 So, yes, I'm aware that this is a summer school and that this is supposed to be a course. And I was sufficiently aware of it that I sent Ching Yu some, like, plan for what I was planning to talk about. And then I sat down to prepare my talk and it looks nothing like what I told her I will talk about. So, now, now, basically, you should think of today, I'm just give some general philosophy. So, you know, depending how much details you want, there may not be any theorems. And then on Wednesday, I will try to go back to the plan. Okay, so the general framework is that FLIR theory takes Lagrangians to chain complexes or maybe they're homology groups. And what you want to say is that if you have Lagrangians parametrized by some space, it's called a space P here, then you will get families of chain complexes or homology groups over P and such a thing is basically what is known as a sheaf. Now, it turns out that there's actually two completely different frameworks or different kinds of sheaves that come out of this idea. And the first one is probably the most familiar one, which is if you allow only Hamiltonian families. So, in that case, there is the familiar statement that FLIR homology is invariant under Hamiltonian isotopes. And that says that the sheaf that you get is a local system on P. And, of course, you know, when you work at the level of chain complexes, it turns out that you don't want to say local system. You want to say derived local system. But regardless, the way you should think about such an object is that it's determined by the value at one point, by the group or chain complex at one point, together with monodromy data. Monodromy data is some kind of representation of the fundamental group or chains on the base loop space or something like that. And what I want to talk about, so this is the subject of kind of another lecture course, which is not this one. So what I will talk about is a different situation. And that's where the flux map from the tangent space P at a point, a parameter space at a point to the first cohomology of L, of your Lagrangian, L, P, fiber P, with the real coefficients, is an isomorphism. So roughly speaking, if you take Lagrangians and you think about Hamiltonian versus Lagrangian isotope, there is a transverse direction to the Hamiltonian leaves. And the transverse direction is exactly controlled by first cohomology. And this is saying you have some kind of slice to that transverse direction. And in that case, the assertion is that we get something that's not local systems, but another familiar kind of class of sheaves and we get coherent sheaves on, well, now you can start modifying things, but essentially on this first cohomology on P, or some associated space. And so that's the subject of this talk to kind of explain this correspondence. So given the audience, I should start by at least giving some intuition by what is a coherent sheave. So now you have to know some algebraic geometry. So let's take their C star, which we will consider as a complex manifold. Well, in fact, for a moment, let's just take it as a complex algebraic variety. So the ring of functions on C star is some familiar object. It's just holomorphic algebraic functions on C star. So that's just Laurent polynomials. And the simplest kind of coherent sheave that I want you to think about. Well, OK, you could think about this. This is a coherent sheave is going to be some special kind sheave corresponds to some finite. In fact, I could just say finitely generated modules over this ring. Let's call this ring something so that I don't have to keep writing it. But the way they will appear in Flur theory, it's better to actually think about complexes. So in fact, perfect complexes are, in this case, again, finite rank free complexes over this ring. And you go from here to here by taking commulg. Exercise one showed that that's actually true. So if you know some algebraic geometry, this is something that you know. And if it's not, then something you should learn. So the complex that I will want you to think about and just keep in your mind is the following. So what does this mean? There is a map from R to itself obtained by multiplying by this polynomial. This polynomial vanishes at one point at one. And it vanishes to degree one. So exercise two, again, if you know some algebraic geometry, this will be easy. Otherwise, you have to think about it. Show that the cohomology of this complex is of rank one over C. So these are the kinds of complexes that show up when you think about Flur, my assertion. The assertion that when we're in this kind of transverse of the Hamiltonian situation, we get coherent sheaves says that we should see complexes of this nature. And that they kind of control Flur theory in the direction of the transverse direction. Now, the reason that I started with C star instead of C, which would have been slightly simpler, is that this picture, well, I'm going to say mirror. But to the symplactic manifold, which is r tons s1, but that's maybe better thought of as t star s1. Now, usually, one thinks about t star s1 in terms of its kind of vibration over s1. But I want to think about it in terms of its vibration over r. So I want to just, again, I'm not going to prove that this is mirror, but I want to show where this ring might appear. And this ring appears when you think about just one Lagrangian, which will be our zero section. And we're going to take this Lagrangian, and we're going to equip it with a base point. And the base point is going to be called q. So this ring r, or r, yes? Yes. So q is a coordinate on t star s1. I mean, I'll still call it q. You can erase the coordinates. I think we won't use that. So the point is that if you take the base lip space, and you take the homology of the base lip space, then this is easily seen to be this ring. So if you go back to your topology, you will see that this is H0 of any space, of the base lip space of any space, is just going to be the group ring of the fundamental group. And in this case, the fundamental group is z, and just gives you this description as Laurent Paul notes. What is the product? Do you mean here or here? You mean, again, so here you know what the product is. You mean here. Well, it's a group ring. Ah, concatenation of loops. The fact that for some reason, I thought it was easier. I don't want to talk about algebraic, arithmetic, algebraic geometry, but here it's perfectly safe to talk about this over z. So if you really want to construct this, you just kind of work with c coefficients. But things are a little bit more delicate if you're working over the integers. You do algebraic geometry over the integers, and you're doing it over z. That's the only. So you could go up there and do things a little bit more carefully, and you'll just work over z as well. OK. So what I want to explain is something very, so every picture I will draw has to, well, you know, it'll be just this s1. But actually, almost everything I will say, it doesn't matter that it's an s1 or not. You could just, you know, the notation will just be q is some manifold. And if I want to stay the theorem later, I will be a little bit more precise about what kind of smooth manifold I want. OK. So what's our goal here? Well, I said I'm going, I should produce for you some coherent sheen. And right now, it's not going to be very clear what that has to do with flux maps, but we'll, that's for maybe for next time. So goal inside x, say, is an exact Lagrangian, l inside x. So for now, another Lagrangian, we want to produce a module. Let's call it, no, that's bad notation. Let's call it m sub l over the chains on the base loop space. And it will turn out that the reason I'm saying this is that, again, if let's just step back a moment, if this q happens to be a circle, OK, so we're in this one-dimensional situation, then what we're saying is we're producing some kind of coherent sheave on c star. And if this Lagrangian here happened to be a torus, then it would happen that we're constructing a coherent sheave on c star to the end. And it turns out, actually, that if this q is anything, then out of this procedure, you're constructing a coherent sheave on a torus of dimension equal to the first co-homology. Now, in that general situation, when q is this general situation, this coherent sheave doesn't know everything that it needs to know about the FLIR theory of l and q. But for tori, there is basically nothing more. And so that's why there is this kind of, here is a little bit of a lie. You do produce this coherent sheave, but it doesn't carry all the information you want, unless you assume you're Lagrangian or tori. So anyway, so let me just explain this story for now. OK, so how are we going to produce this coherent sheave? This is actually fairly old work, but the way you do so is we, as usual, so let's assume that l is transverse to q. And then let's consider, for each intersection point, this space omega qx on q, which is the space of paths from q to x. So this is our base point, and this is an intersection point of x and q. So I said I would produce for you some module over these chains on the base lip space. And all you do is you consider, first, you consider the direct sum over all x, which is an intersection point of these two guys, of these chains, of the chains on the space of paths. And here then you have to shift by something. And I'm going to just write the index of x. And if you don't know what that is, you can just ignore it, because then you can just work. I mean, you just work without any gradings. It doesn't really affect too much. And this is a module over the chains on the base lip space. How is that? If I give you a path from x to q, and I give you a loop from q to itself, then, of course, you can concatenate. As I said, it's a differential gradient. So what kind of chains are these? I mean, are these geometric chains? When I wrote this, I used cubical chains. Is that, I mean, it's not. I used it so that I wouldn't have to worry about. So this module is like, there's a map from this times this to this given by concatenation. And then you're like to say, this induces a map on chains. And if you use singular chains, you have to subdivide. And I didn't feel like reading about subdivision, so I just used cubical chains. But it works with singular chains just as well. I don't want to say the model structure, because if I take something in x, and I concatenate, that's something in q, we get something in x. And if you take a k chain in q and a l chain, And a k plus l. Yes. So you just product the thing when you concatenate? Yes, yes. Which is why cubes are good. Because their products are cubes. Yes. Any other questions? Except for, OK, you get one. But seriously, anybody else should feel free. The paths are inside q. The paths are inside q. This is what Nate corrected. I wrote here x, and I should never have written x. Earlier, your omega q sub x, q is the space of false in q. Yeah, in q, yeah. So we have a module, differential gradient module, over this. But actually, we don't yet have the differential. Because currently, all I use is the intersection points. I didn't use any FLIR theory. So to build in some kind of FLIR theory, you have to consider moduli spaces of critical intersection points. We have a moduli space of strips. And I will just write it instead of, I mean, I'm going to assume that people have some idea what this is. And if you don't, I will just draw this picture. Yes, I question that by the r-action. OK, so if you take this moduli space, then one thing you can do, once you split the r-action, is you can evaluate a long q to get a path. So we have a map from here called evaluation to the space of paths from y to x. OK, so this is whatever. You'll just have to let me just do x to y for now. No, the moduli space itself is a space. It has a map. And that map goes to this actual space of paths from x to y. And the way I order it, so now it's not clear. And my notation is not sufficiently precise for you to know whether it goes this way or that way. But it doesn't matter. It's only one way. It doesn't make sense. OK, so you get this path by evaluation. And now the main point is that we can pick fundamental chains for these moduli spaces. In fact, I should say relative fundamental chains. So with that, I mean, you just ensure that there are manifolds by standard transversality. And then, well, since they are manifolds, you can say triangulate them and then take a sum of all the top dimensional simplices or subdivide them into cubes because we're using cubes. And then take a sum of all top dimensional simplices. So I will write those as these. And this number here is basically the degree of x. Let me just like this. That's the number in which this class lives. So when the difference in degrees is 1, then this is a class in degrees 0. It's just an element. It's just counting. So each one of these moduli spaces gets the number 1 or maybe minus 1, depending on orientations, which we are not discussing. So now I have this evaluation map and I have this class. What I said is, here I said fundamental class. It's not a, I'm saying it's a manifold, every manifold is a fundamental class. If you manifold with boundary, it doesn't have a top class in homology. But it has a class in its chains, such that the image in relative homology generates the class of the manifold real boundary. That's basically what's happening here. That's why I call it relative fundamental class. But it's a chain. You have to pick this chain. Okay. So we have this thing and so we can evaluate it. I'm not gonna keep track of this anymore, except maybe it's supposed to be y, x, let's do x1. Okay, great. So I take some Lagrangian. I look at the intersections with q. I form this big thing, which is a module over the chains on the base loop space. And I also look at these modular spaces and now I get some classes in the chains of now the space of paths from the intersection points to each other, from x to y. Now, this space, we have maps. You know, what I've, we've used already is that we have a map omega q. Let me just now write, let me just write. Let me just do this, even though it's kind of silly, cross omega qx to omega qx. Omega qx. Okay, this is what used, this is what, this is concatenation. So this is a concatenate q. But you also have a map omega qx. There's q's missing everywhere. Now, it'll just happen all the time. Cross this, which is concatenated at the other end. And these operations essentially, I mean they commute because, you know, one of them happens, one of them, well, let me just draw. And then one of them happens here. And the other one happens on the other side. Okay, so you can, you can concatenate with this first or you can concatenate and then this or this and that, it's the same answer, it doesn't matter. Okay, it depends exactly. So if you are actually saying, well, you get some cube of simplicity chains here and then take the product with this cube and the product with this cube then you have to reorder the cubes. But this is not an essential phenomenon. Okay, so we have that and we have that. And this thing here induces a map on chains. So our fundamental class, or rather its image under evaluation defines a map from the chains on the pads from q to x to the chains on the pads from q to y. Ronald, what's going on that you're telling us how did your flow theory of a deferring of the width chains everywhere, is that? Well, you can't do flow theory. You have to, in flow theory you can only, you're only allowed to take homology at the end. Is that okay? I'm telling you how to do flow theory but you're only allowed to take homology at the end. So, okay, so defines this map. But now each one of these is part of my complex M of L. These are just direct sum of these groups. And now I have a map from one to the other. So these give a differential. And that is the construction of the module over the chains on the base, the space of q that is associated to an exact Lagrangian if you want an expression. Basically what we're just saying is you take the sum of multiplication with this thing. And so now there's just a basic exercise. So take L to be this Lagrangian. So I'm sorry, q to be this Lagrangian. And take L to be this one. And now this L has the property that this area here and this area are the same, okay? So when you go through this procedure, so compute this homology of this ML as a module over the homology of the base loop space of S1, which by the way we already decided was this. And the answer that you are trying to get to is the thing that we saw at the beginning, the thing that's over there, one minus z. It's the homology of that complex, it's the same thing. Okay, so so far what I've done is talked only about exact Lagrangians. And so you may want to do things with non-exact Lagrangians. I mean, I said we're gonna try to do some kind of family fleuricomology where we are not gonna stick to a Hamiltonian isotopic class. So surely we're gonna get out of the exact situation even if we're in a situation where it made sense to talk about exact things. Well, the first thing you know, if you've ever thought about Lagrangian Fleur theory in non-exact situation is we have to work over the Novikov field. Maybe there are some cases where you can avoid it. More generally, what is the answer to that question? If you take any zero section and take any Hamiltonian isotopic. You get the simple module over the homology of the base loop space. I mean, you have to say which simple module because in case h1 is non-zero, there is h, pi1 is non-zero, there is more than one. And then you get the thing who's the thing that corresponds to the augmentation ideal of the fundamental group. Okay, so let me say something about the Novikov ring because I wasn't here last week, but I'm pretty sure that nobody talked about it. Did someone talk about it? Okay, so it's best to fix a field K. And now we're gonna talk about the Novikov ring over K, Novikov field over K. So what you want to say, what we really want to do is we want to work over Laurent series over K. So something like K and then you do power series all over it. Let me just put a T here and then you invert T. That's what we want to do. But unfortunately, and this would work if every holomorphic disc we ever encountered had integral area, then we would just, the area would show up as the powers of T. But that's not what happens in FLIR theory. You can have things of arbitrary real area. So instead of doing this, we allow real exponents. So the Novikov field over K, now this subscript I will drop. So it consists of things like this. This is possibly infinite sum. These coefficients live in K, that's why we introduced that. These exponents are real numbers. That's why I was started with the integral case. And they satisfy one condition, which is the condition that is satisfied here anyway. You can write down infinitely many, you can have things here which are infinitely, have infinitely many terms, but only if the power of T goes to infinity. Similarly here, you can have things which have infinitely many terms, but only if the limits of I goes to infinity of lambda I equals plus infinity. So that's the ring that we have to work with. So, and now there is very naive answer, naive expectation, which would be what you would expect if you have thought about the Grunge and Fleur theory and are trying to extend it to the situation, is that we should replace, should work with the chains on the base lip space with coefficient in this. And sometimes it works very well. So let me, let me just draw. So first let me just say one word if you haven't seen this for what it means to instead we're supposed to work over this ring. What it means is over there, I wrote an expression for what the differential should look like. And now you do the same thing as above. The differential is the sum. And what I need to do is I need to break up this modular space into homotopy classes. So the disks may wind around the space in different ways. And as they wind around the space in different ways, they will pick up different areas. So this expression here is just the area of the corresponding disk. And then you multiply this by what we were doing before, except of course we break up and work with only things in class beta. So this is Fleur trajectories or whatever Fleur strips in class beta. This is the standard thing that you do in Fleur theory. When you are not in the exact situation, you still use basically the same modular spaces except you introduce a weight. Just think of this as a weight. Before it was like t equals one. Pardon? Before it was like t equals one. Before it was like t equals one, yeah. So let me just show you or I should never erase that. So again, just think about the simplest situation. It's interesting enough. So take, again, the cylinder. Again, this is Q. But now I want my other Lagrangian to not be symmetric about it. Or in other words, to not be Hamiltonian isotopic to it. This is now going to be some number A. And this is now supposed to be some number B. And what you're supposed to show again, don't know what my numbering scheme is up to, whatever, show that this module corresponds to the cohomology. So I take now lambda. I take Laurent series in lambda. Laurent polynomials in lambda. And then Laurent polynomials in lambda. And now my differential, it has to do something. It's t to the B minus t to the A times z. Okay? So you're supposed to think about this example and see that before we had one minus z because we just set everything equal to that, but now this is what you get. So, and that the cohomology is non-zero. Great. So it looks like we have our extension outside of the exact situation, but I lied to you. This expression is not finite. And in fact, the following, at least to me, amazing fact is true. So let's take this L and this Q, okay? And let's deform L and Q. Let's keep it Q fix. Let's deform L by Hamiltonian isotopes. I think this is four. So the module associated to L in this very special case is invariant under Hamiltonian isotopes if L intersect Q is non-zero, is non-empty. If you look at this picture, there is an easy Hamiltonian isotope which in fact makes L and Q disjoint. You can take L and move it by a Hamiltonian isotope and you will get, okay, maybe I pushed it too far, probably a little bit closer to Q in reality, but it's okay. And then of course, this complex is zero. There's nothing to do, there's no intersections. But up until that point, you can do all the deformations you want of L and you can, as long as there are intersection points, you can compute this complex. And the exercise is to show that they're all actually isomorphic as modules over the Laurent polynomials. Laurent polynomials. So this problem, it's not as, it is a problem, but it's, yes? What could happen if you started with something with more intersection points, well, four intersection points? Well, you would get, again, an isomorphic complex. Even if you eliminate the first two, as long as they intersect, it works. Yeah, okay, yes. So I don't understand why the problem is a problem since you're working with them but why we have high expressions. Well, that's one way to say this, but it's certainly not, by I mean it's not a finite expression, I mean it's not an element of this ring. It's not an element of this chain complex, okay? An element of this chain complex is you have chains with coefficients in this. Now, what you are telling me, and of course, this happens in some of the other theories as well, is that we should actually complete this ring, okay? Is this the next board? Yeah, this is probably the next board. So to get a theory that is behaved in general, in this case, I guess I said it works fine, no problem, but really what we should do when we write such expression or what we think we should do when we write such expressions that you get convergence, we should complete these chains with coefficients in lambda. So let's write that as C hat of, now we're gonna write the whole thing, Q. So what is this? This is gonna consist of expressions of the form sum of, just give me one second, think you can do that. So sum of A i T to the lambda times, sum of T to the lambda i times alpha i, where alpha i's are elements of the chains on the base of space again with coefficients in our ground field, just like before. Before we just put in elements of our ground field and T to the i lambda. Now I just replace these elements of the ground field by chains and lambda i are real and the limit as i goes to infinity of lambda i equals to plus infinity. Okay? So great. So this is the right thing we think and it makes that expression, it makes that differential, differential is well-defined. If we similarly, you have to do a similar thing, similarly use completion on the chains from Q to X and well, that's it. Yeah, that's the module, that's the module. Just to clarify the difference between that one and this one is that this one, the alpha i's not infinitely many different alpha i's but different T, well that one, you only have finitely many alpha i's. Thank you Dussel, yes. Here you can have an infinite series but you can only have finitely many infinite series in the sense that this part can be infinite but you have to take one of these infinite things and multiply it by one element of the chains and take finite sums of those and here you don't. Is this difference analogous or the same as the fact in number theory that we take etalcomology with CL coefficients and take first homology in Z-model and then take the limit instead of taking actually homology with... I don't know anything about etalcomology but I suspect the answer is yes. And I think I want you to... You should explain to me more after the talk. Okay, so we are happy. We think we're happy but because this kind of... The theorem basically says this module over this chains on the base loop space. Well, now I can say it in different ways but let me just do it like this is an invariant quasi-isomorphism type of this module is an invariant of the Hamiltonian isotopic class of L. This thing is... Now, what is it doing? So you may have seen other invariants like there is another invariant which you could have thought about more basic. I thought you just said the opposite. If you leave that to be as joint then it's like... I completed. When you didn't complete... And so what Claude is pointing out is that there is another exercise. Is that you're supposed to show that after completion... Yes. I was gonna come back to this but the group that we were playing with... Well, because that element is invariable. That's exactly why, yes. Okay, so Deuce has done the exercise. I'm not worried about my TA doing the exercise. Okay, so there are other more basic invariants. Can you say... This should be related to... In the mirror there should be sections of O2 or... Sections of O2. No, no, no, we're not doing anything that's sophisticated. One section of the zero section You are taking the zero section and you're restricting it to smaller subsets. You take somehow some smaller chart. That's more centered around that point. When you wrote it... Well, when you... So the... Sorry, let's do it the other way. Q, these lagranges that I'm drawing actually correspond to points. So we have some point and some other point that's some finite distance away. Yeah. Yeah, the wrong section. Yes, that's right. Okay, so, sorry, that... Is it very... Is it obvious that this is well defined? No. That I completed it or before completing it? Yes, it's obvious that it's well defined. Because then now you have an infinite sum of infinite sums. Yes. So you need some kind of... Gromov compactness tells you that the only way that you can have infinitely many, you know, components of your modular spaces is if the areas go to infinity. So it's an element of that ring. One is that for each lambda i, there's a finite number of... The alpha i's have been less finite. Yeah, I understand that. But then when you put it in there, there can also be infinitely many betas that contain holomorphic disks. Yes, but the only way that the beta... So again, there are infinitely many betas, but only finitely many of them have... Yeah, but again, at least... And there are infinitely many alphas here. Yeah, but I mean the map is going to take from the completed thing, cross-completed thing to completed thing, right? So there are like two completions. Okay. We can talk about it. But basically I'm saying that the powers will add up and they will just... You don't have... You only have finitely many things in negative directions. And that's why this is okay. But the fact that the completion makes sense is it a purely algebraic fact or do I have to remember something that's in fact a geometry? No, the fact that this element lies in that completion, you have to remember something that's in fact a geometry. I wrote down a specific element. That's the fact that it lies in that completion is Gromov-Compacti stated. Sir, the homology of this module is again a module with a working chain complex, not the homology of the base loop space. So what... So let me say, what did I say here? Here I said you can take this module over the chains and it's quasi-isomorphism type. Over the chains is an invariant. Yes, but of course it can pass the homology which make your life easier. Then you don't have to say quasi-isomorphism. Okay, sorry, I kind of lost... Okay, there's more basic invariants which are basically the fluoric homology of L, of Q with L, okay? That's another more basic invariant. But there are also other more basic invariants which are you take Q and you equip it with some local system, like a representation of the fundamental group and then you take fluoric homology of L, okay? These are the ones that maybe are most familiar. I mean, I could put CF here instead of HF. And the point is that this group is integrating over all of them. It knows about all of them. E, unitary, I should put here unitary. So the above module knows about all of these. Every representation is a map from pi one. But I actually have the chains so that already has all of pi one. All the group ring of pi one is contained in my. Okay, so that sounds good. So now we have our invariant but it's kind of completely useless for the task of understanding what happens to fluoric homology in families because as we saw above, the fact that this is, as we saw above, as we saw in this exercise, the fact that this vanishes for A, B, if A is not equal to B, basically tells us that you can't use it to extract any information about what happens to Lagrangians when they vary non-trivially in their Hamiltonian isotopic laws. Maybe we can use it for other things but we can't use it for this task. But now, so we have this issue. So if we want to use the ring which actually detects and tells us something non-trivial even when the Lagrangians are not Hamiltonian isotopic, for example, in this case, that's the one we saw at the beginning, then we don't know that it's, that we in general even have a well-behaved, well-defined differential because again, the differential that we write down is infinite and the things that we allow are only finite. But if we use the one that always works, then we get a bunch of groups that are zero and we would like for them not to be zero. So can I ask, well, what do you expect such a thing to exist? It would be nice. No. So the answer is mirror symmetry but I can't somehow. This is not a lecture on mirror symmetry. You just, it's amazing that it's true. How about that? So now, I will explain why you can, how you can salvage this situation. So, so this is Foucaillat's fundamental. So I see this thing was exercise six or? Yes. Yeah, the fact that this group vanishes and before it wasn't zero. Yes. But the relation between exercise four, five and six. So is there at some moment in infinite series that appears, when does that happen? When I completed. So before I said, take the chains on the base loops with coefficients in lambda. And then I said, ah, well, this thing, as Duzer says, only has finitely many chains in it. It may have infinite, you know, the power series that show up and it may be infinite, but each chain only finitely many actual chains show up. One, like I said, is reversible in power series today. The explicit things you wrote down, exercise four and five, there were no infinite series there. There was no infinite series there. So as you vary, at which moment is there becoming infinite series that, I would say like there's this continuous deformation. Now nowhere, no, you don't, you vary that no infinite series will ever show up. But what happens is if you try to prove invariance, you can only prove invariance over the other one. I don't have a better answer to that. I should have a better answer to that question, but I don't have one. Okay, so, yes. It's, having shown you that that this thing is zero, there's still something there. You can still do something. Yes, yes, right? So I've shown you that you would like to work over this ring. And in that example, you get something very nice, which is you get this Lagrangian that knows about the other Lagrangians that are not actually Hamiltonian isotopic to it. It may know, it knows it in a very delicate way, but it still gets you and tells you information about them. But that construction doesn't make sense theoretically because in general, you have to write down, you have to work over the completion. When you work over the completion, then you get zero. And you lose this nice thing that you just had. Yeah, which exercise was the nice thing, five? Four. Four, four is the nice thing. You could get this module, it's non-zero. In fact, you can recover the Flirco-Malgiv-L from that computation, from that module. It knows everything about the Flirco-Malgiv-L, OK? Later, OK, I have to finish my talk. So now I want to talk about Foucaille's fundamental theorem, our family Flirco-Malgiv. Well, I mean, you know, somehow. He didn't say it like this, but it's somehow, it's what? So let's say that we have two Lagrangians which are transverse. Then there exists, so this is a very weak form, but I just, in this language, what I wanted to explain, a completion of the chains on the base loop space of q. So let's call this completion prime, which is intermediate between this one and this one. And both of these are strict containments. Well, this one doesn't have to be strict, but the other one has to be strict. So that, this is this prime thing, so that the module ML prime, which is again obtained by the same procedure we've been doing, except now you do it over this thing, is well-defined. That means roughly, so I don't have time to explain this. And maybe, of course, surely on Wednesday I will explain a version of this, and maybe slightly different language. But this is the main thing that makes family fluorochromology works. This one is, you can't work with in general, but it usually gives you nice answers and non-zero groups. This one you can always work with, but it gives you many zero groups. And this says you can make things slightly better by not completing all the way to here. This module is not necessarily zero. It means that that expression, this expression, sum of t to the pairing of beta with omega, fundamental class of this, gives me an element, again, analogously, in the completion of these spaces, x, y. I guess you want to look for some other property, which you need to look back for, c, x. Well, I can't assert that this will be non-zero because it can't, you know. So right now, it's just you can just, the assertion simply is there is a strict inclusion here. Can you assert it's non-zero and not example four? And which is non-zero in example four, but in example four it was this. In example four, we could have just let this prime be the, we didn't need to use any completion for things to be well-behaved. So the statement is, when you look at these expressions, it looks to you like you have to allow infinite series. But in fact, there is something that you can, you can always cut down your infinite series by a little bit. There's something that you don't, there's some further constraints on these series, which come in from Fleur theory, which means that you don't have to go, you don't always have to go to this thing that is very large. So what better property you get that you didn't, by taking one of the expressions? So that's something I will explain on Wednesday. But roughly speaking, this thing will allow you to, you know, if you now have, say, a foliation by Q, okay? By these, if this Q is just one member of a foliation, then you can try to patch these groups together. It's easier to patch these than it is to patch these. So that's maybe the one thing that, but I don't, I didn't want to. I didn't want to. Sir, the thing that you're going to produce was not the invariant. Under all Hamiltonian examples, it will not be. Will it be an invariant in a small one? Yes, in fact, the size of this completion is directly related to kind of how much hofer energy it's invariant under, okay? So anyway, so that's, I leave it at this very unsatisfactory stopping point and we will continue on Wednesday. Any further questions from Mohamed? Yeah, I still don't know what to do with my paper. Yes. If I don't see it, so I don't know what to do. Well, first of all, it could be Q. And it could also be, as long as, you know, oftentimes there's no reason to work out with complex numbers, I mean, or with characteristics zero. I can work. Sometimes it can be finite fields, yes. If we were working, real example, yes. I mean, yes, yes, yes. If you work on complex tori, for example, there's no reason to work in character six zero. With complex tori. You could do this in over the integers, basically. As I said, but all fields at once not very far from the integers.