 So let me start with some clarification. So last time, I described a module called m hat L over chains on the base of space Q. And I said that this is an invariant. Now, in order for things to actually work, so this is defined if, well, I mean, you can actually have some slightly weaker conditions. Let's say if L and Q bound no disks. So the first thing I want to do is give some context for where this module is coming from, and more generally, what happens if there are disks. And then that's the first part. Hopefully it will only take 20 minutes or so. And then in the next part, I said that there is some kind of, this is some very severe completion. You could do something slightly milder, which is still, well, which is not going to have the invariance properties, but which will no more. And today, I will give a precise statement of what it means that it recovers more information. So first, the context. Did you say anything about the bounding disks you noticed last time? It was going back and forth. At some point, I said it, but I didn't say it at the point where I introduced it. Does that mean L and Q, each of them doesn't bound disks? What, together they don't bound disks? Let me give the context, and then we'll come back to this. OK, so the context is that, so FO cubed have defined a category, which I would call the fukai category of X. X is some closed symplectic manifold, or maybe it has contact boundary. If it's open, it has some property so that polymorphic disks don't escape. And with objects, the grunge in the manifold, let's make them closed. Let's make them embedded. Now, in general, this category is kind of with morphisms, these flirt complexes. Now, in general, this category shouldn't really deserve the name of a category, because these morphisms, these complexes, they're going to have some kind of differential. But d squared is not equal to 0. Instead, we have elements, which we can call m0 of L, that lives inside the flirt co-chains of L with itself, which in this world usually identified with some co-chains on L. And m0 of L prime, it also lives in Cf star L prime L prime. And instead of getting d squared equals 0, you get d squared of x equals something like m0 L times x plus or minus x times m0 L prime. So in fact, this has a generalization. This is kind of already in FO cubed work. More generally, we have as objects the grandians L together with something I will denote E. And this is a unitary local system. System, and I think, essentially, if you look at what's written in the literature, it's only of rank 1. So I'll just impose this here, even though it's fairly straightforward to generalize what they did to the higher rank case. So what does it mean to have a unitary local system? A local system is you have a rank 1 vector bundle over your space. And in addition, you have some data of parallel transport. And parallel transport maps don't depend on the homotopy classes of pads or only depend on the class up to the pads up to their homotopy class. So this is classified as a morphism class of E is classified by the monodromy, which in the rank 1 case is an element of H upper 1 of L with coefficients in the unitary group. Now, if we were working over the complex numbers, I would just write like R mod S1 here. OK, that's basically U1. But if we were working over the novice covering, we can't do that. S1 is the set of things of norm 1. So we need the analogous thing, which is a set of things of norm 1 over the novice covering. And that is just a set of expressions of the form a0 plus sum of ai t to the lambda i, where lambda i, before it was there were arbitrary real numbers going to plus infinity. Now they still go to plus infinity, but they're greater than or equal to 0. And a0 is non-zero. So the wonderful things about these guys is they form a group under multiplication. You can just invert everything. That's your unitary group. And this is the framework that is discussed in FAQ. Now, you can define these Flur complexes. So maybe this is exercise 1. If you've never done this, if you have some idea how to define Lagrangian Flurc homology group, so try to define, just don't worry about the issue of disks bubbling. This is not what you should try to do. So try to define a differential on CF star. So this notation is going to get hideous. So L equipped with the bundle E and L prime equipped with the bundle E prime, which is defined to be the direct sum over all intersection points of L and L prime of hom over lambda from E at x to E prime at x. So the picture you have in mind is, here's my L. Here's my L prime. This L is equipped with some local system over it of rank 1. So this is going around this way. And the fiber at this point called x, that fiber I call EX. And the other Lagrangian is also equipped with its own rank 1 system. And the fiber over it, over this point, I call E prime x. So this vector space is supposed to be just linear morphisms from this to this. Now, we are in the rank 1 case. This is just a rank 1 thing over lambda. And now, you do the same thing as before. You look at your holomorphic disks and you count them appropriately. You just somehow, as I said, if you know how to do Fleur theory, you should be able to do this. But now, you have to, instead of defining a map from lambda to lambda, instead of defining a number, you should define a homomorphism from this rank 1 local system here to this rank 1 local system here. OK, so that's the exercise. So when you have this generalization, the terminology is that a pair E l E is unobstructed if the corresponding m0 group. So again, I'd have to write m, l, e, but I'm not going to vanish this. So to lambda, shouldn't it be the unitary version? No. No, no. You think, again, in the usual case you have a complex vector bundle. And the monodromy is an automorphism, which is an element of S1. I forgot to erase this. So, sorry, let me just do this one. So now, I have my own terminology, which is I say that q is tautologically unobstructed. Sorry, unobstructed. So you actually have to be a little bit careful. You have to also carry data of a complex structure around. If there exists a j, almost complex structure on x, so that q bounds no j holomorphic disk. But I can tell you that it's some count of disks with one mark point to the boundary on l. Pardon? That does not qualify as an explanation. That does not? No, it seems to do. OK, good. So that's what it is. It's some count of disks with one mark point. But again, if you don't know, just do the exercise without thinking about disks. OK, so that's the definition. And the point is that if you want to understand under what generality does this, then can I tell you that this result holds, that so ml is defined if l is unobstructed, which basically means its flare theory is well-defined, and q is tautologically unobstructed. OK, so still, this answer, from a formal point of view, should still not be very satisfactory, because up there I have a whole category, and here I just have a group. Yes? So the same thing you just said would not be true if l were tautologically unobstructed, right? No, you mean if q were only unobstructed, then you'd have a problem. Then this may not be true. But if they're both tautologically unobstructed. Then it's true. Tautologically stronger. But a lot of q means there is nothing to think about. For one particular j is good enough. You just use that j in your construct. You make sure you use the fact that when you define flare theory, when you look at your disks, you don't have to use the same j at all points. So when l and q interact, you just make them use different j's for them. OK, sorry. But that's what we need here. But there is some kind of more general context, which I think is important to realize that this group, it's really just a very small part of a general structure. So in fact, there is an enlargement. So this is now, unfortunately, there's nothing, but this is in literature, so you can just, some year I will write this down. There is an enlargement of the Foucaille category in which objects are pairs l e, where now e is not finite rank anymore. So this is a derived local system, which is, in an appropriate sense, unitary, which is actually the most delicate part. Unitary, it's kind of hard to make sense of what that means, even though I introduced the key notion for that makes it reasonable, unitary, but of arbitrary rank. So you could say, what example should I have in mind? And the example that you should have in mind is something that I already introduced. It's assigned to any point p, assigned to, I'm not calling it a point, it's called x, x in q. OK, so let me turn this into a q. Assigned to x in q, the completion of the chains on the space of pads from the base point to x, coefficients in lambda. So this is this completion that I introduced the last time. You could naively think that you're just going to take the chains on the space of pads from q to x with coefficients in lambda, but as I showed you last time, that's not a very good notion, because when you try to do theory with it, you still run into the problem of not having enough, you don't have enough infinite expressions. So you take this completion that puts them back in. And this is a local system over. So this is some kind of the canonical derived local system. That's exactly what's delicate. So in the end, you end up meaning a module over this corresponding thing. OK? You forgot to rank the word. Now, arbitrary rank. So it's delicate to know what it means for that to be unitary, but you can just define it to be a module over this thing where you just put q, q. Maybe that's an algebra, we discuss that. Then you take a module over that, and that's your notion. This is a derived local system over q. OK. So that's our claim that there is a generalization, and it has the following wonderful feature. You can notion, you can recover it like unitary or finite. We can discuss that. It has to do with, well anyway, we can discuss that. So key fact, basically, you'll need to know that a certain infinite expression converges teatically, because there are so many elements in here. So the key fact, which is the reason for the introduction of that module ML. So if q bounds no disk, so again, it's tautologically unobstructed, then if I take the Flur code chains, so I need the name for this local system, and what I will do, I will do something evil. I will just do base loop space q comma blank. So the Flur code chains of q equipped with this local system, which is the chains completed, omega q blank lambda, with itself, Flur and the morphisms. This is a ring. This is actually isomorphic to this back. Sorry, there was a q here. So now I'm going to put this q, q. So yesterday, I was talking about this guy, and we were building modules over this. Now I'm telling you that this thing appears in some very large focaya category as some endomorphism algebra as of some objective, which is just this Lagrangian equipped with some local system on it. So in particular, if you have any Lagrangian, so given any L, we can produce two modules. We can produce ML hat. This is a module over that guy, C hat base loop space. And we can also produce CF star of L. I'm not going to put any decoration on L with this object. And this is a module over this Flur complex. And the claim is that they're equivalent. These are cause isomorphic modules. So again, the point is I gave you this thing as an isolated thing. It's just some group having to do with L and q. But in reality, it's not. It's just some morphism group in some category. And so there's many additional structures that we won't have time to discuss. Sorry, I just got totally lost. What's quasi-equivalent to what? This module is quasi-equivalent to this module. So first, the equivalent is vector space. In fact, that's an exercise. I gave enough here that you can just prove that as complexes, that the chain complexes, which are this one and this one, and CF star L with these chains on the base loop space completed, are equivalent, quasi-equivalent. In fact, if you stare at them, they're actually isomorphic. The way I define this is really just this group. And you can see it because this local system is base loops from q to something. And what is that something? It's the intersection point. That's exactly how we define this. That's exactly how we define that. OK, so why did I go through this detour of introducing a bunch of nonsense? Just because, yes? This thing on the bottom, I mean, that's the flow complex of L with what? With, no. So the flow complex with this Lagrangian equipped with this local system, which is a local system on this Lagrangian whose values at blank is something like chains on the space of paths from the base point to that point. Is that OK? So why did I do this? I did this because actually, more than one person asked me, what happens if q is not tautologically unobstructed? In general, this construction, this gives an infinity deformation of the chains on the base loop space, of this completion of the chains on the base loop space, as a ring, whatever. And then in general, you don't get modules. So this ML, which we defined as a module over this guy, is not defined as a module over this guy anymore. It's not defined over this deformation. And that's somehow a story that I don't want to go into. But I just want to mention it because it's somehow important to know that this is just the beginning of a very long story. And the other thing that I want to mention is that you should think of this isomorphism here. I didn't have it in my notes and I forgot to say it. Is that this is the analog of there's a result. If you use the trivial local system of rank one, you just get that the comology of q with itself is the ordinary comology. Now I'm telling you, if you don't do that, if you use this huge thing, you get instead what you would expect. So somehow this is just some kind of computation level of this. By that I mean, the only holomorphic curve theory you need to know to prove this is the holomorphic curve theory that goes to this. And the rest is algebra. So now we switch gears. We return to these intermediate completions, to the relation between these kind of CFs, between these kind of modules, an L hat and flux. So my goal is to explain, so in the usual flow theory, if you compute the flow comology between one Lagrangian and another Lagrangian, like let me just point out, just as an aside. I take this Lagrangian here and now I take some other Lagrangian. And let's say that I've arranged things. This is a bad picture. Let's say that we've arranged things so that this area is A and this area is A. Then what we know is that the flow comology is non-zero. Great, we're happy, non-zero flow comology group. But if I take, so this is going to be Q, this is going to be L. Oops, that was L1. So now I take Q and I move it by an infinitesimal amount, some tiny little bit. And my goal is to say something about the flow comology of this Lagrangian that's been moved a little bit with the orange Lagrangian. And in usual flow theory, this is a very delicate point. If you work really hard, you can kind of work out something infinitesimal. But it's really hard to find some statement that tells you what happens to the flow comology of this when you actually move away. The reason is that you can just see over here, orange and white flow comology is non-zero. And if you did the exercise, orange and yellow, the flow comology is zero. That was one of the exercises I gave last time. So we're trying to undo this problem. We're trying to find something involving L and Q that will tell us about L and this Q epsilon. Yes, I said that in the usual setting, you take the flow comology of L with Q. And that tells you essentially nothing about the usual flow theory of L and Q epsilon. Even the Q and Q epsilon are very close to each other. One of them is zero. One of the exercises last time asked you to compute the flow comology between something like this and something like this, where the area here is A and the area here is B. And you basically see that it's zero as long as A is not equal to B. And as soon as you move this away, they're not equal. So you're done. OK. So first I will introduce something that is most natural from the point of view of symplectic topology, even though it will turn out that we will kind of massage it quite a bit to get something that's algebraically convenient. So I want to define, so given sigma, which is a chain. OK, maybe cubicle, but today I'm going to just do singular so that I don't have to worry into the base loops. I want to define the length of sigma to be, I keep worrying whether I get it right or wrong. I think it's the maximum over t in the domain of sigma sub t. So this is a family of loops. They have, with respect, you pick some Romanian metric length with respect to Romanian metric. So what we can do now is we can define, I can finally, like I said, there is some kind of epsilon, some prime completion, which wasn't quite this completion. I'm going to define another completion. So I'm going to define c hat epsilon in it. So this is kind of taking the chains of the base loop space. And I complete them as follows. This is going to consist of series as before. So sigma i is a chain, an unusual chain, coefficients in k, whatever, ground field. And now, before, we said that the limit as lambda i goes to plus infinity, as i goes to infinity of the lambda i is plus infinity. But now, instead, I'm going to require the limit, let me just give myself some room, the limit as i goes to infinity of lambda i minus epsilon times the length of sigma i equals plus infinity. So let's stare at this for a second. So I forgot what I wanted to say. So the first observation is that this is included in here. Because this quantity is always positive. So if this expression goes to infinity, then in particular, if you don't have this negative term, then these lambda i better go to infinity themselves. So that's the first observation. And of course, this includes the usual chains with no completion whatsoever. I mean, basically, because if you have finitely many terms, there is nothing to say. OK, so now comes the second exercise. Let me just erase this one, even though it's not in order. So the second exercise says, show that this is a differential graded algebra. Well, actually, if you try to show that it's a differential graded algebra, you will run into a little bit of an issue, you will run into the issue of understanding why this is a differential graded algebra. So pretend that this is a differential graded algebra and then show this is the case. Or if you know how to make this a differential graded algebra, just use whatever you do there. So the key fact is that the length of alpha, when you multiply two chains, I should have said sigma and tau, you can relate this. I will let you think about, well, OK, this is actually equal to the length of sigma plus the length of tau. And the length of the boundary of sigma. OK, so this is the one that I always get wrong. So it's the maximum. So this is less than or equal to the length of sigma. So this is like a thing you can do for chains on any space with a thought which carries a function? Yeah. People do that thing. Yes, Gromov used it to prove that there are, you know, manifolds with infinitely many contractable geodesics, for example. I mean, not basically things like that. He did it on the free-loop space. So actually, this is a very good question. Working over this ring probably carries a lot of information. But I don't know how to extract. Lots of very delicate information. But I don't know how to extract it. OK, but let me still. OK, so now comes the main thing. So why did I introduce this? Because there is a proposition which is due to Groman and Solomon. Jake, is that the right reference? Or is that your student told me? That's great. OK. And then there is a slightly. So OK, so this is not in literature, but I know that Tchakaya had gave some talks where he was doing Lagrangian-Fleur theory for immersed Lagrangians and explained basically a version of this. And there's also a paper of Duval that does something which is not quite directly applicable, but the right idea, Duval. And it says that so for fixed L and Q, there exists a quant that is an epsilon greater than 0, so that for any holomorphic strip with boundary on L and Q, let me just write it instead of doing that, U in the modular space of strips x and y with x and y intersection points. They had to assume that they were transverse, OK? So that for all such things, we have the following inequality. We have that the length, let me just write length like I wrote up there, length of the boundary of U. I would like to say that it's less than or equal but of course, I don't know that that's the case but if I multiply it by this epsilon, then that will be the case, OK? So this is called the reverse, I don't know where Jake is. He's taking notes, I don't know why he is, but OK. This is a reverse isoparametric inequality. It basically says that, and I'm not going to prove it or even say how it's true, I'll refer you to Jake, it basically says that the, you cannot have very long, you cannot have curves with very long boundary and very small area. Should the length be squared? No, no, it's not squared. You are thinking of the things that go into the proof of monotonicity in which in fact, the inequality goes the other way and then you get squared. But that's like an estimate for things, a very small area. This is an estimate of things, a very large area. So. Can I ask another question on this side? Probably, yeah, sure you can ask it, but if it's short. But it has some like non-compact matter of a finite, a finite dimensional. Yes. It's in function on it. Yes. And it did this, did I gain any information by that? Absolutely, that's what I said. That's what I said Gromov did. I mean, you can use this information to detect. Usually, for example, with the free loop space, you just detect geodesics by saying there are some homology classes. I say finite dimensional manifold. Yeah, you would get some information about how it's, yes, you get some information. Some cycles appear and disappear and that can be useful. OK, great. And sorry, I'm just making sure that I haven't missed everything. So that's the statement. And the next exercise is to show that the expression sum over t to the, remember, sorry, beta omega, this is basically area times the evaluation of the modular space of curves in classes beta and y. In classes beta, with boundaries in x and y, lies in this completion, c hat epsilon of the space of paths from x to y, q, where the epsilon is chosen as in the Gromov and Solomon theorem. So that's the connection between the geometry and the algebra curves here. So last time I said, the fact that these groups are, the fact that this ML hat that I constructed as a module over this, the fact that it's well defined as a version of Gromov compactness. Now, you should think of this as some kind of refined version of Gromov compactness. Not only do I tell you something about there's only finitely compactness for a modular space of curves of a bounded area. But actually, I also tell you something about as the area goes to infinity, what happens to the length. And once I know this refined version of Gromov compactness, I can go down from here to here, to this smaller ring. OK, so now I have to do some gymnastics to extract. Oh, it's here, I see. It's trying not to trip anybody. OK, of course, I didn't do it very intelligently. So now what does this have to do? So in particular, let me just continue, just say one word. Once you know that that element lies in that ring, you just play the same game as before. So in particular, we have something like m epsilon l hat, which is defined over this ring, module defined over this ring. And actually, at this stage, we can start talking about the connection with flux. But it turns out that it's better to use slightly. So as I said, this is an extremely delicate ring to work with. Like, for example, it may have the amount of energy that you may need to represent different homology classes will be different even if the homology classes lie in the same component. So it's really delicate. But so what we're going to do is work with some slightly less refined rings. So let's note, given any inclusion of rings like this, where we have c hat epsilon star omega q lambda inside something else. So I mean, I could do anything. I could just do any r. But I'm just going to make my life easy. Imagine we have some other completion, which is called prime, which I'm about to explain. It'll be a very special kind of use. We can, again, we get a module, get m hat prime l we get a module called m hat prime over this new ring just by tensoring. You just tensor this module with this over this. And it will, again, be perfect. And basically, this is a free module. So then when you tensor, you'll still get a free module. OK. So now the goal is to find such kind of different completions which are slightly easier to describe. So to define these completions, let me start with let me note that, so let sigma be, again, a simplex. And let me just put k here so it doesn't confuse the dimension into some fixed component of the base loop space. This is some kind of component labeled by alpha in pi 1. So then it's easy to see that this expression, the length of sigma, is greater than or equal to the minimal length of a curve gamma appearing in here, which is itself greater than or equal to the minimum of the length of gamma for gamma, which is homologous. So lying in the class of alpha in homology. And so that's what we're going to do. We're going to, instead of using this very refined things where every chain carries the information of its length and we do completion with respect to that, we're doing this slightly less refined thing where every chain carries information of its homotopy class and we kind of complete with respect to that. We forget all that and only pass to homology. So given, I haven't used this one yet, have I? So let's assume that we have a norm on H1. So your length was actually defined by given the maximum of things, right? Yes. Is this intentional? Yes, it's intentional. Yes. It's intentional. So given a norm on H1 of q with r coefficients, well, let me just do the r coefficients. We're going to define a. We're going to define a completion associated to just a norm. The easiest way to do so is to note that we have that there is a dual norm. Actually, this is not the easiest way. This is the way that later on makes the notation simplest norm on H upper 1. And let p inside of here be the set of elements of norm less than or equal to 1. And then I will define now my new completion. And this is now, we're finally somewhere. We're finally kind of at the place where algebra will appear. And the flux will make sense. So I want to define the completion with respect to this. Again, I denote it in terms of this polygon, this convex subset of first homology of this base loop space, coefficients in lambda. This is defined to be the set of expressions, x on t to the lambda i, alpha i. Where? Let me call them sigma i. Where? So sigma i, now I let it be a chain in the space of paths in some class alpha i. Sorry, OK. So far the same as before. But now my condition, instead of being something about the length of sigma i, it's just something about the minimal length of, well, not quite representatives of that homotopy class, but representatives of the corresponding homology class. So that will be limit as i goes to infinity of lambda i minus the pairing of sigma i with p, sorry, of alpha i with p. And let me just say class of alpha i goes to plus infinity for all p in. So again, let me just make sure things are clear. So alpha i, you should think of as an element of pi 1 of q. The class of alpha i is its image in first homology. I have a whole polygon p in first homology. In particular, I have elements with these little p's here, which lie in it. They are elements of first homology. So this is the pairing between homology and homology. So again, one is supposed to show that this is a ring. And in fact, we have an inclusion c hat p omega q lambda. This contains the completion with respect to epsilon. So for any given epsilon, this is true whenever the norm is sufficiently small. So the intuition here is that when the norm is very small, then the set of elements of unit norm inside h lower 1 is very, very large. But in the dual space, the set of elements of unit norm is very, very small. So p is very, very small neighborhood of the origin. And in particular, it is so small that it's smaller than this epsilon neighborhood that we have imposed here. So we have that. Great. So I promised that now finally, we will see something about this picture, which I just erased, something about flux. So if you know anything about flux, it has something to do with h upper 1. So recall that if you have an element of h upper 1, which is sufficiently close to the origin, then there exists a Lagrangian, which is in fact Lagrangian, which I will denote q sub p, which is OK. So I'm going to say slightly abuse of notation, but I want to say c infinity close to q, where how close it is to q depends on how close p is to the origin. c infinity close to q, such that the flux between q and q sub p is given by p. Now I have a defined flux. So what is the flux of such nearby Lagrangian? Well, let's just draw the one-dimensional case. Well, maybe let's not draw the one-dimensional case. Let's draw the one-higher-dimensional case. So let's assume that here was rq and here was rqp. And they're both Lagrangian, and they're kind of close to each other. Maybe some global topology, but nearby there is just some kind of neighborhood of q. So what can I do? Well, I notice that I draw this curve here. And of course, since we're working completely locally, there is a kind of a corresponding curve here. Working completely, but leaving the qp as a section of the cotangular boundary of the q? In the Weinstein neighborhood, yes. So I can connect them by a cylinder. And the cylinder, the homotopy class is essentially unique. And now my number, so I can associate this number. The cylinder has a non-trivial area. Let's call that a1. Or it has an area. Let's call that a1. This is the area of the cylinder. So now we do the same thing with the other curve. We're basically going to do this for basis of homology. There's some other curve. There's other curve here. And then you connect them by cylinder. And then you have some other area. So we have a pair of real numbers. a1, a2 are a given element of h upper 1 in the basis dual to the basis of curves we picked. And this element is invariant. It doesn't matter whether I pick a show. This does not depend on the choice of representatives. And if you get stuck, the main thing you will use is Stokes's Dear. And the fact these are Lagrangian. So that's the Lagrangian that I want to think about. So now we have a bunch of modules. So we have the thing that is like a fluorochromology of L with Q. So this guy has been upgraded to this m hat L. And maybe I should put a Q here, but let me not. And then we have the fluorochromology of L with QP. This guy has been upgraded to, which is by the same construction. Let's call it this. And then we have something which we didn't have before, which has a name. It's called m hat capital P. And now the theorem, so I'll have enough time to give a precise statement, but let me first start with a vague statement. So this is a theorem of mine using Foucaillet's swindle. So the theorem says that this determines both of these. And maybe it's best to write it to just you only say this one. And of course, the point is that this was defined only using, I mean, if P is sufficiently close. So if P is sufficiently small. So this is defined using only Q and L. When I was defining this object, there was no, let's talk about the flux and how you can produce some other Lagrangian close to it by pushing it away in its Weinstein neighborhood in a non-Hamiltonian way. That didn't appear until I wanted to explain to you why this is important. Can I reorder your choice? I mean, if I choose a norm capital P, are you saying I determined this for all small-screwed P? So the correct way to say this is, on the tangent space, there is a norm which is sufficiently small so that in the unit ball, with respect to this norm, you have this determines that. For which small-screwed P? For all small-screwed P in that unit ball. So I'll restate it. So this determines this for all P in P if capital P is sufficiently small. OK, so what do I mean by determines it? Well, in the same way that I here, I suggested that once you've defined this for this epsilon guy, you have it for any prime guy by just tensoring. OK, so what I want to note is that there is a natural map. In fact, there is a natural isomorphism from the chains. I'm not no completion right now, although we could complete for a moment. Chains on the base loop space of Q. Let me drop this space point, it's annoying otherwise, to the chains on the base loop space of Q sub P. This is just saying kind of, I mean, you have locally, you can just identify them. This is a locally trivial picture. So you can base the chain on here, or I'll give you a chain on here and just fix the way of topologically trivializing that picture. OK, so but in fact, I'm sorry, let me just do it the other way. This is an isomorphism, so let me go the other way. So in particular, this induces a map like this. Sorry, things aren't going to come. I'm writing everything the wrong way. It doesn't do that. This induces a map from the completion to this completion with this full completion here. And this is an isomorphism, and it looks like my arrows do indeed want to go the other way. I apologize for switching my maps over and over again. Now, we have inside here, we discussed that we have this milder thing. So the naive thing is you're going to say you're going to produce this from this by using this map. But that's not right. It's almost right, but not right. It's almost right. What you need to do is correct this. And you need to correct this by the flux. You need to correct this by the flux. So the correct map, so the correct map, well the best way to write it is to go back to the components. This is a component pi 0, or if you want it doesn't matter. You can just use the union of components representing a stick pi 1 of q representing, where you can just use all the elements and given homology class. And I want to map this to itself by multiplication by t to the pairing between alpha and our element p in first homology. And then I apply that isomorphism. And this is the map we use. So if you use psi sub alpha for all components, then you get what we wanted. We get the desired map from the capital P completion of the change that was based on q to this guy, current of the P. So let's call this map something. Let's call this psi, no capital letters. Maybe I'll call it psi hat. And now, finally, I can tell you what the precise version of the theorem says. Precise version of the theorem says, if capital P is sufficiently small, then there is an equivalence between m hat curly P L and the tensor product of m hat P L with the chains in the base lip space omega P q over this kind of free thing. This is just some kind of tensor product. If you do it in practice, you may have to do some derived version, but that's OK. And so that says that this now solves the problem that I started with. I told you that my goal was, here is my q. Here is my, let's use a different color. Here is my, I called it q epsilon, but now let's call it qp. I would like to know, I'd like to recover the Fleurcomology Fleur theory of L with qp, knowing only something about the Fleurcomology of L with q. And this does it. This is the formula that does it. OK, so once again, I failed to say anything that I promised you I would talk about. But next time, I will try to go down in abstractness and only talk about q, which are tori. So thank you. In this case that you've drawn, is it just giving the answer 0? Absolutely not. I thought that was the exercise from yesterday. Oh yes, 0, because the areas are the same. Yes, yes. So it tells you that this thing is 0 because of, because it's given better. But actually, what it can do this time is that you can be in a situation where the Fleurcomology of purple, of orange and q in the usual definition is actually 0. But orange and yellow is non-zero. Non-zero out of 0. You can now get non-zero out of 0. And just to understand this, the nature of this ring, how to say, I only allow long loops to be involved in very high order perturbations. Something like that, yes. You're trying to control how long the loops are that appear when you allow infinite series. This is what we're doing here. But maybe it'll make more sense tomorrow. So these q over script p, are they only defined up to C0 small and ultimately nice? No, they're defined up to, I mean, the ones that I'm talking about are C infinity small. Just do the ones that in a Weinstein neighborhood come from the graph construction. We can still return my C infinity small and ultimately nice to it. Yeah, the ones, all such things are C infinity close to each other. So in particular, that doesn't, that module is not dependent. In the end, in the, no, but in the end, the module, the quasi equivalence class in the module depends on nothing. It only depends on the Hamiltonian isotopic class of this guy. But in the construction, you use the representative that is C infinity close. I mean, in the proof of this, you don't, you can't prove this for, for a Lagrangian which is only C0 small. You have to use the one that is actually C infinity close. Who did Fakaya swindle? Well, anybody who would have tried to prove this by careful estimates. So I will explain tomorrow what the swindle is, what I think I was being the swindle. You would have expected that what I should do is go back to Groman Solomon and do some very careful estimate, additional estimate of kind of how does this affect transversality of some modular spaces. That's what you would expect. As the energy grows up, you'd like to use this to control the quantitative transversality. But in fact, you don't have to do this. You can just use some trick about swindle, about tamedness of the space of almost complex structure, openness of the space of tame, almost complex structure. And then you just don't have to do any estimates. In fact, you don't even have to, that essentially implies this. With no, with no, with no numbers, with no quantities, but it implies the existence of that epsilon.