 Tack så mycket. Jag kommer att fortsätta den här seriösa talen. I dag blir det lite olika. Jag vill tala om hur det här inte-kontaktomologi- är related till lite fysisk. Jag ska tala lite om fysisk. Det blir en sketchy, imprecise och så vidare. Det kanske inte i spiriten av helmets polyfols. Jag hoppas att det blir intressant och att det blir britt. Först vill jag fortsätta från senaste gången. Rekomna att vi gjorde en kalkylation. Om du tog annot, så var det just annot- och så kunde vi se att det är kond normalt. Lambda U sittar i unitkotangent bundle av R3- som vi observade var bara equal till 1 jet space av S2. Det här var bara den ena kalkylation som var verkligen kallad. Men om du remerter kan vi dra framtiden. Det var en typ av torres. Ops, det var en bra bild, men det kunde vara så. Det var såna här fyra disker som... Det är bara rekollektioner, så det är inte underståndigt. Vi kunde finna de fyra holomorphiska disker- och använda flotter. Det var två generater för den här... Jag sker... Jag måste... Om hon inte är här är det okej. Det var två generater C av grad 1 och E av grad 2. Algebra var bara... De följde att D, E var equal till 0 och D, C var equal till 1- minus E till X, minus E till P, plus Q, E till X, E till P. Man kan se att de följde att de följde E till X, E till P och Q- som kodde den... X är den långtidiga tors, P är marigin och Q är den kallad av S2. Det är en relativ homologisk kallad. Nu... Vi först kände... Det var också homologisk kallad. Hormologisk kallad av S2 var bara equal till denna. Nu ska vi ta Q equal till 1. Om vi restriker denna homologisk kallad av Q equal till 1... Vi får 1 minus E till X, minus E till P, plus E till X, E till P som är equal till 1 minus E till X, 1 minus E till P. Det är inget accident. Det turneds out that these two factors divides the augmentation polynomial of any note. I want to try to show that. It gives you... This is somehow using some standard property of this SFT package. This is always when you specialize Q equals 1 for the other dots. Ja, exakt. Jag skulle ställa det som en theorem. Så för any k... The augmentation polynomial of k, specialised at Q equals 1, is divisible by this 1 minus E till X, 1 minus E till P. Why is that? In fact, this comes from... The proof comes from Kubotism maps in this Lechandrian... I'll draw the picture right here. Much like we defined the differential in this contact model GDJ, you can also define the chain map. In this case we have lambda k sitting inside this U star R3. And it bounds, it's the ideal boundary of the, say, full conormal, where the conormal sits inside T star R3. And this guy is exact, and that's somehow important for this argument. And then this Lk actually induces a chain map epsilon from the algebra of the un-naught, or of the knot. Maybe this was the notation. Into integers that counts where... And I'm drawing the picture. If I want to count epsilon over Rapecord A, so this is equal to a count of modular spaces, which I schematically draw like this, so I have a positive puncture at the Rapecord A, and then this goes down and the boundary lands on Lk. So now what I would want to do is I want to check that this is a chain map. The chain map comes from this SFT compactness result, namely, and gluing, but why is it chain map? So what do you need to do? I should say I can't read it, so I'll write this later. But this is zero. Maybe I'll write D equals zero. So dimension is equal zero, so I can't read it in such planes. But now if you think about instead looking at the modular space of D equals one such thing, so then it's boundary. So the boundary of this thing is exactly two-level buildings, which on the top has this D equals one disk, and on the bottom the second level is D equals zero disks like that. And now that means indeed that this is a chain map, because the boundary of this one-dimensional modular space now consists of such things, and they are exactly the things that contribute to this kind of fire composed with D. This is there for zero, and this is a chain map. So this is an augmentation, and what does it do on the homology variables? Well, on the homology variables, so the only homology variables surviving is P and X, and on the homology variables is just the inclusion map. So these kills, so these maps e to the x, sorry, e to the P, the meridian is map to one, and e to the x, you can use the homology variable remaining in this lambda k and map into kind of e to the x. So this shows that this factor is in the augmentation polynomial. And so we would like to also see that the other factor is in the augmentation, yes? Can you see that again, like I just said, you would be there is a reason why this factor is augmentation. Yeah, so this map if you want to, so how do you count these disks, right? So remember you had some kind of capping here, so you count them by counting how much they wrap around the generator of, this guy is just S1 times R2, right? And the R2 or D2 fills the P variable, right? So the map does not see if this disk, it kills the P variable, right? It's somehow just inclusion map on homology from torus to the solid torus is identity on A, it takes x to x and P to zero. And that's what this map does as well, right? So if that means that if I put e to the P equals 1 and I define my chain map like this, I have a chain map. So therefore e to the P equals 1 is in the augmentation righty and this factor sits in the polynomial, okay? Right, so we'd like to understand why this thing is here and it's therefore a similar reason because there is another feeling that fills out x and let's try, there are many ways to see this, but one way of seeing it is, maybe here it's better to use S3 than R3. So I haven't this yet, but let me anyway use S3 rather than R3. So here is S3 and here is R0. And then I have the conormal and going out like this. I don't have to draw it, but you see if the conormal intersects the zero section along the knot and I can there, so basically Lagrangius looks like that and I can now surger them so that I get this thing and what do I do? So what I do is I take a zero section and I join the conormal and the topology of what I get is actually the knot complement, right? So I have another feeling which is slightly more complicated, it kind of goes down and then spreads over S3, but that's the topology, so I can sort of do, it is exact. Or if you want to see that from the beginning you can take a function, Morse function that sort of explodes just along the knot, right? And take the graph of that. So I have another feeling, let's say Mk, which topologically is just S3 minus the knot and that, if I just care about, it's kind of, of course, interesting topology here, but if I care about the homology, it's not so interesting, it exactly kills the longitude and keeps the median. So therefore I also have this factor explained by counting disks on this other Lagrangian. So these two exact Lagrangians, they somehow give me these factors and that's kind of proof of this there. Just because I wanted to pay attention, what was the first Lagrangian? The first was just the Lagrangian co-normally itself, going down to the zero section, this one going down to zero section. So now, and as you see, this works great for this Q equals 1. And remember that Q was somehow, Q was related, it's the E to the T where T is the class of this S2. So here we feel the S2 and it's very hard to recover Q. Och så, what now I want to explain is what's with augmentation polynomial when Q is not equal to 1 and how can we see it geometrically there. And that turns out to be related to a lot of physics, so I want to kind of give just a very brief description of this physics so that otherwise it's somehow very hard to guess where things come from. So, and it's sort of more like overview, advertisement or something. And it's a long story in physics. But anyway, the starting point for this story is Charon Simon's theory. And that's, it's been, from physical point of view, is you have a three manifold and then you have Charon Simon's actions. So say S over UN connection, so three manifold M. And then, and A maybe I should say UN connection. And then you take this action form. So this action form is somehow not quite gauge invariant, but it changes, it comes from some characteristic class in four dimension. So it changes by multiples of four pi if you change the gauge thing. And then, and then in physics you would write down the path integral, which is a function of the underlying manifold. And since it's a path integral, it's hard to understand. So you integrate overall gauge equivalence classes of connections and you take this e to the i k over four pi, where k is an integer called the level of this Charon Simon's action. So this object from mathematical point does not make sense, but you can treat it with Feynman perturbation theory and get something in it. And basically, if you do that, you get an expansion in terms of, so maybe an expansion in terms of k, which is this level, and N, which is the UN. So, okay. And then the starting point for our relation is actually Witten's old result from 1992. You're saying this path integral doesn't always work fine, but you can use some Feynman perturbation to approximate it? Right, you can expand it. So the critical points of this action are the flat connections. So you can kind of expand it around these flat connections in some kind of Feynman expansion. And I'm not going to... There were a lot of things done about that, and I'm not going to talk about it, but... So Witten in 1992 saw that this Charon Simon's is actually related to topological strings here, which in our world would be some grammar-Witten theory. So what he says is that this Charon Simon's partition function is equal to a certain count. So it's really like coming from topological strings here, but when it localizes, it's a grammar-Witten invariant, and it's a little bit strange grammar-Witten invariant from our point of view, but it's the count of polymorphic curves living in the continental bundle on M with boundary on the zero-section, but you have to take n-copies. So somehow you imagine it's like this, your multi-sections, but not perturbed, and the total weight is n. So you have exactly n-copies of it lying on top of each other. And these such, if you... So the kind of... String-coupling constant. This is the genus parameter that you use when you expand this just like in SFT. So that's equal to this change of variable. So basically what Witten is saying that this is supposed to count polymorphic curves in T star M with boundary on M. Now, being kind of well-educated symplectic geometry, you see that there are only constants of such maps, but if you look at formally the dimension, you see that this is a three-dimensional Calabja manifold, and the zero-section has mass of index zero. So formally any such disk is rigid, so there should be a count. And he invented something called string-field theory to relate it directly to the Schern-Simons count. But I think this would be a great polyfold project to make sense out of this count, I think. But... So if I set k equal to zero, is it somehow clear that the right-hand side is not making sense? Which k? The k... The level of the Schern-Simons? Or... Right, then the Schern-Simons tries to just integrate over the whole space of connections. Yes. Right, so then it should not really make sense, right? Right, so then the right-hand side also shouldn't make sense. The right-hand side? It's sense, okay, sorry. Are we excluding k for some reason, k equal to zero? No, no, no. I think not. It's supposed to say for k equal to zero. This quantity is the Humphrey polynomial. The right-hand side. Wait, wait, what? So is there some specialization? Is there some idea of how I should get that just from the space, from the function one on the space of connections? Well... Or what particular thing about the space of connections? If I set k equal to zero. So you want to compute somehow the volume of this space, right? That's what you're saying, yeah. Yeah, okay, I don't know. So I didn't think about it much in these terms, but in terms of the kind of coefficients in the expansion. Of course these coefficients should somehow mean something if you go back. In... On that side, yes, but maybe perhaps not on this side, right? Right? No, I don't have a good answer, so I need to think. But this, of course, is a kind of normalization of this volume. It will come out as you're saying, this is not quite the Humphrey polynomial yet, because we have no not, right? But it will become, it will become, and we will normalize it. Is it when there's no not in my three-manifold? What is the term assignments at the coefficient, the constant coefficient? The coefficient function is not really meaningful, it's only like other expectation value divided by a partition function, which is meaningful. The z of m is all along some sort of normalization constant. Sure, but maybe let's take n equal to 1, k equal to zero. And so it's totally okay for that normalization constant to be infinity, which it might be. But still, when I evaluate at some other not, and I divide that by this infinity, I get meaningful answers. That statement there, should be enough. Let's save this for discussion anyway. I think this is a good discussion topic, it will happen after lunch. Yeah, so, right. Let me kind of go one step further, and perhaps I can, perhaps this will answer your question. So, then the next step in this story is somehow due to Guri Baffa, and they want to somehow, so they use what's called this conifold transition, and it's a way to relate various string theories, open and closed string theory. So, it's also not understood, I would say well understood mathematically, but there is a physical proof of this strange correspondence. So here, this is a picture of T star, this picture of T star S3, with n copies of the zero section. And then, T star S3 you can view as a quadric in C4, and you can pinch it to a cone, where I basically pinched the zero section, but now you can resolve it the other way. So, instead of, this indicates that I fill in the S2 by disk topologically, instead I can fill in the S3 by disk topologically. So, what I get topologically is somehow just S2 times R4, but it appears nicely if I use this algebraic geometry language as the O minus 1 bundle over Cp1. And now, what this physics prediction says, or the physics theorem, is that if I do the open, this grommawitten over here, then it's related to closed theory of closed curves here. I'm opening out the black one, maybe here. And I have to require some relation between the areas here, and I take the area of this Cp1, and that's a kind of scalar parameter on this manifold, it's T, so that's finite, and that's equal to the coupling constant, the string coupling constant, times n. So I'm sort of letting n go to infinity, and this is going to zero in a controlled way so that it ends up at T. And then the open grommawitten theory in this T star S3 with n copies on S3 should be equal to the closed grommawitten theory, so kind of grommawitten count in, let's call this x, this manifold, with this identification of parameters. I've got the area of Cp1 is called Q. I think Q is e to the T, so this is sort of logarithm of Q. Right, so now... Is there an easy interpretation of n on the right-hand side, or do I have to sum this over all n to extract it from the right side? I think you have to sum over all n, so it's not... It's not a sum, you can specialize it. It goes to infinity. Yeah, true only if n goes to infinity, I think you have to do... So now maybe we can recover this k equals zero in some understandable manner here. So what can we say? So somehow this curve count can be done, so we're supposed to do the count curve here, and you can do it... I don't know, pande pande, maybe you can do it. And this, and I'm sorry I don't answer your question, but we can also express these turn assignments. And they are in fact, so mathematically it's a sort of check that they actually agree, right? So although there is no mathematical proof of the mechanism, in this case there is a mathematical check of the prediction of this thing. Sorry, are you saying... Oh, so the check is between the two grammar points. No, no, I mean these grammar points I think we cannot quite do, but we can expand, and I'm sorry exactly as I couldn't answer your question, but there is a way to write that somebody else probably knows well to write this partition function for S3 in some rather understandable manner. And there the answers check out. So what is the k equal to zero? Ja, that's what I didn't answer, and I still cannot answer. I'm sorry about that, but I will try to come up with a good answer. Ja, so that was what I might suggest. Maybe you can, from here, recover what it would be. Because this is a curve count that we are more familiar with. What I would like to hear is some kind of completely fuzzy physics explanation of what I'm supposed to do with the space of U1 connections to get this thing. Yes, and I'm not going to give it to you, but I can certainly try later. Okay, so now, we still want to include notes in this story, and let us include notes. Sorry, just a quick question. So you're saying that Q is e to the t, right? Q is e to the t, and before we specialize Q to 1, we're just setting t to zero. And that's exactly the code, when the area of this is zero, then you're basically in the cotangent bump, basically here. So you don't care about these things, no curves. Okay, so there is a way to include notes in this story, and this is due to, I said, yeah, this is probably not a guruvaffa. What I said is a Gopakumarvaffa, but here it's a guruvaffa. So because I missed the note, so in order to include notes. So what we do is we take this t-store s3, and remember this is s3, so that's the n-copies of this we have on each other, and then we add the conormal of the note. And now we want to count holomorphic curves in the same spirit with several boundary, they can look whatever they want. But there is just one copy of this LK and many copies here. And that can be shown on the Schern-Simon side to correspond to some insertion of something in the path integral. And basically what happens is that you're supposed to take this path integral and you insert one minus into the x, into the minus x, actually. So this is s1 times r2 again, and this x basically corresponds indeed to u1 connection, so it's a monodromy around this generator of LK times the holonomy of our connection a along k. So this is supposed to take determinant and inverse. So it's the same type of argument as written used from the beginning, so I'm not going to give it. But what this then comes down to, and of course, so now this is why I don't answer your question primarily, because the thing that we're interested in now would divide out by all the, I mean, expectation value. So we're somehow, you know, this i4k and so on. So here also, I'm a little bit sloppy. But anyway, so we're normalizing this by the s3 partition function. So what comes out, in fact, here is the colored homfly polynomial. So you expand this as a determinant, and then you get symmetric traces in symmetric representation. So after all, this is just the sum of the case, the homfly polynomial case symmetric, maybe I put kh, it's not too many case, e to the kx, or perhaps minus. But anyway, so what is this? This is something fairly possible to calculate. So homfly polynomial is not polynomial when you just use this sU2, which is derived from some, so you can compute it for any note by some relation with coefficient like this. So polynomial for this guy, polynomial for that guy, plus polynomial for that guy is zero and some coefficient. So it's somehow iteratively computable, and the colored homfly is not quite computed from that, but you need to take also cables of the note. But it's fairly, fairly computable thing. So now what we see is, we see this picture, and what we would like to do is we would like to apply the guppakumovaffa trick to just to go from the kohent band of s3 to the resolved conifold instead. And we can certainly do it, and believing in the first strange statement, then it is clear, as mathematicians, why do we want to do it? We want to do it because this is somewhere where it's possible to count curves, so that's what I'm aiming for. So this is kind of quite nice manifold, right? You have, it's a compact, it's not a compact manifold, but it has a kind of a compact port and a positive infinity. So I want to have somewhere to kind of make a little bit sense of such things like this curve count. This is not too bad, right? Such counts. Right. Is it better because it doesn't force all the curves to be generated? Yes, yes, that's right. So that's right. So somehow this count we understand, I think this count we don't really understand as well. So, but what would happen is somehow exactly the same. So here it's provided that, let's assume for a moment that the cone normal does not intersect S3. In fact, you can always shift it off a little bit. Maybe I'll skip the explanation of that for now. And then imagine that we have these curves, somehow this holomorphic curves, which used to look like this, they end on the S3 zero section. So now, in the transition, what I should imagine is that all the holes of these curves, they kind of just shrink to points and then it's clear this is in T store S3. And if we believe this is clear what's going to happen in X. So in X, this boundary just stays and then you get somehow this disk instead, right? So basically what all this thing, this dictionary is telling us is that this function should be equal to the Gromer-witten invariance of Lk in X divided by all the closed curves in X. So the homfly polynomial encodes the holomorphic curves with boundary on Lk sitting in X, okay? Do you get some kind of mark point conditions on the? No mark point condition, just for illustrational purposes. The boundary did shrink. It's not really mark point, in fact it's... In the physics proof, there's a whole disk sitting there somehow. So, okay, but this is the picture. So now, in order to still cut some... I'm skipping a little bit of this. Maybe I shouldn't. Yeah, I'm saying two words. So let's call this function somehow the wave function of K. And then one can now try to take out the contributions from kind of small disks, indeed doing more or less what Katrin was asking for. So I'm going to do sort of GL1... GL1, churned simons on this S1 times R2. And when you write down the form of the action, you see that what you're doing... So here at infinity sits some sort of torus. And the connection has kind of periods P and X. And what you're doing is just when you write up the path integral, you see that you're just doing quantum mechanics in variables P and X. So the action somehow periods P and X. So therefore what you find when you quantize is that you should replace. Okay, so if you do... So you want to do... Let's say GL1 or U1, GL1, churned simons theory on... Why do you want to do that? So I said I'm supposed to count holomorphic curves. So you're going to play wind game in reverse now? Yeah, just one second. So I'm going to try to count curves starting on LK and then kind of closing up. But there will be these small curves that just kind of constants in some sense that go from LK to itself. And I want to in some sense get rid of them. And then I play Witton's game in reverse because I know by Witton's argument that they would be churned simons theory here. And in words what happens is that to churned simons theory on this guy just looks like... The path integral looks like quantum mechanics in variables X and P. And then we know what this quantization from some kind of... I don't know, whatever, high school. And so that means that we should replace P by this G as d dx. So P acting, and that's a kind of multiplying by these short little strings. P acting on the wave function should be G as d dx, somehow. So I can, it's awesome, so I can. And then also remembering maybe next year or some other year, I don't know. So remember that you can then express the wave function in these short wave asymptotics, how you derive... When you have this Hamilton Jacobi theory, how you kind of could guess quantum mechanics, something like that. So this is somehow the standard expansion. And then there comes higher order, kind of zero, Gs, Gs squared and so on. But remember what this was, so this was just... It's still up there, it was just a count of holomorphic curves. And Gs was the genus parameter, so this is here, by our other interpretation, supposed to be just the disk potential. So I'm going to call it wk of x. So this is disk potential, disk count. So we're counting ridded holomorphic disks on lambda, so we forget about all things with some topology in it and just count disks. Och så, this thing is predicting that we would have P is dwk dx. And in fact, looking at the homfly, somehow dequantising the homfly polynomial or its recursion relation, what you find is that this thing should... So I'm really saying very briefly what happens. Men anyway, it should give a local parameterization of an algebraic curve. In other words, if that's given by a polynomial, I will write that polynomial, maybe there are many a's, but maybe I'll anyway write a of k e to the x e to the p, and I'm keeping this Q, and I think of it just as a variable. So this is somehow the thing given by this. And basically you can compute this a from the homfly, colored homfly, or a piece of it. And then what was observed, and this is the starting point of this talk really, and it's joint work by myself and Lenning and Aghanagikenvaffa. So what was observed was that this a k is equal to the augmentation polynomial in all examples. So I'm going to try to explain now why that is the case. So there's somehow my next call. So now we can in some sense forget about all of this physics background, but still, so what we are going to try to relate is, now we're going to try to relate augmentations with Q not equal one, so more global things, with curve counts in x. So let me just kind of redraw this picture. If what? It's not the one that you have, but the one that you don't need. Right, if you have many components. So then you have Lagrangian given by actually many equations. It's still Lagrangian in the space, and basically this part of the story is more or less the same. It will again give various branches of this augmentation variety, but again, the disk count, it's a little bit, right. So basically if you have many component link, you can also have other connected feelings of those components. So you cannot just look at the co-normals, but there have to be other Lagrangians which allow you to map mixed chords, the chords going from one to another down to something non-trivial as well. But basically all this, they give a variety, which should be the characteristic variety of the D-module of the Humphly again. So maybe I'll reach that to the last. But indeed, when you have many components, it's a higher dimensional variety that replaces this guy. So you have... The question is, if you quantize it, it kills you as if you quantize it. If I quantize the AK... It kills the partition of parts. It kills the... Exactly, yes, yes, yes. And that's if I kind of, if I'm fast enough to get there, I would want to explain what that looks like from the point of view of contact tomology. So it's a more complex, like an SFT version of this contact tomology. You should know this quantization of this guy. Okej, så... Let me look at what we have. So we have this X, and inside this X it's LK. And I didn't explain how to shift it off the zero section, but I'm just going to now say that it happened. So, and we have... This maybe is LK. And inside here we have a count of disks, which I will call W sub K of X. That's a count of disks. Counts, holomorphic disks. And again, we're in this clubby-outsetting, so they're all rigid. With boundary on LK. And now the thing that we're setting out to prove. So I would say theorem in quotes, but maybe I'm still saying theorem. So this is... One can question how much of this is really proved in terms of perturbation theorem. But I'm not going to enter that discussion, so I'm calling it theorem anyway. So the theorem is that if you take this P, it's equal to the WDX, is a branch of the augmentation variety. Just to understand what this is saying. So, DWDX is a polynomial in X. Ja, right. Maybe I should write it. It's a kind of polynomial in E to the X, or power series. So some count occurs that wraps K times around X. So this is an equation of the variable X and P, and we're saying that is a branch. That's locally a branch of the augmentation variety. Is WKVX a thing that's defined in mathematics today? I would say so, yeah. I mean, maybe somebody else would contest it. Maybe no, I mean, I will hint at how it would be defined. But it's a borderline defined. But not just... What about it makes it not... So you're supposed to count this open grammar-wittning variant. So it's some kind of framing contribution that's maybe not complete. Definitely you need all this kind of abstract perturbation machinery. I don't know, maybe... Is it defined? Okay. But note that what is defined, so defined or not. The augmentation variety is very rigorously defined. So it's transporting some things to infinity where you can actually deal with them, perhaps easier. Okay. There's Q. Q is here. Sorry, yes. So we're keeping track of the full... So this has a homologi class. It wraps around here on the boundary. But also there is the CP1, the S2 inside. That also carries homologi. I'm sorry, there's a Q here. I thought it was on the question. Augmentation is about the fillings in the simplification. And you are filling inside somewhere else. Why is that okay? I'm going to explain that. So that's the kind of the key. So now let's first try... How would we go about proving something like this? Well, we would just try head-on with our old scheme. We tried to count holomorphic curves like this to define the augmentation. So kind of rigid things, D equals zero like that. And then we tried to prove the chain map equations. We have to look at one-dimensional modellispass. But now there's a kind of huge difference between the previous thing. And that's that what can happen is that some kind of boundary bubble starts forming. And then splits off... This is somehow the relative version of what Helmut... Just these pictures of what space. So this boundary lives in LK. And the whole disk lives in X. And now I have... It's a one-dimensional modellispass, so somehow I'm following... This is also in X with boundary on LK. This is in X with boundary on LK. So what I'm trying to explain is that there could be one parameter family that splits into a rigid disk from above and a rigid disk formed on L. So this is somehow the obstruction. So that would say that our previous nice thing that said that this chain map phi composed with D is zero, that's not true anymore. So we kind of... This is the bad... LK is not exact. LK is not exact, right? That's exactly. Even the manifold is not quite exact. But in particular there could be such disks and they obstruct. And this here, unlike in Helmut's lectures, this is called dimension one phenomena. So it really matters. But now we somehow learn from Foucaia how to overcome this thing. So that's what Mohamed's talking about. So you have to find these bounding co-chains. And here, of course, there is no bounding co-chain because what one would like to do. So here everything is rigid, so it's not so much higher modellispass as you have to care about. But what is the scheme is that you would take this disk. You take this disk and you try to find... So here is your disk and you try to find the chain that its boundary bounds inside S1 times R3. But of course you cannot because S1 has some homology. But you can do almost the same thing. You can pull out its homology to infinity. So instead of finding bounding co-chains, we find sort of... I don't know what to call them, but we take for each such disk... For each rigid disk D, now sitting inside X with boundary on LK. We take a chain which at infinity looks like some... So this is a chain inside LK. So the boundary is an S1. It's a circle sitting inside LK. And we take a bounding co-chain, a bounding chain. I don't know what to call it, a co-chain bounding chain. Det börjar med den här och ender på K times X. I infinity har vi den här länge. I infinity har vi den här cycle som bara ser ut som X. Så vad som är multiplis är kvar för den här. Vad är X? X är den länge variable, så jag kanske skulle kolla det som X. Det är en specifik mål. Och nu, hur är det att kura den här problemet? Det kurser det i en månad att när du... Du kan nu någonstans passa det här med att skapa... Ställa disker som den här med inserterna av de bounding-chains. Så när du hittar den här, så är det modelleringen. Du går in till den bounderingen, och så har du den här bounding-co-chain. Och du fortsätter... Du fortsätter med en familj som den här. Det här är en whole idea av... Nej, kanske inte whole, men det här är... Det här är Foucaia and Company's idé. Och så, i det här sättet kan du nu... ...avgöra den här inserterna... ...på grund av att uppgradera din disk för att skapa... ...inte bara dessa disker som är D equals 0 disker... ...men du måste skapa... ...särskilt... ...det är svårt att dra den här liten bilden. Jag skapar skematiskt en liten del. Så att skapa disker med inserterna av alla dessa saker. Så nu... ...vad skulle kompagniseren vara? Vi tog ut den bästa delen av den bounderingen. Det kompagniseren nu är att vi inte har intervju splittning... ...så vi har splittning av denna som helst. Det är 1 och det är 0, 0, 0. Men vi behöver förstås hålla på med alla inserterna... ...och det är ingen förväntning att inserterna... ...levar bara i kompagniserna. De kan säkert lägga ut till infinity. Men i infinity är de väldigt särskilt gällande. Om jag har en av dessa disker som nu är disk i differensen... ...och i infinity... ...den kan säkert lägga ut till infinity. Om jag vill se hur många gånger den här disker... ...hur många inserter kan jag göra... ...så måste jag hålla på med hur många gånger... ...det intervjuar med den standarda delen. Om du tänker på det... ...så i infinity... ...det chans som kommer ut från dessa... ...det ser ut som... ...det ser ut som... ...det var Ckl e kxql... ...det är bara kopior av denna delen. Så du har den här. I andra ordet... ...skärta denna inserterna... ...är det samma... ...för att om jag lägger kx runt... ...jag kan... ...jag kan inserta kx... ...så det är exakt... ...sättning P är equal till dwk dx. Så... ...om jag vill ställa... ...om jag vill säga att jag har bara en kurv här... ...det ser ut som e kx... ...och nu har jag en annan kurv här... ...som går... ...och efter att jag klockar den... ...går runt... ...det här cirka. Så hur många kurv ska jag ställa? Om jag kan inserta den här... ...jag kan välja en av de k-klarna... ...men jag har k-klarna... ...så min kurv ska vara kx e kx... ...och därför... ...jag har en kopplad av kopiorer och så vidare... ...men den här mechanismen är exakt samma. Så om jag lägger P... ...är det equal till dwk dx... ...då den här uppe klockan... ...körs med de här inserterna... ...och den här matchen... ...det är en prov av den här... ...för att den här bilden är klart. Om jag drar med inserterna... ...då är det det... ...klarna-klarna håller... ...och... ...och... ...och gör den här uppe klockan... ...det är exakt... ...det som vi ska göra. Så vad har vi gickat i enda? Vi har sagt två olika klockor... ...klarna och polomofiken matchar med varandra. Well... ...för exempel, i den ena och ena exempel som vi vet... ...kan du nu kolla... ...om du vet den här... ...klarna och polomofiken... ...kan du nu kolla... ...klarna och körs med den. Så vi vet hur det är... ...det är polomofiken... ...och nu kan du lösa... ...lösa för P i t.ex. Och då integrerar du ens... ...och får du den här potentialen... ...på den här klarna och polomofiken... ...så... ...så att vi vet polomofiken... ...som du kan komputera... ...på elimineringen. Så för polomofiken är det en mer komplicerat sak. Du kan nu kolla polomofiken... ...med kvarna på den här klarna... ...därför... ...på X. P i t.ex. Hur är Q-sektorn... ...och stäcker... ...klarna? Kapital Q? Ja, så det spelar inte en roll. Jag menar att det bara... ...klarna på hur många gånger... ...det raps runt... ...det centrala CP1. Så det inte... ...det inte interfererar med den här klarna. Så det bara raps runt... ...det är som... ...så... ...så den här W... ...det var W. Och då dW... ...det som du vill... ...det är bara... ...Q är bara... ...det är bara K, C, K, L... ...U, K, X, Q, L... ...så det bara raps runt. För nåt som du har sagt... ...det kan vara att... ...det gransk av det argumentation som du har... ...det är alltid ett stort gransk av argumentation... ...det är ett stort gransk. Nej, det är inte det. Det är så att det är så att det inte är så. Jag vet inte. Men... ...men i exempel är det inte. Det är inte... ...men i en speciell... ...det är... I konferens vill du faktiskt veta den diskpotentialen... ...det är att veta vilka gransk du ska ta. Ja, men... ...så här kan du... ...så det här... ...det här kan vara att... ...kalkalera när P equals 0. Då vet du att du ska ta den. Men det finns många frågor här. I en del av att röra hela... ...när det är lättare, så är det okej. Och det händer ibland. Men inte alltid. När det är lättare... ...så vill du veta mer gransk... ...för att röra allt av det. Och jag vet inte hur att göra det. Så jag vet att jag inte har tid. Men låt mig kanske säga... ...förlåt och lägga det för diskussion. Så... ...så... ...indeedas... ...jag var säga... ...så det är... ...det är suppost att det är... ...en liv av det här... ...så det... ...det uppgräns... ...så vi har den här ordentation-pollinomium... ...det uppgräns... ...det här till en operatoräckvation... ...med samma symbol... ...men där... ...där faktiskt... ...det här P... ...P nu är equal till... ...GS... ...DDX... ...i linje med vad jag drar för vägfunktion. Och... ...och det som händer här... ...det här ska få till... ...det här ska få till generator av ideal... ...korspåning till den här ordentation-pollinomium... ...eller vad det är som heter... ...D-model-ideal. Och då... ...vad... ...vad vi jobbar med... ...det är väldigt mycket jobb i progränsen... ...är att se hur man ska använda SFT. Så att komma till en del av kontaktpollinomium... ...men... ...äst att använda... ...i något sätt... ...all... ...körs... ...up here... ...and count them, and recover this relation. Så det är... ...and we manage somehow... ...it's an interesting story, I think it's far from finished... ...but it's looking good, so we could... ...we are not somehow trivial, but we manage to do it for hopeful link... ...and for truthful notes. It looks kind of interesting, and it certainly has a lot of challenging... ...holomorphic curve problems, among them... ...obstruction bundle gluing... ...and also I would say some new phenomena where you... ...where you cross kind of multiple times and so on. So maybe, but I think I'd better leave this to discussion... ...and stop at this. Any questions? I think that... ...so the count of the object in the middle... ...where I think it's sort of like hybrid of... ...objuntation and finding co-chain... ...which essentially makes the sum of the object in the hand... ...and also linearize the homologi. Yes, it's that story, that's exactly right. And you can, of course, this trick applies, of course, fairly generally. I mean, now here is a very good case when all the curves are rigid... ...but if you want to go into applying this, you could... ...you see that when the homology... ...when you have these Lagrangians with some infinity... ...and the homology of the Lagrangian generated from infinity... ...then you can always pull out cycles like this. They're kind of as good as bounding co-chains... ...and then you get similar type equations in the non-exact case. But the cycles that sit inside purely... ...and bound holomorphic curve, they are behaving exactly like in this standard... ...compact Lagrangian fluoromology story. But some of them you can get rid of by transporting to infinity. The lesson maybe is that this is... ...it's actually kind of quite efficient way of computing. So I would say that compared to what's... ...so this orientation polynomial, maybe this one is a little bit difficult... ...but this guy is just... ...as I said, it's just counted by elimination theory... ...from something very concrete. So this calculation is comparatively very simple. Whereas this other computing directly inside is a kind of... ...difficult task, I would say. So you mentioned the... ...what do you call it? Flaming ambiguity at infinity? How does that enter into the... I'll tell you. So of course I'm drawing pictures sufficiently fast... ...that I'm kind of doing the opposite to Helmut's. I'm doing so slow, you can find the mistake. If you draw really fast... ...so the question is really... ...what is this going with the potential? Remember what we suppose... ...so we start from something and we realize we have to count... ...so it splits like that so we would have to count... ...insertions of this bounding co-chain for this guy here. Men nu... ...of course, there could be some splitting... ...which somehow splits... ...you cannot say when things split, right? So when you think about this... ...it could be that some of the insertions of some other guy... ...actually ends up here, right? So the count of the disk potential... ...which I'm saying just count disks... ...does not quite just count disks... ...if it's supposed to work with this formula... ...because it will have to count disks with this bounding co-chain. So if I have one disk... ...and then I have already kind of some lower energy disks... ...I have to count that disk... ...so all possible trees... ...of insertions of other disks in some sense, right? Så, egentligen... ...det gråmavitten count... ...depends on somehow the choice of these bounding co-chains... ...and that's a little bit like... ...if you frame the boundaries in some different way... ...you would get some different count, right? So that's some kind of bounding... ...framing ambiguity, I would say... ...or you can maybe borrow the framing from a framing on... ...on the solid torus, something like that. Admittedly, I mean, this is easy to see... ...I didn't work out the details in any sense... ...but indeed, this... I mean, the level of formula... ...can you say how to... ...compute the different... ...I mean, that physicists have predicted... ...what the dependence on the framing should be? Ja, there are a couple of different framing. So one framing that they talk about... ...is somehow just the choice of basis... ...ex compared to P. ...and basically, this guy should know everything. You can make a change of variables and compute things. But realizing it again is somehow that... ...he prefers to hear, somehow... ...my feeling here kills the P variable. If you do something in some other framing... ...preferably, you would like to kill... ...the variable that you solve for in this potential formula. And prior, I don't know how to get all the suitable Lagrangians for that. But basically, this polynomial should know everything. Det är så, jag tror. Men att skriva ett geometriskt sätning... ...för den som vet att det inte är så. Hur är det att köra dig själv... ...på grund av det här genuskördningen? Ja, så... Det är inte så, det är exakt... ...att köra den här... ...men tänka på alla den här genuskördningen... ...som du kan bygga med insersioner. Om du har den här disk... ...och du vill insera den här... ...det bygger dig en disk. Men nu, när du gör den här genuskördningen... ...får du två insersioner. Du kan välja båda och så vidare. Och du vet att det är precis vad du får... ...för att ta den här... ...att äta den här... ...men kanske kan jag öppna i en diskussion. Så det är en typ av kurvkont-derivation av det här.