 Tack så mycket. Vi tittar på det här polifallt dränkingsgården. Vi ska introducera det i eftermiddag. Så idag vill jag tala om hur man kommer att komputera det- –när det inte är kontakt och homologi i flotterna. Och då beskriver jag några konsekvenser av den här kalkuleringen. Så... Låt mig recalla vad vi gör. Vi har en nott i R3. Och då tar vi den konormala... Unikonorma lyft, som är en set av unit. Vi tar poängen... Det sitter... Jag vet inte hur jag säger det. Det sitter i unit kulten i R3. Det är en set av poängen som ligger över nott– –och som har kovektorer som är perpendicular till kulten. Det är en tropologisk torus. Det är en lösandrig torus. Vi vill komputera det som jag kallar lösandrig kontakt och homologi– –av den här lambda k. Det är en algebra. I den lösandriga kring är det en paus– –som finns där på R3. Det är en passaget om svår kvar cilja– –fer tomma料 i rörelsen. Vi tar den på Y4. Nu har vi en kväll– There are basically three variables in this second commodity. So one is the, so here is the longitude x, here is the meridian p in the torus. So we think of the torus as a boundary of a tubular neighborhood of the knot. And then there is also kind of a, so topologically this is just R3 times S2, so there is an S2 class as well, so this is actually a three dimensional space. And so the coefficient ring, maybe I know where to write it, I write it up here. It's actually equal to the polynomial ring on three variables, e to the x, to the p, and then I'll write the last one q. You can write, think of it as e to the t. But this, so q is related to the S2 factor and x and p related to the longitude and meridian on the conormal of the torus. You haven't written plus minus one of the q, just the meridian? Yeah, I should, thank you. Okay, so now the theory that we are going to compute was just again a reminder. So the, if we compute the differential of a rape chord A counts holomorphic curves in the simpletization of the following shapes, it has a positive puncture at A and then several negative punctures at B1, B2, B3 and basically homologivariables encodes the relative homologi class of this disk capped off suitably. So I won't kind of talk so much about these caps, but if you close it off, then you get a disk with a boundary on this lambda k inside here, so it has a second homologi class. Right, so and basically, so this is a holomorphic disk with punctures, one positive, several negatives. And so basically the first goal today is to tell you how one could actually count in this setup such disk. So the first observation is that this unit contendent bundle in fact is contactomorphic to the one jet space of S2. So the map is just, if you take say X in the base here, Y in the fibers, Y is the unit vector, at X then you map it just to Y in the first and then take the component of X which is perpendicular to Y and last you take this X dot Y. So this is somehow, this is now sort of Q, P and Z. So this is contactomorphism, so in fact we can compute this by calculating homologi, Legendary contact homologi in one jet space of a surface and so let's kind of study this a little bit. So the idea was to use flow trees and let me now try to describe what they actually are. So if we're given, so this is just, this is T star S2 times R. So if we're given lambda inside this J1 of S2, we can project it into the zero jet space of S2 which is called, this is called the front projection. And what is it? Well, so the front then, the front of lambda has generically two types of singularities. Cusp edge, they have some smooth points where it looks like a graph of a function. And this is S2 times R, I should say. And then it has the following singularities only. So there are these cusp edges where it looks like that. So here is somehow it's the zero section and here is the R direction. So it's a, okay. So cusp edge and the other singularity is the swallow tail. So looking like that. So it's two cusp edges coming together and the kernel sort of converting to this cusp direction. And swallow tail. So in other words if you have a lechonry and then you can draw it in three-dimensional world, so kind of in S2 times R, you would draw some multi-graph. And that determines for you the lechonry. Again, remember that locally, so locally here is the zero section and here somehow is your pi of F. Then, no, pi of F, pi of lambda. So then this gives you, so locally you have Z of, say, Q1, Q2. And then you can solve for the peak coordinates in terms of the function and the Q coordinates by taking derivatives. So the front determines for you the lechonry. Okay, so, but in particular if you're at this, in general of course you may have more than one such sheet and they even may be singular, but let's forget about the singularity for a bit. So we stay outside the singular locus. Then over this point you have locally a number here three of functions, graphs of functions. So your lechonry in four you locally defines some finite number of functions unless you're kind of at the point of the cusp edge where it's a little bit difficult to say what this function is. But for generic points in the basis of open dense outside codementation one subset, you have these functions. And so now I want to tell you what is a flow tree. And the flow tree, and the flow tree is, first I should say why, I'm going to tell you what the flow tree is. The flow tree is... Outside codementation one or outside codementation two? No, outside codementation one in the base. So outside, you know, outside the image of this cusp edge, right? Cusp edge is one dimensional, right? So once you're outside, you really have sheets. And when you're at the cusp edge you have a couple of sheets and then this bent thing. Below crossing, I just told you that. That's fine. That doesn't matter if they're close. Just I need these functions. So I am going to explain to you what are these Morse flow trees that we need to count. And they will exactly... First of all, they are rather finite dimensions or combinatorial. You can actually find them, which is good. But then it will turn out that they correspond exactly to holomorphic curves. So it gives you a calculation of this differential in terms of Morse theory data, basically. So a flow tree is a map, let's say u, from a tree. Let's call it gamma, so this kind of abstract tree. Into, in this case, s2. So into the base. And if I draw it, so okay. And we require that along each edge. The flow tree, the image of the tree agrees with... Vi fixar också Remanian metric on this s2, so we can talk about gradients. So it follows the gradient, maybe with... This is of course stupid to write something like that. It follows the gradient of these local function differences. And so you sort of have pieces of gradient segments. And then at the vertices, you have certain matching conditions. So you always have a cyclic order of these things. And if you look at this thing, you can actually lift it naturally to the Lagrange. So here is this minus gradient fi minus fj. But above that, there are these sheets. So this somehow is fi and fj. And of course you can just take this line and lift it twice. So you can lift it to somehow to the sheet where it belongs. So you just lift it straight up. And there is a kind of orientation rule, which I'm not going to bother you with. But it's supposed to look like a holomorphic curve. So on the one of them, I think you always orient by minus this gradient. Can you put some more axes on that picture? I could. So this thing here lives in the base. So this sits in s2. And over s2, so somehow I have to draw many more axes. So this is a big axis containing the fibers of t star s2 and r actually. I can think of it as I lift all the way to the Lechandrian. So the Lechandrian lies. So this is a sheet of the Lechandrian. This is another sheet. So it's somehow just here projects down and I sort of lift them up. And as long as I'm outside the singularity locus, I'm fine. And I haven't sort of told you about this. I'm not quite there yet. But let me draw in one dimensional example. It would be kind of maybe illustrate something. So just in dimension one, this is easy to make precisely. Here we have some kind of x axis. And then here is c axis. So we have this c minus y dx. So I would have maybe one line like that and one line like this. So this is the front. This is my lambda. And if I go over to this xy plane, then maybe it's better to do that. And I have y is equal to this c dx. So this zero sheet, one sheet. The other sheet lying here and one sheet is lying above. And my flow line is just kind of following. Of course, this is stupid thing. It's just in one dimension, but it's lying in the base. But over this thing in the base, I could lift it to either this sheet or that sheet. That's because they're just above here. Now if there are more dimensions, there's kind of no big deal. It's just locally difumorphism, so I can do it. Okay. And I now require that when I come to this vertices, I want the lift to match. So somehow if I look, so now I'm drawing one of these sheets into which I lift. So I have this incoming thing from, let's say, fi here. Then I should continue in the same sheet and I should be able to patch them here. So that the total lift is a continuous curve. So basically it would look like boundary homomorphic. That's what we want. But instead of, so basically at this junction, there is one thing looking like that. And there is one another sheet where it looks something like that. And the third sheet where it also looks like this. So there are three local pictures, right? So somehow I'm trying to, so this is the picture of some homomorphic. This is one dimensional picture where I, in one, so there is no, in dimension one it's going to straight. But in dimension two it's making a little bend here. This one is this one and then last I have this one. So there are these one, two, three sheets where the three boundary components live. Or this homomorphic disk and so that's what it's supposed to look like. Okay. But now for our calculation, this is some sort of theorem I proved many years ago. So somehow the first people to explore this tree thing was, I think Fukaya O wrote a paper about this where you have, here would correspond to having Lagrangian without any singularities at all. So now what I did was I extended to the case when there are singularities and maybe not by now I guess. So in the paper there was some kind of restriction of singularities but certainly it was fine to do this in demand for two dimensions where there are only these two singularities. So let me tell you what are the counterparts of these rigid disks that we count. So a rigid... Sir, do the vertices of the trees, are they meant to stay away from the projection of the singularities? Yeah, actually. So the vertices of the trees lie in the good part where you have these? Yeah, I'll tell you kind of exactly where vertices is everything now. So indeed the vertices need not stay away from the bad part so they can be there but somehow the tree always begins and ends either at this cusp edge or at critical point. But I'll tell you. So a rigid tree has only the following vertices. So they can have valency one and there it could be a... Rabe chord in some sense. So somehow... I don't know how to draw like this. So here you see... This is a front picture. Here you see kind of a critical point. So this would be a rabe chord where you have a critical point. And a flow line can certainly start there just like in Morse theory. So this is sort of a Morse vertex, Morse critical point. Or maybe I should say critical. And it can be of two flavors just like before. So when you orient this flow tree... See either you go up, up the rabe chord or down the rabe chord. Up the rabe chord is positive puncture, down the rabe chord is negative puncture just like for the holomorphic curves. And we look at flow trees with only one positive puncture. So they have critical vertices and then there is another type of vertex where the tree just goes right into the cusp edge. So that's an end. So you see... Of course you could have like a flow line coming this way. And then it's a positive function. Difference just shrinking to zero at this. So it can go there, right? So I should draw one more picture here to make you... These pictures live in S2 or the other front? This lives in S2 and this is somehow a picture of the front over where this S2 picture is. And if I draw the picture in this other XY thing then this is just some kind of holomorphic disk like this. And this is a holomorphic disk like that. This is one valent, yeah. And then I have some two valent vertices. So again I have two valent critical points. So that's just in some sense a little bit special case of this. But that's when I want to have for example a positive puncture at a minimum. So then I would try to do flow out of this minimum but I cannot. It's just constant flow, right? It doesn't flow out. But if I have some other shading between here then what I can do is I can take this thing and I can split it. And then the things flow in different directions. So that looks like that. So it's a sort of degenerate version of some combination of vertices but it needs to be included. It happens only if this guy is a maximum minimum. Can you label the lines and the sheets on these pictures? Yeah, here. I have two numbers attached. Ja, so here it goes between flow line two and one. And here it goes between flow line two. No sorry, one and zero. This guy is between one and zero. And this one and maybe it's between one and two. This is this flow line and this is this flow line. What's happening to this thing? You've got a red chord which is an isolated thing between one and zero. And then you're flowing. So when you flow, what do you flow? I flow with this. So in this picture it looks like something like this that you just pass by this. So if I draw it, in the limit it will be extremely flat. So the flow line that just goes passing by, that would be this one that just continues but it can decide to smoke sheets here, right? That's kind of how one should view it here. What is the limit you're taking? I haven't taken the limit yet. But what I'm doing is I'm scaling the fibers down. So I have a Lechandrian sitting somewhere and I just scale the fiber so that it's coming closer and closer and closer to zero section. And for sufficiently small, when it's sufficiently close there will be one one correspondence between trees and disks. So that's, I mean in dimension one there's maybe always but in high dimensions it's not. Okay, so this is the easy two valent and then there is the difficult two valent which is called the switch. So the switch is the flow line. So here is the caspe, I shall call it sigma. It's a flow line between, and I should again label this sheet with a, so this is in the bottom but maybe I'll draw it here. So this is some other sheet somewhere. This is difficult to draw. So here I have a flow line that going between this third sheet, no this top sheet and the lower one and it starts out creeping up along the lower one and then it can go up until it's on the upper one. So it can sneak by the caspe. So it's a flow line that is tangent to the caspe here and then it starts out. So here it's on the lower one and it's moving up to the upper one. So this is unfortunately invisible in dimension one. So this is the only true high dimensional input. Maybe. Okay. And finally there are three valent vertices. What is going on in the base in that picture? In the base you see. So in the base I should draw the base maybe more properly. So here is the base, this is S2 and there is somewhere, it's the image of this caspe, right? So this is sigma. So here for this sheet the multiplicity is 0 here and 2 here, right? So you cross this caspe thing. And then you have the flow line and then you have the flow line which actually is somehow defined only up to this point because then it stops but if you continue it the flow line is tangent to the fold and at this tangency it switches from one to the other. Okay. So if you think you have a family of flow lines then this is picking a specific one that gets stuck onto the sigma. Right. And then the trivalent things. So they are a little bit easier. First there is something I call Y0 where they are just, so that was the one that I drew up there. So just three smooth sheets and then the flow comes in between two and decides to split into two. It can happen anywhere somehow. Yeah, so I think we could do something like this. So basically when you lift it it will look like this. So this is on sheet 0, this is on sheet 2 and this is on sheet 1. So it's like that. And there is a similar one which I call Y1 which involves also the caspe edge and that somehow the picture is the same but here is the sigma and it splits and right over the caspe edge. So this is Y1, 3. And the picture indeed is the following in terms of fronts that they have a flow line that comes in and then decides to split into two flow lines between the between the two newborn sheets. So it splits and there are sort of two ways to go there. This Y1 vertex. Okay. And then and now as you see, so now our trees they are in some sense so they kind of complete. So they start at critical point and they close up at either critical points or at these ends. So it will be a curve exactly as a boundary of holomorphic curves that goes from end points of course to end points of course. It's somehow exactly looking like boundary. Holomorphic curve. And the main theorem and reason for saying this. So why don't we have end range? No, so because I look at rigid trees. So you can certainly imagine something like this happening but that would be in one parameter family. But when I look at for generic data which is somehow all I care about they will have only trivalent, maximally trivalent vertices. And if you would look at five parameter families then you would have to look at some deeper tendencies etc. So this is the calculation for rigid trees. If you have two flow lines coming in and meeting each other, do you demand that they interact with three relevant vertex or can they just cross? They could cross. They could cross. That's fine. It's a map locally looking like. So they can cross. It need not be embedded this boundary. I'm a little puzzled by this particular because zero and two look completely symmetric to me. The way you are translating it into a disc is not symmetric because one is playing a different role. Your disc has got boundaries on zero and two. It also has a boundary on one. So they are playing different roles. I have here always cyclic ordering. I know which one comes after. I need to match that order. You've got an R action. You could say that. I just have a local numbering. When I lift this, it ends up in some sheet. Actually, I have this orientation convention. One of this lifts is oriented towards the puncture and the other one is oriented away from it. When I come and I go towards the puncture, I know which sheet I should continue on and I require that it's oriented so that it matches there. I see. So you're orienting sort of coming in on zero, two spoken and then going out on the other two. That's why it makes an oriented curve. That's why I can lift it and it's not similar. That's right. So this is some kind of theory and it's a lot of work to prove that they actually correspond to allomorphic curves. So the Morse function I suppose I think of is the Hamiltonian that I was looking at when I was degenerating my curves. No, the Morse function, the functions that you're supposed to think about or the functions locally defined by the Lechandrian itself. So the Lechandrian gives you locally some number of functions and the gradient lines are of the differences of those functions. So somehow in order to understand this theory, one need only to do one exercise actually. So this exercise is the following. It's in the end of my paper on flow trace. So you take R2. So this is the zero section. And then you take some other sheet which is sloped say this way and then you would get flow lines going from left to right say. And then you leave the upper sheet but on the lower sheet you somehow isotope it a little bit so that you introduce a ring of singularities and actually maybe I should try to draw it. So the slice here is this. So this is sort of the first rather nice to move in Lechandrian knots here. So you take this ring and now you just check what happens with this family. So you will find all these singularities playing a role and maybe for this polyfoil minded people it's a kind of interesting exercise because it shows that these trees so here rigid trees are rigid so they kind of are sitting there. But here you will find trees somehow looking some way that have the property that they live in a one parameter family but this piece of the tree completely fixed but that piece is moving. So it's somehow these flow trees are in some sense desperately non elliptic. So this is I think I could explain this more. This will take up all my time so I will leave to discussion this. But just kind of reminder of elliptic series so somehow a flow tree living in say one or high dimensional family it can be completely fixed some places and other place move. So this is somehow anti elliptic behavior. So I think this is this squeezing thing that I have not yet stated my theorem I should but it's doing something rather non trivial. It's killing some is adiabatic limit so maybe it's not surprising but okay. So the main theorem in this business then is that so the Lechondien differential can be so this theorem is actually stronger but let me just state it like this now so it can be computed by counting flow trees instead of disks so we need actually one more one small extension of this maybe not so small but one extension of this theory which is the following Do you just never have to speak about the swallowtales? Do you never have to mention the swallowtales? No that's right. I mean of course when you try to calculate they will play some role but the reason is basically this that the trees that we are looking at disks we are looking that they are rigid and they have one dimensional boundary and this guy has co-dimension too so they should never pass this point and indeed they don't. I try to calculate they will play some role so these are some kind of extension of this which we will use so now imagine that you have your some Lechondien inside this J1 of S2 so this will be actually the connormal lift of the unnote in our case and assume that you have so this Lechondien looking here and then you have some other Lechondien which lying lambda which lies in a small neighborhood of this lambda zero so lambda zero has a neighborhood which looks like the one jet space of lambda zero this is a kind of two blue neighborhood in this business so we take lambda inside the one jet space of lambda zero inside one jet space of S2 and now we are interested in counting disks on lambda zero and then what we do again we take lambda and we do this pinching thing which allows me to kind of relate disks and trees and now so I am stating this kind of very slowly but the theorem is that what you should look for is you should look at disks on lambda zero with flow trees growing out on the boundary so basically you can have a big disk so let me draw so you can have a big disk and somewhere in this narrow region grows out these narrow pieces of the holomorphic disk so basically if you pretend that you know everything for some reason about holomorphic disks on lambda zero and you know the flow trees between lambda and lambda zero then you know all the holomorphic disks you need for lambda as well and the theorem says that in order to count the disks on lambda then you can count instead what they call naturally quantum quantum flow trees they appear in many other guys called cascades and what not it's okay right so now let us actually let us actually try to apply this machinery to calculate the not-contact homology of the unnot so the plan the plan is actually to calculate the not-contact homology for the unnot and then it will actually be quite easy to calculate for any not so what's the strategy so strategy is as follows so in the theorem is something that is allowed to have two disks on lambda not connected by a flow tree no so we because we have just one positive puncture so such a guy would have to have two positive puncture in this case but in general if you count everything then I mean if you look at all orbital holomorphic curve such things would happen but here it won't so it's much easier in some sense the action can only flow down okay thanks right so so the plan for calculation is as follows so we take the unnot and that defines that gives us this the unnot which is a torus and now if you take any other not you can braid it around the unnot so this is kind of k and that gives you this lambda k but then in fact if you think about it this lambda k sits now inside some one yet small neighborhood of lambda u so in order to calculate the thing we need to understand all the holomorphic disks on lambda u and then all the trace between lambda k and lambda u and this is actually kind of one can do and let's start with lambda u so first we need to draw the thing so and we take this aha we take the u to be just a exactly like round circle in the in the x1 x2 plane and so we are trying to draw its front so remember what what was somehow this here is a zero strangely but anyway so what was this last the last coordinate so there was somehow c over in in one jet space of s2 zero jet space of s2 that was c was equal to position dot covektor so somehow now we have in the co-normal we need to understand what's happening to this this little circle of covectors and then somehow we would be done so I'm gonna draw it for you it's very easy so here is the s2 this is the zero section and I will draw the r axis to be just like you know starting at the order and going to infinity so the lift of this circle so here when I'm here the why dot the position vector is positive so I would and over at the poles it is zero so I would get some kind of circle like this right so so somehow it's the circle that I see and here the function is positive and here it is negative so that's it but now we have symmetry so we can just swing it around so the actual lift the front of the co-normal lift is this strange torus mapped with two singularities which looks like cones over the poles and where are the rape quartz well the rape quartz so this is unfortunately a little bit non-generic so I will perturb it in a second but the quartz first actually they come in in an S1 family right over the equator there is a whole S1 family of quartz which corresponds to actually binaural geodesics in here so one should have said but I didn't so so anyway you have these quartz and and if you perturb it a little bit so you can make it short on one side you get the chord C and long on the other side and then remember this grading formula we have and the grading formula says that the grading of C is 1 and the grading of E is 2 so in order to calculate the differential we only have to care about what happens to C and you see it's pretty clear what happens right we have two flow lines going out from C one goes up towards the pole and one goes down towards the pole just a little bit degenerate so we need to perturb it and then to see what's really going on in terms of flow trace maybe I shouldn't use color short so the yellow the yellow thing here is the zero section okay the white thing are graphs of that's the front so they define a function graphs of function so here if I'm standing at this point I have one function difference the difference between this sheet and that sheet and if I follow the gradient flow that decreases kind of function then I just flow straight up to the pole so there are sort of the flow lines that we see there are two such flow lines one going up, one going down here it's a little bit unknown what happens because it's too non-generic but I will perturb it so it becomes generic right this is also a nice exercise in cone when you perturb it one can say many things but it's somehow in the middle is the middle stage of a circular version but anyway if I draw this so here is the cone and if I draw it from the top it somehow looks like this so you have two sheets and there is a dot in the middle now I'm going to try to draw for you what it looks like when I resolve it and I will draw the projection because it's simpler so there will be four cusps and two double lines and outside here it looks exactly the same so this may be kind of slightly difficult to see the first time you think about it but let me draw the movie maybe yes can I get this picture by photographing this absolutely yes so if you perturb this yes you can so you would have to roll it up a little bit along your cylinder instead of having it in the plane so you would find four inflexion points which are these these guys here but this you cannot get by perturbing this guy actually you can never get two rab course two by normal course because they count with orientation so you always have four if you do it so you're sort of thinking in this picture that you have a top circle which which goes to the two top faces of that tetrahedron and you have a bottom disc which you've got on the back two faces and then you attach them along the tetrahedron think of removing the interior of that tetrahedron yes and two of them two smoothed like this and the other one smoothed yeah you can imagine right so maybe I don't have to draw the movie ok maybe I'm not drawing the movie so anyway if you look so this is in the front and the Lagrangian lift of this is just a cylinder like this and somehow the middle circle is completely killed in the projection so ok and now we need to figure out what's going on with these flow trees up here and in fact one of them looks like this I should draw it with some other color one of them looks like this and the other one looks exactly the same in the beginning but I won't draw it exactly the same and then it splits over here and in this picture it corresponds to the disc coming from here and coming from here and it can go either this way or that way so there's somehow these two options so one the eye shaped disc maybe is this one what is that cylinder this cylinder is the pre-image of this in the Lagrangian itself right so in the Lagrangian this is almost isomorphism exactly it crushes the middle circle and when I draw the boundary of this the lifts of these trees it basically looks like this one goes down one way and the other one goes down the back so there are these two curves and when you lift them they kind of differ by this so in fact we have four discs and we find that this DC is equal to one and there are some signs here which I won't discuss right now minus e to the x minus e to the p plus q e to the x e to the p so I'm also not because I'm sort of running out of time so I'm not going to tell you about the q right now it's somehow basically I am going to tell you about the q so basically you have to count intersection numbers of these discs which are just lying above the flow lines with some fiber and if you take this fiber I think to lie here then you get exactly this calculation if it's up on the north pole this is a choice again of capping path and so on you see that there are four discs and they live in in four different home topic class now I'm telling you there are two flow lines that we see starting from here right but then what's going to happen up here that's not clear from this picture and we have to see how would they continue so I have exactly two flow lines one flow line going up up to this point but then there are different histories one going straight and one going and again kind of even here illustrating this not so super elliptic super elliptic the two maps actually agree here right not so elliptic to then not agree ok so so this is our calculation and and I'm somehow desperately much behind time behind schedule so let me try to do something with this calculation so now the one key thing that will appear in the talk tomorrow is what's called the augmentation I should also say what is this D of E D of E is easy it's actually easy to see that this D of E is just a Morse difference so it's C minus C so that's zero so I want to talk about the augmentation augmentation variety which is probably the most important invariant derived from this contact model so we have this we have our algebra this A of lambda k and we can look at the locus where you have a chain map into integers so maybe I will switch to some point to some other coefficients what doesn't matter I'm sorry to ask this you've got two E's in that picture you've got the E which is a generation and then you've got E's separately and D C C is one dimensional and D does that go down in dimension the differential decreases grading by one so that should be multiplied by E that's just an coefficient E is stupid let me change to something else A what about A but then D C if it goes D C is one dimensional so this should be a multiplying by that generator A has degree 2 and C appears in its boundary so that's degree one so the differential decreases degree by one and the constants all are degree zero goes into the ring which is degree zero thanks right so the orientation variety is the locus so I should say is the locus of in the space e to the x e to the p and q where there is a chain map like that let's take a look at this one and only example what is this locus chain map where this has the trivial differential and it lies in degree zero so here for the unnote this is extremely simple because the only thing that it requires the only thing that it requires is that the image of the differential degree one is killed by this map so this we try to define maybe epsilon and the image the differential has to be killed so we see the equation so the equation for the orientation variety is equal to one minus e to the x minus e to the p plus q e to the x e to the p ok and now in the last few minutes let me somehow just very fast explain what happens with a general note so I'm going to follow this maybe I'll keep the orientation variety follow this strategy and tell you sort of how it goes you mean the variety is where that van is the variety is that's right the variety is the zero set of this thing which is the orientation polynomial thank you so so if you have a general note then as indicated on the top board we will braid it along the unnote and in fact so here I'm drawing strangely the unnote as a straight line and then the braid will consist of some strands here but in fact what you can do is you can have these strands all kind of increase along the knot except you have to close them up when you're here so somehow because what is a braid kind of turning but you can certainly keep on going outwards and turn and just kind of have the outwards motion dominate so that means that there will be no binormal chords at all here rather all of them will be here so if you have a brand on n strands then you have chords sort of aij here and bij here and they go both ways so you have something like 2n squared from here all twisting takes place in the point where they're going up yeah, it's completely straight here so the twisting does not really interact with the chords at all, that's right then if you look at it from the top you see, now I should draw this thing extremely close to the original one so you see that there will also be these chords that correspond to the chords on the unnote so you will have some kind of chords cij and let's call the other ones not e because I call them a no, not a, you see I call them aij anyway with the twill so there are remnants of the long chords on the unnote that goes between different sheets and now what we need to do, we need to understand what are the flow trees on this torus and in fact so this lambda k inside j1 of lambda 0 is given by by the function so let me sort of introduce some more chords so here I have my circle I have a coordinate s along the circle then I have some then I have some coordinate cij along the along the this is a d2 neighborhood is s1 times d2 and this one is given by the and if I have my kind of braid here so that defines for you a vector fj the j strand in this disk so somehow as a function of s the point is moving in the disk as I go around and the function of this guy is just fj of s dot cij so this somehow is the function that generates this front and it makes it not so hard to find the flow lines etc so let me just tell you very briefly what happens so here I am now drawing this is the front of the this is the front of the the 0 it needs a torus and I am drawing it like a square and the yellow lines or what's going to the pose so this somehow is the preimage of the cone point and then when I if my strands my aijs and bijs they will be here aij bij and this is actually aji i and bji it doesn't matter so much how I include this so if my strands just go completely straight then the relevant stable and stable manifolds here just go completely straight like this all of them so this is the only stable and stable manifolds where the function difference stay positive if there are also flow lines falling down here very rapidly the function difference become negative and I cannot find any trees so my trees have to follow these things so for the unlink calculation is simple what happens when I introduce a twist when I introduce a twist I change the flow line like this instead like that and this one is falling down here and coming back up so now in order to count the trees you just need to keep track of how many times did it intersect something here so it's not a kind of terribly difficult to do it takes some combinatorial skills to be able to organize this but then basically you can find all the trees by composing these things so one tree we look like that and then you feed this to the next thing and there will maybe be another tree like this and then eventually you collect them up here so this is one piece of the I'm gonna say something for two more minutes this is one piece of the calculation but note that in this picture if I draw this kind of carefully I can also draw the images of all these other big holomorphic disks so they look something like this so I need to make sense out of these big disks with flow trees kind of attached on them and this can also be read off from this picture so basically this picture tells you everything and gives you a similar expression so let me just very schematically finish by saying what how you deal with the orientation polynomial so so what you get is some sort of and I will so you will have in degree in degree one you have cijs and bijs and in degree zero you will have aijs okay and basically so you find something like this that d of cij and constants no there is short aijs these guys the a till the aijs has degree two so they would have been eij but no they degree two so I don't have to care so much so this dcij and equal this bij that's if you think about this just some kind of polynomial I don't know PCij in the aijs and this is some other polynomial pbij in the aijs so and now what we want to do we are sort of obsessing about this orientation variety or polynomial and how to find it well we need to find values where all these polynomial have common roots so basically you can find this orientation polynomial orientation theory so you have to take your note you sort of write it down like this draw this picture and find it there is a formula from a braid presentation and take this thing and then you do elimination theory you get the polynomial and this polynomial is called orientation polynomial and in this case the elimination theory extra simple because there is nothing to eliminate so maybe let me finish by stating what I wanted to say and then and maybe one exercise so the ideal exercise here is maybe the simplest exercise is a calculate for hopflink and so how the q is somehow related to having something on this from this disk so calculate or try fall if you wish just a little bit more difficult so calculate for hopflink and find this orientation polynomial or orientation variety here it's more complicated ok so what I intended to say and actually also to prove was the following that the the theorem is due to Lenny and there is also some other myself and Silibach and Larchev this is the original proof and then the other proof is some other by string topology somehow but that's showing that in fact the orientation so one could say that the note contact homology knows the unnote so that one can prove directly using some kind relation to fundamental group or actually grouping of fundamental group and that's how we go about it in this paper that is in the makings is soon ready but Lenny proved something maybe stronger so he proved that and that's not so hard to see from this string topological thing so he proved that this e to the 2 minus p times the a polynomial of the note which is a polynomial in e to the x and e to the p divides the orientation polynomial at e to the x and e to the p to p and q is equal to 1 and the a polynomial by work of Kronan-Rovka no by work of other people using Kronan-Rovka I don't know it recognizes the unnote so there is somehow if you are not you have some interesting representation of fundamental group with a complement in the SU2 which gives you some, so a polynomial is something derived from SL2 representation of the not group so in fact this orientation polynomial strictly contains the a polynomial and the starting point from for tomorrow's talk is that from this this orientation polynomial therefore gives you kind of a deformation of the a polynomial that was also found by other means from from physics and then the tomorrow's talk is to kind of relate it to and talk about what's coming after this relation has been established so I'll stop at this So in higher dimensions it's so true that the cost of locus as for dimension 1 and higher singularities like lower dimensions so why do you need restrictions on the singularities the answer is that you don't but it requires proof because if you take a holomorphic curve then certainly you can say that it ridges a holomorphic curve it's boundary does not pass through any of these bad things but now you're taking a rather bad limit you're squeezing the Lagrangian if you wish down towards the zero section where basically everything you like about holomorphic curve theory like this estimate between the internal metric and external metric everything is blowing up in the wrong way so someone has to prove that also in the limit you can preserve this this property so it is true but we wrote something that we never really finished but it's a non-trivial statement because it's very hard to control this transalty in the limit so what restrictions do you have do you need so in the paper that I wrote I basically do exactly these two type singularities but in general you can do it generally and you don't have to know anything about these other singularities except that they have a higher code dimension so it is a theorem in the writing but there is something to prove you cannot just state that they won't go there because in the limit you don't really have your favorite elliptic theory anymore In polynomial is somehow SL2 so there is SLN element of the polynomial so in fact what turns out that this divides but there are some additional factors here so this is not equality it's just divides and these additional factors come actually from other such representation like GLN so this knows in some sense about all these flat connections on not code that's not true imagine about the ones with how to say it that looks like that looks like you can use but you can take high rank representations of this augmentation of this algebra and it's somewhere else to get you always rank one representation of this algebra and that knows the other and then you're saying that it's a Lagrangian subarik it should replace the equal and one of the subarik that's right and which would correspond to taking parallel copies of this torus so indeed one thing we should wrap up the discussion is something interesting ok