 OK. Thanks to the organizers, right. So anyway, so I'm supposed to talk about obstruction bundle gluing. And the references are so that these two unreadable papers by myself and Tobs gluing something, something, something, parts one and two. And the fun part begins in the second paper, section five. So there it explains the gory details for obstruction bundle gluing for holomorphic curves in four dimensions. But I'm going to start with the simplest non-trivial example I could think of, which is for Morse theory. And that example is on my blog, flurhomology.wordpress.com, in a posting from July of last year. And I'm not going to talk about this at all, but there's another nice example of similar techniques in a paper by Urkel Bau in Kohanda on contact homology, 1412.0276. So you might find that interesting also. So what is obstruction bundle gluing? So obstruction bundle gluing is a way of gluing things when transversality fails, but it doesn't fail catastrophically. So you might not need polyfolds, but it kind of fails, so you need something. And it'll be interesting later to compare this with the polyfold stuff. So Helmut next week is going to talk about the definition of symplectic field theory by polyfolds. So the polyfolds say, well, you can get some numbers. And these numbers satisfy some properties so that you define invariance. But maybe you don't know much about these numbers. But in some situations, you actually want to know what these numbers are. You want to know what's going on. For example, if you have a situation where you know what all the holomorphic curves are, but transversality fails, you want to know what are the numbers with which you're counting these things. An obstruction bundle gluing is relevant to this question, as we'll see. But here's the Morse theory example I want to start with, because it's easier. And then I'll talk about the holomorphic curve stuff later. So I want to consider circle-valued Morse theory and circle-value just to allow this phenomenon that I want to tell you about to occur. So we have a finding-dimensional smooth manifold. And we have a Morse function. And you choose a metric that allows you to define the gradient vector field. So Morse means, I mean, if you have a circle-valued function, then locally, its derivative is the derivative of a real-valued function. So locally, it looks like a real-valued function. So the same definition of what's the Morse function. And then you can define a Morse complex where you do the usual thing. You count gradient flow lines between critical points. The only issues you have to be a little careful, because there can actually be infinitely many flow lines between any two critical points. So I draw my manifold x like this. And f is this sort of circle direction in the picture. Maybe there's a critical point over here and the flow line is some critical point over there. The flow lines can sort of go around in the circle direction many times. So there can be infinitely many of them. So to get a finite count, you pick some level set that doesn't contain any critical points. And say it has no critical points in it. And this theory is to find over a nova covering, which in this case is power series in a form of variable t with coefficients in z. And the differential of a critical point, so p is an index i critical point, is the sum over critical points q of index i minus 1 of. And then we have the sum from k equals 0 to infinity in the form of variable t to the k times some count of flow lines from p to q. Let's put a k here. And these are flow lines that cross the level set sigma k times. So the flow line like this is counted with t to the 0. If it crosses this three times, it's counted with t cubed. And once you do that, you get finite counts. So then you get circle value more theory. And it's kind of fun in ways which are completely relevant to my talk. Now, you could say, well, is this an invariant? If I do a homotopy of the function f, and if I change the metric g, replace it with some other metric, will I get the same homology? And the answer is yes. And you can prove this by following the usual strategy of defining continuation maps. But for certain obscure purposes, which are beyond the scope of this course, sometimes you want to know in more detail what the continuation chain map actually is. You want to define an explicit chain map. So you want to deform f and g in a homotopy. And maybe at some times during this homotopy, it will fail to be generic. And you want to know, as you cross that non-generic time, what exactly happens to the chain complex? Stay? Is it staying? I can't see. Yes. OK, good. So you have a? So about this circle value more theory, why not to unwrap the circle? Yeah, you can do that. Although you still have to account with the Nova Covrain, because then you'll have a non-compact manifold. But then by now, our usual most functions change. Ah, but this is, you're about to see why it's not that simple. OK, so now I have a homotopy. And let's say it's not generic at time t equals 0. And there are various ways it can fail to be generic. But the one I'm interested in is where I have a critical point q of index i, say. And I have a flow line from q to itself. Let's call this gamma 0. Or actually, let's call it u0. So u0 is a flow line from q to itself. So the reason why I'm doing circle value more theory is for this to be possible. This can't happen in real value more theory. In circle value more theory, this can happen. Let's suppose that this intersects sigma just once. I want to know what's going to happen to the chain complex. And so you could have some other critical point p of index i plus 1. So here's p. And there could be a flow line. Let's call this u plus from p to q. So this thing has index equals 1, this flow line. And then unwrapping the circle as Vivek suggested to draw the picture, we could have a bunch of copies of this flow line u0 from q to itself. So then we could say, well, here's this pretty non-generic thing. It's a broken thing with many levels. And then if I perturb t a little bit away from 0, what's going to happen to this? So it could happen that for t slightly negative or slightly positive, there's a flow line like this obtained by gluing all this together. So the question is, so how many ways to glue this to a flow line for t not equal to 0? And that's going to tell us how the chain complex changes. Maybe there's no flow line here when t is negative, and there is suddenly a flow line appears when t is positive. So then the differential of p is getting added to it t to the something times q. So this is the simplest example of a problem I could think of for which we want to do obstruction bundle gluing. And I sort of know what the answer is for obscure reasons. So I'll tell you what the answer is, but then we're going to try to actually understand what's going on. So there's a lemma, which is that there exists a power series a equals 1 plus or minus t, and then I don't know what, such that. So the differential after t is slightly positive, so d plus, and it's a differential for t is slightly positive, and the differential for d is slightly negative are obtained by conjugation by a map aq, where aq of q is this power series a times q, and aq fixes all other critical points. And this a basically is a count of sort of flow lines that are created or destroyed in this bifurcation. And then it's a fact which follows by combining various obscure forgotten theorems that in fact this power series is 1 plus or minus t to the plus or minus 1 with four possibilities. These signs are not related. And I want to understand this directly by obstruction bundle gluing. So you're saying you take any generic path that goes through this, and then each such path will be one of those matrices for the path. Because I thought you were just going to ask you, what's your model for going through? Because obviously, if you could go through from t minus to t plus, you could reverse t and go through in the other direction. Opposite power series, yeah. So we're going to see that this depends on some very subtle data. But the nice thing about the obstruction bundle gluing is the analysis, if we look at the analysis carefully, you'll tell us exactly what's happening. It's very precise. But we have to look at it carefully. First of all, is it correct that t is used in two different ways? Oh, shit. Right. Let's call this q. No, use q already. Ah. OK, well, we're just going to have to learn to tolerate some ambiguity. But let's make this a capital T. How about that? Is that better? All right. OK, sorry about that. Also, this lemma, this is a fact I know from the usual more series, and it's only involved with two terms. This lemma? Yeah, so this lemma, it's proved by doing what you suggested and sort of unwrapping the circle and looking at what happens there. OK, other questions about this? Yeah. Is there a straightforward criterion for which version of A is applied? It's a non-straightforward thing. So we're going to, there are four or five different signs and putting them all together. It's going to tell us what A is. Oh, and by the way, from now on, so to avoid my head completely exploding, I'm going to replace this with Z mod 2 coefficients. So normally you have to count flow lines with signs, but I think it's sort of indecent to discuss that in public. So this sign no longer matters. This sign still matters. So if using Z mod 2 coefficients, A is either 1 plus T or 1 plus T plus T squared plus dot, dot, dot. OK, so how do we glue this stuff? Well, so let's do a warm-up before we get to this. Where is the hook? I'm only impressed when we say audience numbers. Not yet. Oh, wait. How do I get that one? Oh, gosh. Here we go. I'm sort of afraid that I'm going to be hanging on to the boards and then lift it up. OK, so what's my warm-up problem? So the warm-up is let's look at the gluing needed to prove that D squared equals 0. So this is something which is supposed to be easy. I'm going to make it difficult, OK? So I'm going to review this in a slightly strange way because the instruction bundle gluing setup looks a little different from the way people usually do gluing. But so I'm going to do this in a strange way, which is then going to generalize conveniently to the non-transverse situations. So we have a flow line p of index i minus, index i plus 1, sorry, the flow line q. So let's call this u plus to q, which is index i. I have another flow line u minus to critical point r of index i minus 1. I want to glue these things to get an end of the modular space of curves from p to r. So let's write v for the gradient of f. I think v is the upward gradient. And then a flow line u will satisfy the equation ds u plus v of u equals 0. So I'm thinking I really should draw the arrows go in the other way so they're parameterized like this. So my flow lines actually go up, OK? So u is a map from r to x. And I want to glue these things. Also, there's a, sorry, these signs are going to kill me. OK. All right, and there's also a linearization of this equation, which I'll write as du. So this goes from sections of the, sorry, of u star tx. And we probably want to use some monoch space completion. It doesn't really matter which one, as long as you have some decay conditions. So we could take l21 sections. So this is the derivative of this equation with respect to the deformation of u. So du of some c is the covariant derivative of c in the direction of the flow line minus the covariant derivative of the vector field v in the direction c, with respect to, say, the Levy-Chavita connection. All right. OK. So now, so how am I going to glue these things? So I'm going to choose a very large value of r. And I'm going to translate u plus up and u minus down. The total translation distance will be r. So I pull these things apart by distance r. And then I'm going to have some cutoff functions. So I guess u plus up by r over 2, let's say. And u minus down by r over 2. Are you translating in the source r or the target x? Well, it's sort of both. But when u is a map, we'll define an r. So I'm going to compose that with a translation of r. So I need to choose some cutoff functions. So there'll be a cutoff function beta minus, which will look like this. So here's r over 2. Here's minus r over 2. So it's 1 over here. And somewhere between 0 and r over 2, it dies. And then there'll be a cutoff function beta plus like this. I'll call this parameter s. And the size of the derivative of these cutoff functions is on the order of 1 over r. And then what? So now this translation by r is making them over that more or less? So basically, you're pulling them apart so that sort of at time s equals 0, they're sort of close to the critical point q. And then we choose a coordinate chart on x in a neighborhood of q. And then we can pre-glue these things. So this is a curve which is, well, let me not put the s in there. So it's beta minus times u minus plus beta plus times u plus. So for s less than minus r over 2, it's just u minus. For s bigger than r over 2, it's just u plus. And in between, we use these cutoff functions to interpolate between u minus and u plus. OK. Now the usual approach would be you start with this thing, and then you perturb it. And you argue that you can perturb it to get an actual flow line. Wait a minute. You haven't shifted that. Shouldn't you put the shift in by r? U plus were shifted. I guess I'm using the same notation. When you say u minus is the, we shouldn't be shifted both. Yeah, so these have already been shifted. So when I said shift, when I said translate them up and down, I didn't use a notation for that. So these are the shifted, the translated u minus and u plus. Sorry about that. And the betas also have r translation, right? Does it change the r or the beta plus and minus? No, I'm not doing anything to the betas. Betas stay the way they are. The betas do depend on r. Yes? Supposing I have understood the two gluing from yesterday, is this like the same kind of thing, one dimension down? Yeah, it's the same kind of thing. So I mean the picture, if you want a picture, then add a picture, and it's gone. So this is u plus, and this is u minus. Then our flow line, so as s increases, so first we're following u minus. Because u minus has been translated, we're following it for a very long time. So we're very close to q. And then these cut-off functions kick in, and we start interpolating to u plus. And then we follow u plus like that. And I have some coordinate chart here. So all the cut-off function stuff is taking place inside this coordinate chart, so it makes sense. Yes, so as s increases, so we're following u minus, until we're very close to q. And then these cut-off functions kick in, and then we're interpolating to u plus, and then we're following u plus. You cannot require that there's a relation between u minus and u plus, like u minus equals 1 minus u plus. I don't need that. But it could be, well, that you have two times. So what is the, you assume that the image of its coordinate point is 0? Right, yes. So this addition is defined with respect to my coordinate chart in which the critical point corresponds to 0. If I wanted to say this a little more correctly, I would talk about exponential maps and so on. But I'm trying to make it as simple as possible. OK, so that's the pre-glue curve. And the usual way that this is presented is to say you would now try to, you'd argue that this can be perturbed to an actual flow line. Then I'm going to do something a little different, which is I'm going to perturb the things before pre-glueing them and then pre-glue them. So it's going to look like this. So we're going to choose or let psi plus or minus be small sections, c0 small sections of pullback tangent bundle. And we're going to look at a curve. So let's consider beta minus times u minus plus psi minus plus beta plus times u plus plus psi plus. So this is my pre-glue curve, but it's modified using sections of the, well, it's modified both on u minus and u plus. So on u minus, this is perturbed by psi minus. And up here on u plus, this is perturbed by psi plus. And in the middle is perturbed by some combination of psi minus and psi plus. OK, yes. So u plus and u minus are always shifted. And I want to solve for this to be a flow line. So we're going to solve. So when is this a flow line? So I will just write the equation. You've done this before, but the answer has been never. If the size is 0, then it won't be a flow line. So I want to choose the size to make it a flow line. So I want to solve for that. I'm going to write down the equation for this to be a flow line and then solve it. So this is a little messy, but this is really the very simplest case I could think of. And also, to make my life even easier, let's assume that v is linear in this coordinate chart. I don't assume that there are some extra terms. In general, in this business, I wrote these gluing papers with TOBS, and so this is my analysis training. So what I've learned is basically you write down an equation, then the stuff you want, and then there's a whole bunch of crap. And you have to estimate all that crap to show that it doesn't matter. But I'm trying to make this so there's as little crap as possible. So let's just write the equation. So I have ds minus v of this thing beta minus times u minus plus psi minus plus beta plus times u plus plus psi plus. So what is this? So there's ds beta minus times u minus plus psi minus plus beta minus times, I probably should be writing the covariant derivatives here, but I'm just going to not worry about that. So ds u minus plus ds psi minus. And there's a similar thing with plus. And if you don't mind, I'm going to take the liberty of multiplying this term by beta plus and this term by beta minus, which I can do because the support of the derivative of beta minus is contained in the region where beta plus is equal to 1. So I'm allowed to do that. Doesn't change anything. Then I have to put in the vector field. So I have minus beta minus times v of u minus plus the derivative of v in the direction psi minus. And then in general, there's going to be some additional error term. I'll write this as q minus of psi minus. So basically, this is the Taylor expansion of the vector field. So it's the vector field plus its first derivative plus some quadratic term. And then minus beta plus times v of u plus. I thought we agreed that the vector field is linear. Yes. So near the origin, this only comes up away from outside of my coordinate chart. Outside of your coordinate chart, you just take exponentials instead of sums in here. So you put something by sum, doesn't it only make sense in this coordinate chart? So this is to be interpreted as the exponential map of u minus evaluated on psi minus. So I'm sorry for my sloppy notation. I'm just trying to make it simple. OK, and then we can cancel some stuff out because, so u minus is a flow line. So ds u minus plus v of u minus equals 0. So I can cross this out. But likewise, u plus is a flow line. So I can cancel this ds u plus with this v of u plus. And what do I get? So I get beta minus times. So another nice thing is this term, I really should make this a covariant derivative here. Let me do that to be a little more honest. So this nabla s of psi minus, minus nabla psi minus of v, is the deformation operator applied to psi minus. So these two terms here are d minus of psi minus. Well, this is d minus is the deformation, the linearized equation for u minus. And then what else is there? So then there's this q psi minus. And then there's this ds beta plus of times u plus plus psi plus. And then there's an analogous thing times beta plus. So beta plus times d plus psi plus plus q plus psi plus plus ds beta minus. I guess these q's I have to subtract, doesn't matter. And then this times u minus plus psi minus. So this thing is the failure of my thing to be a flow line. So I want to make this thing equal to 0. In the more general situations, like in the papers with tobs, there's a bunch of extra crap in here, which you then have to estimate. What time did I start? Or how much time do I have left? 22 minutes? OK, sweet. I can actually get somewhere then. All right. So I'm going to rewrite this as beta minus times theta minus of psi minus psi plus beta plus times theta plus of psi minus psi plus. So this is my equation for the thing to be the flow line. And then the first lemma you need is that if r is sufficiently large, then there exists a unique pair, psi minus psi plus, such that I guess these are in what you have to assume some decay conditions, like L21. So it's a unique pair of psi minus psi plus, such that psi plus or minus is perpendicular to the kernel of d plus or minus. So this is perpendicular in the L2 sense. And I'm assuming that we're in a situation where d is defined, so these things are cut out transversely. So in this situation, d plus or minus is subjective. And it's kernel is one-dimensional given by the derivative of the r translation of the flow line. So there's this unique pair, such that both theta minus and theta plus are equal to 0. And then if we have this lemma, then we've glued, because if both of the theaters are 0, then certainly their sum is 0, and so it's a flow line. And the idea of the proof, so I'm not going to go into too many analytic details because it will be too messy and take too long, or to be a little more honest, because I don't remember all the stuff from this paper, which was written eight years ago. It's one of the reasons why the invention of writing was such a milestone in human culture is that you no longer have to keep things in your head. You can write them to have them and forget them. Anyway, the idea of the proof is to let me first rewrite the equation a little bit. So let pi plus or minus, is this really what I want? Sorry, hold on a sec. No, I actually don't need this yet. OK, sorry, never mind. Which term is guaranteeing that after the term, the end of 1 goes to the start of the other one? I mean, these are decaying as you go to the ends. So these go to 0, so you're not going to escape this coordinate chart, and it's going to be OK. I think the initial set of guaranteeing that is 0 and it's 0, if you have the ansatz. So there's an inverse, so d plus or minus inverse, which is going to go to the target, use your favorite bonnet space completion, which will give you an isomorphism to the orthogonal complement of the kernel of d plus or minus. So I'm going to rewrite this equation. So the equation, so what is theta? Let's look at the equation theta minus equals 0. So this says that d minus psi minus is equal to q of psi minus plus ds beta plus times u plus plus psi plus. So then I can apply this inverse. So that's equivalent to saying that psi minus, remember, I assume that psi minus is orthogonal to the kernel. So psi minus is equal to d minus inverse of q psi minus plus ds beta plus u plus plus psi plus. And likewise, theta plus is equal to 0 if and only if psi plus is equal to d plus inverse of q plus of psi plus plus ds beta minus u minus plus psi minus. So these are the equations we need to solve. And then you want to use the contraction mapping theorem in your favorite bonnet space completion. So you pick your favorite bonnet space completion, and then you want to show that the right side of this equation is a contraction mapping on the set of pairs psi minus and psi plus. If you look at this, well, this term with the u plus and u minus is a constant, so that's fine. And I have the derivative of beta plus times psi plus. The derivative of beta plus is on the order of 1 over r. And I don't know what the operator norm of d minus inverse is, but if I choose r to be much larger than the operator norm of d minus inverse, then this part of the equation will be a contraction. And also, this quadratic term, if you set things up so that psi minus is c0 small, then this part will also be a contraction. So you get a unique solution. So ta-da, we've glued. And then you can fact to check. So this construction gives you a homeomorphism from the set of r sufficiently large to the set of gluings. So if you choose two different r's, you're going to get two different glued curves this way. And any curve which is sort of close in the sense of the usual compactness to this broken curve is actually obtained by this construction. And it's a slightly weird construction because if you look at this thing here, so if you look at the psi minus, then when s is large, like up here in the picture, then the psi minus doesn't matter at all. So what we're doing to the curve up here doesn't depend on psi minus at all. And what we're doing to the curve down there doesn't depend on psi plus at all. But still, I'm solving for psi minus and psi plus, which are defined in the whole real line. So it's a little bit weird because the psi minus and psi plus contain various additional information which I don't care about. So it's a little weird, but as we'll see, this is a useful way to do things because it allows us to use the analysis of these operators, d plus and d minus. Yeah. So where I can't currently see why it matters if v is linear not in this? Oh, I just used it in this equation. So I had this, I had v evaluated on this whole expression, and then I expanded that linearly, which I couldn't do if you were nonlinear. If it's not linear, then they're just some extra error terms, and it's not a big deal. Question? It seems like we solved a stronger question. So we pushed it, we solved it to talk in my own sense, because it's not linear, it's only the combination. That's right. Or we understand the combination of these two? Well, I don't know what to say. Anyway, there's a unique psi minus and psi plus satisfying these conditions. If I didn't put all these conditions on, it wouldn't be unique, and then it would be more confusing. OK, so now let's do an example where there's an obstruction. But are you saying that it would be possible maybe to prove using different theta minus and theta plus to get that equation to be 0, even though individually they're not 0? I mean, you can change psi minus and psi plus. So in the region where both of the CODA functions are nonzero, you could add something to psi minus and subtract something to psi plus, and you get the same curve. However, then you'd no longer satisfy the equations theta minus and theta plus equals 0. OK, now here's warmup number 2, and it's going to get a little more interesting. So warmup number 2, I have a one parameter family of functions, and at time t equals 0, there is an index 0 flow line. So this goes from, say, q to r. So these both have index equal to i. And then maybe there's another flow critical point, p, of index i plus 1, and the flow line, u plus. And then I want to glue these to an actual flow line from p to r for t not equal to 0. Now the thing is, you can do this only for one sign of t. So either you can do it for t positive and not for t negative or vice versa. And the analysis is actually going to tell us which. So how does this work? So what's the construction? So again, we choose r very large. We translate this before. And then what's the data or the inputs? So there's no assumption that we have t, g, t not for t to 0. It's generic. Yeah, I do need generic. I do need generic. So I'm going to assume that this is a co-index with the code and the cooker would be run and then it would be filled out by v. Right. What if the fact that you walked into a situation and here you treat this? What? When you four walls have two flow lines and you are gluing them, what is the distance between this situation and you? So what's different in this situation is that the linearized operator for d0 is no longer surjective. So this has a one-dimensional co-kernel. And we can't glue these for fixing t. We have to change t to be able to glue. So the inputs for the gluing construction are a psi minus and psi plus as before plus t. So I'm going to change to a different small time t and perturb by psi minus and psi plus. And then from these inputs, we get here. And once they've been larger than all of this discussion, we're trying to show that something about homotopy invariance and this construction in the start? That is one thing you could do with this. Right now I'm just doing it as an example of gluing construction. But yeah, if you want to understand how Morse complex has changed under bifurcations, then this is part of what you have to do. OK, so then this, you get a flow line, if and only if again, beta minus times theta minus of psi minus psi plus t plus beta plus times theta plus, excuse me, there's no minus. This is 0. So you can do a similar calculation to this. There's going to be an extra term because when you change t, you change the vector field. So this theta 0 of psi minus psi plus t is equal to d0 of psi 0 plus, there's going to be that ds beta plus of u plus psi plus. And then an additional term, which we'll write as t, t times v prime, where v prime is the derivative of the vector field with respect to t. And then there may be some additional quadratic terms which I'm going to ignore. And a similar expression for theta plus. So theta plus is d plus psi plus plus ds beta 0 u0 plus psi 0 plus t times the derivative of v plus and other stuff. You said we used the index differences being one for subjectivity. We don't actually need subjectivity. We don't need the v plus minus 2. So we could take orthogonal component. Right, so we're going to do something like that shortly. So here, this inverse operator was defined in the whole target space. In general, it would just be defined on the image of the operator. No, well, it will work up to an error, which I've got to show you. So here's the lemma. Let me state the lemma, and then we can discuss it. So the lemma is that for t sufficiently small and r sufficiently large, there exists a unique pair, psi 0 psi plus, such that. So as before, psi 0 is perpendicular to the kernel of d0. Psi plus is perpendicular to the kernel of d plus. Theta plus of psi 0 psi plus t is equal to 0. And as for theta 0, I can't necessarily get it to equal 0. All I can get is that this is perpendicular to the image of d0. And the idea of the proof, well, I shouldn't have erased this. But the idea of the proof is, again, use the contraction mapping theorem. So when you do that, the inverse operator is only going to be defined on the image. So let me just write it down. Let's let pi 0 be the projection to the image of d0. So then we have a d0 inverse. Goes from the image of d0 to the orthogonal complement of the kernel of d0. And we're going to solve an equation that looks like, well, what I want is that pi 0 of theta 0 is equal to 0. That's the equation I want to solve. So I can write this as pi 0 of d0 psi 0 equals blah, blah, blah. So now I can apply the operator d0 inverse to it. So I can write this as psi 0 equals d0 inverse of blah, blah, blah. I guess there's a pi 0 here also. So d0 inverse pi 0 of blah, blah, blah. And then you can do the contraction mapping theorems before. But what's the punch line? So I didn't quite get as far as I was expecting to get. So I don't think I'm going to be able to completely explain this example today. But let me tell you what happens. So to glue, I really want theta plus and theta 0 to both equal 0. And I can't get that. Well, I can get theta plus to equal 0. But theta 0, it lives in a one dimensional vector space. Lives in the complement, in the co-cernel of d0. And it's a one dimensional vector space. I can't get that to equal 0. So now we have this picture over here. So I have m will be the set of all pairs, t and r. And over this, I have an obstruction bundle, o. And so this is a trivial bundle. The fiber over any point is the co-cernel of d0. And I have a section, s. And what is the section, s? So s of t comma r is equal to the state of 0. So the gluing construction says, I can almost get the gluing to work except the state of 0 might not equal to 0. The state of 0 is the obstruction to gluing. So when this is equal to 0, I can actually glue. So this is some function of t and r. When this function is 0, I get a gluing. And I run out of time. So in the next class, I'll tell you how to actually compute what this function is and how to figure out whether you can glue for t negative or t positive. And then I'll do the harder example. And then I'll do the homomorphic curve stuff. But I'll try to make stuff modular. Like the Titanic had this compartment. So if one part of it fills with water, they just seal it off so the ship won't sink. So I'm going to try to do that. So if you didn't understand anything, it'll sort of start over. As long as if too many compartments fill with water, we'll still sink. But all right. So I'll be hanging around if anyone wants to talk about this stuff. Thanks. And Chris, I mean, office hours tomorrow for this, officially? I have no idea. Anyway, I'll be hanging around after all of my talks to answer any questions. Call it office hours. In this example, the D0 doesn't depend on where you are in this space tent. It's a very trivial adjustment. Other examples will be less trivial. But this one is, as I said, I try to make the easiest one I could think of.