 As Mohammed said, I had really little time to prepare this. But I would just like in this discussion to go over some of the exercises he gave during the lecture. And then I think also there are some common questions. People has been asking me, so I'll just go over that. So if you don't mind, let me first start. So let me just recall. In his lecture, we define three possible rings that you can look at or three possible chains on the base loop space. You take either not complete these chains on the base loop space or take some special completion. But then for each of them, you can define a module. So in his notation before, it was called ML. If you deduct the exercise quasi-isomorphic to four chains of L. So let me start. Remember, we have a Q. Let me just, Q and L, they don't. Sorry, let's just say they are totalologically obstructed, which is something he defined, means it doesn't bound any disk and in some closed manifold. Then we can define three modules. So one of them is just take floor coaching complex, Q, with this special local system. Take chains on the base loop space. You fix a base point, Q. So Q is the base point. And by this thought, we mean this local system at x takes the path space from Q to that point in Q. Do I need to recall this or we are fine? So this is one of the module. This is a module over. So if I'm being precise, there should be a Navico of field where consider chains in Navico field. So this is the module. Obviously, you can just act by loops on this chain complex on this local system. And then, I mean, I will do an example later, but the claim is this, which this group, you should think of since the elements in this base loop space, the chains on it, if we were writing this notation, we are taking a finite sum. So this is a finite sum of possible things coming from the Navico of field. And then where this sigma i is some chain is a preferred ground field, if you want. OK, so then similarly, you can define m hat and l hat. Also, you're supposed to think there's a quasi-asomorphic. So I mean, I'm just translating everything to flour theory now. But then now you look at a completed ring. So this is a module over this ring, which is also appeared in his lecture. But the difference between the first one and the second one is the second one is possible to be an infinite sum. I'm just going to be sloppy with things like this. And, well, oh, sorry. Of course, you will have the usual condition that this is the tiatic convergence. So that means that this exponent eventually goes to infinity. So i goes to infinity. OK, so this is the usual completion you will look at. But then in the lecture, there also appear to be a special kind. Let me just say, so let me first take this epsilon guy. So it all depends on what kind of convergence you're allowing. So this is a module over a special ring like this, where the sum still can be infinite. But you want, so this is due to the result of Grofman and Solomon that if you look at this, so this is the only condition here is that the sequence goes to infinity. But in this ring, it says that the sequence goes to infinity in a per-square rate that it has to be larger than the length of the generators. So we also define length here. So this goes has to be faster than this, so that all together it still goes to infinity. Yeah, so this is the, well, where sigma is still something. So obviously, this is a special kind of completion because we are looking at some special kind of rates that's going to infinity, OK? So let's look at all this in a special example, which he gave. So in his lecture one, we had this cylinder thing. We're fixing S1. So you're supposed to think this is a Q. And another guy, L, wrap around like this, where area of this is A, area of this is B. So let's assume, well, we want to do the case with null coefficient. So let's assume A is bigger than B for now. So the area is not the same. So we had the exercise to compute what happened in the first and the second case. That's what he gave in the lecture. So you're supposed to compute this ML, which, OK. So let's first look at the first case. What happened to this chain complex? If I look at this kind of Lagrangian in T star S1, you can think of, OK. So as a chain complex, this thing is just generated by, so its definition is that it's generated by x, which lies in the intersection point. So I take chains from this per square base point. Let's fix the base point. For example, just by convenience, let me see base point is here. To the intersection point, OK. So really, in this example, we only just have two intersection points. x is the front intersection, y is the back intersection. So this chain complex will reduce to the case that you just have two direct sum of q in our coefficient, where q is now equal to S1. So this is our chain complex, where we want to compute its homology. OK, to do that, you first have to think about. So obviously, these two things are modules over chains on the base loop space of q. And there is a way to think this differential graded algebra. The way you do this is that you first change your original definition of base loop space to be the more loop space. So that means that instead of looking at maps from per square interval 0 to 1, you look for any length to q. So that if you do things like this, you don't have a concatenation to be infinity. You just have a strict product on chain level. If I use a special chain complex later. And while he write this chains on the base loop space, a correct model for this is really taking cubical chains and normalize it. But this way means that it's n, n-symplex, just going to purely be generated by maps from interval to q. Modulo by normalized way meant that modulo maps that degenerates. So if gamma is modulo of sigma, which is degenerate. And sigma is degenerate means that it factors through a lower dimensional parameter recipes. So it's less than for this. Do you mean that the loop space is there? Oh, yeah, yeah, yeah. Sorry. I want to define this for any manifold. But here, I'm in loop space. So if you use this model to compute your chains. So it gives you a model for the chains on the base loop space of q. And if I fix a parametrization of these loops, then essentially you can see that. So I want to claim what I want to claim is that in the case of S1, you compute this homology. So I didn't tell you the differential, but differential is basically just restrict to the boundary map. Like this is just a version of a singular chains where you're using cubes. So then chains. So if I take homology for S1, what I get should be with Navikov coefficient. So this is something easily can be checked. It's just the polynomial of Navikov coefficient. Because basically everything is supported in degree 0. If you think about it, I mean, you have a base point over there. I'm mapping an undimensional family into this. If I fix parametrization, then really what's left is only the degree 0 guys survives if I model degeneracy. So this z and the inverse corresponding to if I fix orientation on S1, the wrapping number around each of the, like how many times you are wrapping around this. So that's why somehow in this example, he gave this range showed up eventually. If you're computing the homology of the base loop space, is this always true for the base loop space of S1, then it's also true of do you have to impose this moving out by the general simulation? Oh, no, no. I'm doing this because at chain level I get a DGA. That's a nice structure. But homology is always this. Chain model, you can compute this by singular homology, then you still get this result. But I want a nice chain level structure. So the claim is that if you take this normalized cubical chains, then you think about the product. Product, originally if you just take singular chains, for different simplices, you have to do subdivision. But take a cubical chains, you can just take domain, you just take the product and map these two maps in each vector. And then there's normalizing because if you don't normalize, you can compute cubical chains for the point. When you find this cubical homology for the point, then you find this is even infinite generated. Because there are some non-if you think about the homology, I mean the way it's defined, it will have like a generator in each every degree. So you don't want such a thing to appear, so you have to take normalize it. So that you see, when you take normalize, if the target manifold is just a point, it has to be a point, right? So you're saying each star here, but what you really mean is that C star, if you look at the chain complex, it doesn't have anything in dimension limit. If the chain complex would only have something in dimension, see if it's degenerate. Well, otherwise it would have chains in arbitrary dimensions, but actually, of course, it should be- Right, you don't want, right. Right. So I mean, on the chain level, if I model the degenerate systems, this is a nice model for base-loop space. So if I take this, this is the claim is this is always differential-graded algebra, and given by the Pontragon product on the base-loop space. Sorry, but I don't understand. Why don't you want to change it at every level, at every degree for the point, but that happens for single chain systems? No, I think actually, if you do this cubicle chains just, even for homology, it might be have some- Yeah, what you said was correct. Single or homology, if you don't model the degenerate cubes, you don't get the homology of a point. Yeah, I mean- It gets z and each even agree. Right, I mean, on chain, you have a higher support tool, but then on homology level, you also have homology in higher degree. So then this is something that you don't want, you want to prescribe. You have one thing in each dimension, yeah, and then you have a D which might whatever it is. So, you should change the A chains. Let's see. Oh, okay. Going back and forth. My correction was wrong. Correct, the first time you said it. My point is, okay. My point is, okay. This is a chain level supported at each degree, but on homology is also, if you don't take normalized chain, it will also support it at some degree that you don't want. So, infinitely, yeah. I think it even has infinite rank. I might be wrong. I did this exercise at one point. Because basically, you will have some alternating things show up. So, this is explaining why you take normalized here. Well, anyway, so the point is we want this. Then we want to compute this homology. Sorry, I'm going so slow. But then this means I have to redraw the picture. So, we already have a chain complex over there. Yeah, did I write it down? Yeah. So, if you write down this chain complex, so y, basically, you will say that it has to be of this form. So, let me change twice one. So, it's like easy to see. Shift by the index of y. There's some map here, which we don't know yet. Sorry, my nose is a little bit. I want to go from y to x. But basically, you have to count this J-holomorphic strip contribution. So, since I didn't want to talk about index, let me fix the index of y to be one, and zero index of x to be zero. I mean, up to renaming them, you'll have this, then you want to count the contribution of this disk and that disk. So, if the holomorphic strip should have some orientation given by the gradient flow line, let me fix their orientation like that. Then you want to, so basically, this map in Mohamad's talk is given by the evaluation of this modular space to the path space between x and y. So, that means that if you look at the contribution from this one, first of all, it's an area is a, so t to the a, then you have to consider. So, this map is given by concatenation of evaluation of the fundamental chain of strips between, so if I have a modular space for q and l, and in this case is if I fix orientation here y and x, you want to evaluate, so this is the modular space yx. But it has two contribution, so first of all, the first one will give you, so if it's, the first one will give you evaluation on this loop. So, it has a prescribed direction here, so let me call, we don't have color chalk or, let me call this loop gamma a, the other evaluation to be gamma b, because I give you all the direction, you just evaluate them. So, this is the first one is gamma a, and should be some orientation here. Let me just claim to be this one. So, this is concatenation here. So, I want to, so this is just a path space, but I want to identify this to be, so this is in the case of S1, so it's just a rank one module over the base loop space of S1. This map is just, I mean, everything is now we call co-efficient here. This map is just by concatenating a path. So, here you would like to, since we have chosen to prescribe the base point before, this map here is just given by, I choose a path from y to q and concatenate that. So, from this intersection point to q, and so then any path here, I can complete it to be a loop to be well-defined. Also, here I choose another one from intersection point of x to q. So, in this situation, I just choose this two paths, but it doesn't depend on your choice, the event choice result. So, then under this isomorphism, we want to say, what is this map? Okay. So, you go around. You will see that, so we basically have to do certain path concatenations. So, you first, let me say, go from q to y, and then this map here is like concatenation of its inverse. Can you concatenate this gamma BTB? So, my claim is that, if you want to go into detail, you can do that. My claim is that one of these paths going to wander around this S1 once, the other one gives you a trivial contribution. Maybe you can see that here. Let me see which one is running around. Let me check this one. So, supposed to go from q to y, and then- I think it's impossible to do that. Yeah. Yeah. I think I have my connotations. The one is once and the other is your, you should just say that. Okay. I check this. One is a one-day around once, the other one is one-day around zero times. So, that's how you see, there's a contribution of Z here. So, then if you take homology here, I mean, this is basically, this complex have some certain differential. So, this is another complex. So, this is a bi-complex. You can first take homology here, then you will see that this is exactly what I claimed before, and this map descends to homology like this. Because basically, I've already identified the positive direction of wrapping with Z. So, whatever the non-trivial contribution will give me a component of Z here. So, in my calculation is this. Anyway, you guys can check. So, there's a Z inverse here. Anyway, so, this is what you get. If you pass homology, then you can compute homology of this complex. So, okay. So, you will see that since this element, okay, up to multiplication by Z, this is, let me multiply by Z on both sides. This is just that, right? Then, from this complex, you can see that homology of, so, let me denote homology of each of them to be HF, okay, maybe with the head added later. So, this thing, this local system thing, basically, it's going to be non-trivial because you are now to invert. So, what you can do is not write this to be this, right? If this is isomorphism, then that means this element is invertible in this way. But basically, if you want to invert this element, you start to see infinite series, right? Just use the geometric series expansion. Then you will see that it's not possible to invert this element in the first ring so that homology is not zero. But then in the second ring, just like what I just said, because you can invert, this is isomorphism. Notation is so complicated, sorry. This has to be zero because you can write this makes sense in this completed ring. So, what do we get from this point of view? Because I have assumed this A is greater than B, then that means that I could have Hamiltonian deform this Lagrangian L to one that's Hamiltonian so chop it to it with a nearby S1. So, this S1 doesn't have to intersect this S1 here. So, what you want is that this module you ask, is quasi-asomorphism type this module here? Is quasi-asomorphism type? Is it invariant and Hamiltonian entropy? The answer is that, so also in his lecture, he stated that if you look at this head version, so quasi-asomorphism type of this is invariant under Hamiltonian isotope. But you can see from the other one, because obviously one is zero, one is not zero because they don't intersect. So, this one is not. Okay. So, this is what you got from this exercise. No, it's not the Hamiltonian, it's not the non-Hamiltonian isotope. They're all invariant of the Hamiltonian isotope, is that the point is? No, no, no, no, no, it's not invariant of the Hamiltonian isotope. So, ML, no, it's not, because I can Hamiltonian isotope this L, because there are some positive error here, right? I can without, I mean. Oh, I see, you can invariant. Yeah, I mean, isotope, that's it's disjoint, so. Yeah, okay, okay, sorry. But also, the first motive which turned out to be no zero was defined in this case, only because we were really lucky and otherwise this was really small. Yeah, yeah, yeah. I mean, I'm just. Otherwise, this thing didn't exist. Yeah, sure, sure. I mean, I could have a look at this finite sum thing. Sorry, and you can still define it, I think it just not invariant of Hamiltonian isotope. But your map, which you define, which is a concatenation, this evaluation of that, that may involve anything in the name. But, I mean, starting this complex because you're only taking finite sum, because the other component is only finite, the chains by writing this, I mean, there are some ambiguity about how he defined this. The finite sum, so, yeah, I mean, the T's can be infinite series, they're just elements of the notebook, right? Yeah, but then if you look at this sum. Finitely many signals. Yeah, so this is my point. T can be infinite, but simplex can only be finite in my name. So when you take this sum, you're not gonna get infinite sum. Right? It's always well-defined. It's not like we are lucky. But what is your map? Do you have a single differential and an L? Yeah. And that was defined as sum over all classes A in relative pi two of T to the A evaluation of the mobilized space in that class. When I evaluate, I basically got some element in the base loop space or past loop space. Yeah. So then. So I think, but there is a mistake that you should allow the coefficients to be arbitrary elements of notebook. Yeah, no, no, I'm allowing that. When you write it like that. You should, I mean, there can be infinite sums of T's. So the first C star. If you put an F instead of the T, where F is some notebook for any of them, no, no, no, no, no, no, no, no, no, no, no, no, no. I mean, T is there's something in the Navier-Covac, like a polynomial, Navier-Covac field. Yeah, but like you're, I mean, by definition, that's the linear combinations with kind of coefficients belonging to the field, Navier-Covac field. But when you write it like that, since you have your own finite things, it doesn't cover everything. Oh, I see what you mean. Yeah, yeah, yeah. Right, right, right. FI is seamless. So. Yeah, sure, the second one should also be FI and then you're asking that its valuation goes to infinity. Yeah, yeah, yeah, sorry. Well, yeah. Questions? Yeah, so the rifle ML is a direct sum of two things. Right. What is the differential of the R and the T? No, no, this, the differential is as a DGM. So this complex is the ML. The differential is this map between the two summand. Right? I can, I can design. I don't think that Muhammad mentioned it in the lecture, but in the differential, you know, when you have the sum and you can calculate the valuation of the fundamental class of the marginalized space problem, you also have to differentiate in the chains, right? I mean, yeah, I already differentiate chains over that, over here because I first take the homology of this chain complex. Don't lie, don't lie. Right, I was saying this thing on the board is a double complex. Is that making sense? I think for the last two, it would be okay. Yeah, yeah, yeah. To write T to the alpha. Yeah, I also think so. So it's just, because this T to alpha is notation in his lecture, so I got confused. He didn't write this out, so I did this thing and then got confused right now. Anyway, but in this way, you don't have ambiguity. But then, because this can be infinite in both components, it doesn't matter if I take lambda or not. But now there's this slight confusion of, I mean, I'm just algebraically confused. Yeah. Because, so this is the full completion where the homology vanished and we lost information. Yeah. But then the intermediate completion was the one where we got this. And so I haven't got, yeah, we, you mean this is the intermediate thing that we have? I think that the way you've written it now with the valuation, it looks to me like the, this Hakenode thing. Yeah, yeah, yeah. It is nothing. Yeah, because I'm basically, so in his lecture, there are only three types of things showing up. Later, he used something about epsilon. So basically there's P. So epsilon is some kind of convergent radius of some analytic function. And his P is taking in epsilon ball, the polytope. So later he changed his notation again because he want to set up the, he want to define this rigid analytic space. And that space is only nice if you look at a polytope. But now we're just looking at any epsilon now, which also, this will also show up in the first lecture. Okay. Any questions on this? But I think the point is that if A minus B is bigger than epsilon, then one over T to the A minus B Z is in the ring. And therefore that, the same balance. But if A minus B is less than epsilon. Right, right. Yeah, yeah, exactly. I was gonna say that. I mean, I'm just gonna remark on that because I also have other stuff. So you can also look at what happened, right? What happened on this chain complex I'm writing now with this ring here. So that exactly depends on, so it depends on the relationship of alpha, like you said, with this difference, then in one case one actually, then the other case doesn't work. I mean, you can do the same exercise here. Right, so if A minus B is bigger than epsilon, then the coefficients of Z to the n are actually 32 faster for it to be in this ring, therefore. Yeah, so that's why he's saying he gets a little bit of stability where if you just move a tiny bit, then you can use essentially the ring as in say, so to say, right? Yeah, yeah. Okay. Even if it's not the torus, you still get alfinoid domains, I think maybe. Oh, no, no, no, because I was talking in S1 case, really. Even only in torus case. So I haven't changed my notation to P yet. Okay, okay, okay. I'm just saying that later this P showing up when we look at the Lagrangian torus vibration and want to build a mirror. But up to this point, nothing about that yet. Everything can be defined using this epsilon without polytope. Okay, and also, okay. So this is later, we're gonna look at the case that, so if your Q is TN, so before nothing is TN, yeah? So before nothing is TN, yeah? Then, you know, basically from this computation, we did four, it's already behind the board. I mean, we did the computation for homology of S1, or base loop space of S1. So using this fact here, you can get the Lagrangian polynomial when you look at base loop space of TN. Let me just use this notation. So it means that the Laurent monomial or, I mean, it's a n-tuple, okay? Just by projecting to each factor, you can get this. So if I pass to homology, so basically, everything here is chain level, right? So I can pass to homology if I have TN. Then, because of this thing, you will see that homology of, okay. Let me maybe write it here. So you can pass to homology for each of them. In the case as restrict to TN that we're interested, really, you know what I mean. I don't want to waste time. And this one has a prescribed as epsilon, is it? And this thing is a module over the homology. So now the homology here is just Laurent polynomials in unvariable. So this is what Muhammad meant that these are the algebraic function on a punctured complex torus. Since each of them give you, well, you're supposed to think in this situation where you have Navic offering instead of the usual mirrored symmetry where you have C star, but really you have to, that's analogous to that. But this one, what are we doing is basically a completion so that you complete in Navical parameter direction, but you also complete in the Zi direction. So they are homology classes for this. So later, so this is his notation in class, but really these are Laurent series. So they are allowed to go into infinity by this bracket, formal Laurent series. So these are the algebraic formal neighborhood of what you have before. But then this one is saying that we are taking something, a module. Actually later, we're gonna show this thing even as a coherent module. So at each point for each specialization, the rank of this stays the same, but it's only true when you look at the homology of this kind. So in his notation, I think he changed to P here. But let me just, I just meant the homology of this. So this thing show up in the last lecture. Where is bad notation? Okay, so if we have this, the point of this discussion is that if I look at, really okay. So we want to compare a flour chain complex of one fiber, which is given by Q in this case with a nearby fiber. So to compare that, you have to consider, so if this family of Lagrangian isn't Hamiltonian, then there's some fluxed contribution from this variation. So, and this is gonna be important because that's exactly capturing the area difference between different J holomorphic disks. So what I meant is that, so I have something, so for example, I have a Lagrangian like this and another Lagrangian like this, which is L, so this is just S1 now. For each element alpha in S1 of R, S1 is just fiber. I'm specializing to the case of S1. Then you can define the flux to be, take a representative of this alpha, then look at, so graph of alpha. So that alpha, for example, be some harmonic one form. Then graph of alpha is some well-defined nearby fiber in the cotangent bundle of S1, okay? Then I can look at the path from zero to alpha interpolate by T alpha. For these family of forms, you can represent it again by harmonic forms. I mean, these are homology classes, maybe I should do this. Then represent by some one form, graph of them of T to the alpha, okay? Then this flux thing is just given by, you evaluate the area swiped out by, so these nearby fibers. So this point is alpha, q alpha. So you can evaluate alpha on any whole form homology class in H lower one to be the error shaded here. And this is important for us because, so basically what I want to say is that we want to look at what happened if we move this Lagrangian bit little bit and then how much is exactly controlled by this alpha in this H upper one. And then you want to look at the nearby fiber. This alpha is controlled by this. And then obviously, so in his lecture, he has this, well, Foucaille's trick, which basically saying that you can choose some J holomorphic, sorry, choose some almost complex structure and some diffeomorphism so that when you pull back that diffeomorphism of the original almost complex structure, this disc contribution here is identified with the original disc contribution here. But then, so these contributes to the flourishing complex between alpha, S1 alpha with L. So D alpha counts U alpha, which is this shaded thing, but then, where are the color chalk? Oh, okay, no, I mean no color chalk. It's probably under the table. No, I don't say anything. Anyway, so just read. But then there's another disc like here. So this U contributes to the differential in this flourishing complex, okay? In this picture. Okay, right? Oh, okay. Well, anyway, let me, oh, yeah, yeah, I mean, sure, sure, sure, okay. But I want to continue with my discussion to save time, sorry. I think everybody know what I mean. Well, anyway, so this thing counts that, but then the differential also counts area, right? So the key point where flux comes in is that you have to estimate this arrow. This arrow between, the arrow between one. So this thing in green, but it shows that in yellow is the difference between the energy of U and different energy of U alpha. And to first the approximation, this is, so Mohan would actually estimate a precise form for this, which I can talk about more, but then the most important contribution of this is actually just the flux. So you're supposed to evaluate this alpha to this generator. So to the boundary of U. So that means that although in my picture, this U boundary of U only go from here to here, right? But in principle, what could happen is that this this might wrap around this fiber in multiple times. So then you actually have to count this flux multiple times. So this, so this is like roughly, I can tell you that plus some arrow. So this pair is exactly given by the flux, okay? And then, but then this contribution may happen in every single direction. We're just doing this for one as one. So that's why eventually you're adding all this the ice. So basically what happened is that you're supposed to think of this ZI is some formal deformation of the original chain complex, such that if I plug in Z to be T to the alpha, and this alpha is the one from here, then if I evaluate the original differential with this evaluation, then I will correct my error by this amount, okay? And then, well, then the error here, what I mean by approximately this is because really, for example, in this picture, the wrapping number is zero, right? So you don't get any contribution from here. But then you can actually choose a zero section of base point and the written out, so think of this intersection as a sheet over this zero section and write this as a graph. And think of this intersection as another sheet over this zero section and written this as a graph. Then this shaded region is exactly the integral of this one form, subtract the integral of this one form, because it's exactly the area from here to here, subtract the area from here to here. So this area you can actually control by the flux and choosing some base point in this one. So this choosing base point is in the end of his lecture where he's saying that when you do actually like a change of chart, you have to consider like in different chart how these errors behavior will behave. So, I mean, so the upshot of this discussion is that using this flux variable and then the focaya streak, you can actually define isomorphism. So if you look at another completion in the end, this completion gave you a module over this string and later you can specialize to different Z value to get the corresponding fiber, which represent the fluorocoating complex with a nearby fiber. So if I want to look at the origin L with a nearby alpha, q alpha, then in this ring, okay, let me do everything with epsilon for now, since I didn't talk about the q yet. So, okay. So this ring, sorry. Oh, right, right, this is correct. So I can look at the original ring, which define just for q and then I can tensor this over this ring and then take a special fiber nearby fiber like this. And where this map is the map he introduced in the talk. So basically you need a map to specialize this fiber. So this is like a ring of function on a neighborhood, but now you're just looking at one point. I can say something about it, but maybe I might already. Yeah, I know, I know. Yeah, but this, basically this map is exactly your evaluate Z at t to the alpha because we want to go to alpha and then the flux of looking at the fiber at q and q alpha is exactly given by evaluation, like I said, and then compose the Foucaille's trick map that you can actually identify the disc contribution with this chosen difumofism for each nearby fiber. Your isomorphism sign there, this is again positive. Yeah, I mean, everything, I mean, this one, no, I mean, this is isomorphism, this is not, this is isomorphism, but if you want to, so this is, okay, so there is, in his lecture, he basically stated this result using the ML that I used at the beginning. Everything here is quasi-isomorphic. Actually, so this result, I think, he might say it in the end, but I mean, we didn't really have it in this form, but I basically want to change this notation to flow theory and tell you what happened in this case, but then in each case, this isomorphism is just a strict isomorphism. Questions on this? Or short question for anything? You don't say anything more? I am down up for here, but I don't know. Questions?