 Thank you very much. So I promised to you I would talk about something involving tori this time. And actually, this is going to happen. But first, we have to start with a correction. So let me first say, so let's q of the grongian in x. So if q is exact, I explain that you can basically build some sort of complex of q equipped with a local system of the chains of the kind of space of pads from the base point to any point. And here we can work with z coefficients if we want. This is Cf star of this with itself. And there is a natural map from the chains on the base loop space at q. I'm going to put another q here just to be clear. And this is in equivalence. And this equivalence, as I said last time, this is just a direct generalization of Fleur's isomorphism between ordinary cohomology and Fleur's cohomology in the exact case. But I asserted incorrectly last time that this is true when you pass the completion, then it's not true. So even if q bounds no homomorphic disc, it is not true. We have an isomorphism from this completion that I talked about last time. It's Cf star of q equipped with these large local systems. Unless you basically redefine the morphisms using, say, for example, the topology on these spaces using teatic topologies, which you may or may not want to do. So anyway, the statement is true. And I asserted this and this one is not true. And I think at the end of this talk, I may give some indication for why this thing fails. But I think it'll be better to just wait until the end of the talk. OK, so now that I paid my debts, yeah, there is a lambda. Here's the Roudabin lambda. Chain here with coefficients lambda. And here there's not a z, there's a lambda. Thanks. OK, so now for the rest of this talk of the day, we're going to set q to be a torus. And I'm going to just unwind some of what we've been doing in that case. And hopefully we will understand things better in this special case. So let me just observe that the homology of the baselip space of the torus, I'm going to just do things. Let me do it with lambda coefficients, uncompleted. This is supported in degree 0, because the torus, the universal cover of the torus is Rn. So the only thing that basically exists is the fundamental group. That's the only non-trivial homotopy group. So this is isomorphic to the grouping of the fundamental group of the torus. But the fundamental group of the torus is abelian. So this is the same thing as the grouping the first homology of the torus with z coefficients. And for much of the talk, it would be convenient for me to just choose a basis of each one. And in that case, you can write this as Laurent polynomials with Laurent polynomials and coefficients in lambda. And I will just, I don't want to be writing all of this. I'll write this as my notation for this will be just lambda, Zi plus or minus, plus or minus. That's my notation for the Laurent polynomials. So an element of this ring can be expressed as sum over alpha in Zn of C alpha Z to the alpha. And the notation is that Z to the alpha is equal to Z1 to the alpha 1 Zn to the alpha n. These are the Laurent polynomials. And that's a finite sum. And so if you know any algebraic geometry, you're supposed to think of this. These are functions on an algebraic torus. And since I'm not an algebraic geometry, I just say, i.e. I take lambda minus 0 to the n. And I try to think of it functions on this space, algebraic functions. Here, I'm going to put algebraic functions, which the notation is that this thing is called lambda star. So now, let's think about that. This is what showed up when we thought about the exact case. But if q is not exact, then we know that we should instead use, should consider instead this completion, which I will write as lambda. And then here, I put brackets Zi plus or minus 1. And what does this notation mean? Now what I'm going to do is I'm going to just unwind our definition of our completion. So an element of this ring is now, again, expressible as a sum of Z to the alpha. What I didn't say here is that, of course, Z alpha lies in lambda. But now, there's some additional condition. I'm sorry. Well, anyway, it's all the way up there. Alpha lies in Zn. I'm a little bit confused by what you're doing doesn't seem to be very homotopy. What you did last time didn't seem to be a very homotopy invariant in the sense that maybe it depends on the choice of model, like chain. There's chains on the, I'm saying is the chain, the completion at the chain level, is it equivalent to this H0 thing? It's supported in degrees 0, so it doesn't have any deformations. So I think, let's talk about that. Maybe if we want to talk about it, we'll talk about that later. But I think right now, you should imagine, we used to work at the chain level and we're going to pass to homology. Now, maybe we lose some information when we pass to homology. But I'm saying in this case, we don't. But even if we do, let's just lose this information and see what happens. OK, so alpha is in Zn and C alpha lies in lambda. But as we said, now instead of having polynomials, we have series. So there is a condition. And I'm going to write it now the limit in terms of an expression which I haven't defined yet as alpha goes to infinity of the valuation of C alpha equals plus infinity. So there is two undefined things. The first thing is, what do I mean by the norm of alpha goes to infinity? And here, I'm just saying, use any norm on Zn. For example, you could just take the sum of the absolute value of alpha i, just the box norm. Doesn't matter what you read. No, this is the one I want to use. OK, so the next thing I have to define is what do I mean by this valuation? So C alpha. So now I'm going to define the value. So the valuation is a map from lambda star to r. And what it is is that it takes the valuation of an expression of the form ai t to the lambda i. i goes from 0 to infinity is equal to lambda 0, where I have assumed a 0 is not 0. And all the other lambda i's are strictly bigger than lambda 0. In other words, this lambda 0, the valuation, is the leading exponent of an element of the null recovery. So that's the definition of that ring. So this valuation of C alpha, I could have introduced it earlier. Because remember, I just introduced right now this lambda star. And I said there is a map to r. Let's think about the inverse image of 0. So these are the sets of non-zero elements of the null recovery so that the leading order term has exponent 0. Well, we already talked about that. That's u lambda. That's the unitary elements. Are there any questions so far? So it's actually, I mean, I claimed that every element of this completion of the homology of the baselove space has this form. And you can take that as an exercise, so kind of exercise. Prove that this h hat of the baselove space of the torus is in fact isomorphic to these. And the proof is basically just a matter of rearranging things. When we define the homologies groups here, we define them as I take an infinite sum consisting of chains in the loop space with coefficients in our ground field, which you can imagine to be like k, like these AIs, and then some t to the lambda. And what you do in this isomorphism is you just collect together all the terms which correspond to a given component of the loop space. And that's how you re-express things in this way. OK, so we understood that Laurent polynomials correspond to functions on lambda star to the n. And so now the question is, what are these functions on? Expression. Sorry, is this rather the ring of functions on a space? So usually, this question might be kind of subtle or whatever. Or if it is a space, it's not obvious which space it is. But in this case, it's kind of completely easy to answer, because it's already given to you as something and then some expression z. So the correct way to answer this question is we just need to figure out which expressions can be, let's say, plugged into the zis so that we get convergence. In other words, if we start with given, let me just now write b1 all the way to bn in lambda star to the n, we want to know. And so when does, what conditions, the coefficients bi imply that let's write f of b is theoretically convergent whenever f lies in this series, in this ring. And the answer is actually pretty straightforward. So the first thing you observe that if b lies, well, not in this lambda star to the n, but in the smaller subsets, you land that to the n in the unitary element, then the valuation by definition, as we said, the valuation of bi is equal to 0. And since the inverse of a series of with leading order term constant is also a series with leading or nonzero constant is also a series with leading order term constant, you also get that the valuation of bi inverse is 0. So the valuation of bi of b to the alpha, by that I mean b1 to the alpha 1 all the way to bn to the alpha n, is also 0. This is just a statement that u and, so u lambda, which was described as the inverse image of the valuation, I also explained last time that it's a group under multiplication. So in particular, if f equals sum of c alpha z to the alpha, and I plug in f of b, this is now going to be sum of c to the alpha b to the alpha. And I want to understand, does this expression, which lives in lambda, does it actually live in lambda? Which means, does it theoretically converge? Which means, can I check that the valuation of this expression goes to infinity as alpha goes to infinity? But this is easy to see because the valuation of c alpha b to the alpha is just sum plus the other guy, but the other guy is 0. And our assumption was that this valuation goes to infinity. So what we have worked out is that u lambda to the n is contained in what you would call the domain of convergence of all functions in this thing. That's basically some kind of convergent power series. And the exercise is to prove the converse. If you take a point, which is not in u lambda, then you can find one of these series and you can find an f in here so that this expression, this evaluation, doesn't make sense and you get infinitely many terms whose valuation is, infinitely many terms whose valuation doesn't go to infinity. So then you can't add them up. Questions? OK, so what have we done? So we have, I mean, in the world of algebraic geometry, we have a geometric interpretation of these rings. So the summary is that this homology of the base loop space of the torus with coefficients in lambda corresponds to algebraic functions on the torus. And this completion corresponds to, well, I'm going to say, analytic functions on u lambda to the n. And so the next thing that I want to do is, so in particular, last time we constructed some modules associated to Lagrangians. So if l is a Lagrangian, then this guy, m of l, that I constructed, which was a module over this, is a coherent chief on lambda star to the n. And this completion is a coherent, when this guy with the hat, a coherent chief on u lambda to the n. So this is fine. It looks good. You have to remember that, again, this construction basically only makes sense in the exact case. Maybe it also makes sense in the monotone case. I haven't really thought about it. Maybe I'm going to put here. Yeah, it probably will make sense in monotone case. But I'll just put it in question mark. But this one makes sense in general. If you have to impose these kind of unobstructedness assumptions, which I was a little bit sloppy about at some point, but beyond that, that's what happens. So at some point, you'd like to say, OK, well, that's all we have. That's life. But this is unsatisfactory. And unsatisfactory, because u lambda to the n, or in fact, I'm just going to put u lambda inside lambda star, is very thin. It's like a unit circle inside S1. I mean, this is basically what I'm talking about. We've constructed the coherent chief a complex analytic coherent chief on S1. Basically, we think S1 has an infinitesimally thin annulus living in C star. And the expectation coming from mirror symmetry is that if you have some situation where there is any sense in which mirror symmetry holds, then even in the general case, you should still have a coherent chief over something that's about as big as lambda star, something which is locally and about as thick as lambda star. And so that I'll come back to that. Well, I'll explain a little bit what that means. We expect to get larger domains over which our coherent chiefs are defined. But here we have to be careful. So when we constructed, when I said the ring of functions module over the ring of functions on lambda star, is the same thing as a coherent chief? Well, it's like there is an equivalence, but it's not exactly the same thing. Because over there, we're working in the algebraic category. And now we already see that we are instead should be working in some kind of analytic category. So we will produce such a coherent chief by a local to global procedure. Instead of trying to build one thing that's very large, that is as large as lambda star to the end, let's just build something near the unit circle. Just imagine that we're trying to build something on C star, and somehow we're having trouble controlling our convergence. But we know that near the circle, we can ensure convergence. So the first thing we do is we try to thicken the circle a little bit. And if we can do that, and we know that we have some kind of non-zero radius around the circle where we do get convergence, then there'll be another chart over here. And then we will get convergence maybe using a slightly different model in that chart as well. And if you can do this near circle of every radius, then you can cover all your entire punctured plane with such an anuli and produce an analytic coherent chief on C star using this local to global procedure. And that's what's going to happen, except, well, we won't have time to talk about the details. But the main exception is that instead of working over the complex number, we have to work over lambda. OK? So the next thing I want to do is basically justify the description, or maybe give an alternative description, of these intermediate completions. So now we consider, let's for simplicity make it a polytope P contained in H1 of our Lagrangian, which is going to be, might as well be just be a torus, with our coefficient. Sorry, give me one second. And for convenience, I want to make it integral affine, which means that we have a collection. We have alpha i lying in H lower 1 of the torus with coefficients in Zn. And then we have some numbers. Let's call them ai lying in H, lying in R. And then our polytope P is given by the set of points P, satisfying that the pairing of P with alpha i is greater than or equal to ai for all i. You should just imagine that this is H upper 1. You mean with coefficients in Z? Sorry? You mean Z in the coefficient of the form? Here I meant to put Z, and that's why here it's integral affine. Yes, sorry? Thank you. This is sometimes slightly technical point, but so now what we can do is we can take, so we have P inside here. Fortunately, I didn't give myself enough room, so I'll just rewrite it again. P is contained in H1 of Tn with R coefficients. And remember, there is this valuation from lambda star, lambda star to R. It induces a natural map on cohomology groups. And then, so let's call this map valuation. So let's define yp to be that. And really, this is the analog of some kind of thickening of the real torus inside C star to the n. And the reason is that, oh, I should have said, containing the origin. So containing the origin. No, actually, no, that's automatic from this. It has to be compact, yeah. I guess that's not implied by polytope. I'll put it just for safety. What? I would like for it to be. So it's like a thickening of S1 to the n in C star to the n, because this S1 to the n, remember, correspond to u lambda. And u lambda is the inverse image of the valuation, is the 0 valuation of 0 under the valuation. So that lives inside here. U lambda to the n. OK, you lambda to the n. So that's the notation we're using. So last time I introduced, I'm going to regret this, but it's OK. So last time I introduced some, again, completion of the chains on the base loop space that corresponds to these things. But now I'm just going to just rewrite. So I'm going to get like HPTN lambda. And this is in the new notation is denoted by lambda. And again, convergent series plus or minus, like a P here. And you can describe this as before, as the set of Laurent series. So you can write it differently. So the easiest way to do this actually is the limit as the norm of alpha goes to plus infinity of the valuation of C alpha plus the pairing of alpha with P is equal to plus infinity for all P in our politics. Yes. So these areas are negative, right? So you're right. Yes. Yes, in my, yes. In this one. Yes. And maybe I should make them strictly, strictly negative so that they don't. Here if you want to put that in the middle. I would rather 0 or not be in there. OK. So this is just, I mean, again, if you plug in, if you put P equals 0, you get the condition for that. You get the condition for being in this H hat lambda, OK? Without a P. But this is basically saying you impose the same condition not only at 0, but at every point along the polytope. So this should be interpreted. This implies convergence. Well, actually, maybe this is an exercise. This, these series, sorry. This corresponds to the subring what might have been called a convergence series without a P consisting of functions, functions f, such that f of y converges for all y living in this yp. OK? yp was this, no, was this subset of the first homology. But the first homology is just lambda star to the n. Just, I mean, if you choose a basis, lambda star to the n, you look at all the series which happened to converge here, and you get this expression, which then is this completed, p-completed co-homology group, shouldn't say p, that I introduced last time. So this subset yp has a name. yp is called an affinoid domain. It's a very special kind, but the notion was introduced by Tate. And this ring of functions is unsurprisingly an example of an affinoid ring. And Tate proved that it has good properties. For example, it's noetherian. So yes? How do we use it as a protocol for p? In fact, when we prove this, if it were just some arbitrary norm, then as far as I know, it's still well-behaved. But I think you lose this noetherian property. The theorem property is basically saying it's the convex hull of finitely many things. You can just use these finitely many points to control everything that's happening. OK, so what I'm trying to say is that these names should be suggestive because affinoid is basically saying, if algebraic geometry is like, maybe I should say this. So the kind of thing to keep in mind, so algebraic geometry is basically you glue affine varieties by algebraic maps. And then analytic geometry, maybe you should say rigid analytic geometry, over the Novikov ring, Novikov field, I should say, is you glue affinoid domains by affinoid maps, basically by maps preserving their analytic structure. And so now, just to, oh, I see. It's time for me to raise the boards. The thing is topology of a race is historical. Yes, I don't want to talk about it. I don't want to talk about growth in the topologies here. Otherwise, it's a bit misleading, but it's a completely similar issue. I was intentionally misleading. Anyway, so that's a theory which you can use to study these spaces. And the last thing that I want to recall from the last lecture is that we can, given L and Lagrangian, we constructed something called m hat p, which is a module over this ring, which we now can interpret coherent chief on yp. Last time, I explained that this epsilon thing, we can do something less refined, which is the thing that we are doing here. Intuitively, the epsilon thing defines some kind of unit wall in H1, but with respect to some very, very complicated norm. And, well, we could just simplify our life and just work with polygons. Not as, poly carries less information, but it's closer to the world of algebraic geometry. OK, so you should think of this. So this is the construction of some kind of, it's a completely local construction right now. So it would be nice, as I said before, that we have some expectation that we can build sheaves over bigger, bigger spaces. But we need a natural source of these bigger spaces. They need to come from some natural symplactic. Can I ask a question? Yes. So besides satisfying our mirror symmetric expectations, why do we want our sheaves to be defined on these bigger domains? Because in examples, they will be. So give me a second. So let me assume. I guess I should maybe notice, but the epsilon in that theorem that you used the previous time, did that depend on the almost complex structure we used? Absolutely. Everything depends on the choice. So OK. So we will not actually get a P that is kind of independent of which chain we use. It's not dependent of epsilon. No, it's not independent of the chain. So we'll actually have to define it by saying there exists and, yeah. We'll get there, yeah. So let me try to give an example of a situation in which you expect to have this larger thing. So let me, let X be, now again, let X be symplactic. And let's say that I have a family that, OK, I have to change the notation. I'm sorry. So F is a B, be a family of the Grange and Torai, parametrized by B in some parameter space. This is some kind of smooth manifold. And let me assume that there is a condition on the flux. Remember, I defined the flux homomorphism, some version of flux. Anyway, I defined some version of flux last time. And following two conditions. Well, first, the natural map from the tangent space of B at any point into the first cohomology of the fiber at that point, Lagrangian at that point with coefficients in R is an isomorphism. And two, this is going to be some strong version of not having any holomorphic disks. There exists a J so that F, B, OK, so that for all B and B, just to be clear, F, B does not bound any holomorphic disk. So this is a condition. It may not be, this one especially may not be the most natural thing in the world, but let's just impose it. This one, the easiest way to get it, B is the base of a torus vibration. If you have a base of a Lagrangian torus vibration and you try to figure out how the flux of the Lagrangians vary, it has to vary non-trivial because it's a vibration. They don't intersect. And then you can see that this is actually an isomorphism. So now, I think many people, including, for example, Mike here, have thought about a similar problem, but where your family is some kind of Hamiltonian family. So in the Hamiltonian family, it's known that you can build some kind of family flare cohomology, which detects some non-trivial information about how the flare theories can be integrated over your manifold. But as I said in the first two minutes of my first talk, the Hamiltonian setting what you get are local systems, are locally constant shears. The one reasonable question is to ask, what is the analog of family flare cohomology in the non-Hamiltonian setting? And the big machine that I described yesterday, these kind of ML and ML hats and so on and so forth, they are the answer in complete generality, at least the local answer, in the generality where the cues don't bound any holomorphic disks. That's what these guys are. We're doing non-Hamiltonian flare homology. But if you think about tori, you can make these things much more explicit, because if you believe that algebraic geometry is geometry, then you can give them a geometric interpretation in terms of these spaces. So let me explain a little bit more. So assumption one allows us to construct space, which I will call y sub b by gluing together y sub p's for p contained in h1 of fb with coefficients of r. In fact, this is actually quite straightforward, because it's basically a version. I said that the flux map defines a map from the tangent space to h upper 1. But in fact, if that map is an isomorphism, then there is a canonical identification. So I mean, I don't know who this goes back to. I would probably say that this is some version of Arnold Yuval's theorem. Although that's the case for torus vibrations, and you don't need torus vibrations for this, but OK. So again, assumption one is in place. There is a neighborhood of any point in b, which is canonically identified with a neighborhood of 0 in the first commonsch. If you don't know that you have a simple something coming from some type of biology, you would not really expect that there is such an identification, but it wouldn't be canonical. But in fact, it is canonical. And the simple reason is, if this here is a b, here is a b prime, I will now give you an element of h upper 1. I take this b prime. I think of it as, here's my b, here's my fb, here's my fb prime. And then I compute the areas of those cylinders that I talked about last time. Oops, no, one, one. OK, anyway. So compute areas of cylinders and then get your class. So if you have this neighborhood, which is identified with neighborhood of 0, we might as well assume that the image is a polygon by just shrinking it a little bit. It's an integral affine polygon. What guarantees that the different branches near by core I don't intersect? They could understand. So it's not a base of a Lagrangian porous polygon. That's an example. I should say eg, sorry. That's a sharp eyesight. Yes, it's not. I'm not saying it is. I think that is an example. It could be more general. So that tells you this. We might as well assume that it's a final polygon. So if this is b, now you have some kind of cover by small. OK, I don't know why I'm doing this. So we get a cover pi of b by integral affine polygon. In fact, the reason that we shouldn't even be doing this because we're going to have to refine this cover, but it's OK. Mohamed, can I ask a silly question? Yes. If it's pretty small, can you really find a integral polygon or do you want a rational polygon? The coefficients are the slopes are integral, but these quantities are real. That's a good question. OK, so you get this cover. And in fact, if it's sufficiently fine, then all intersections, let me call them p sub i's, which are intersections of these guys, p i 1 intersect p i k, are again integral affine. In fact, I think I only need the double intersections for what I'm about to do, maybe the triple intersections too. So now recall as in like 15 minutes ago, there we go. I constructed that I said that there is some kind of affinoid domain associated to any integral affine polygon in first cohomology. So now we define y b to be the union of all the y p i's. Modulo the equivalence relation, this is induced by y p i j mapping into both. If you take an intersection, you just glue it. So maybe you have to think a little bit about why both of these maps are actually kind of maps of affinoid domains, but they're basically inclusions. There's just some kind of change of coordinate in passing from one to the other. So it's quite straightforward to do this. So just as a, sorry, what's going on? Have I written on the board on the top recently? OK, let's assume I haven't. Let me just move on. So just as an aside, so if this x over b is a torus vibration, and you can make sure that property 2 holds, so property 1 holds, and 1 and 2 hold, and in addition, b is spin, which is kind of a small, delicate point, then this y sub b is, in fact, the mirror of x. And if you don't know anything about the mirror symmetry, you just ignore the sentence. But this is just somehow what we're doing. OK, you could say what we're doing is we're doing family flow homology in the non-Hamiltonian setting. But another thing you could say is we're just doing mercenaries. Are you making it so that y sub b is something that you can get from an honest, complex object like the idea of a mirror symmetry? No. It won't be, in general. I don't know what are the cases in this. The cases are nice. In one spot I have a mirror symmetry map that you should get. Yes, but it's not clear. I guess now we've extended it. And it's not clear. OK, let's talk about that later. But the answer is it's not clear how often it's supposed to be. OK, so now we would like a coherent chief on yb. And one way to get this, we'd like a cover y sub pi, possibly not this original cover that I started drawing on the board. And coherent chiefs on the p i's. But these are just modules over these kind of ring of functions. And then one more piece of data. If all we want are coherent chiefs, then we only want one more piece of data. And isomorphisms, I guess I should have given the name to the modules. I wonder what I'm going to call them. Let's call them ML pi. So these are modules that are associated to a Lagrange. Modules are called ML pi. And isomorphisms of these coherent chiefs over the double intersections. What does that mean algebraically? ML pi, you can tensor that over this ring and z. Where did the p go? With the ring associated to pij. And this should be isomorphic to the one built from pj. So currently I am lying, because of course if this will be a construction of just a coherent chief. And then when I'm about to end the talk, I will say, well, really what you need to do is do all this at the chain level. But you can certainly get a coherent chief from this. It won't be the correct answer. It'll be some kind of you take some complex and you just take the homology and don't worry about how what the extensions are. So we already have a DG module like yesterday? We have a DG module over each one of these. But the description of how you would go from this local thing to a global thing is not as simple as what I'm saying here. So here we are working at the homological level. OK, so as I said, we already know how to produce these. So let me just briefly say. So when I said that there is some cover using this embedding, using this kind of Arnold, this classical symplectic topology, I said, well, I can build a cover from it. Now we're going to build some cover that maybe much, much finer. Notice that this one only depends on the symplectic topology of the vibration. The one I'm about to build will depend on L as well. So for all B and B, there exists some Hamiltonian difumorphism. Diffumorphism, let's call this phi sub B, such that phi sub B of L is transverse to FB. Now this is an open condition. Transversality is an open condition. So we can find the neighborhood such that of B. Let's call it now, such that phi B L is transverse to FB prime for all B prime in this neighborhood. So you would think that this would be enough. This is kind of the hope if they're transverse, then you can define flow-comology between them. And they look very much the same. Because if you have B here and B prime here, and you try to keep track of the intersections between L and the fibers, then you can just parallel transport it canonically, identify the intersections here and the intersections here. Unfortunately, flow-comology is not just defined in terms of intersections. It's defined in terms of the intersections and the holomorphic disks. And so this neighborhood which satisfies this geometric property is not going to be good enough. Instead, we have to pick some kind of J. It's really going to be a family of J parameterized by the interval 0, 1. Because anyway, we have some issues with our Js, but I'm not going to worry about that later. Let's just hide that, and OK, I'll write something. Family of almost complex structures to define CF star flow-comology of this phi B L with FB. Here I'm just saying so that the right transversality and then comes over there, you really have a plus and minus. Yeah, plus and minus everywhere. I'm just that would take too long to write. Yes, sorry, is there anywhere else it's missing? Here, I think I wrote it. OK, so now comes Fakaya's swindle. So Fakaya says, well, so as I said, in principle, you would think that there is a very delicate way of trying to compare the flow theories here and at the nearby points. But in fact, we just pick the neighborhood of B so that for all B prime and B, in fact, in the end, you want to do things slightly more controlled, but let me just say it like this. There exists defumorphism. Let's call it psi. Let's put a B prime here. It's a defumorphism of X. Let me repeat it again. It's a defumorphism. It is not a symplectomorphism. Such that following properties hold. The first thing you want to do is make sure that you don't destroy this transversality. So let's just make sure that this defumorphism takes our Lagrangian to its. So is it for B prime in the neighborhood? So for all B prime in the neighborhood, thank you. There exists this psi B, which preserves our Lagrangian. Except our Lagrangian, the one we have in mind, is not L, but its image under this Hamiltonian defumorphism, which made it transverse to the fiber. Just preserve it. And two, the next condition is that you want to take FB to FB prime. And finally, if you pull back your J or your family's of J, psi B prime pulls back. Let me just write it differently. Preserves the tameness of J. Well, these two conditions are open. Well, no, not this condition. This condition is open. And these conditions are E. You can easily see that you can achieve them in a neighborhood of the identity. And if given such a J, given such a psi B, psi B prime. Can we really achieve it if we want it to be good? There's intersection point. Yes, that's why they're transverse over the entire neighborhood. That means over the entire neighborhood. We can talk about it later. But basically, you're drawing this picture. You're just going to slide this guy here, keeping out where it is. So first, you build it near your neighborhood of FBL. They're both transverse to it, and then you extend that. So when you have this, then what you can do is you can, as I said, pull back J. And this is, again, this is almost complex structure by assumption. So this can be used to define this Flur group of L, FBL, sorry, with FB prime. But actually, holomorphic curves for this and holomorphic curves and J-holomorphic curves are identified by applying psi B, psi B prime. They're not the same. Because all the data is compatible with applying psi B prime, the one you used here is the one that is pushed forward or pulled back. I don't remember. I guess here I pushed forward. From your original J, you just take a solution to the equation. You apply your diffeomorphism. You get a solution to the equation with now these different boundary conditions. That's all that's happening. And this is what proves the result that I explained last time. So this implies that not only that m hat Lp, well, now let's call this neighborhood p, not just that m L hat p is defined, not only that, but that it also, it recovers m hat. OK, this is now going to be a little bit delicate. Let me just put little b here, little Fb here. And hat Fb prime, I'm also running out of time. As I explained, I mean, last time I said what you basically do is you take this thing and you tensor it with a ring and you'll get this. But if you tensor it with basically the same ring, but over a different map, you'll get this one. So but now what we have done is we have constructed. So I think I'm just going to give myself five minutes. So we have constructed coherent chiefs over an affinoid cover of y sub b. There was this, did I call it y sub b? Yeah, y sub b. There was this big analytic space, this analytic space. Since you can now cover the base by neighborhood satisfying all these conditions, for each such thing, you produce a module, which we now can think of as a coherent chief on this affinoid domain. And then the only thing left to do, so we must glue these together. IE, we must compare them to each other when we restrict to the double intersections. And I don't have any time to say anything about, but only to say that what you should keep in mind is now that I have a b0 here. And now I have a b1. And over this b0, OK, they're not polygons anymore, but it doesn't. Over this b0, we chose this Hamiltonian diffeomorphism, phi b0 of L. And over this b1, we chose this other Hamiltonian diffeomorphism, phi b1 of L. So the Lagrangians that we are considering over this polygon and the ones over this polygon, they're not the same. So when you want to compare over the double intersection, you have to do something. But we know what you are supposed to do. You're supposed to do continuation maps. So that's what you would do. You would do a continuation map. You kind of have some kind of moving family of Lagrangians. And that would give you a chain map from one filler complex to the other over every possible point you could build here. But that's not good enough because you need to have the equivalence as modules over these big rings, OK? So then you need to actually make sure that the continuation maps converge. So need convergence of continuation, OK? So you step back and you refine the cover. So there are many technical things which I decided to overlook. And there's only one of them which I want to point out, which is that in reality, when you go through this construction and you try to check whether you get coherent sheaf or not, this refining works perfectly fine and you do get these maps. But they need to satisfy the co-cycle condition. And it turns out that satisfying the co-cycle condition is an additional constraint which we didn't talk about at all. One way to see that it's not. So one issue has to do with signs, OK? So let me just erase and say out 30 seconds. So there is two issues that are left. One issue is signs. And to resolve the issues of signs, you assume that B is spin. And the other issue has to do with some kind of action which really has to do with choosing base points. And in the case of vibrations, you want to assume that there exists a section. I haven't thought about what happens in general. OK, thank you. OK, we have time for one question. Yes. We can see quickly what we need for the time of three minutes of the almost complex structure. OK, so if you didn't have the timeness of the almost complex structure. Well, you can, no, we're sure you can see that. So if you want the timeness of the open structure, it's here. So here I said you can use the push forward of J to define the Flurc homology of this thing with Fb prime. OK, well, that wouldn't work if it wasn't tamed. Right? If you don't have a tamed almost complex structure, then you can't define this Flurc complex. So you need it in order to actually compare the Flurc theory at the nearby point. Otherwise, you don't have. Basically, another way to say that if you don't have tamedness, you don't have convergence. If you don't have tamedness, you don't have Gromov compactness. And if you don't have Gromov compactness, you don't know that your series that you write down, they actually convert. Yes, but their areas change. OK. I would like to make a remark. Just a short remark, the name of Borkovich, I suggest to mention the one with the name of T, because it gives better Borkovich structure on the intermediate geometry. In particular, B can be identified with what is known as the Borkovich skeleton. OK, like Jan says, B is the Borkovich skeleton.