 Thank you so much. Thank you for the invitation. I'm super excited to talk about this and for organizing this. I know it's, yeah, especially difficult nowadays. So thank you so much. So I'm gonna be sharing screen and then having the video here. Let me know if there's any technical difficulties because I probably won't be able to tell. I'll mostly be highlighting this document I have on one note because it glitches if I try to write. But hopefully there's like a small element of it being interactive. So this is a project that was started at WISCON which was hosted at ICERM and it was joint with Bajaraku, Agnes Gablid, Alexandra Meringovic, Mimi Murphy, Laura Starksson, and Angela Wu. And I'm gonna give a bunch of background before delving into the problem. So if you don't know much about symplectic geometry or contactology, hopefully this is somewhat approachable. So we're gonna start with the basic, which is what is the definition of a symplectic manifold? Well, it's going to be a two-dimensional, I should say an even-dimensional manifold equipped with a non-degenerate closed true form omega. And this will be a symplectic manifold. So two examples, just off the back, are R2N with a symplectic form DXI wedge DYI or CN with a symplectic form DCI wedge DCI bar. CP2 then has an induced symplectic form from the symplectic form C2. Wait, on CN. And it's called the Fubini study symplectic form. As a non-example, S4 is not a symplectic manifold. And so our goal is gonna be to study closed symplectic manifolds. And we're gonna do that by first considering a co-dimension to some manifold with an induced symplectic form. This is gonna be called a divisor. And so then we're gonna split our closed symplectic form into two pieces, the complement of the divisor and the neighborhood of the divisor. And the complement will then be symplectic manifold with boundary. Yeah. And there's a result by Donaldson that tells us that for any symplectic manifold, there exists a divisor. So a co-dimension to some manifold with an induced symplectic form. And then Juru tells us that you can actually choose the divisor so that the complement has a Weinstein structure. So it's a Weinstein domain. This is a very specific type of symplectic structure, which will play nicely with a Morse function. And we're gonna be able to produce handle the compositions from it. So the goal of this project was to produce explicit Weinstein handle the compositions for kind of the nicest beginning examples. So this would be complements of smooth torque divisors and torque form manifolds. So we're gonna see CP2 showing up, CP1 times CP1, blowups of CP2 and blowups of CP1 times CP1. And so those will be the manifolds we look at. Great. So what is a handle decomposition? Let's start with the basics in case this is new. So a handle decomposition is essentially a thickened cell decomposition. And it gives you a way of describing smooth manifolds. So you can start with by building a bunch of zero handles. So this will be a D0 times DN and you build up N0. And then you're going to attach one handles, which will be a D1 times DN minus one along an S0 times DN minus one to the M0 and so on and so forth. And so a K handle will be a DK times DN minus K and it's going to be attached to the MK minus one along SK minus one times DN minus K. And so very briefly, this is a K handle. And so a local diagram of it. And we're attaching along these green boundary components SK minus one times DN minus K and we're going to call the SK minus one the attaching sphere and the DN minus K is really just the thickening that we're doing to get something that we can then smooth and get the appropriate dimension. And just very quickly, Morse function on M will give you a handle decomposition. So in the example, in the case of a surface, for instance, a zero handle will just be a disc. One handle would be a strip and it's attached along an S0 times D1, which is pictured in, I guess the S0 is pictured in red and you can see the D1 strip. And the two handle would be a disc attached along an S1 times D2. And we're going to be working with four manifolds. So let's just go into it. In a four manifold, a zero handle will be a D0 times D4, so just a four ball. A one handle will be a D1 times D3 and it's attached along an S0 times D3. So I've pictured two, three balls and they're going to be identified across the vertical component. And you can imagine them as, I always imagine them as like the feet of a bridge that's spreading out in four dimensions. And then with a two handle, that's going to be a D2 times D2. You can no longer picture that, but the attaching sphere is, it's attached along an S1 times D2 and that's just going to be a knot with a thickening, right? And the thickening is given by an integer N which we'll call the framing. And then you have what a three handle is and a four handle. Those actually are going to turn out to be relevant for Weinstein manifolds. I'm going to put them here for completeness. So let's look at some examples. Well, if I take a four ball and I just attach a single one handle, I'm going to get a four manifold whose boundary is a S1 times S2. If I attach another one handle, say here, then I get a manifold whose boundary is a connect sum to S1 times S2, it's denoted that way. And yeah, and so then the two handle will be attached essentially as a knot with a framing to the boundary of connect sums of S1 times S2. And then kind of going backwards for a surface, right? We can see a handle decomposition on the Taurus really nicely. So you could have started with the zero handle, which is a desk attached to one handles. I always get this trying wrong some times quickly. And the two handles will be these, the pink strips, the A1 and A2 are identified together or the B1 and B2, it's a single one handle. And you can also then attach a two handle, kind of running attached all along this green curve, right? And so then you'll get a Taurus. And in order to get the discotentant bundle of the Taurus, all you actually have to do is thicken this handle decomposition on the Taurus. So you'll start off with a zero handle, which I pictured in purple. And then you have to add in one handles, the black one is the one with the, which we're gonna label A with attaching spheres A1 and A2 and another one handle in orange with attaching spheres B1 and B2 here. And then the two handle is attached along the screen curve that runs through the one handles. And that's actually like a smooth handle decomposition of the discotentant bundle on the Taurus. And we're gonna see that this is gonna be like the main protagonist to some extent of this talk. So now if we wanna put in some symplectic structure in, I first have to give you a few definitions. So let's start off with, we have a symplectic manifold with boundary that's compact and exact, exact meaning that the symplectic form is D of a one form alpha. And we're gonna ask that it has a vector field Z, which we're gonna call a Leoville vector field. And that means that along the flow of this vector field, the symplectic form is symplectically expanding. So the lead derivative of Z with respect to omega is equal to omega. So that's one important that gives you like the definition of a Leoville vector field. And we're also gonna ask that it's transverse to the boundary. And what this is gonna give us is that it induces a contact structure in the boundary. The boundary is gonna be an odd dimensional manifold. And the contact structure here will be given as the kernel of the one form restricted to the boundary. And if you take all of this together, you get a Leoville domain. And so if you've never seen the definition of a contact structure, that's going to be a hyperplane, it's glitching. Okay, the hyperplane distribution Psi in the tangent space of some odd dimensional manifold, call it Y. And Psi can be given as the kernel of a one form. And this will be a contact structure if the one form satisfies that lambda wedge d lambda to n minus one is non-zero. So here's a quick example, symmages from Wikipedia of R3 with the standard contact structure. So to each point I've assigned a plane and the contact structure is given by the kernel of dz minus y dx. You can see that the planes, if you move up and see, you'll get the same array of planes and they're kind of twisting one way and then twisting the other way. And one geometric kind of fact about contact structures is that at least for, say if you're in a contact three manifold, you can't put in a surface whose tangent space is contained at every point in the contact planes to the contact structure. So that's just not possible. But, and that's part of the definition of a contact structure, but you actually could put in like a knot and whose tangent space is contained in the contact structure. And we're gonna see those repeatedly and those are gonna be called Legendre and knots. And I will redefine them in a second. So what did this have to do with handle decompositions? Well, Weinstein defined basically on a K-handle a symplectic form in a Liouville vector field so that the Liouville vector field was transverse inward along the screen boundary, which will be the one along which we're attaching and transverse outwards along this blue boundary. And this then means that the green piece, SK minus one times DN minus K has a contact structure. So, yeah, so that's really nice. And then actually it turns out that the attaching sphere, SK minus one is going to be an isotropic some manifold of the contact manifold. And this is gonna mean that the tangent space at every point of S is gonna be contained in the contact structure of this boundary. Great. And so Weinstein domain comes from essentially taking say a Weinstein zero handle, attaching a Weinstein one handle so that the Liouville vector fields like match up and then attaching to handle and handle and so on, right? And so it's another way of saying this is that a Weinstein domain is a Liouville domain with a Morse function five so that the Liouville vector field is actually gradient like with respect to this Morse function. So the Morse handle decomposition and the symplectic structure playing together really nicely. And these are like an important class of examples of symplectic manifolds and there's like a lot of cool results about them. And some basic results about them were given by Eliashberg who showed that Weinstein two-end manifold is built from Weinstein key handles but the K the index of the handle this K has to actually be less than or equal to N for two-end manifold. And if K is equal to N, it's a critical handle and it's gonna be attached along a Legendre. So this is an isotropic sphere of like kind of the largest possible dimension that a semantifold could be isotropic and that's why it gets this special name Legendre. And there's like a lot of results which I won't mention about like the subcritical handle attachments kind of being different from critical handle attachments but there is this distinction. And yeah, and that's all I'll say. So again, here concretely, a semantifold and a contact manifold is a Legendre and if it's tangent spaces are contained in the contact structure for all points on lambda and if two times the dimension of lambda plus one is equal to the dimension of the contact manifold. So in the case of a four manifold which is really all we're gonna concentrate on, the critical handles are two handles and there will be attached on the Legendre knot and it's going to be when you've taken zero handles plus one handles, you get this boundary connects some of us one times this too, right? And then the framing actually is determined by the Legendre knot. So it's not just any integer, it's going to be an integer given by the Thurston-Benekin number of lambda and this is a classical Legendre invariant. You also add in a minus one. And this was shown by Lee Ashford. So to some extent, if I draw you a Legendre so say if my manifold was just, my handle decomposition was just this guy, a zero handle plus this two handle attached along some Legendre lambda. And I can tell you that TB of this minus one is gonna be, wait, TB, it's gonna be minus one, minus two I think. Then I've given you all the information. And what I've just drawn, I kind of jumped ahead was the front projection of the Legendre knot or one of the Legendre knots to be more precise. So if you take R3 with the standard contact structure, you can project down to X and Z, so forgetting the Y coordinate and you can try to draw what the Legendre knots look like. And so here's one of the Legendre none knots. And it turns out that because the tangent vector has to lie in the kernel of DZ minus YDX that the slope recovers the Y coordinate. So you don't have any points that have infinite slope, which is why you get these cusps. And you also actually don't have to even tell someone like what the crossing looks like because it's gonna canonically just be crossing in the following form. And so here's a Legendre and Truff oil. This actually is a different Legendre and unknot. So these are not Legendre and isotopic even though they're both topologically the unknot. And there's infinitely many Legendre and representatives of the unknot that aren't Legendre and isotopic to each other. So you can just essentially add it and see the Zags and I'll make them distinct. And if there's some Dominican number, actually we'll detect this for the unknot. Okay, so a handle decomposition gives you basically a diagram of a smooth manifold. And you can then ask when are these different handle decomposition diagrams equivalent? And so it's under three categories and moves. You can have handle slides, handle cancellations and isotope of attaching spheres. The isotope of attaching spheres I think should be fairly clear why it's true for handle cancellation. What's happening is that like say if you attach a one handle and then a two handle kind of running through the one handle then it's like you just added in a bump. And so that picture for the four manifold would be the following. And you can just like take away both handles. And this means that like you have a K handle and a K plus one handle that runs through it once and just take it away. And a handle slide means that a K handle will just like morph over another K handle. And you can see here the, this is like a diagram for a two handle and the pink one moves over the blue one and they'll do it just kind of slowly. And then these two spaces should be the same, right? And in the case of when we're talking about four manifolds all you would see is like a not K one and not K two. And that that should be the same as attaching a not K one and then a not K two that kind of runs around the K, a push off of K one and connects them with K two. So you'll really be seeing the feed of like these E glues. And I really love the fact that like Kirby did a lot of work on this. So it's often called the Kirby calculus and like Kirby Ray, like the video game, like Kirby eats things. So yeah, small aside. So now in the Weinstein category you also are gonna get that Weinstein handle bodies will differ by handle slides, handle cancellation. And now isotropic isotope of attaching spheres will give Weinstein homotopic manifolds. So that means that it's not just the same smooth manifold but that you're deforming the Weinstein structure from one to the other. And Gauff gives a really nice description of this point for Weinstein four manifolds and Casal and Murphy give nice descriptions for what the front of like a Legendre and handle slide would be and handle cancellation. And that's all I'm gonna have time to do but hopefully I'll have some time to do it kind of in action. We also put out a conference paper on archive giving an introduction to all of this and with the hopes that it would be like a nice introduction for grad students who wanna learn about Weinstein manifolds. So that's available on the archive and gives it kind of a sketch of this algorithm. Okay, so now we're gonna get into Toric manifolds which are gonna be the source of our examples. So a Toric four manifold is a symplectic four manifold and it's also going to be equipped with an effective Hamiltonian torus action. And because of this, it's going to have a moment map. So it's gonna be some map to the plane where in each coordinate you get that it's a Hamiltonian function. And I'm gonna kind of run through this a little bit but one way to think about them is through the image of their moment maps. So in this example, we have C2 and it's gonna have actually at the standard torus action whose moment map is as follows and it's like really encoded by the moment map. So that's why I don't actually have to actually write it down. And the image is gonna be the following upper quadrant of the plane. And the pre-image over any of the purple points is gonna be a torus. That's actually gonna be the torus, the orbit space for any point there. And the pre-image over one of the edges is gonna be an S1, which again is the orbit space of any of the points in that S1. And the pre-image of a corner is gonna be a point. And so those are the fixed actions, the fixed points of this torus action. And in the case of C2, there's only one. Great. So CP2 is actually an example. It's like one of the nicest examples, first example of a torus for manifold and it has the following moment map image. And again, the pre-image of any of the purple points is T2 over any of the edges. The pre-image is S1 and it has three fixed points. And Atiyah, Guillemin and Sternberg showed that the image of a moment map is gonna be a convex polytope in the plane, which satisfies a few constraints and they called it a delzant polytope. I'm not gonna define it further for the purpose of this talk, but they're pretty cool. And essentially, if you kind of draw a delzant polytope, you have a torque for manifold, at least in the plane. And you can do this for torque and manifolds in delzant polytope, so then we're well understood. And something to note and we're gonna use throughout is that the pre-image of the facets of the delzant polytope is a symplectic divisor. We're actually gonna call it a torque divisor because it's also invariant under the torus action. So it's really, really a really nice surface in these four manifolds with an induced symplectic structure. And we can actually see them pretty nicely because we know that the pre-image over any of the edges is a circle or it should be this guy. And as we get close to one of the nodes, it kind of pinches down. And then here we have like another family of circles. And so we get three spheres that are pinched together. And you'll see this pattern like for any, like if I had, say if I had a hexagon was a delzant polytope, then you would see six spheres pinched together in a necklace, right? And you wanna notice actually that the complement of these torque divisors is actually fairly straightforward. It's just gonna be a disc cross the torus, which is the discotangent bundle of the torus, right? Because we know that the pre-image of any of the purple points is the torus. So I guess if that were it, that would be the end of it. But what if actually we smoothed the torque divisor? And so now we have to think, well actually what is happening at these singularities? These will be normal crossing singularities. And so locally, that means that they're given by, let's say in this chart Z1, Z2, they're given by Z1 times Z2 is equal to zero. And when we smooth it, we wanna replace it with the following local model, the Z1 times Z2 is equal to epsilon, right? And we wanna ask what's happening to the complement? So this shaded purple region. And well what's happening here, let me also draw Z1 times Z2 is equal to, so epsilon and Z1 times Z2 is equal to minus epsilon. If epsilon goes to zero, right? We get the Z1 times Z2 is equal to zero. And you can see that like if I somehow attached a bridge here and connected these two shaded regions, I would then get what I wanted, right? And what does that mean to attach a bridge here? It actually is going to mean that we wanna attach a two handle to the disco tangent bundle of T2, right? And a two handle is attached along a thicken knot and S1 times T2 and it's going to be attached along the boundary of D star T2, which is the unit at cost for your bundle of T2. And you know that the cost for your bundle of T2 can project down to T2, right? And so then you could have also projected down this thicken knot and that's gonna project down to some circle, some gamma AB for AB coat prime in the torus, right? And then just by definition, the Legendrian curve, this S1 times T2 is going to be the conormal lift of this curve. And with a contact structure on S star T2, you know that this conormal lift is Legendrian, right? And so this seems kind of like a complicated way to say like attach a two handle along some curve, map it down to the circle on the torus and bring it back. But it's because we can find what this curve is on the torus and then we're gonna need to lift it to the conormal lift because a priori we don't know what this Legendrian curve is. So our first theorem is that the complement of a divisor smooth at one node has a Weinstein structure given by attaching a two handle to the discotentium bundle of T2. Along some Legendrian, we're gonna call it lambda AB and AB is gonna be given by the difference of the inward normals at the node. So if I'm smoothing at this node, for instance, the inward normals of the Delzap polytope are zero one and one zero and the difference will be one minus one. We're always kind of taking this direction for taking the difference. And so now we wanna ask, well, if I know this curve, do I know the front projection? I want actually the front projection of the conormal lift. Right? I really want like an explicit diagram of this Legendrian. And so first, actually, let's even understand the discotentium bundle of T2. I'm gonna do a time check. Okay, so this is one of the trickier parts. So I'm happy to pause now. Are there any questions? Okay, so let's recall that you can build T2 with the following handle decomposition and that we had like basically built the following smooth handle decomposition of the discotentium bundle on T2. And gonf shows that you can give the following Weinstein handle decomposition of T-star T2. So this is now the Legendrian curve along which the two handle attaches. You can see the two one handles and B, right? And it looks markedly different from the smooth one. And he essentially plays around with the smooth one and then adds in a twist to get the correct framing. And so that's like one argument for it, but we really need to understand a little more in detail how to obtain this handle decomposition for the purposes of this problem. And so what we're gonna do is we're gonna look at the following like Legendrian curve plus kind of strips on the one handles and we're going to pinch, we're going to cone them off. They're still being, this is still a torus, but now it's gonna be a singular torus because we're coning off these purple strips. And then I'm gonna, this is now actually, if you look at it, this is a Legendrian knot. And you're going to do a Legendrian isotope by adding in a twist. You add in another two sets of twists and you cone off for the orange handles and then you get this guy. And now when you see that, that's more reminiscent of the two handle in the following handled body, right? And what's really happening here is we're giving a homotopy of the Weinstein structures from the canonical Weinstein structure on the cotangent bundle of the torus. The canonical one has the delieval vector field is just in the fiber direction. And so the zero section is the entire torus, right? And if I contracted, or if I flowed backwards along the delieval vector field, I would hit that torus and the symplectic form on that would vanish. And so that's actually going to be a smooth Lagrange and skeleton of the cotangent bundle. So just, this is the delieval vector field and this is T2, right? And omega restricted to T2 is equal to zero. And T2 here is smooth. In this example, if I look at the Lagrange and skeleton, so if I flow backwards along the delieval vector field, I'm not gonna hit a smooth semana fold. I'm gonna hit something with singularities. And it's gonna be given by essentially the core of the two handle plus you cone off at the one handles as follows and then you're going to, you have the two handle lying over it and then you're gonna cone off everything else into the one handle. And so you get a singular torus. And these diagrams give essentially a homotopy from the smooth torus to the singular one. And because the Lagrange and skeleton, so this whenever you flow backwards along the delieval vector field and hit a semana fold where the symplectic form vanishes, that encodes like the information of the Weinstein manifold, you actually can then build a Weinstein homotopy from one to the other. And so we first showed that if you have a liable manifold where the Lagrange and skeleton is a smooth closed Lagrange and L, then the cotangent bundle is exactly that manifold. So it's like a straightforward thing that people knew of as we've just written it out. And then this was also known but not written out this way. This comp handle body of the cotangent bundle of a surface is Weinstein homotopic to the canonical Stein structure on the cotangent bundle on the surface. So maybe instead of saying induces, I wanna say Weinstein homotopic. Great. So this all seems like really complicated and maybe you wanna take a deep breath because it's about to like maybe get a little worse. But yeah. So part of what's happening is that, so this is actually, if you just took the following, you took T-star T2 with D of P1, DQ1 plus P2, DQ2, that would give you the canonical symplectic structure on the cotangent bundle. Great. We're gonna add in essentially a morcification and it's gonna be D of P1 partial of phi over DQ1 plus P2 of partial phi over P2 where phi is a Morse function on T2. Q1 and Q2 are the coordinates on T2 and P1 and P2 are the fiber coordinates. And then we're gonna look at some subdomain where we're looking at a level set of phi. So phi is equal to C and P1 squared plus P2 squared is less than or equal to one. And this actually will turn out to have contact boundary. And all of this like symplectic manifold with this like morcified Weinstein structure is gonna be giving us essentially this handle decomposition because this handle decomposition was given by looking at a Morse function on the torus, right? And so if you say we're to add in a T here that's ranging from zero to one you would be getting this homotopy as well. And now when we're looking at the subdomain well, why do we wanna look at this because we wanna understand the contact boundary, right? Of this new guy and we then wanna look at the genre and curves on it. And so it's gonna turn out actually that if you look at a conormal lift of a level set of this Morse function that that is the same as a red push-off of this other Legendre which is when phi is equal to C and P1 and P2 are equal to zero. And so if all of that was confusing essentially we're able to identify the following curve on the torus with the following Legendre and SRT2. And if we have a curve that's really close to it that kind of goes like this, for instance from A1 to B1 we actually know that it's gonna do something like this. It's gonna go from A1 to B1 like that go through the handle. So when it goes over a handle it kind of cusps upwards and we know how it cusps and if it goes into a handle just goes in. And we can also use the fact kind of the relative order of the strands here and see that in the you will also see a relative order in this picture. And so our strategy is to isotope this curve close to the blue curve K just where the two handle is attached along and use a contactomorphism. So an identification of a neighborhood of the two handle and SRT2 with the one just space of S1. And so let's do this in practice. So if I took say the one zero curve I would have it here and then I'm going to have to all of this also has like orientations which I'm sweeping under the rug but I would have to push it close to that. If I had the minus one zero curve I would have to I should pick a different color. Sorry to say orange. I would have to push it down here. And then I can like I done in practice you can imagine like cutting the square along that purple this purple point and kind of opening it up and getting a strip which will be J1 of S1. And so you'll get the following. You will go from A2 back to A1 with this orange curve and from A2 to A1 with the following curve. And then you wanna use this contactomorphism to map it to the front. And so if I do the green curve I start at A2 I go over B1 and into A1, right? And now if I do the orange curve I start at A2. So I wanna start at A1, go over B2 and into A2. And so it really tells you kind of which part of this curve maps to which part of here because you're identifying like this, like I say this segment is going to be identified with some segment over here for instance. Any questions? How am I doing on time? Okay, I'm speaking until 1250. Great. Yeah. All right, so let's do one of the examples which is when we just smooth once. And actually this is gonna be kind of a canonical example because it turns out that for any delzant polychope if you take a corner of it there's a SL2Z transformation so that that corner looks like the standard corner and other stuff happens to the other corners, right? But essentially if I'm taking any torque divisor in any torque format, I fold in a smooth once I should just pick that corner after an SL2Z transformation look like the standard one. And when I smooth at the standard corner I know that the difference of the inward normals is one minus one. So I'm looking for lambda of one minus one. One minus one here is in blue. I'm gonna have to push it kind of close to the K curves as follows. So they're kind of going like this, like this. And then when I map it to the J1S1 diagram I'm mapping, so I'm gonna go from A1, sorry, I wanna go with the one that's closest to the kind of black boundary of the torus even though that doesn't actually exist, right? Or it's all identified together, sorry. I'm going from B1 to A2 is this curve. And then I'm gonna go from A1 to B2 here. And then when I map it into, I use the contactomorphism going from A, sorry, that's wrong. I'm going from A2 to B1, yes. So I go from here to here, right? Cause this is A2 and B1. And then if I go from A1 to B2, but going over A2 and B1, I'm doing this guy, right? So I start at A1, go over B1, go over A2 and get to B2, great. Now, after doing these Weinstein handle slides, handle cancellations and Legendre and isotope, you're gonna end up with the following Weinstein manifold. And this turns out to be just, you take T star of S2 and like plummet to itself, you're gonna get this, so that's what it is topologically. And some of the moves you have to do to get to that, this diagram, for instance, you could do a Legendre and Rheidemeister move that essentially pulls this cusp upwards and then you can get rid of it. And so then you have the following curves and then you can do some more Legendre and Rheidemeister moves and you're gonna wanna cancel, I think, this guy. Great, so something else to look at, excuse me, is what happens when you smooth twice. And here we're actually going to smooth twice at a particular type of corner. So if I take CP2 and I blow it up at this point, that means that in the Delzat polytope image, I've like chopped off in a certain way and I've added in two new nodes, two new fixed points and they will have the following inward normals. And so if I wanna smooth at this point in this node, the torque divisor and look at the compliment of the smooth torque divisor, that means I wanna add in lambda one zero and lambda zero minus one. And so here they are in blue, because I was worried I wouldn't have time to draw them, but the green one is lambda one zero. I push it upwards in lambda zero minus one and I'm gonna push it to the left and then that maps that the green one's going from A2 to A1 and then the blue one is going from B2 to B1 and over it and then you map it into the SRT2, right? And so you go from A1, pass over B1 to A2 for the green one, for the blue one, you go from B2 over A2 to B1. So this becomes really algorithmic and yeah, it's really neat. And so here a few of the Legendre and Isotope nodes you would do and this is overwhelming it's because just throwing a bunch of diagrammatic moves at you, but you would do a right of my sturt, set of two and three moves to move this caspa, then you would get rid of it, then you would have the following diagram, then you would hear, well that once is drawn, how you're going to cancel the pink one handle with the green one and then once it's taken away, sorry, first I should say you're gonna handle slide over the green one and that means that you're gonna like to get the following diagram and then you're going to get rid of the green and pink handle and you're gonna get the following picture and simultaneously I was doing some right of my sturt moves that get the light blue curve into the following shape, kind of curves around and then once you have that here, you're going to slide over the blue. So the black one say is gonna do something like this at one point and then the other one will have to do something over it and you'll get something that looks like this after canceling the light blue with the orange one handles and now you just stuck with a genre and isotope and like what happens in the end is that you get the truffle, which is pretty neat and so that means that you get a compliment also that is simply connected. The moment you're smoothing two notes that are coming from a blow up. So let me just, oh, I'm at time and yeah, so can I say a few of the like further questions? Yeah, I was gonna say yeah, we started a few minutes late so you could go for a few more minutes if you'd like. See, thanks. So I haven't drawn some of the more complicated examples but you can really like, we look at like CP2 with all of the notes smooth or CP1, which has a following, the slide polychope the square with say two opposite corners or all four notes smooth. You can do it for more and more complicated ones in your Weinstein Handle Decompositions will look wackier and wackier but what's nice is that you're getting a different description of them. You can start asking a few questions about say some manifolds in these spaces. So for instance, this is the diagram for CP1 times CP1 where you smooth that opposite notes and one thing to note is that if you take like the core of the two handle and then you look, it's the end knot, it's going to have a filling, so a hemisphere that lives inside of this four ball and you're gonna get a sphere and actually in this case, it's gonna be a Lagrangian sphere. And so you can start to actually find Lagrangian some manifolds really easily from these Weinstein Handle Decompositions and one neat thing is that we found in this example that the purple region would give you one Lagrangian sphere and the yellow one would give you another and they're in different homology classes. So you know they can't be isotopic to each other but there's actually a result that in CP1 times CP1, all Lagrangian spheres are Hamiltonian isotopic to each other. Like there's only one at Hamiltonian isotope. This is a result of Heim. So in this implies that actually the isotope is passing through the divisor, right? For this particular one, which is pretty neat, I think. And we haven't yet been able to find, say, a genus to Lagrangian cementifold in these complements. So so far we've only found Torian spheres. You can also ask like what happens when you smooth different sets of nodes, do you get the same simplectomorphic complements? So say if I have some space and I've done these three nodes versus another space and I do a different set of three, do I get the same complement? And it turns out that like if you write them out where the PIQI are those like, sorry, the AIBI coordinates of the curves, that if there's an also to Z transformation between these matrices, but yes, the complements will be simplectomorphic. And finally, it would be really nice to be able to use invariance of these Legendrians to like understand better questions about Torque for manifolds, such as like Merisimetry ones, but that's like a question. And I should end there. Yeah. Thank you.