3/3 Lagrangian Floer cohomology in families
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Transcript: English(auto-generated)
00:00
Thank you very much. So I promised you I would talk about something involving
00:24
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 Lagrangian in x.
00:42
So if Q is exact, I explained that you can build some other complex of Q equipped with a local system of the chains of the space of paths
01:03
from the base point to any point. 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. And I'm going to put another Q here just to be clear.
01:20
And this is an equivalence. And this equivalence is, as I said last time, this is just a direct generalization of Fleur's isomorphism between Fleur cohomology,
01:41
between ordinary cohomology and Fleur 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 holomorphic disk, it is not true.
02:12
We have an isomorphism from this completion that I talked about last time to CF star of Q
02:21
equipped with these large local systems. Unless you redefine the morphisms using, say,
02:42
for example, the topology on these spaces using t-adic topologies, which you may or may not want to do. So anyway, this 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.
03:02
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 there would have been a lambda. Chain here with coefficients and lambda. And here there's not a z, there's a lambda. Thanks.
03:20
OK, so now for the rest of this talk of the day, we're going to set Q to be a torus. OK, and I'm going to just unwind some of what we've been doing in that case. And hopefully we will understand things
03:42
better in this special case. So let me just observe that the homology of the base-lip space of the torus. I'm going to just do things. Let me do it with lambda coefficients, uncompleted. This is supported in degree zero, OK,
04:01
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 group ring of the fundamental group of the torus.
04:23
But the fundamental group of the torus is abelian. So this is the same thing as the group ring, the first homology of the torus with z coefficients. And for much of the talk, it will be convenient for me to just choose a basis of each one.
04:45
And in that case, you can write this as Laurent polynomials,
05:02
with Laurent polynomials and coefficients in lambda. And I will just, you know, I don't want to be writing all of this. I'll write this as, you know, my notation for this will be just lambda z i plus or minus, plus or minus, OK? That's my notation for Laurent polynomials.
05:22
So an element of this ring can be expressed as sum over alpha
05:42
in z n of c alpha, z to the alpha. And the notation is that z to the alpha is equal to z one to the alpha one, z n to the alpha n. These are the Laurent polynomials.
06:00
And that's a finite sum. And so if you know any algebraic geometry, you're supposed to think of this.
06:21
These are functions on an algebraic torus. And since I'm not an algebraic geometer, I just say, i.e., I take lambda minus zero to the n. And I try to think about functions on this space,
06:41
algebraic functions. Here I'm gonna put algebraic functions, which the notation is that this thing is called lambda star, OK? So now, let's think about that. This is the guy, this is what showed up
07:00
when we thought about the exact case, OK? But if Q is not exact, then we know that we should instead use, should consider instead this completion,
07:25
which I will write as lambda, and then here I'll put brackets z i plus or minus one. And what does this notation mean?
07:40
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 c alpha z to the alpha. What I didn't say here is that, of course,
08:00
c alpha lies in lambda. But now there is some additional condition. Oh, no, so now, I'm sorry. Oops, well, anyway, it's all the way up there. Alpha lies in z n. So I'm a little bit confused by,
08:21
what you're doing doesn't seem to be very Hamatabi. What you did last time didn't seem to be like a very Hamatabi invariant in the sense that maybe it depends on the choice of model, like chain. There's chains on the, so I'm saying is the chain, the complete addition at the chain level, it's quasi-equivalent to this H zero thing?
08:43
It's supported in degree zero, so it doesn't have any deformations. Okay, so I think the, well, let's get, 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're just walking, we used to work at the chain level and we're going to pass to homology. Now, maybe we lose some information
09:00
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. Okay, 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,
09:22
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's 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, any norm on Zn.
09:47
For example, you could just take the sum of the absolute value of alpha i, okay? The box norm, doesn't matter what you use. It's, okay, no, this is the one I want to use.
10:05
Okay, so the next thing I have to define with is, what do I mean by this valuation? So C alpha, so instead, so now I'm going to define, so the valuation is a map from lambda star to R.
10:23
And what it is, is that it takes, so the valuation of an expression of the form, ai t to the lambda i, i goes from zero to infinity, is equal to lambda zero, okay?
10:40
Where I have assumed a zero is not zero, and all the other lambda i's are strictly bigger than lambda zero. In other words, this lambda zero, the valuation is the leading exponent,
11:04
so the leading exponent of an element of the null recovery. So, okay, so that's the definition of that ring.
11:25
So this valuation of C alpha, I could have introduced it earlier, because remember, I just introduced right now this lambda star, okay? And I said there is a map to R. Let's think about the inverse image of zero.
11:44
So these are the set of non-zero elements of the null recovery, so that the leading order term has exponent zero. Well, we already talked about that, that's U lambda.
12:06
That's the unitary, that's the unitary element. Okay, are there any questions so far?
12:28
So it's actually, you know, if you, I mean, I claimed that every element of this completion of the homology of the base loop space has this form, and you can take that as an exercise.
12:42
So kind of exercise, you know, prove that this H hat of the base loop 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 homology groups here,
13:02
we define them as, you know, I take an infinite sum consisting of, you know, 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,
13:22
which correspond to a given component of the loop space. And that's how you re-express things in this way. Okay, so we understood that Laurent polynomials correspond to functions on lambda star to the N. And so now the question is, what is this,
13:40
what are these functions on? R is expression, sorry, is this rather, the ring of functions on a space.
14:03
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 answer to that, the correct way to answer this question is, we just need to find, we need to figure out
14:30
which expressions can be, let's say, plugged into the Zi's so that we get convergence.
14:50
In other words, if we start with given,
15:00
let me just now write B1 all the way to Bn in lambda star to the N, okay, we want to know. And so when does, what conditions,
15:24
coefficients Bi imply that, let's write F of B is theatically convergent whenever F lies in this series,
15:43
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,
16:01
but in the smaller subsets, you lend that to the N in the unitary element, okay? Then the valuation, by definition, as we said, the valuation of Bi is equal to zero. And since the inverse of a series
16:20
with leading order term, a constant is also a series with leading, non-zero constant is also a series with leading order term, a constant, you also get that the valuation of Bi inverse is zero, okay? So the valuation of Bi, of B to the alpha, by that I mean B1 to the alpha one
16:42
all the way to Bn to the alpha N is also zero. This is just the 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. Okay, so in particular, if F equals sum of C alpha Z to the alpha,
17:05
and I plug in F of B, this is now going to be sum of C to the alpha B to the alpha, okay? And I want to understand, does this expression, which lives in lambda, does this, well, no, does it actually live in lambda?
17:21
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 the sum plus the other guy, but the other guy is zero,
17:41
and our assumption was that this valuation goes to infinity, okay? 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
18:05
in this thing that's basically some kind of convergent power series. And the exercise is to prove the converse,
18:27
i.e., if you take a point which is not in U lambda, then you can find one of these series, you can find an F in here, so that this expression,
18:41
this evaluation doesn't make sense and you get infinitely many terms whose valuation is, you get infinitely many terms whose valuation doesn't go to infinity, so then you can't add them up. Questions? Okay, so what have we done?
19:00
So we have, I mean, in the world of algebraic geometry, we have a geometric interpretation of these rings, okay? 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,
19:30
and this completion corresponds to, well, I'm going to say analytic functions
19:44
on U lambda to the N. And so the next thing that I want to do is, so in particular, last time we constructed
20:02
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.
20:30
And this completion is the guy with the hat, a coherent chief on U lambda to the N.
20:47
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.
21:01
Maybe I'm just gonna put here, yeah, it probably makes sense in the 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.
21:22
So at some point you'd like to say, okay, well, that's all we have, that's life. But this is unsatisfactory. And unsatisfactory because the U lambda to the N, or in fact, I'm just gonna put U lambda inside lambda star
21:41
is very thin, okay? It doesn't, it's kind of, it's like a unit circle inside S1. I mean, this is basically what I'm talking about. We've constructed a coherent chief, a comp, pardon? C star, sorry. It's like, yeah, let me, S1 inside C star,
22:01
maybe that's gonna be C star. Okay, we've constructed an analytic, a complex analytic coherent chief on S1. Basically, we think thought of as like an infinitesimally thin annulus living in C star. And the expectation coming from mirror symmetry is that if you have some situation
22:20
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 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. So we expect to get larger
22:51
domains over which our coherent chiefs are defined.
23:07
But here we have to be careful, okay? So when we constructed, when I said the ring of functions, a module over the ring of functions on lambda star,
23:21
is the same thing as a coherent chief. Well, that's, 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're essentially working in some kind of analytic category. So we will produce
23:41
such a coherent chief by a local to global procedure. Instead of trying to build one thing
24:02
that's very large, that is as large as lambda star to the n, 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.
24:21
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, another chart over here.
24:41
And then we will get convergence maybe using a slightly different model in that chart as well. And if you can do this near every, near circle of every radius, then you can cover all your entire punctured plane with such annuli and produce a analytic coherent chief on C star using, you know,
25:01
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, the main exception is that instead of working over the complex number, we have to work over, we have to work over lambda. Okay? So, so the next thing I want to do
25:31
is basically justify the, the, the description, or maybe give an alternative description of these intermediate completions.
25:44
So now we consider, let's, let's for simplicity, make it a polytope P contained in H1
26:02
of our Lagrangian, which is going to be our, might as well be just be a torus with our coefficients. Sorry, give me one second. And, and for, for convenience,
26:20
I want to make it integral affine, which means that, that we have a collection, we have alpha I lying in H lower one of the torus with coefficients in Zn. And then we have some numbers, let's call them Ai,
26:41
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
27:01
is greater than or equal to Ai for all I. You should just imagine that, you know, this is H upper one. You mean, you mean with coefficients in Z? Sorry, you mean Z in the coefficients of H1? Here I meant to put Z and that's why here it's integral affine. Z. Yes. So not Z. Sorry?
27:20
Not Z to the nth order. Thank you. This is sometimes slightly technical point, but. So, so now what we can do is we can take, so we have P inside here. Fortunately, I didn't give myself enough room,
27:41
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.
28:00
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 I mean, really, this is the analog of some kind of thickening of the real torus
28:25
inside C star to the n. And the reason is that 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.
28:41
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, like in C star to the n, because this S1 to the n, remember, correspond to u lambda,
29:00
and u lambda is the inverse image of the valuation, is the zero valuation of zero under the valuation. So that lives inside here, u lambda to the n. Okay, u lambda to the n. So that's the notation we're using.
29:20
So last time I introduced, I'm gonna regret this, but it's okay. 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. Try something like HPTN.
29:46
And this, in the new notation, is denoted by lambda, and again, convergence series plus or minus, like a p here. And you can describe this as before
30:01
as the set of Laurent series. So you can write it differently. So you can, well, the easiest way to do this, actually, the limit as the norm of alpha goes to plus infinity of the valuation of C alpha
30:24
plus the pairing of alpha with p is equal to plus infinity for all p in our pot.
30:42
Yes. So these alpha, these ai's are negative, right? So you're- Yes. Yes, in my, yes. In this way. Yes. And maybe I should make them strictly, strictly negative so that they don't-
31:00
Here, if you wanted to put that in the middle of it, yeah. I would rather zero not be in there. So this is just, I mean, again, if you plug in, if you put p equals zero,
31:23
you get the condition for that, you get the condition for being in this H hat lambda, okay, without a p. But this is like, this is basically saying the same, you impose the same condition,
31:41
not only at zero, 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,
32:13
what might have been called the convergence series without a p, consisting of functions, functions f,
32:23
such that f of y converges for all y living in this y p, okay? y p was this, no, was this subset of the first cohomology, but the first cohomology is just lambda star to the n.
32:42
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 cohomology group, I shouldn't say p, that I introduced last time.
33:07
So this subset y p has a name. y p is called an affinoid domain. It's a very special kind, but the notion was introduced by Tate,
33:31
and this ring of functions is unsurprisingly an example of an affinoid ring.
33:50
And Tate proved that it has good properties. For example, it's Noetherian.
34:02
So, yes? You're trying to be lucid as a protocol for p. In fact, when we proved 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.
34:22
The Noetherian 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. Okay, so all I'm trying to say is that, well, maybe what I'm trying to say, what I'm trying to say is that these names should be suggestive because, you know, affinoid is basically saying,
34:42
you know, if you, if algebraic geometry is like, maybe I should say this. So, so the, the, the kind of thing to keep in mind, so algebraic geometry is basically you glue affine varieties by,
35:07
by, by, by, by, by algebraic maps, and then analytic geometry, maybe you should say rigid analytic geometry,
35:23
over the Novikov ring, Novikov field, I should say, is you glue affinoid domains by, by affinoid maps, by basically by maps preserving their analytic structure.
35:54
And, and so now, just to, oh, I see, it's time for me to raise the boards.
36:24
Yes, I don't want to talk about, I don't want to talk about growth in the topologies here. I was intentionally misleading.
36:41
So, anyway, so that's, that's a theory which you can use to study these spaces. And the last thing, at least the last thing that I want to recall from the last lecture is that we can, given L a Lagrangian, we constructed something called M hat L p,
37:06
which is a module over, over this ring,
37:23
which we now interpret, we can interpret coherent chief on y p. Last time I explained that this epsilon thing,
37:42
we didn't need, we can do something less refined, which is the thing that we are doing here. Intuitively, the epsilon thing defines some kind of unit ball in each one, 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,
38:00
but it's closer to the world of algebraic geometry. Okay, 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.
38:22
But we need a natural source of these bigger spaces. They need to come from some natural symplectic. 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.
38:42
So give me a second. So let me assume. I guess I should maybe notice, but the epsilon in that theorem that we used the previous time, did that depend on the almost complex structure we used? Absolutely. Everything depends on the truth. So, okay, so we will not actually get a P that is kind of independent of which chain we use?
39:02
It's not dependent of epsilon. No, it's not independent of the chain. So we will actually have to define it by saying there exists and yeah. We'll get there, yeah. Okay. So let me try to give an example of the situation in which you expect to have this larger thing, okay?
39:21
So let X be, now again, let X be symplectic. And let's say that I have a family that, okay, I have to change notation. I'm sorry. So F sub B be a family
39:42
of Lagrangian tori parameterized by B in some parameter space. This is some kind of smooth manifold.
40:03
And let me assume, 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. Hold first, the natural map from the tangent space
40:23
of B at any point into the first cohomology of the fiber at that point, Lagrangian at that point with coefficient in R is an isomorphism, okay? And two, this is gonna be some strong version of not having any holomorphic disks.
40:42
There exists a J so that F, B, okay, so that's, for all B in B, just to be clear, F, B does not bound any holomorphic disk, okay?
41:09
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,
41:26
B is the base of a torus vibration, okay? If you have a base of a Lagrangian torus vibration and you try to figure out
41:42
how the flux of the Lagrangians vary, it has to vary non-trivially because it's a vibration, they don't intersect, and then you can see that this is actually an isomorphism, okay? So now, many, I think many people, including, for example, Mike here, have thought about a similar problem,
42:01
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 FLIR cohomology, which detects some non-trivial information about how the FLIR theories can kind of be integrated over your manifold. But as I said in kind of the first two minutes
42:22
of my first talk, in the Hamiltonian setting, what you get are local systems, are locally constant shears, okay? The one reasonable question is to ask what is the analog of family FLIR cohomology in the non-Hamiltonian setting? And the big machine that I described yesterday,
42:43
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, complete generality, in the generality where the cues don't bound any holomorphic disks. Okay, that's what these guys are. So we're doing non-Hamiltonian FLIR homology. But if you think about tori,
43:01
you can make these things much more explicit because you can, 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,
43:34
which I will call Y sub B, okay, by gluing together Y sub Ps
43:47
for P contained in H1 of FB with coefficients R. In fact, this is actually quite straightforward because it's basically a version,
44:00
you know, I said that the flux map defines a map from the tangent space to H upper one. 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-Youville's theorem.
44:24
Although that's the case for torus vibrations and we don't need torus vibrations for this, but okay. So again, assumption one is in place. There is a neighborhood of any point in B
44:44
which is canonically identified with a neighborhood of zero in the first cohomology.
45:03
If you don't know that you have something coming from some type of topology, 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 I, this here's a B, here's a B prime, okay, I will now give you an element of H upper one.
45:20
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, wrong one. Okay, anyway. So compute areas of cylinders
45:44
and then get your class, okay? So if you have this neighborhood which is identified with neighborhood of zero, we might as well assume that the image is polygon by just shrinking it a little bit.
46:01
Is an integral affine polygon. What guarantees that the different the graduate nearby tori don't intersect? They could intersect. So it's not a base of the Lagrangian torus.
46:22
That's an example. Yeah. I should say EG, sorry. That's some sharp eyesight. Yes, it's not, I'm not saying it is. I'm saying that is an example.
46:40
Could be more general. So that tells you this, so I might as well assume that it's polygon, so if this is B, now you have some kind of cover by small, okay, I don't know why I'm doing this. So we get a cover, P I of B by integral affine polygon.
47:10
In fact, and the reason I shouldn't even be doing this because we're gonna have to refine this cover, but it's okay. Mahama, can I ask a silly question? If it's pretty small, can you really find
47:20
an 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. Okay, so you get this cover, and in fact, if it's sufficiently fine,
47:44
then all intersections, let me call them P sub I's, which are intersections of these guys, P I one intersect P I K, are again integral affine.
48:01
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, I mean recall as in like 15 minutes ago, there we go, I constructed, I said that there is some kind of affinoid domain associated to any integral affine polygon
48:20
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.
48:42
If you take an intersection, you just glue. So maybe you have to think a little bit about why both of these maps are actually kind of maps of, you know, domains, but they're basically inclusions. There's no, there's just some kind of change of coordinate in passing from one to the other. So it's quite straightforward to do this.
49:00
So just as a, sorry, what's going on? Have I written on the board on the top recently? Okay, let's assume I haven't, let me just move on. So, so just as an aside,
49:27
so if this X over B is a fibration, is a Taurus fibration, and you can make sure that property two holds.
49:43
So property one holds, and you know, one and two hold. And in addition, B is spin. This is kind of a small, delicate point. Then this Y sub B is in fact the mirror of X.
50:02
And if you don't know anything about mirror symmetry, just ignore the sentence. And this is just somehow what we're doing, you know, okay, you could say what we're doing is we're doing family floor cohomology in the non-Hamiltonian setting. But another thing you could say is we're just doing mirror symmetry. Are you making the assertion that Y sub B is something that you can get from an honest complex algebraic variety
50:21
like you're doing with this? No, it won't be, in general. I don't know what are the cases, the cases are, I mean, what's the point of mirror symmetry meant, that you shouldn't get? Yes, but it's not clear. I mean, I guess now we've extended it,
50:42
and it's not, okay, let's talk about that later. I mean, the answer is it's not clear how often it's supposed to be. Okay, so now we would like, so we would like a coherent sheaf on Y.
51:03
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 sheaves on the PIs,
51:22
but these are just modules over these kind of ring functions. And then one more piece of data. If all we want are coherent sheaves, then we only want one more piece of data.
51:46
And isomorphisms, I guess I should have given the name, the name to the modules. I wonder what I'm gonna call them. Let's call them MLPI, okay?
52:01
So these are modules that are associated to, associated to a Lagrange. Modules are called MLPI. And isomorphisms of these coherent sheaves over the double intersections,
52:25
what does that mean algebraically? MLPI, you can tensor that over this ring, lambda Z, where did the P go? PI, with the ring associated to PIJ.
52:47
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
53:02
of just a coherent sheaf, okay? And then when I, 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 sheaf from this. It won't be the correct answer. It'll be some kind of, you take some complex and you just take cohomology and don't worry about how, what the extensions are.
53:20
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 on, we're working at the cohomological level.
53:46
Okay, so as I said, we already know how to produce these. So let me just briefly say,
54:03
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 gonna build some cover that may be even much, that may be much, much finer. Notice that this one only depends on the symplectic topology of the vibration.
54:21
The one I'm about to build will depend on L as well. So for all B and B, there exists some Hamiltonian diffeomorphism.
54:41
Diffeomorphism, let's call this phi sub B, such that phi sub B of L is transverse to F B. Now this is an open condition. Transversality is an open condition. So we can find the neighborhood such that,
55:01
of B, let's call it now, such that phi B L is transverse to F B prime for all B prime in this neighborhood. So you would think that this would be enough.
55:21
This is kind of the hope. If they're transverse, then you can define flow cohomology 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
55:40
and canonically identify the intersections here and the intersections here. Unfortunately, flow cohomology 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.
56:04
It's really gonna be a family of Js parametrized by the interval zero one because anyway, we have some issues with our Js, but we'll worry about that later. Let's just hide that. Okay, I'll write something. Family of almost complex structures
56:23
to define CF star flow cohomology of this. Phi B L with F B. Here, I'm just saying, you know,
56:40
so that the right transversality holds. And then comes- Yeah, yeah, plus or 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.
57:03
Okay, so now comes Fukaya's swindle. So Fukaya says, well, so as I said, in principle, you would think that there is a very delicate way of trying to compare the FLIR theories here and at the nearby points. But in fact, we just pick the neighborhood of,
57:28
our 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.
57:41
There exists diffeomorphism, let's call it psi, let's put a B prime here. It's a diffeomorphism of X. Okay, let me repeat again. It's a diffeomorphism, it is not a symplectomorphism, such that following properties hold.
58:01
First thing you want to do is make sure that you don't destroy this transversality. So let's just take, let's just make sure that this diffeomorphism 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, psi B prime,
58:23
which preserves our Lagrangian, except our Lagrangian, the one we have in mind, is not L, but it's image under this Hamiltonian diffeomorphism, which made it transverse to the fiber. Just preserve it. And two, the next condition is that you want to take
58:45
FB to FB prime, okay? And finally, if you pull back your J, or your families of J, psi B prime pulls back,
59:02
let me just write it differently, preserves the tameness of J. Each, well, these two conditions are open.
59:23
Well, no, not in these conditions. 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.
59:46
Can we really achieve one and two together? There's an 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.
01:00:00
But basically you're drawing this picture, okay? You're just gonna slide this guy here keeping L where it is. So first you build it near a neighborhood of phi b L They're both transverse to it and then you extend that
01:00:24
Okay, 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
01:00:43
This Flir group of L phi b L. Sorry with f b prime, but actually holomorphic curves for this and holomorphic curves and J holomorphic curves
01:01:01
are identified by applying psi b, psi b prime Okay, there is this I mean, well, they're not the same but because all the data is compatible with applying psi b prime
01:01:21
The one you used here is the one that is pushed forward or pulled back. I don't remember. I guess here I read pushed forward From the your original J. You just take a solution to the equation you apply your diffeomorphism You get another 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 proves
01:01:44
so this implies that not only that m hat L P 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
01:02:07
M Hat, okay. This is now going to be a little bit delicate. Let me just put little b here little f b here and m hat f b prime. I'm also running out of time As I explained I mean last time I said, you know
01:02:21
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 Okay, so but now what we have done is we have constructed So I think I'm just gonna give myself five minutes So we have constructed constructed
01:02:49
Coherent sheaves over an An affinoid cover of
01:03:00
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 The base by neighborhood satisfying all these conditions For each such thing you produce you produce a module which we now can think of as a as a coherent sheaf on this affinoid domain and
01:03:24
Then the only thing left to do so we must glue these together Ie We must compare them To each other
01:03:40
when we restrict to the double intersections and that I don't have any time to say anything about but only To say that you know, what you should you should keep in mind is now that now I have like a b0 here
01:04:03
Okay, and now I have a b1 Over this b0. Okay, mate, they're not polygons anymore, but it doesn't This 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
01:04:20
So the Lagrangians that we are considering over this this this polygon and the ones over this polygon They're not the same. Okay 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
01:04:40
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 floor complex to the other over every possible point you could build here
01:05:01
But that's not good enough because you need to have the equivalence as modules over these big rings Okay, so then you need to actually make sure that the that the continuation maps converge so we need convergence of Continuation, okay
01:05:24
So you step back and you refine the cover so there's kind of many technical things which I which I decided to overlook and there's only one of them which I want to point out which is that in
01:05:41
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 they're satisfying the coach cycle condition is An additional constraint which we didn't talk about at all
01:06:01
One way to see that I that it's not so one issue has to do with signs Okay, so let me just erase and say 30 seconds So there is two issues that are left One issue is signs and to resolve the issues of signs you assume
01:06:22
That B is spin and the other issue has to do with some kind of action Which really has to do with choosing a 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. Okay. Thank you
01:07:00
Yes
01:07:07
Okay, so if you didn't have the tameness of the almost complex structure Well, you can you can know sure you can see that So if you want the tameness of the open stress, so the tameness of the open structure, it's a here So here I said you can use the push forward of J to define holomorph to define the floor cohomology of this thing
01:07:24
with FB prime Okay, well that wouldn't work if it wasn't a Right, if you don't have a tame almost complex structure, then you can't define this for the complex So you need it in order to actually compare the Fleur theory at the nearby point
01:07:45
Otherwise, you don't have basically another way to say that if you don't have tameness you don't have convergence If you don't have tameness, 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 converge Yes, but their areas change, okay
01:08:08
I suggest to mention along with the name of K, because it gives a better approach to normal community and geometry In particular, B can be identified with what is known as the Brokovich skeleton
01:08:22
Okay, I think Jan says B is the Brokovich skeleton
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