 So indeed, so the subject of my lecture series is not contact tomology. And so this is some subject that we studied for a number of years. And in some sense, it can be viewed as applications of this symplectic field theory package to not theory or more generally to differential topology. So the series will have three parts. So today, I'm going to give some kind of introduction, defining not contact tomology in particular, but starting by talking about the Lausandre in contact tomology, which is part of this symplectic field theory package. And then try to get to the point where you can actually get the feeling for how you would compute these things more concretely, not just showing that the theory exists, but actually compute it. And then tomorrow, no, maybe not tomorrow, but this second lecture, I will talk about, probably continue with that, and then talk about relation to string topology. And finally, in the third lecture, I'm going to talk about relations to more physical theory. So like Cern Simons and the topological string theory, there was more recent developments of the subject over the last few years. But anyway, so let me start by giving some introductory remarks about the relation between differential topology and symplectic geometry or contact geometry. So there is somehow this basic thing that if you have a smooth manifold M, then it has a cotangent bundle, which is a symplectic manifold. In fact, Weinstein manifold, so exact symplectic manifold. And one may now ask the question, how much of the smooth topology or just the topology first of M does the symplectic geometry of this T star M remember? So the first, so I am not going to explain these results. I'm just going to state them so they see where what we are going to do fit in. So see if you have a symplectomorphism between T star M and some other T star or some other manifold N. Then, if you look at the zero section in here, M, that sits in there as a Lagrangian submanifold. And the image of that would be an exact Lagrangian submanifold in T star N. And there is a theorem which has some history, but somehow this full version is due to a Bosaid and Krog. And it says that if L is an exact Lagrangian submanifold in T star M, then there is a simple homotopy equivalence between L and the zero section. So in fact, the homotopy type of the zero section is remembered by the symplectic topology. And this proof goes via flow homology with local coefficients. So there is some history to this theorem that I want to tell you about, but there certainly were lots of contributions. Compact, yes, sorry, L, M, and L are closed manifolds. So that's right, so compact, no boundary. So the next question is, does this go beyond? Would the symplectic geometry feel anything more? Then the homotopy type. And the answer, again, is yes. So maybe the easiest starting point for such theorem is to start when M, in fact, sort of has no homotopy theories. So that would mean that M is a homotopy sphere. And then there is a theorem that builds on the first theorem in this kind of spirit. This is due to Abu Sayyid again, but this theorem is in a paper by myself and Prague and Smith, and it says the following that if sigma and sigma prime are all-dimensional homotopy spheres, and if the cotangent bundle of sigma is symplectomorphic to the cotangent bundle of sigma prime, then sigma is the class of sigma in the class of homotopy spheres. There are equal modulo boundary parallelizable spheres. So in this dimension, 2k minus 1, perhaps. All-dimensional, 2k. So when you look at this, this is some kind of classical differential topology, which I won't talk much about. But if you have a homotopy spheres, they form this age-corporealism group. And inside there, there's a subgroup, the boundary parallelizable manifolds. And we know that their cotangent bundle can be equal, symplective manifold, only if they're equal modulo of this boundary parallelizable one. So thank you, which corresponds to changing the orientation of sigma. So this is sort of not so serious. OK, anyway, so this theorem tells you that in fact the symplectic geometry of the cotangent bundle is pretty sensitive. And I would say that the proof that I'm not going to talk about, but goes beyond, you know, Fleur theory and SFT and the like. It's kind of different version where you use a modular space of allomorphic curves, not just to get some fundamental cycle or something like that, but it's actually a k theory class of it or something like that. So basically, you're using the modular space more than you're using things derived from it. So it's somehow a very interesting way of using modular spaces. But here it's a little bit ad hoc used. So it's very hard to use it directly. And so I encourage everybody to think hard about this and see if you can find maybe the expectation of what is the hoc curve, the sharpen. Yeah, so the sharpest expectation of this is this nearby Lagrangian conjecture will say that any exact Lagrangian sitting inside here, actually it's Hamiltonian isotopic to zero section. Do you still believe in this? Do you still believe in this? I'm not so sure. But then somehow you should ask Thomas Krog. So he's kind of the leading disbeliever, right? But anyway, I don't know. I mean, I don't know. So for example, yeah, I will mention something in one second in this direction. So now what we are going to talk about is not this, but we're going to talk about a relative version of that. So if you have a sub manifold, K inside the manifold, so again, closed, everything is closed, then we can associate the symplectic geometric object also to that. So and the natural thing is to take the Lagrangian conormal inside the cotangent bundle of M. And what we are going to do is related to that, but we're also going to look at the Lagrangian conormal. So I'll define this in one second. Inside the unit cotangent bundle of M viewed as a contact manifold. So the Lagrangian conormal is the set of points in the cotangent bundle. So Q is the coordinate along the base. And P is the corresponding fiber coordinate. So that Q is in the sub manifold. And P, when you restrict it to the tangent space at this Q, is equal to 0. So it's somehow exactly the normal vectors of K, but thought of as in a line in the fiber direction. So this is obviously Lagrangian sub manifold. And the Lagrangian thing is this just LK intersected with, say, the unit cotangent bundle if it's the Riemannian metric. And that's the Lagrangian sub manifold. And what we are going to do is actually we're going to specialize quite a bit more. So we are going to, for this lecture series, we take K to be a 0 in M, which is equal to R3 most of the time. And sometimes, perhaps S3, it doesn't make much difference. So we will study a very special case of this. But before we go to this special case and I define the holomorphic curve theories, et cetera, that we are going to use, it's worth mentioning some results about this stuff and see how well it works with respect to things like this theorem. And there, so one should say, and that that's maybe kind of a good reason for talking about this, that this theory, so when you apply the holomorphic, the SFT package or some part of it to this lambda K, then it's quite successful. So for example, as we will see, the theory detects the unnot. So it will be, I'll explain this, so there will be a DJ associated to any note. And this DJ looks like that of the unnote, exactly only when it is the unnote and for no other note. And it is a pretty nice not invariant that knows about things like a polynomial of the note and et cetera, et cetera. But let's first consider somehow this counterpart of this in high dimensions, where there was some observations recently made by Dimitro Grurizel and then another student of mine, Ericsson Östmann, and it's the following. So if you start in R6, then in R6 you can embed three-sphere. So there is the standard three-sphere and the note theory here, in fact, is pretty simple. So if you appeal topologist, or C0 topologist, or maybe no, Lipchitz topologist maybe, or if you're Lipchitz topologist, then there is only kind of one note class. But there is an integer worth of smooth notes. And the invariant can be seen as follows. You take this embedding of the S3 in R6 and then you would find the four-dimensional cypher surface embedded, and you take its signature. So the signature is some kind of multiple of 16, and then you can realize all these notes. So there are these notes, three-sphere notes in five-space, which are actually differentially noted, but PL standard. So one might ask, would this lambda k, so k is k sitting? So what about this lambda k, which is now the conormal of S3? And topologically, so it's the boundary of tubular neighbors. It's S2 times S3. Does this lambda k, remember, smooth not in S? Is that 6 or 5? This is 6. So the boundaries would be in S2, right? So lambda k is if you morph it to S3 times S2, right? So it's a trivial, normal bundle. And the answer is no, because in fact, it does not. And why is that? So the observation is that actually, so let me draw the note and then take this tubular neighborhood, boundary of tubular neighborhood. Now the conormal lift can also be somehow, it's clearly isotopic to the conormal lift with the outward normal of this boundary of a tubular neighborhood, right? It doesn't matter if you lift right on the note, or you just fatten it up and lift a little bit along other vectors. And in fact, the boundary of a tubular neighborhood of the note k and of the standard one, even though the notes are not isotopic, the boundaries of tubular neighborhoods are isotopic. So this is some kind of just consequence of H-Cobardson theorem, or Smale's Handled Cancellation. So here you can define a function, Morse function on S, if you put S6 instead of R6. So you have a minimum here and an index-free handle here. And so the complement of that, you can cancel all the critical points. So that means that the complement looks like an S2 times D4. So somehow the complement retracts into an S2 sitting here. And the S2 is completely free in R6, right? Because it's 2 plus 2, it's too little. So that means that you can isotope this tubular neighborhood to that tubular neighborhood. And therefore, you can isotope lambda k to lambda S3. On the other hand, and this is somehow interesting open question. So the secret to this is that this weird note, the Hefliger note, it actually lives here somehow in a tubular, on the boundary of a tubular neighborhood. And one could, for example, ask the question that if you pick one of these Lachandrian isotopes, anyone, would it have to take, could it take a section of this kind of normal bond into a section of the other one? And probably, I would guess the answer is no. So that it still knows about this, but it doesn't, somehow it's some parametrization question, right? And does the Lachandrian isotope have to really mess up the parametrization? OK. So if that's not true, does it break this nearby Lachandrian conductor? No, no. It doesn't break really anything. But for the nearby Lachandrian conjecture, I think great question is if, so now this is in R6. And if you think about this T star S3, that's also a six-dimensional thing. And you have more or less the counterparts of this. You have the zero section. You have a kind of Hefliger counterpart to zero section and so on. And I think it's unknown if they are represented by Lachandrian embeddings. So I think this is a reasonable question. Can you find Lachandrian S3 in the isotopic class of the Hefliger sphere, in T star S3? And that would certainly break the conjecture, but maybe it's not that easy. I don't know. Well, OK. So this was end of the introduction, almost. But one could maybe say here that one of the things that little bit go wrong here is that the co-dimension is too big. So when you're in co-dimension two, then certainly this non-contact tomology, exactly as in the case of knots in R3, will be pretty powerful, because it fills a lot of properties of the fundamental group of the complement. It's OK. But that's enough of introduction to the general area. So we will now study this non-contact tomology. Should we think of the fact that lambda k doesn't remember smooth not in this as a shadow of the fact that there's no kL not in this? Yeah, I mean, so here it is. But so what I suspect is that lambda k, not as a submanifold, but as a parametrized submanifold, as a parametrized Lashornian submanifold does remember smooth not in this. That would be my guess. And that's, of course, I don't know how to prove. Otherwise, I'd give some other talk. OK, so. Do you think that the smooth morning is distinguished by the signature of the Cypher surface? So if you take those Cypher surfaces and put those in the co-dimension component, then what? Yeah, so here you wouldn't quite have a Cypher surface, but you would have a surface going. So if you have, so here's your S3, and then you have your strange not. So there you would have a four-manifold interpolating between the two. And you could take the signature of that. So that would be the counterpart of this Hefliger signature, I think. But this is just, you're just assuming now the knots are smooth. And then you're asking, can you actually find a Lagrangian representative so that this cubordism has non-trivial signature? And that's, if you can, you're famous, I think. Yeah. OK, so what I'm going to do next is I'm going to describe for you Lechondrian contact homology in a setting which is sufficient for what we are going to do later, but not completely in this unit called tangent bundle of R3. But I will start out a little bit more generally. So y is a contact, contact 2n minus 1 manifold. And the contact form alpha is the contact 1 form. And inside this y, we have a Lechondrian, so manifold. And what is this Lechondrian dj? So we'll define Lechondrian dj, it's part of SFT. So if you want to, just before I go into details, what is it? Well, it is the attempt. It is the answer to the following question. You try to do some kind of flow homology on the path space. So gamma, so here is lambda, and here is lambda, and this is a path gamma taking lambda to itself. And you try to do the flow homology for the action functional, which sends such path gamma to the action. And then you cannot quite do it, and somehow these SFT type splitings that you see forces you to use some other algebraic structure than just a chain complex. And that's what I'm going to explain. So since this is a lot of kind of technical business going on, so let me be precise. So we will restrict to, so we assume that there are no closed orbits in Y, which is the case, for the unit container bundle of R3, maybe not for unit container bundle S3, but anyway. And we will also assume that the first-gen class of the contact plane field is 0. And I think that I will assume that this Y is simply connected. And what else? I'll probably assume that the mass, huh? It was about to ask whether no contractable closed orbit could be enough, but since you're talking about Y, 0. It would be, but I want to somehow to get the thing sorry, maybe it's 0. OK, what do you prefer? OK, so and then, and I will also, I'll talk about this mass log class, but I'll take the mass log class of lambda to be 0, to have the theory graded. Are these assumptions really necessary or just to make it easier to explain? It's just to make it easier to explain. So they are not necessary, and some of them are slightly serious. So these two things affects the grading. So you'd have to do something about the grading. Right, so this is somehow is the serious. The serious assumption, and what would happen is I will now define for you an algebra which just generated by quarts, and it will be an algebra over some coefficient ring. But if there are orbits, then the Legendrian algebra would be an algebra over the contactomology orbit algebra that I'm not going to talk about. And that orbit algebra certainly does require a lot more of machinery of the nature that's being discussed here about polypholes or something like that, so abstract perturbations. Whereas this is what we are doing here, we can actually get away with the classical perturbation theory. OK, thank you. So I'm probably forgetting half of the more assumption, but it doesn't matter. So what is this thing? So what we are going to do is first we have the rape vector field. Probably you already discussed this, right? So this is somehow in the kernel of the alpha and normalizing, so this is equal to 1. And so the generators of our algebra will be rape quarts. So they will be quarts with n points on lambda. And the algebra itself, so a of lambda, will be the following. So it will be an algebra over the group ring of the second homology of y relative lambda generated by rape quarts. So this is a unit algebra generated by rape quarts. And these algebra is not commutative in any way. So rape code A times rape code B is not rape code B times rape code A. It's just like that. And this is a. The H2Y lambda classes, they commute with rape quarts? Yeah, so right now, for the first words, I wanted to say this is a kind of simple version. So I will first take them to commute. It's not really necessary. So you can soup this up, and we will do that tomorrow. But for now, I will take the coefficients just to commute with all the rape quarts. But it's not really necessary. So this could be like some kind of module where you can put these things in between, and they don't necessarily commute. And we'll see geometric reason why you can do such things in a little while. So we are defining a dj, and we just defined the a kind of. So now the g, so the grading. So grading. So here is a chord C. And so what we do, we pick a path inside the Lagrangian. So this is some path inside lambda. And now, since y was simply connected, we could somehow extend. We can fill it in with a disk, right? And now, on the boundary of this disk, we have the tangent spaces of lambda giving a field of Lagrangian planes in the contact plane field. And when you have that path, you somehow you can take the tangent space of lambda at the endpoint and transport it to the starting point with the linearized rape flow. And you get almost a closed path. You have to close it up somehow by positive rotation. And you take the mass of index of that or the con disender index. So let me, I'll explain in examples how you compute this later. So anyway, the grading of C is equal to the con disender index of this C, which is this mass of index. I was talking about minus 1. So I'll leave it at this. It's just you have to think about this is somehow related to the homotopy class of this Lagrangian plane fields. And this is what normally enters into grading when you compute this indices. So OK. So we have an idea about the grading at least. And now what is the differential? This is, of course, the most important piece of the algebra. So first we fix an, so first we consider as was done before here. We consider the simpletization r times y. And here we take the symplectic form d e to the t alpha. I guess everybody saw this before. And we fix an almost complex structure, j, which is translation invariant, and takes the contact plane field to itself. And somehow this j of the extra t direction is equal to the ray direction. So this is probably everybody. You probably talked about this, right? Yeah? Oh, no? Oh, yeah. Oh, yeah. Oh, yeah. This was multiply-covered. This was multiply-covered. That's good. So yeah, I don't know if you should divide by one factorial or just divide by how many times it was mentioned. But anyway, so it's OK. OK, so you can recognize this thing. And then we have this notion, again, probably multiple-covered of finite energy. So we have the energy or omega energy. So it's an area in the contact planes. And then we have this cut-off area in the alpha energy, but maybe lambda energy in some other torque. So going in this other direction where we have to take a cut-off function. And then, as probably was discussed, at least in the case of orbits, we know that finite energy curves. So I'm just taking this a little bit in-precise. So finite energy curves, holomorphic disks is enough or asymptotic to ray chord strips at punctures. So indeed, note that by this requirement, if you take the just ray chord C times R, then this is a holomorphic strip. These are the trivial strips which counterparts the trivial cylinders from Helmholtz's talk. And now, if you have any such finite energy disk, let's say, mapping in there, then it has a number of positive punctures and a number of negative punctures where it's exponentially asymptotic to this disk provided that all the ray chords are non-degenerate. OK, and the modular spaces that we will use, they are modular spaces of a specific kind of disks with one positive and several negative punctures. A ray chord A and the word of ray chords B underline, which is B1 up to BM, we consider the modular space M sub A, which is this is, I'll explain what A is in a minute, A B is the modular space of holomorphic disks of the following form. So this disk has one positive puncture at A. And it has several negative punctures which appear when you go around the disk as B1, B2, B3, and then finally, BM, right? So that's determining this word. And finally, remember that we have these capping disks. So we can somehow take a little disk here, a little disk here, here. They're just, we just choose them, it's up there. If for each chord, we choose such a disk. So when we fill it in by these capping disks, then this creature, the holomorphic disks and the capping disks, they define a homology class inside H2, right? Relative. So that's what this guy is. So this is the modular space we'll use. And the dimension, the formal dimension of this MA in our case is simply equal to the grading of A minus the grading of B, which is the sum of the gradings of the factors in B. So the differential that we are going to define is simply defined by counting such things in one-dimensional modular spaces. So the kind of minimal dimension. Something interesting happens. So because, as you've heard about this Orin variant, somehow the minimal dimension of an interesting disk is 1. And so the differential from this algebra to itself, maybe I have lambda, satisfies Leibniz's rule. And it's linear on coefficients. So coefficients just go through. And on generators, it's defined as d of A is the sum over A minus B equals 1. And then number of R families in this modular space times the word B. And maybe I put here each of the A. So indicating we're in the group ring. And then, so this is the definition of the differential. And then we have the expected theorems. So d squared is equal to 0. And A d, let me not say too much about this, is invariant up to homotopy under deformations. Let's say under Le Chondrin isotope of lambda. So in the case that we needed, so in the case when the ambient space is a one jet manifold, one jet space, as we'll see the unit container bundle of R3 is, this theorem was actually sort of fully proved in work of myself and Etner and Sullivan. And that you can do without using abstract perturbations for some reasons. And maybe I'll see if I get to explain it. Otherwise, you can explain maybe during discussion. So what it says is that if you pick this j generic, then d squared equals 0. And also when you deform the thing, you get a homotopy. One would need to say what is this homotopy. But in particular, the homology of the thing is invariant on this. OK. Sorry, Tobias, A is the energy. A, sorry, A is the, this A is the algebra. No. In the formula for the differential, we have E to the A. E to the A, A here, yeah, right. This is just stupid notation for, so A is in H2. And then I write somehow, E to the A for the same thing in the group ring, right. So this is some integer number. And then times this thing, so it lives in the grouping. So it's just, you can think, if you want, you can think about it, it's just A. This is, OK. Right. OK, so let's consider the simplest example of this theory that was also the first somehow was discovered. It's called the Chekhanov-Iliashburg Algebra of a note, the Schrodinger note in R3. So consider the ambient manifold to be R3 with the standard contact sector dc minus ydx. So here, a Schrodinger note is a note which lies in the kernel of this thing. And so in particular, when you project it, so I should say first that the rabe, the rabe is just here, the vertical vector field, the z direction. So when you project the Lechonry note into xy space, then it projects to some, if you take it generic, it has double points. And it bounds zero area. If you kind of take the ydx integral, it's zero. And the rabe cores are very easy to spot. So they just correspond to double points, right? Because it's in z direction. And now one can pick the almost complex structure to be the pullback. So the contact plane field projects onto this plane so that there's nothing, I don't know what this is, bijection, I don't know. Funny thing is beautiful pictures on his website. Great. But anyway, so it's isomorphism on this plane, right? So therefore, you can take the standard complex structure and pull it back to the contact planes. And therefore, you can read off holomorphic curves in the simplectization by projecting them down and see that they project to curves in here. So in fact, it's possible to compute here by the following. So we need to distinguish what is a positive puncture and a negative puncture. So the rule is that at positive puncture, the holomorphic curve goes up the rabe coordinate, negative puncture goes down. So here you have such decoration. And then the differential counts holomorphic polygons with convex corners. So that's somehow this index, index, index requirement, if you wish, with one positive and several negative corners. So we need to say kind of few more words to complete this description. So here in this case, what is this H2 relative? That's just H1 of the knot. So there will be kind of one variable in the group ring. And so we fix a point on the knot. And what more? So the other thing is the sign. So here signs is the following. So remember that these chords, they have a grading. And in this case, I should say what the grading is. So the grading is the following. You start at the top. You go along the knot. So I assume the knot has mass loving. So you go along the knot, you end up here, and you close it up with a positive rotation. So the grading of A is twice, so that's the mass love, times the rotation number of the positive close up, minus 1. And so there will be negative at the even, so let's see. So right. The polygons have to be embedded, or could they be immersed? They could be immersed. As long as they can certainly cover themselves, but they need no branch points. But they are immersed, and they have convex corners. And then now I'm kind of getting slightly confused. But let's see. So now what you do in order to get the, sorry, so I'm sorry. So somehow, I'm sorry, I should have gone a better picture. So you somehow, you shade, I'm going to explain. Signs, right. So you shade at each even, even index crossing, you shade somehow half of the corners like this. And then in the disk, so we're going to have to look at papers to get this right. But in the disk, you count the number of shaded corners with a minus 1 to something. So for example, for the unknot, there is nothing here. So what is the differential here, d of a? So this is a. First, let's see, what is the grading of a? So grading of a is 2. It rotates once, right, when you do it. So it's a 2 minus 1. So that's 1. And there are two disks, two rigid disks, this one and that one. And one of them does not pass through this chosen base point, so that's a 1. And the other one, there are no shaded corners because there are no even corners at all. And the other one does pass through this base point, which means that we kind of view it as going once around the knot. So it is 1 plus, I know what to call it, maybe mu, which is the homology class of the knot, right? So you see somehow that the DGA of this unknot is sometimes, I mean, most often is acyclic actually. But if you take mu equals minus 1, then it is not acyclic, so it has something. So that's a. Mu was the mass loss. Mu, sorry, mu is the generator of h1 of lambda. So this is coefficient, right? So the algebra. You want q? Mu was mass loss. OK, maybe, I don't know, t is sometimes used. Let's put t. So t, t is the generator of h1 of lambda, which is here the relative coefficient, OK? So I was kind of, I'm now, the time is flying here. I still don't know this. But anyway, so I was asked to give some exercises. So let me kind of give exercise. So exercise, of course, is to compute this for some other knot. So compute is for the trifoil. If I can do it correctly, I'm not sure. I think that's correct. So compute is something wrong? Yeah, it will go. It's topological term. Sorry, yeah. This should go down, right? Down, under, over. Like that, right? I think that looks better. So compute dj for this. And also, think a little bit, what is the significance here of t? So this, of course, is a harder question. But yt is equal to minus 1. It's something, a kind of special thing here, so on. OK, good. So the next step, so we see that this theory is actually very computable in the simple case that we are in R3, or maybe surface times, or something like that. We can compute this thing. But we are actually aiming for something a little bit higher dimensional. So we're aiming for the unit-potented bundle of R3, which, in fact, is the one jet space of S2. So the next thing I want to explain is how this thing can be done in one jet space. So one jet space. Oops. So if you have a manifold m, then the one jet space of m is just a cotangent bundle, thought of as differential of the function times R, which is the function value. And on here, there is a nice contact structure, which is dc minus the action form pdq. And pretty much as in the case of the plane, where this is actually a kind of special case, when we take this m to be R, and this is one jet space of R. So we can also here use, we can find, so pull back the almost complex structure from an almost complex structure on t star m. So in other words, we take an almost complex structure on t star m, the contact planes, maps again isomorphically onto the tangent place of t star m, and we pull it back to the contact planes and extend it in the usual way in the other direction. And then we have a similar description of the holomorphic curve. So again, so holomorphic disks now correspond to holomorphic polygons of the same nature, but now with boundary on the projection of lambda, which is now a Lagrangian manifold. Inside this t star m. So basically, it's the same thing as four nodes. But now our Lagrangian is higher dimensional, and it still has some number of double points. And we have to find out how to count the polygons. But first thing to observe is that wherever it went here, that this notion of positive negative corner is exactly the same. So you have a crossing, and at the positive puncture, it goes up the rape chord, and at the negative it goes down. So we have somehow exactly the same dictionary. Now the question is, how can one possibly count these things? So here we are somehow we're lucky in the sense that we're basically using Riemann mapping theorem. But when the ambient space has higher dimension, there is no such easy way to find the holomorphic curves. But the idea is that one can degenerate the Lagrangian towards the zero section. So let me try to explain this. So just could you raise the board up behind this? Sure. So now we have some projection of lambda sitting inside t star m. And the idea is the following, that if here we have the zero section, vm, and then this lambda somehow is exactly Lagrangian submanifold here over here. And now what we want to do is we want to scale. So we take the, we have some almost complex structure, and then we scale. So now you take q comma p, and you map it to q comma, maybe I'll take sigma times p, where sigma is tending to zero. So what happens is, of course, that this thing is getting skinnier and skinnier, and gets closer and closer to zero section. And then one can prove that the holomorphic curves, so the basic case of this is very old. And the flora, that the holomorphic curves, when you have a graphical one, they correspond indeed to gradient flow lines of the function difference that this graphical thing defines. So when you have more complicated things like here, then you cannot say that they correspond to gradient flow lines, but rather they would correspond to gradient flow trees, which locally have, locally have the, looks like a gradient flow line, but then it's some sort of tree that is being drawn. So let me try to explain what those are. So I'm probably not going to finish that explanation today, but anyway I can start. So that's somehow the next main subject of what I'm going to try to explain. So this is called gradient flow trees. And I will have to wait till next time, but what I can explain is at least the grading formula. So grading. So when you have this Lagrangian inside the one jet space of M, we projected it to T star M, which was somehow holomorphically important. But we can also project it to the zero jet space of M, which is just M times R. And there it looks, what happens when you project the Lagrangian is that you get the front. And so here is R, and here is this M. And for example, the unnote that we drew before has the following front. And the intersection points, self-intersection points, well, how to reconstruct. So here we reconstruct the p coordinate, pi, is equal to the partial derivative of c with respect to qi, where you have a graph, then you have certain singularities, which I'm not going to talk too much about right now. But at least you can see that the double points, they correspond to points on the front. This is a little bit too good. But anyway, points on the front where the tangent planes are parallel. So here there's one chord in the middle. And now I want to give you the grading formula. So the grading of such a rave chord A is equal to, I should say, one more thing. That the singularities, singularities of this front, they are not so hard to describe in low co-dimension. The highest strata of the singularities are so-called cusp edges. So they look like that. It's a semi-cubical cusp. So c squared is equal to x cubed is the local thing here. And then you just multiply it by r in the other direction. And those are all the singularities in co-dimension 1. So if I pick a path from top to bottom, then I will hit only these type singularities, no kind of this co-dimension 2 or anything. I don't have to describe. And the grading of A is the number of down cusps. So that means I go down the cusp in the C direction, minus the number of up cusps. Is it the opposite way with the arrows? It's the opposite way with the arrows. So I go up in the C direction, plus the Morse index. So the Morse index here is that I have a critical point. If I take the difference, the top minus the bottom, I have a critical point of some function. And the degeneracy condition is that this is a Morse function. I take the number of negative eigenvalues of the Hessian. And then I subtract 1. So for the un-note here in the picture, you see I start here, I go down one cusp, that's 1. And I have 1 for the Morse index, that's 2. And so then I subtract 1, I get my 1 back. OK, so then we can expand. I will stop at this, but we can expand. You can now do an next exercise without any kind of calculation. So you can compute the higher dimensional un-note. It's this thing, which is a, now here this is an, oops. Here you have n dimensions, right? x1 up to xn, and this is z. This is the front. So this is an sn minus 1, cuspital edge, and so compute. And so here you can just compute, because you don't have to know anything about trees or anything at all. So you can compute the dj very easily. So I will stop at this and continue on Wednesday. And then I will continue first with the trees and then somehow try to cover moment here. Thank you. Just for Tobias, can you just say where the Morse index comes from in the picture again? Yeah, so the Morse, so in general here, so when you have, when you're at the point in the base where you have two parallel tangent planes, if you look at the function which is the difference of the upper graph and the lower graph, then you have a function which has a critical point at that point. And that's a Morse critical point, and you can take the Morse index of that with the positive function difference, so the top of the rate minus 1. I don't think the right most double point should really change that example, the real exercise. But do you think that exercise, the trifolding? The trifolding. The right most is wrong. The right most is wrong. Thank you. Yeah, you should go down when you go post. Thank you. Yes. Sorry. Somehow. Some convention. That needs to be the same at both ends. OK. Any other questions? So like, the theory, it's, if you have a, how to say it, the hyper-circuit is an m cross r, and then you tell me something about it. But if I took that whole thing and I was multiplied by a trivial factor, I'd go up a dimension by taking that hyper-circuit as 1, so how does the theory change? The same theory, or what's the other one? So if you multiply it by s1, then the theory changes. It's actually one of my exercises I want to give you. But indeed, the theory changes. So basically, over this s1, you have, so when you just multiply, you get the front with very degenerate things. So you have these s1 families, right? So you have to perturb it away. So you get one, something sitting over the minimum, something sitting over the maximum. Over the minimum, you have exactly a copy of what you had before. And over the maximum, you get, so here, here somehow, I have a sort of a generator A hat, corresponding generator A here. And when I want to compute this DA hat, then I get, let's say, gamma of DA. So this, what is this DA? DA is equal to some sum of monomials, b1, b2, up to bm. And the gamma of a monomial up to size. So let me forget about signs. So it's doing the following. So it's b1 hat, b2, bm, plus b1, b2 hat, b3, bm. Plus, and then so it goes. So I just move the hat once. It's like some kind of homotopy operator, if you wish. So that's what it does when you multiply by s1. And when you multiply by r, you kind of have to choose whether you want to live here or you want to live here. So it's not so clear. I have to say what you do at infinity. OK. Thank you for this again.