at all of these replicas He should not be able to compress them and store them in states significantly smaller than the size of the original replicas And this has some uses and the decentralized file storage More formally we can define a replica encoding scheme as a set of three algorithms There is a setup algorithm, which takes in your security parameter along with and and an a priori found on the number of replicas this Parameter set is able to produce and outputs a public key and a secret key There's an encode algorithm which takes in a message and produces an encoding and Corresponding decode algorithm which goes from an encoding back to the message itself Now there are a couple properties we want of such a replica encoding scheme Perhaps most obviously we would like it to be correct. So if you Run a message through encode and then run it through decode, you know, hopefully you would end up with the same message again We'd also like for the there to be some efficiency properties Where the replica shouldn't be too much larger than the original message itself in the case of the schemes that we look at This ends up being only an O of security parameter additive factor larger than the original message Finally to capture the Incompressibility property that we want that and we discussed earlier There's we define this soundness game where a challenger takes a message Runs encode to produce several replicas of the message And gives it to an adversary and this adversary takes a look at these replicas and Outputs a state file which can be anything it wants to a second adversary and the second adversary Only has access to the state file along with the original message and its goal is to produce Reproduce as many of the original replicas as possible So to judge the success of an adversary in this game We compare the size of the state file and produce with the size of the replicas. It was able to successfully reproduce And we say we say a scheme is s sound if we're all poly time adversaries the ratio of this Of this storage use with the replicas it was able to reproduce is Almost always a larger than us And ideally we want this asset s factor to be as close to one as possible So in the dgo construct 19 construction, they use a couple of ingredients They use capital permutations Random oracles as well as invertible random oracles, which are just random or random oracles to permutations Which also had a publicly computable inverse So how the construction actually goes is that well, there's a message m we break it up into B blocks of length capital And for each individual block We start by XORing it with the random oracle output of this random string row concatenated with the block index i and then We take this whole thing We run it through the invertible random oracle Then we run it through the copper permutation inverse And then we do run it through the random Invertible random oracle again and so on and we keep on doing this until we are until we're I guess essentially satisfied after our rounds and At that point the actual encoding is just going to be yr for all of the individual blocks along with the row Random string from earlier So we can see that this is At the very least correct in that we should we can recover the message from the encoding by Just using the tropical permutation public key Since to work We to work our way from yr to y0 We just need to evaluate the tropical permutation in the forward direction Along with the inverse of the invertible random oracle which is also public But we could hope that this is secure Because we'll take the because going the other way would require directly at least requires using the secret key So for the adversary, which doesn't have the secret key We'd hope that they need to commit a lot of space to remembering these tropical permutation inverses And so In the work last year, they They show that for rounds equal to omega of capital times n This is a pure replicating coding scheme for any constant s less than one And what our first contribution is it's actually a lower bound on the efficiency of this This construction Where we show that for rounds in little o of b times n This construct this construction is not an s sound replicating coding scheme Uh, but if you look closely at these two theorems, you can see that they're actually Uh, kind of contradictory in that For our lower bound, we we claim that the number of rounds needs to scale with the number of blocks Uh b But this this factor is not present in the actual construction of replicating codings in the original theorem So let's take a let's take a closer look at these two results That's we can start with our lower bound results and the way we show Our lower bound is by constructing a specific trapper permutation And along with an adversary that has an attack strategy on the replica encoding schemes When the when instantiated with this trapper permutation and so How we do this is in addition We take any trapper permutation and in addition to the public key and secret key We also we also attack a public obfuscated program to the trapper permutation where Which assists the adversary in inverting inverting it Now to ensure that this is still a secure trapper permutation. We also require that they That when in order to use this obfuscated program You need to certify that you've already that you already know the in some sense know the outputs that you're trying to request So in particular we require that they give the XOR of all the random oracle outputs. It's trying to compute So the actual paper this is slightly more complicated and uses indistinguishability obfuscation To show security But for the sake of this talk We'll assume we have this sort of heuristically obfuscated program here And of course the idea here for the attack being that You only Need to store this small XOR in the state in order to get all of these trapper permutation inverses Now concretely for the adversary of the scheme What it'll put in its state is the random string row As well as the what the XORs of these y of the y Blocks for every where will it also are one of these for every single round that the construction has to go through As a reminder on the right, this is what the actual replica encoding is and what then this is what the adversary will try to reproduce So we can simply run through the steps of the construction and see that the adversary Using the state and compute all of these steps So to start off you we take m we split it into these blocks Because it has row it can compute the XOR with the random oracle with no issue To get the y zero blocks And now to go from the y's to these this just requires an evaluation of the invertible random oracle, which is public There's no problem for the adversary And now to go from the z's to the y's. This is normally where you use the trapper permutation inverse But all that all that the adversary needs to do is just take one of these Blocks one of these quantities that it's remembered in state To just like that invert all of these blocks and get its y i's And the important thing to note here is that this XOR quantity doesn't scale with the number of replicas Sorry, it doesn't scale with the number of blocks in each replica But of course the size of the replica itself does scale with the number of blocks it contains So in order to make sure in order for an adversary to still be as spatially Inefficient in order to make it Unsound in the soundless game you would need The only way to make this adversary in particular unsound is to increase the number of these XORs which it has to Store and these XORs only scale with the number of rounds in this game. So that's Oh, that's how we get our lower bound and so if we go back to the DGO theorem this actually I think highlights an oversight in their analysis, which was that they implicitly assume that The storage of the adversary and sort of the bits of the storage of the adversary can sort of be attributed to individual random oracle queries, which it's made um And that between different random oracle queries these sort of accumulate additively um but from the From our previous construction we can see that this is not always the case Uh, so just taking taking our first result. This leaves open a pretty I would say important open question, which is can replicating coatings actually be constructed? uh, unfortunately Uh, we also show that just by reparameterizing the scheme to include this to include this b factor this This is this same construction is still a secure replicating coating scheme Uh, I'd also like to mention that there was some concurrent work by moran and wix, which also Notice the same issue That the original DGO construction has but they they constructed replica encodings, which they call Uh incompressible encodings From entirely new they they propose an entirely new construction of this from different assumptions using different techniques uh, and so But uh back to our result so Since the prior work, you know heavily relied on this implicit assumption that the entropy of queries sort of Was separated we needed to We needed to implement a new proof strategy to show the security of our reparameterized scheme So our security proof roughly breaks down into two main components which we call sequentiality and switching So for the first step of sequentiality What what we click what we show is that That there's some block in which any adversary information theoretically has to query In in sequence where they have to query y zero first before they query The random oracle on y1 before they query on y2 and so on all the way to yr And now once we've shown this We can argue the next step which is that Uh given the limited amount of space which the adversary has If an adversary is successful in the on the normal encoding There's also must be some location on this uh sequential path Which if you replace the i plus one with a random With sort of a random element then this on this other like alternate encoding the adversary should still be successful And using and once we arrive at this point, uh we can sort of We're able to take this adversary And if it's able to if it's able to succeed on this in this random version Then it should also just by Just by inspecting the random oracle queries, which it makes we should also be able to just invert the trapper permutation Which of course should not be possible Uh for the sake of this talk, we'll focus primarily on this first step of sequentiality and how to show this um Yeah, so this is done through what through a fairly simple observation which is So supposedly have a random permutation here Uh where we've highlighted the We've highlighted the matching pairs with the same color And if you're only if you can only make a few few queries to this, uh random permutation Then the probability that you output a pair that you haven't queried is actually very low um in particular for the Problem that we care about we say that okay if the domain is of size two to the kappa And you can only query this random permutation in polynomially many spots then The probability that you output this matching pair That you output any matching pair in fact is going to be negligible in kappa times two to the negative c times kappa for any c less than one And so another way to interpret this actually is that even if you're given c times kappa's State we can treat each of the possible state configurations as its own adversary and sort of and just union bound the Probability that any such adversary succeeds with negligible probability And In the actual okay in the actual proof this also scales up to outputting multiple pairs so Correspondingly say if you want to output three matching pairs Then even if you had three times c times kappa state It should you should this adversary should still only have negligible probability in dune so How does this correspond to the original notion of sequentiality? Well, if you imagine that at some point you have an adversary which is non sequential And go like they queries y zero and then queries y two without querying y one then We can reduce this to the matching pairs game before by By after they query y two we can compute We can compute z one from y two and we compute Why we can compute y one from y zero and now we have this y one z one pair Which is a matching pair of the random oracle who we've never queried their random oracle on y one itself and so That's how you show sequentiality And for and finally we have the third result Which is that using sort of the same techniques We can construct the a replica encoding schemes in the random oracle model using by replacing the Invertible random oracle with a feistel network structure And it turns out that using a very similar Using a very similar sequentiality and switching framework with a bit more casework involved We can show this the security of this construction as well Thank you


So, today we continue the investigation of the relation with the fundamental group and we will come to what are called as regular cover lines. Before that, we have to consolidate the result that we have proved last time and improve upon them a little bit. Let us go back a little bit here. This theorem it says that induced homomorphism P check is injective. The second part says that there is a surjection theta from pi 1 of x to the fiber P check P inverse of x which is a constant on each right cosets of this k which is subgroup P check of pi 1 of x bar and hence defines a bijection of the right cosets of k with the set f. So, this is purely algebra now. Only hypothesis topology is that P from x bar to x is a covering projection, x bar is path connected, x is path connected, I have chosen a point x bar on top of x. Once you fix this notation, there is a homomorphism P check from the fundamental group of the top space to the fundamental group of the bottom space and that homomorphism is injective, it is a monomorphism. So, the subgroup can be identified under P check with the sub with the group pi 1 of x bar x. Of course, identification has to take place under P check, this map homomorphism is important. Moreover, if you take the right cosets of the subgroup, they correspond in a nice fashion to the points of the fiber. So, this is the statement. Proof is not very difficult, once you have done lemma 1 most important, lemma 2 is purely geometry, everything follows very easily. So, let us look at this one. Let omega bar be a loop at x bar, I am proving the injectivity. Let H be homotopy of P composite omega bar when you go down to a constant loop, it just means that P check of this point, this element in the fundamental group is the trivial element. I want to show that this is service trivial in pi 1 of x bar x, which means that omega bar must be homotopy to constant path as a loop in x bar. When you go down in x, it is homotopy to constant loop. All these homotopy must be relative to end points. So, start with a homotopy such that this homotopy is the starting point is P composite omega bar and there is a constant path. The homotopy is remember it is relative to the end points, these two end points 0 and 1 are constant all the time. Now H bar with a lift of H such that the starting path is omega bar. I do not know anything about that one, this is homotopy lifting property of P I have used that is all for everything. Then if you look at the H bar 0 as just like as what we have seen before, it is inside P inverse of x because this is a constant path below. So, this is inside the fiber therefore as seen above by discreteness of the fiber and connectivity of the interval H bar of 0 x 0 cross i, it follows that this entire thing is one single x bar because that is where you have lifted this whole thing. So, same reason it follows that H bar of 1, S is x bar because it is a loop right. So, all of them are at this point the H bar of T 1 for everything T and S belong to S. Therefore, H bar itself is a homotopy of the of the what of omega bar relative to 0 and 1 to a constant loop. The problem is constant loop the top will be inside the fiber but the fiber is what discrete all the time I am using the same thing therefore it is constant ok. So, P check is injective by this one. The next thing I have defined a set theoretic function from the right cosets of what of k, k is the image of this fundamental group ok to the fiber of P at the point x. So, this is done by defining a function on the group itself and then restricted to cosets observe that it is a constant. So, it can you get a after all cosets set on the set of cosets is a quotient set of of the group by the group by the subgroup right. So, this is what I am going to given a loop omega which represent an element here after fundamental group has to be represented by loops ok. Lift this omega to a path at x bar this x bar is fixed for all this investigation ok and let the function theta be defined theta of omega equal to omega 1 the end point of 1. Any lift this this end point is independent of what class I what loop I take in the class it depends only on the class therefore theta of the class omega is well defined ok. By the first lemma that we proved theta is well defined well defined means what if I change omega to omega prime here omega prime 1 will be also equal to 1. Now take any z inside this fiber x bar is path connected join this one you will like tau be a path from x bar to z in x bar ok. Then see that p composite tau is a loop at x see all these we have seen already I am just repeating it and theta of p composite tau how is got by this you have to lift it you have to lift it at x bar and look at the end point but the end point is there it is started like that there is only one lift remember and it is already a path there and you have taken the image of that path there will be given path. So it is tau itself and end point of tau is z so theta of this one will be z theta of p composite tau will be z. So this shows that this theta is surjective now I have to show that this theta takes same value on the right cosets of k ok. Now I want to say that if it only so that automatically it will be an injective mapping and surjectivity we have already proved. So theta omega is equal to theta lambda what are omega and lambda they are elements of pi 1 of xx ok some loops at x and they represent some class if this happens if and only if the lifts of omega and lambda at x bar at the same end point that is the definition of theta ok. So this is if and only this one but this is the same thing if and only if omega star lambda inverse ok lifts to a loop at x bar. So this also you have seen right this was lambda 2 actually if and only if omega class of omega into class of omega inverse is inside k because this is a loop at x bar ok this what is this k remember k is pi check of pi 1 of xx bar. So this is if and only if this is an element of k k omega is k lambda or k lambda inverse belongs to k it is right cosets. So this is just group theory. So this complete the proof completely analyzing ok whatever happens to the geometry of the slopes lifts and so on converting it into algebra ok. So we shall now investigate the effect of changing the base point in x bar ok of course without changing the base point at x that means you look at fibers of x fiber of x and change the base point there what happens is what we are we are we are interested in now ok and once again x bar is path connected that is important ok. The isomorphism class of pi 1 of x bar being x bar path connected is not affected this part we have seen long back ok change the base point above. But the isomorphism class changes does not mean that p check is the same ok. So can you also say the subgroup does not change that is not clear right. In fact it does supposed to be not true also and this is where whatever we did in the previous theorem comes into picture again. So that has the answer already there. So the regular coverings is the title of this one. For various points x bar in p inverse of x not there may be finite there may be infinite you do not know p inverse of x not is a discrete set right this make we know. Look at all the subgroups p check of pi 1 of x bar x bar ok they are all subgroups that is what we have proved subgroups of pi 1 of x x not and all these subgroups are conjugate to each other verifying relation the best relation that you can assume ok they are conjugate to each other ok. Once again the previous theorem is they contains this proof let us go through that one little bit ok. Take a path which joins x 1 to x 2 x 1 bar to x 2 bar ok its image will be a loop at x.  So this will be a conjugate of alpha right sorry this is this I am taking directly I have not applied p here. So I am in x bar itself I am doing. So this will be a loop starting with alpha if alpha is in x x 2 ok from x alpha is in x 1 it will it will take this one to x 2 is a loop at x 2 start with alpha and x 1 first omega inverse. So omega inverse from x 2 to x 1 then alpha as a loop and then omega from x 1 to x 2. So this will become a loop at x 2 bar. So this defines an isomorphism pi 1 of x bar 2 on to pi 1 this we have seen already. So this is the isomorphism between any two fundamental groups at any two different points of path connected space ok this will depend upon omega of course if I choose another path this may be different this is an isomorphism isomorphism. Observe that when you take p on omega that is a loop. So let tau equal to p composite omega in pi 1 of x when you p of this p of this one becomes tau inverse p omega p alpha something from element into this tau. So tau inverse tau now tau is a loop so its class is an element of pi 1 of x.  When you pass on to the base space the above isomorphism becomes a conjugation by an element tau inverse here tau inverse alpha tau ok is that is that clear I mean there any doubt in this one. So pi 1 of x bar x 1 bar p check of this is some k 1 p check of this is some k 2 right take a loop here that will look like p of some alpha where alpha is a loop here look at this this map it will give you a map it will give you isomorphism of this one with another elements of pi 1 of x bar x. But p check of that will be k 2 so but this will give you tau k 1 tau inverse k 1 tau 2 is equal to k 2 when you come down. So let us say the conjugation is tau inverse ok. So various groups k 1 k 2 etcetera all of them their image is of p check when you take the different points inside p inverse of x ok they are all conjugates. So this definition and this theorem automatically leads us to study of normal subgroups because for a normal subgroup all the conjugates are the same conjugating element may be any element from the group right that is by definition right. So this leads to the notion of normal subgroup here. So let us make a definition a covering projection is called a normal covering this normal over it is borrowed from from group theory. I am not sure whether whether group theory has borrowed it from the covering space theories or the other way around ok because group theory was developed by Galois much later than Poincaré ok Poincaré had already studied these things. So one is not sure whether this normal covering was borrowed from group theory too here or the other way around ok. It is also called Galois covering that is definitely for Galois and there is other word regular covering also all three different wordings are used by different authors ok. What is it if the subgroup p check of pi 1 of x bar x is normal in pi 1 of x. This x bar is some point in p inverse of x. If it is normal for one x bar for all other points for all other points in x it will be normal ok alright. So it is immediate from definition that if the group itself is abelian then the conjugation is identity no matter which element you conjugate then every covering projection is normal or you can call it Galois or you can call it regular covering ok. However we shall soon see that there are many interesting spaces with pi 1 of x not abelian. If all pi 1 of x were abelian this would not have been a useless definition because everything is normal after all.  This is the definition of normal covering in terms of algebra. So purely in terms of topology what does it mean in terms of geometry what does it mean that is what we want to investigate. But this is already the more or less we have done the basic thing all that I have to do is combine propositions 7.1 and 7.5 immediately ok. So this 7.1 it says this one conjugation and the other one is already there for us alright to this theorem. So we will just we will just put them together ok regular covering ok. So this is the theorem the covering projection is normal if we turn down leaf given any loop attacks inside x all of its loops all loops means what you have to only change the starting point that is all we have no other freedom. If you specify starting point the lift is very different. So starting point you take in any at any point of the fiber all of them will be either loops or none of none of them is a loop. So this is purely in terms of topology if this happens for any one p then p will be a normal covering why can you see why this is true if it is something is normal what does it mean k is a normal subgroup that is the definition right. So what is the meaning of normal subgroup conjugating by any element is again inside that element take a loop omega ok in x if its conjugate some conjugate inside k then only it has a loop as a lift right yes or no. Yes sir. Right if it only if that is what we are seeing if this omega does not belong to no conjugate belongs to this one then none of its lifts will be a loop. So as soon as it is a loop you can conjugate by that element you will get into inside k that is what we are seeing. So because of that theorem this is the precisely the meaning of this one some element some conjugate belongs to k as soon as that happens all conjugates will be also inside k that is a group theory if no if even a one conjugate does not belong then no conjugate will belong that belonging is converted into lifting into a loop or lifting not into not a loop which is does not belong. So group theory part is converted into this one but this is purely in terms of topology ok yeah. So next time I will stop here next time we shall use this one to to illustrate you know not to not to develop any theory but to illustrate that the fundamental group of wedge of two circles this is a figure eight it is fundamental group is non-abelian we will use that non-abelian therefore there must be some subgroup which is not normal everything will be normal if it is so abelian. So we will try to do that first I think and then use a covering ok and then show that some loops will lift to a loop some other loops the same thing some other lift will not be a loop same one single loop below in the covering at different points you lift them at one place it will be a loop another place will not be a loop that will show that the covering is not normal that means the fundamental group the image of pi one of the above space inside the pi one of x is not normal if you have a subgroup which is not normal the group cannot be abelian so that is the way we will do but we will do it next time thank you