 All right, thank you. So I guess this is my last chance to try to get my point across. So before getting buried in calculations again, let me try to explain what's the general idea of obstruction bundled gluing. So suppose we want to glue a bunch of things together. Usually, we'll talk about holomorphic curves, although my example's a far worse theory. So suppose I have a bunch of holomorphic curves to UI to want to glue them together. Now, so there's some moduli space M. So this moduli space M consists of configurations of the UI together with some gluing parameters. So in general, these holomorphic curves might themselves live in some moduli spaces. So for any collection of holomorphic curves in their respective moduli spaces, and then some gluing parameters, saying like how much we translate things, et cetera, we can try to glue. Now, the best situation is where all of these linearized operators, so DI is the linearized equation for UI, the best situation is when these operators are surjective. And then there's no obstruction to gluing, and you can just use the usual construction. Everything's fine. The second best situation is where these operators are maybe not surjective, but their co-kernels have constant dimension. So as you move around in the moduli space, the dimension of the co-kernel of this operator doesn't jump so that I have an actual vector bundle over the space M of configurations, where the fiber over a point in this vector bundle is the sum of the co-kernels of all of the operators for all of the pieces. And this is when we have a well-defined obstruction bundle. The worst case is where these co-kernels dimensions may jump. Then you're in Kerinishi land or polyfold land. So that's when you really need to do something fancy. So these obstruction bundle cases sort of in between the classical case or everything works fine, and the general case where you need really high technology to solve the problems. So now let's consider this obstruction bundle case. So then you can try to glue, and you write down the gluing equations, and there's an obstruction. So you can glue up to elements of the co-kernels of all of these operators. And so you can then define an obstruction section S. So S inputs a gluing configuration and gluing parameters and outputs the obstruction to gluing. And whenever S is equal to 0, you can glue. And in general, you can prove that this construction gives you a bijection from the 0 set of S to the set of possible gluings. And then you want to understand what are the zeros of S. And the section S is defined in some strange, indirect way from the gluing equations. So you want to approximate S by something easier to understand. So S0 is this linearized section, which is obtained basically from taking the leading order terms in S and throwing everything else away. And then you argue, so if you're trying to do some count of the number of zeros of S, then in general, this moduli space M is going to have some boundary. So the gluing parameters can only be in some certain range. And you argue that on the boundary of this moduli space, the terms in S0 are much bigger than all of the other terms in S. And so if you linearly deform S to S0, zeros never cross the boundary. So any kind of count of zeros that you can make will be the same for S and S0. So then you just have to understand the zeros of S0. And then the idea of S0, I think I didn't explain this so well last time, so let me try again. So for each of these holomorphic curves or whatever that you try in the glu, there's a component of S0 in the co-colonel of that operator. And the contribution of S0 for one of these curves has contributions for all of the things that are adjacent to it, for all of the things that are being glued to it. So if I look at this U5, I want to say what is the U5 component of this section? So for each of these curves, U1, U2, and U7, there is a contribution which comes from the asymptotics of the end of that curve that's getting glued to U5. So there's a contribution here from the asymptotics of this end of U1, a contribution from the asymptotics of this end of U2, and a contribution from the asymptotics of this end of U7. So basically, you look at the asymptotic expansion of this curve, and you can write a kind of Fourier expansion of it in terms of eigenvalues of the linearized operator, and you look at the leading order term of that. Excuse me, eigenvalues of the asymptotic operator. And you look at the leading order term, it's going to have some exponential decay. So this contribution is going to look like e to the minus r lambda, where lambda is the eigenvalue, and r is the amount that you translate this curve. So that's the basic idea. And then if all goes well, you can then actually count the zeros of S0 to figure out how many gluings there are. So are there any questions about this general picture before I plunge into my next example? Yeah. So when we do the gluing, for example, U7 part dominates in that end, and we do the gluing construction using S0 corresponding to that whatever highest eigenvalue it is. And then we do it for U2, U1, but I'm confused about how we do the gluing in between these, like, in the interior of U5. Like, how do we glue the part coming from U2 and the part coming from U7? How does that happen automatically? I'm confused about that. How does what glue to what? So we glue near the connection with U7. So up here? Yeah. And then we glue near the connection with U2. But those two things should come together to. Well, these don't really interact. I mean, these are sort of far away from each other. But the levels of the gluing, I guess, I mean, there should still be a U5 giving those two gluings, no? So we're sort of doing the whole gluing at once. So we're sort of pre-gluing. So I guess in this example, I've got three levels. So I translate these guys up by some large amount. They can move by different amounts up. I translate this one down by some amount. And actually, these guys can also translate up and down with respect to each other. And I pre-glue using some Cotto functions. And then I deform using sections of the tangent bundle over each of these curves. And then I solve this whole system of equations for all these statuses to be 0, or at least in the co-kernel of the operators. So somehow, in the glute homomorphic curve, in the interior of U5, nothing seems to change. Oh, change the whole thing. But how does that a little bit change happen? I'm missing that point. I mean, there's a section psi 5, which describes how much I'm moving U5 in the gluing. But then we multiply those by Cotto functions, which just die up. I mean, there's a Cotto function beta 5 for U5, which is supported on U5. And it dies on the ends of U5. And then there's a function psi 5, which tells me how much I'm moving U5. So I am moving the whole thing. And OK, so the abstraction, OK, I understand now. There is no equation for psi 5 which contributes to this abstraction gluing procedure. That part is already solved. That part is 0 on the nodes. And then for the other, like this psi 2 part, then I get some condition. Something should vanish in the of this kernel of the i. Some number, that number should be 0. But for the other one, number is already 0. Yeah, I don't really quite understand. OK, all right. If you're happy, I'm happy. OK, other questions? If you could ask another question. How many equations are you going to write down? Seven. So what's so funny? So philosophically, say, if you look at the top floor, you look at maps, model, or common r, actually, that is on each level. In principle, the object is, if you look at asomorphisms, it is a map model of the image shifted by r in r cross v. Well, it depends what you're trying to do. I mean, you could shift the different components. But this are different asomorphism classes of objects. OK, OK. But that would, of course, in the modernized space be a different point. So from my perspective, I think basically you have a blue asomorphism class. It's essentially a blue asomorphism class. And then you have two blue parameters in this case, which is sort of, if you look at the average value, at some points distributed here, the average value here is how they go up. I guess it depends on exactly what problem you're trying to solve, whether you want to glue things. Well, exactly what kind of equivalence class do you want to glue? So am I right in conclusion, in this picture, there are two gluing parameters? Well, for Helmut's point of view, yes. For my point of view, there would be more, because there would be one, two, three, four, five, six gluing parameters. But he can't sort of moving the thing on a level against each other as the same, where I don't. So then I have more objects there. That's where the parameters then come. Maybe, I think the definition of this, if you multiply space, also counts those as different things. OK. All right, that's fine. So let's just not find. From that point, if you're trying to solve that problem, then there are just two gluing parameters. Yes? I think the more or less space would be the same. So I guess the tangential question is, again, you're assuming the dimensions of all the co-carnals are the same, is that right? So then what about, I guess you also need that dimension of the co-carnal on the interior of the moduli space is also the same? Well, I'm not sure what that means. I mean, each of these pieces lives in some moduli space, which is not breaking or anything. So this U5 lives in some moduli space of unbroken things. Right, so presumably that's a piece of the boundary of some larger moduli space. Yeah, so this whole configuration is part of the boundary of some moduli space. And we're trying to understand how many ways can we glue this boundary or corner configuration to something in the interior of that moduli space. All right, so let's do an example. So I'm going to talk about a non-degenerate contact three manifold. So we have the ray vector field R. We have J, a generic almost complex structure on R cross Y satisfying the usual conditions, which I wrote down last time. We want to glue holomorphic curves in R cross Y. So let me talk a little bit about branched covers of trivial cylinders. So suppose gamma is an elliptic ray border, elliptic simple ray border. So what does elliptic mean? Well, here's my orbit gamma. So it's a periodic orbit of the vector field R. And you can look at the normal bundle to this ray of orbit, which is actually sort of identified with the contact plane field restricted to this ray of orbit. And the derivative of this flow defines a map from this normal plane at a given point in gamma to itself. So you just look at the derivative of the flow like this. Defines a linear map. So linearized return map from the contact plane at, say, gamma of 0. This has a symplectic form on a defined by d lambda. The linearized return map preserves this symplectic form. So it's a 2 by 2 symplectic matrix. Non-degenerate means it does not have 1 in its spectrum. And elliptic means that this map is conjugate to a rotation by some angle 2 pi theta. So theta is sort of like a rotation number of this orbit. And the non-degeneracy of the orbit and all of its multiple covers tells us that theta is irrational. And then we have the trivial cylinder R cross gamma and R cross Y. So this is a holomorphic cylinder because of the usual conditions on J. Then I want to talk about branched covers of this trivial cylinder. It's lost. Where's this took? So you go down more. By the time I figure out really, by the time I master these chalkboards, my course will be over. All right, so suppose I have a branched cover. So this is a degree D branched cover of the cylinder where sigma, let's say, has genus G. It's connected in genus G. And let's say it has positive ends of multiplicity a1 through ak. So in other words, so here's R cross gamma. And I have this surface that's covering it. So each end of the surface is going to give me a covering of the sort of end of the cylinder of some multiplicity. So the positive ends of multiplicity is ai. The negative ends have multiplicity b1 through bl. And the sum of the ai's and the sum of the bj's must equal D, because it's the whole thing is degree D. So this has a complex structure on it, pulled back from the complex structure on the cylinder. This is a perfectly good holomorphic curve. It's a nice example of failure of transversality. So Chris Wendell did the sort of a closed version of this for closed curves in his talks. And this is a sort of relative version of that. So why does the transversality fail? What's the Fredholm index of this curve? So the index of this map u0, well, in general, for the index of any curve in a R cross contact three manifold is minus the Euler characteristic of the domain plus twice some relative first turn class. So this notation is, so tau is a trivialization of the contact plane over all of the ray orbits. And this c tau is the relative first turn class of the contact plane field restricted to this curve u. Then there's a sum of conlead-sander terms. So you have the sum of the positive ends of the conlead-sander index of the ray of orbit with respect to the trivialization tau minus the sum over the negative ends of the conlead-sander index of the negative ends. The same formula works in any number of dimensions, except the minus chi gets replaced by n minus 3 times chi. Now, in this particular case, I can just use the same trivialization for everything coming from some given trivialization of gamma. And then the trivialization is xi trivialized along with the gamma. So tau is a trivialization of xi over the ray orbits. So the index of u0 is minus Euler characteristic of sigma. And this first turn class goes away if you use the same trivialization of gamma. So then I have plus the conlead-sander stuff. So what that comes out to be, so it's 2g minus 2 plus k plus l. So that's minus the Euler characteristic of the curve. And then, so if I fix some trivialization, then the linearized return map for this ray orbit gamma is rotation by 2 pi theta. And the conlead-sander index is a, so if I have the conlead-sander index of gamma itself is 2 times the floor of theta plus 1. And then if I'm looking at an ai-fold multiple cover, then that has the linearized return map for that as a rotation by ai times theta. So I get 2 times the floor of ai theta plus 1. So that's the sum of the conlead-sander index terms. For the positive ends and then for the negative ends, I have minus sum from j equals 1 to l of 2 times the floor of bj theta plus 1. And for convenience, I'm going to write this little differently. I'll write this as 2 times the ceiling of bj theta minus 1. Could you say again why the relative turn-class transition? Because the trivialization, if I use the same trivialization coming from a given trivialization of gamma, then that trivialization extends over the entire curve. And so the first turn-class is 0. So the first, relative first turn-class is sort of the obstruction to extending the boundary trivialization over the whole curve. Is it a branch point? Yeah, yeah, it's still OK. But it's because you have a cylinder, right? Otherwise, you add some ding-a-ding or a cylinder. Yeah, I mean, everything's just mapping to this one ray of orbit. I trivialized the contact point field over it. OK, so now I can, oh, I did that. Sorry, that's not the way I wanted to do it. Excuse me. Let's make this, let's keep this the floor. I actually wanted to change this to the ceiling. Because now I can cancel some stuff out. So I can cancel this plus k with this minus 1. I can cancel this plus l with this plus 1 over here. So this whole thing comes out to be 2 times g minus 1 plus sum over i of ceiling of ai theta minus sum over j of floor of bj theta. And then I always get confused by which way it goes. But the sum of the ceilings is greater than or equal to the ceiling of the sum. So this thing is greater than or equal to the ceiling of d theta. And this thing is less than or equal to the floor of d theta. So this whole thing is greater than or equal to 2 times g minus 1 plus ceiling of d theta minus floor of d theta. And because d theta is not an integer, the ceiling minus the floor is equal to 1. So this is equal to 2g. So what does that tell us? So it tells us that the index of this multiple cover is greater than or equal to 0. That's a nice fact. It's not true in higher dimensions, unfortunately. But in three dimensions, this branch cover trivial cylinder always has index greater than or equal to 0. And it can be equal to 0. So the index of u0 is equal to 0 if and only if the genus is equal to 0. And the sum over i of ceiling of ai theta is actually equal to ceiling of d theta. And the sum over j of floor of bj theta is equal to the floor of d theta. So this is sometimes true. So in some cases, the index is 0. So little combinatorics just for comic relief here. Because we need some jokes in between all the analysis. OK. Great. So why did I tell you this? So what's the situation we can have? As we could have a Fredholm index 1 curve here. Let's call this thing u plus, which has a bunch of negative ends at covers of gamma. So this u plus could have negative ends at gamma of multiplicities a1 through ak. Then I could have some curve u minus of also Fredholm index 1, which has positive ends at covers of gamma of multiplicities bj. And then in between them, I could have this curve u0, which sometimes has index 0. So suppose we're in a situation where this thing actually has index 0. So then in principle, maybe I could glue this configuration. And I want to know how many gluings there are. So the reason why I looked at this problem, or Cliff Tobbs and I worked on this problem in our paper, was to prove that the differential in embedded contact tomology satisfies d squared equals 0. So for that application, you actually need to glue stuff like this. Although in our paper, we actually did something much more general. So we glued more or less anything of this kind, even much more generally than what's needed for embedded contact tomology. We also did orientations very carefully, if you ever want to read about orientations. Very carefully. Is it? Yeah, question? Does it have to be a cover of a trivial cylinder? Maybe that goes more like different pieces that go to different parts of gamma. I guess that it wouldn't. Well, you could certainly do this kind of thing more generally for branch covers is something which is not a trivial cylinder. We didn't do that, although that's of interest when you're trying to look at cowardice maps in embedded contact tomology. So Chris Garry gives thinking about that. It's also an example of the Bowen-Honda paper. That's true, yes. The Bowen-Honda paper looks at covers of cylinders, non-trivial cylinders. They're not branched, but they're non-trivial, which come up in Shane Hometopes for invariance of cylindrical contact tomology. It's a very nice example of obstruction bundles that they do. Why do you think the same thing with more general? So I mean, do you see any dimension of 3 or is it many? We just did dimension 3, yeah. But the kinds of configurations that we glued were more general than what you need for your CH. I think one advantage of dimension 3 here is there's sort of low dimensional phenomena related to automatic countisality that will help you prove that the obstruction bubble is well defined. Yeah, that's coming in a second. Right, so in this situation, a situation like this, we have an obstruction bundle. So I say I want to try to glue this thing. So what is this M? Well, it's basically a modular space of branched covers. So the reason why this is an example of failure of transversality is because the dimension, so the modular space of all such curves U0 is a manifold of branched covers of the cylinder. And the dimension is 2 times the number of branch points. And the number of branch points is k plus l minus 2. So whenever this thing is not a cylinder, there are some branch points. So it's a modular space of positive dimension, manifestly. But its index might be 0. So then transversality just doesn't work. There's absolutely no way you can fix that by choosing a generic J. Life sucks. OK, so now in this problem, we could sort of fix the gluing parameters so we could fix how much we're going to translate the stuff up. And then we have a modular space of branched covers. And then over each point in this modular space, we have the co-kernel of the operator D0 for the curve U0. So we'll assume that the curves U plus and U minus are cut out transversally. So we'll assume that the operators D plus and minus are surjective, which you can do if at least these things are somewhere injective. You can assume that for generic J. However, D0 definitely has a co-kernel. And as Chris just pointed out, the dimension of the co-kernel never varies. So the lemma, while I have all these combinatorics up here, one more lemma. I'm just certainly doing this badly. It's not the right way to do this. Yeah, in general, you could look at non-disconnected branch covers, too, on how to use the board. Well, they should have sent us an instruction manual. What do you see in the restroom? There's this big instruction sheet on how to wash your hands. But you just have to be like that for the boards. Then you'll have 25 steps. OK. So the lemma is that the kernel of D0 is always equal to 0. So this is Elkris Wendell's talk. He talked about unbranched covers of closed surfaces, improving that those had kernel equal to 0. And here I'm talking about branched covers of trivial cylinders. So it's a slightly different thing. So this is just a simple calculation. And it's related to automatic transversality things. Right. In fact, I forgot to tell you what D0 is. So D0 is not what you think it is. I'm very sorry. I'm getting scattered here. So first of all, this is a sort of, I want a normal deformation operator. So this operator, D0n, was a go-to. Or you could take sort of L21 sections of U0 pullback of the normal bundle to the trivial cylinder. So these are basically sections ascribing how to move my curve U0 in the direction normal to the trivial cylinder. So you could think of this as deformations that don't move the branch points, roughly speaking. And then this maps to L2 sections of, well, 0, 1 forms on sigma with values in this bundle. So we're measuring deformations that don't move the branch points. So the index, right? So this thing, the dimension of m is 2 times the number of branch points. And the index of this operator, D0n, is minus 2b. So the curve U0 itself is index 0. So the sort of usual operator for that curve would have index 0. But because I'm sort of restricting the domain of this operator by not allowing the branch points to move, that decreases the index by the number of constraints I put on, which is 2 times the number of branch points. So that's why the index of this operator is minus 2b. And my lemma is that the kernel is 0 so that this co-kernel has dimension equal to 2b again. So I have a rank 2b bundle over a 2b dimensional base. And I can expect to get some finite count of 0s, which would be the number of ways to glue. And what's the proof of this? Well, so you let psi be a non-zero element of the kernel, suppose such a thing exists. You want to count how many 0s does it have. So by this sort of Coralman similarity principle, these have isolated 0s, which look like the 0s of a holomorphic function. So the algebraic count of 0s is greater than or equal to 0 with equality if and only if it's non-vanishing. Now, it will be non-vanishing sort of on the ends. So at each end, it has a kind of winding number around 0. So as you go around one of these circles in the ends, you can count the winding number of this section around 0. Then the number of 0s in the interior is going to be the difference between the winding numbers and the ends. So it's going to be the sum over i, sum from i equals 1 to k, of the winding number with respect to this trivialization tau of psi around the i-th positive end minus sum from j equals 1 to l of the winding number of respect to tau of psi over the j's negative end. And then you can put bounds on these winding numbers. So basically, the asymptotic behavior of psi is described in terms of asymptotic eigenvalue of an asymptotic operator. I certainly don't have time to explain that in detail. But you find that the winding number is less than or equal to the floor of ai theta, while one of these winding numbers is greater than or equal to the ceiling of bj theta. So then this whole thing is less than or equal to, you're right, thank you. So this whole thing is less than or equal to the floor of d theta minus the ceiling of d theta, which is minus 1. So I just show that 0's, less than or equal to minus 1. That can't happen. Well, whoever's reading this is unhappy, but this nonsense. All right, so we have a well-defined instruction bundle. This didn't actually disuse the index to be for 0 assumption? No, the kernel's always 0. I actually suspicious as to whether this really is a low-dimensional phenomenon. It seems like it should be related to the fact that the operator for the trivial cylinder is always invertible, which is very general. All right, let's talk about that later and see if it generalizes. By the way, how much time is left? 20? Really? Well, 17. 17 minutes? OK. Wow, OK. Time flies. OK, well anyway, all right, great. I don't know what I'm saying. All right, so we have an instruction bundle. So this instruction bundle is good. So we're happy over here. All right, so now we can do obstruction bundle gluing. So what's the picture? Is that thing as happy or sad as the question? Well, it makes us happy that this method is applicable. And lucky you, you didn't have to do it. So I think people greeted my papers with tobs with a mixture of horror and shod and froida. OK, so let me do the simplest non-trivial example. So simplest non-trivial example. So this is where k equals 1 and a1 equals 2 and l equals 2 and b1 equals b2 equals 1. So my curve u0 just looks like this, just a pair of pants. The top has degree 2. The bottom has degree 1. This is mapping to r cross gamma. And let's assume, we should assume, that theta is between 0 and 1 half. And this is what you need for the index. Well, up to shifting it by an integer, this is what you need for the index to be 0. If theta is between 1 half and 1, the index will be 2. OK. And there's how many branch points are there going to be? There's going to be one branch point. So it's going to be somewhere over here, mapping to somewhere over there. So the modular space of branch covers, we can identify this with a cylinder. Basically, you just need to keep track of where this branch point is. It's actually, the modular space is actually a double cover of the cylinder, because you need to keep track of asymptotic markers. Well, let's not go into that. Now, when you're actually gluing, you're not looking at all branch covers, because I've only shifted u plus and u minus down by some finite amount. So there's only sort of a finite range in which this branch point can be. So this really is actually something, some interval, say minus r, r cross s1. Then we try to glue these things. So this is going to be an instruction section. So can you say, actually, why you trumpeted r at that point? Because I shift u plus up by some finite amount. I shift u minus down by some finite amount. And there's only this finite interval in which I can put this branch cover. So the branch point can't be way up there, because then I'm far away from gamma. So when you're writing this, it's really kind of annoying to sort of sort this out and figure out where exactly the branch point can be, and so on. Then there's another simplifying thing you can do, which is you can arrange that the linearized rate flow around gamma is j linear. In other words, you can choose your j in such a way that the rate flow from one point in gamma to another point in gamma is actually a complex linear map. When you do that, this causes the normal operator d0n to be actually a complex linear operator, and not just a real linear operator. It simplifies the calculations a little bit. So the co-kernel, well, it's a two-dimensional real vector space or one-dimensional complex vector space. So now I have a complex line bundle over m. So this m is an interval process 1. I have a complex line bundle over it. But this is more interesting, because now in my Morse theory example, the bundle was trivial, because you had the same co-kernel over every point. But now these curves use zero in varying, so they have different co-kernels. So it's a more interesting bundle. Then I have some instruction section s, and I have no idea what it is, but I have an approximate section s0. So I fix the gluing parameters. If I say let's just translate u plus up a lot, u minus down a lot, and just fix it. So the gluing parameters, like some constant times this are. OK. So they have an s0. So s0 has contributions from the negative end of u plus and the two positive ends of u minus. And you can actually sort of figure out what these are. So I'll spare you the whole story, but I'll show you what it comes out to. So you get that s0, well, let's give these things names. So this is, let's call this s and t. So s, oh, not this again. I have some of your s0 and s0 sections. No, I have. OK. Just make a little R. Let's make this stigma entail. When I was taking undergraduate quantum mechanics, they would write equations on the board like this. It looked like this. p, p equals p, p. The letter p would mean three different things. So there's three different things in this equation all represented by the letter p. I sort of couldn't. I found this very difficult to deal with. But then on the exam, when I had to write my answers, I would just sort of write the same letter everywhere. And then I got an A. Let's put a hat on this. How about that? Is it your trivialization fix so you can just use tau? Or are you going to change it? Like dynamically scoped variables or something. I forgot this. All right. Anyway, just s0 hat. So s0 hat of s t, what's it going to be? So there's going to be a contribution from the positive end, which is going to be some constant, say, alpha times, let me get this right, e to the lambda s plus i t. Or lambda is an eigenvalue, some operator. So basically, this e to the lambda s means that when the branch point is higher up, you're sort of closer to u plus, so you're seeing more of a contribution. And this i t reflects the fact that as the branch point moves around, well, so you have to compare these co-current elements for different elements in the moduli space. But to do that, you basically need to understand the winding numbers of these co-current elements as you go around the end. So the fact that this term is here represents the fact that the co-current element has winding number 1 as you go around the end. And then there's going to be some, from the negative end, there's going to be some contribution, let's call this, I don't know, lambda plus, the form e to the minus lambda minus s. And these, on the negative end, the co-current elements have winding number 0, so there's actually no t term up here. So I don't really have time to explain this in full detail, but you get an expression like this. So this is actually quite simple. The general case where you have lots of branch points, it becomes very challenging to calculate the number of zeros of the section. You'll search the tricks. But this case is simple enough that you can see directly what's going on. So what's going on? So if s is large, then this term is much bigger than this term. So in the complex plane, when s is large, as t goes around s1, the section's going like this. So let me label this circle here. If s is small, then this term is very small. And then this term is big, but this is, as t goes around the circle, it's a constant. So if s is small, it's some little circle like this. And then on that finite cylinder, you're interpolating between these two circles. So one of these circles has winding number 1 around 0, and the other one has winding number 0. So the number of zeros is equal to 1. There's one way to glue, up to science. In the general case, we calculated the number of ways to glue something like this. We found kind of remarkably that the number of ways to glue is equal to 1, and only if you have exactly the kind of data that comes up in embedded contact tomology. Otherwise, you get some number other than 1. Isn't that crazy? All right. So in my remaining three minutes, let me just close with a question, which we can think about next week. But next week, how am I going to tell us the definition of s of t? So s of t is supposed to count index 1 things, but not just index 1 curves, but any sort of index 1 building. So you could have an index 1 building, which looks like this. So there's some nice, let's say, we can make it embedded, if you like, embedded index equals 1 piece. Perfectly good curve. This is the kind of thing we want to count. And then maybe attached to it are some branch covers of trivial cylinders. So they can be attached above it. So these have some branch points in them or below it. They could also, in sort of worst cases, this thing could have some additional ends. The covers are the same orbit, and there could be some further branch covers of trivial cylinders on the side. So you have some building where 1 piece is an embedded index 1 curve, and then there are a bunch of other pieces either above it, below it, or on the side, which are index 0 branch covers of trivial cylinders. And something like this should make some contribution to the s of t differential. And the question is, how much? So my question is, what is the contribution to the s of t differential? So I have some conjectures about this, but it's not really well-defined question until we've seen the definition of the s of t differential. I think it might, depending on, I mean, so you have solutions on the boundary, on corners, and you have an algorithm kicking certain things in, systematic way, kicking certain things in and out. And there are some conventions. So depending on the conventions, I mean, they lead to the same algebraic theory. So it's conceivable that, actually, depending on the convention, the answer to the question would be different. Yeah, that's exactly what I expect. So I expect this to depend on some choices. So just to draw the picture for the simple example that I did. So we did this example where you glue a pair of pants in between two index 1 things, like this. So this is index 1. This is index 0. This is index 1. We found that the number of ways to glue this equals 1. Now, this building consisting of the upper piece plus the middle piece makes some contribution to the s of t differential. I don't know what. Let's call it c plus. And this lower piece, obtained by gluing the lower piece to the middle, makes some other contribution. Let's call it c minus. And this forces us, modulo. I guess, in general, you need some combinatorial factors in the s of t differential. But ignoring that, the fact that there's one way to glue basically forces that c plus plus c minus equals 1 in order to get d squared equals 0. And I would expect that you could choose one of these to be whatever you want. And then the other one is forced by this equation. So I have some more general conjectures along these lines. Can you just say it again? Because the number of gluings is 1. So there's an end of the index 2 moduli space converging to this thing. So it has to be somehow counted by d squared. So either this thing, it could be counted as c plus times this thing, which is contributing 1. I'm ignoring the signs here. Or as this thing, c minus times 1. So the total contribution to the differential squared is c plus plus c minus. That has to be 1. It'd be nice if I could say that you could make the choices such that whenever there's some junk on the top, it counts 0, for example. Although I'm a little confused with what to do when there's stuff on the side. I don't really understand that. I was actually thinking about this question 10 years ago. And I think I gave a completely incomprehensible talk about this 10 years ago. But I think maybe now we have the language in place where we can actually answer it. All right, thanks. What if I saw on this, in the needle part, some further picture, in the embedded part, do you assume like a standard ECH assumption? No multiple powers of the type of voltage? No, I'm not assuming that. Doesn't have to be one of the things counted by ECH. Although that's, of course, an interesting case. But those are the assumptions that need to be answered if I'm correct. Right, well, this is a different question. This is not a question about how many gluings. It's a question about how do you want to count this? So SFT says there are some numbers, maybe depending on some choices, that have the properties to give you an invariant. And silly people like me say, what are these numbers exactly in these examples? So if you had a slightly more complicated thing where you had, say, three times winding, just one end at the top and three times winding, but then you have three separate disconnected things, order one at the bottom. And I seem to remember you had this partition. I mean, it's a one-wave to do that, or do you have to partition it into two things? And I seem to remember these trajectory. I think in that case, the number of gluings was also one. But it's, in general, it's a kind of complicated combinatorial formula, which I don't remember off the top of my head. It's in the beginning of part one of the two-part paper with top. So there's some complicated combinatorial formula for the number of gluings. Right, but when you split things, when you sort of stretch along the torus and you have a small orbit, you have something sort of that the ECH thing that you get is sort of one multi-covered thing at the top, the generic one, and then lots of single ends at the bottom. Right, yeah, so these partition conditions would say that. Yeah, so in this picture, if you're actually looking at, if your U plus and U minus are counted by ECH, then the numbers A, I, and B, J are determined by D and the angle theta. So if the angle theta is positive and very small, then at the top, as you said, you have just one end. And at the bottom, you have D ends, well, with multiplicity. And it's just splitting into one thing in the middle. Or would it split into lots of things in the middle? Just one thing. So I don't really think of splitting the middle into levels. We don't really look at it that way. Yeah, so Chris Garry will answer all of your technical questions. And he also, if you don't have technical questions, he has a couple of sort of other examples of this phenomenon prepared to tell you about. Guys, it was brought to my attention that the next speaker wants to have plenty of time for a leisurely coffee after lunch before we reconvene at 2.15 in the stink microwave.