1/3 Obstruction Bundle Gluing
Hutchings, Michael
Obstruction bundle gluing is a method of calculating the number of ways of gluing certain configurations in which transversality fails, but not too badly. We will introduce this technique and show how it works in simple examples from Morse theory, contact homology, and embedded contact homology.
 Okay, thanks to the organizers. So I'm supposed to talk about obstruction bundle gluing and the references are, so these two unreadable papers by myself and Taubs, 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. And 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 and 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. And obstruction bundle gluing is relevant to this question, as we'll see. Okay, so here's the Morse theory example I want to start with, because it's easier. And then we'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. And then, so Morse means, I mean, if you have a circle-valued function, then locally it's 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. Okay. And then you can define a Morse complex where you do the usual thing. You count gradient flow lines between critical points. The only issue is 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 defined 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 one. And then we have the sum from K equals zero to infinity. 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 across the level set sigma K times. So the flow line like this is counted with T to the zero. If it goes across 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. Yeah. 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. So you want to define an explicit chain map. So you want to know sort of what, so you want to sort of 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. Okay, good. Okay. So you have a... So about this circle value, why not just unwrap the circle? Yeah, you can do that. Although you still have to account with the nova covering because you'll have a non-compact manifold. And then if I know how usual most functions change. Ah, this is... You're about to see why it's not that simple. Okay. So now I have a homotopy. And let's say it's not generic at time t equals zero. 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 zero. Or actually, you'll call it U zero. Okay, so U zero is a flow line from Q to itself. And so this... The reason why I'm doing circle-valued Morse theories for this to be possible, you can't... This can't happen in real-valued Morse theory. In circle-valued Morse 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 one. 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 one, this fault line. And then sort of unwrapping the circle, as Vivek suggested, to draw the picture, we could have a bunch of copies of this flow line U zero 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 zero, 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, how many ways to glue this to a flow line for T not equal to zero? And that's going to tell us how the chain complex changes. Like, if there's 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 my, 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 equals one plus or minus T and then I don't know what. Such that, so the differential after slight, where T is slightly positive, so D plus and it's a differential for T slightly positive. And the differential for D 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 one plus or minus T to the plus or minus one 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 parts that goes through this and then each such part will be one of those matrices which is part. 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 Q plus you could reverse T across in the other direction. Opposite power series, yeah. 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 will, 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. I just wanted to make sure that that was correct. Yeah. Right. Let's call this Q. No, use Q already. Okay, well, we're just going to have to learn to tolerate some ambiguity. We're never going to, but let's make this a capital T. How about that? Is that better? Yeah. All right. Any more? Okay. Sorry about that. Also this lemma, this is a fact I know of in the usual moist 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. Okay. Other questions about this? Yeah. Is there a straightforward criterion for which version A is applied? There's a non-straightforward. So we're going to, there are like 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 count, 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. Okay. So how do we glue this stuff? Well, so I'm going to, let's do a warm up before we get to this. Let's see. Where is the hook? Ah. I'm really impressed with what it says audience numbers. Not yet. Wait. How do I get that one? Oh gosh. Here we go. I'm sort of afraid that I'm going to be like hanging onto the boards and then like lift it up. Okay. Right. 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. Okay. So this is something which is supposed to be easy. I'm going to make it difficult. Okay. 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. Okay. 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 moduli space of curves from P to R. So these, let's write V for the gradient of F. I think V is the upward gradient and then these, these flow line U will satisfy the equation DSU plus V of U equals 0. So I'm thinking of, I really should draw the arrows going the other way. So they're parameterized like this. So my flow lines actually go up. Okay. So U is a map from R to X. And I want to glue these things. Also there's a, there's a, sorry. These signs are going to kill me. Okay. All right. And there's also a, there's 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 monochromatic 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 sum 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. Okay. So that's, 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. Okay. 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. Okay. What does that mean? Okay. Mark, 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, right, and I'm going to, 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. Okay. And then what? So now, this translation by R is making them overlap more or less? So basically, you're pulling them apart so that sort of at time, at S equals 0, they're sort of close to the critic, they're both close to the critical point Q. Oh, no. And then we choose a coordinate chart on X in a neighborhood of Q. Okay. And then we can preglue 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. Okay. 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 though. Shouldn't you put the shift in by R? The U plus were shifted. I guess I'm using the same notation. Well, U minus is the, we're supposed to be shifted rather than be close, isn't it? 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 have, these are the shifted, the translated U minus and U plus. Sorry about that. And the betas also have R. Is it changed the R to 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 understood the two gluing from yesterday. Is this like the same kind of thing in one dimension down? Yeah, it's the same kind of thing. So I mean the picture, if you want a picture, then I had 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, or 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 Cotto functions kick in and we start interpolating to U plus and then we follow U plus like that. And we have some coordinate chart here. So all the Cotto function stuff is taking place inside this coordinate chart. So it makes sense. Can you do the understanding on that side? Yes. So as S increases, so we're following U minus until we're very close to Q. And then these, these Cotto 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 beta minus and beta plus like beta minus equals 1 minus. I don't, I don't need that. It could be, well, that you have two times. So what is the, you assume that the image of it could be 0? Right. Yes. So, 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. Okay. So that's the pre-glue curve and the usual, the usual way that's presented is to say you would now try to, you'd argue that this can be perturbed to an actual flowline. Then I'm going to do something a little different, which is I'm going to perturb the things before, before pre-glueing them and then pre-glue them. So it's going to look like this. So we're going to, 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, 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, it's perturbed by some, some combination of psi minus and psi plus. Okay. Yes. So U plus and U minus are always shifted. Okay. And I want to say what, I want to solve for this to be a flowline. So we're going to solve, so, so when is this a flowline? So I'll just write the equation. And if you've done this before, the perturbation answer would have been, yeah. If the size is zero, then it won't be a flowline. So I want to choose the size to make it a flowline. So I want to solve for that. I'm going to write down the equation for this to be a flowline and then solve it. Okay. So this is a little messy, but this is really the very simplest case I could think of. And also, let's, to make my life even easier, let's assume that, 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, well, I mean, so I wrote these gluing papers with TOVs and so this is like my analysis training. So what I've learned is basically you sort of write down an equation, then there's 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. That's, I tried to make, but I'm trying to make this so there's as little crap as possible. Okay. So let's just write the equation. So I have ds minus V of this thing, beta minus times u minus plus i minus plus beta plus times u plus plus i plus. So what is this? So there's a ds beta minus times u minus plus i 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 side 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 one. So I can, I can, I'm allowed to do that. It 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 side minus. And then in general, there's going to be some additional error term. I'll write this as q minus of side minus. So this is, 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 is all, this only comes up away for, away from outside of my coordinate chart. Outside of your coordinate chart, you just take exponentials instead of sums. So you put them by sums and it only makes sense in this coordinate chart? Yeah, yeah. So this is to be interpreted as the exponential map of u minus evaluated on side minus. So I'm sorry for my sloppy notation. I just tried to make it simple. Okay. 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 zero. So I can cross this out. And 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, so I get beta minus times, so, 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 nav s of psi minus, minus nav of 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, it 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 zero. In the more general, the more general situations like in the papers with the tobs, there's a bunch of extra crap in here, which you then have to estimate. Okay. How much time did I start? Or how much time do I have left? Okay. 22 minutes? Okay, 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 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, 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 rejected. And its 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 zero. And then now if we have this lemma, then we've glued because if both of the theaters are zero, then certainly their sum is zero 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 this stuff from this paper which was written eight years ago. It's one of the reason 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, well let me first rewrite the equation a little bit. So let, what? You were ravens that were tying. Uh-oh. So what's my notation? Right. 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. Okay. Sorry, never mind. Which term is guaranteeing that after the term, the end of one goes to the start of the other one. I mean these are decaying as you go to the ends. So these go to zero. So you don't, you're not going to escape this coordinate chart and it's going to be okay. I think the initial set of guarantees will tell us the zero and it's zero. So, I think so there's a, so there's an inverse. So d plus or minus inverse, which is going to go to the target. Use your favorite, favorite bonnet space completion. So we'll give you an isomorphism to the orthogonal complement of the kernel of d plus or minus. Okay. So I'm going to rewrite this equation. So the equation, so what is theta? Let's look at the equation theta minus equals zero. 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 zero 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, these are the equations we need to solve. And then you want to use, use the contraction mapping theorem in your favorite bonnet space completion. So you pick, 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, this, this term of 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 one 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, 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 c zero small then this, then this part will also be a contraction. So you get a unique solution. So tada, we've got to include. And then you can, the 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, 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, 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 the, 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 it is linear or not. Well 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 there's just some extra error terms and it's not a big deal. Question? That's right. 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. Okay so now let's do an example where there's an obstruction. But are you saying that it would be possible maybe to do using different theta minus and theta plus to get that equation to be zero even though individually they're not zero? I mean you can change psi minus and psi plus. So in the region where they were both of the cota 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 no longer satisfy the equations theta minus and theta plus equals zero. Okay now here's warmup number two and it's going to get a little more interesting. So warmup number two I have a one parameter family of functions and at time t equals zero there's a flow line, an index zero 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 one and the flow line u plus. Okay and then I want to glue these to an actual flow line from p to r for t not equal to zero. Now the thing is you actually 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. The analysis is actually going to tell us which. Okay so how does this work? So what's the construction? So again we choose r very large we translate as before. And then what's the data for the inputs? So there's no assumption that we have t, g, t for t to zero it's generic. Yeah I do need generic. I do need generic. So this I'm going to assume that this. Okay so the co-index of this would be the co-index of the co-index of the co-index of this would be one that you would prefer to go for. Where is the fact that you want in this situation appear in the previous. What? In four walls that you could align when you were gluing them. So what's different in this situation is that the linearized operator for d zero is no longer subjective. 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. Okay and then from these inputs we get. And what's the larger one? All of this discussion. We're trying to show that something about the homotopy invariance and this construction in the start or? That is one thing you could do with this. Right now I'm just doing it as an example of the gluing construction. But yeah if you want to understand how Morse complex has changed under bifurcations and this is part of what you have to do. Okay so then this you get a flow line if and only if again beta minus of 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 zero. Okay 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 is zero of psi minus psi plus t is equal to d zero, ah, zero. So it's equal to d zero of psi zero 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 zero u zero plus psi zero plus t times the derivative of v plus and other stuff. You said we used the index differences being one for surjectivity but we don't actually need surjectivity. We don't need the v plus minus two. 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 will just be defined on the image of the operator. No, well it will work up to an error which I'm about 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 zero psi plus such that so as before psi zero is perpendicular to the kernel of d zero. Psi plus is perpendicular to the kernel of d plus. Theta plus of psi zero psi plus t is equal to zero. And as for theta zero I can't necessarily get it to equal zero. All I can get is that this is perpendicular to the image of d zero. 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, well maybe I should write it down. So let's let pi zero be the projection to the image of d zero. So then we have a d zero inverse goes from the image of d zero to the orthogonal complement of the kernel of d zero. Now we're going to solve an equation that looks like well what I want is that pi zero of theta zero is equal to zero. That's the equation I want to solve. So I can write this as pi zero of d zero psi zero equals blah blah blah. And then I can write this as so now I can apply the operator d zero inverse to it. So I can write this as psi zero equals d zero inverse of blah blah blah. I guess there's a pi zero here also. So d zero inverse pi zero of blah blah blah. And then you can do the contraction mapping theorems before. 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 zero to both equal zero. And I can't get that. Well I can get theta plus to equal zero but theta zero it lives in a one dimensional vector space. It lives in the complement in the co-kernel of d zero. It's a one dimensional vector space. I can't get that to equal zero. 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. This is a trivial bundle. The fiber over any point is the co-kernel of d zero. And I have a section s. And what is the section s? So s of t comma r is equal to the state of zero. So the gluing construction says I can almost get the gluing to work except the state of zero might not equal to zero. The state of zero is the obstruction to gluing. So when this is equal to zero I can actually glue. Okay? So this is some function of t and r. When this function is zero 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, you know, the Titanic if it has compartments. 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. Are you and Chris having 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. All at office hours. Any questions? In this example the D0 doesn't depend on where you are in the space. It's a very trivial adjustment. Other examples will be less trivial. But this one is, as I said I tried to make the easiest one I could think of. Any questions? Let's thank Michael again.
 Okay, thanks to the organizers.  So I'm supposed to talk about obstruction bundle gluing and the references are, so these two  unreadable papers by myself and Taubs, 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.  And 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 and 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. And obstruction bundle  gluing is relevant to this question, as we'll see. Okay, so here's the Morse theory example  I want to start with, because it's easier. And then we'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. And then, so Morse means, I mean, if you have a circle-valued  function, then locally it's 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. Okay. And then you can define a Morse complex where you do the usual thing.  You count gradient flow lines between critical points. The only issue is 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 defined 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 one.  And then we have the sum from K equals zero to infinity. 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  across the level set sigma K times. So the flow line like this is counted with T to the  zero. If it goes across 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.  Yeah. 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. So you want to define an explicit chain map. So you want to know sort of what, so  you want to sort of 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. Okay, good.  Okay. So you have a... So about this circle value, why not just unwrap the circle? Yeah,  you can do that. Although you still have to account with the nova covering because you'll  have a non-compact manifold. And then if I know how usual most functions change. Ah,  this is... You're about to see why it's not that simple. Okay. So now I have a homotopy.  And let's say it's not generic at time t equals zero. 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 zero. Or actually,  you'll call it U zero. Okay, so U zero is a flow line from Q to itself. And so this...  The reason why I'm doing circle-valued Morse theories for this to be possible, you can't...  This can't happen in real-valued Morse theory. In circle-valued Morse 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  one. 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 one, this fault line. And then sort of unwrapping the circle,  as Vivek suggested, to draw the picture, we could have a bunch of copies of this flow  line U zero 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 zero, 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, how many ways to glue this to a flow line for T not equal to zero?  And that's going to tell us how the chain complex changes. Like, if there's 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 my, 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  equals one plus or minus T and then I don't know what. Such that, so the differential  after slight, where T is slightly positive, so D plus and it's a differential for T slightly  positive. And the differential for D 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 one plus or minus T to the plus  or minus one 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 parts that goes through this and then each such  part will be one of those matrices which is part. 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 Q plus you could reverse T across in the other direction.  Opposite power series, yeah. 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  will, 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.  I just wanted to make sure that that was correct.  Yeah.  Right. Let's call this Q. No, use Q already.  Okay, well, we're just going to have to learn to tolerate some ambiguity. We're never going  to, but let's make this a capital T. How about that? Is that better?  Yeah.  All right. Any more? Okay. Sorry about that.  Also this lemma, this is a fact I know of in the usual moist 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. Okay. Other questions about this?  Yeah.  Is there a straightforward criterion for which version A is applied?  There's a non-straightforward. So we're going to, there are like 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 count, 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. Okay. So how do we glue this stuff? Well, so I'm  going to, let's do a warm up before we get to this. Let's see. Where is the hook? Ah.  I'm really impressed with what it says audience numbers. Not yet. Wait. How do I get that one?  Oh gosh. Here we go. I'm sort of afraid that I'm going to be like hanging onto the boards  and then like lift it up. Okay. Right. 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. Okay. So this is  something which is supposed to be easy. I'm going to make it difficult. Okay. 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.  Okay. 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 moduli space of curves from P to R. So these, let's write V for the gradient of F.  I think V is the upward gradient and then these, these flow line U will satisfy the  equation DSU plus V of U equals 0. So I'm thinking of, I really should draw the arrows  going the other way. So they're parameterized like this. So my flow lines actually go up.  Okay. So U is a map from R to X. And I want to glue these things. Also there's a, there's  a, sorry. These signs are going to kill me. Okay. All right. And there's also a, there's  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 monochromatic 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 sum 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. Okay. So that's, 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. Okay. 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. Okay.  What does that mean? Okay. Mark, 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, right, and  I'm going to, 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. Okay. And then what?  So now, this translation by R is making them overlap more or less?  So basically, you're pulling them apart so that sort of at time, at S equals 0, they're  sort of close to the critic, they're both close to the critical point Q.  Oh, no. And then we choose a coordinate chart on X  in a neighborhood of Q. Okay. And then we can preglue 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. Okay. 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 though. Shouldn't you put the shift in by R?  The U plus were shifted. I guess I'm using the same notation.  Well, U minus is the, we're supposed to be shifted rather than be close, isn't it?  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 have, these are the shifted,  the translated U minus and U plus. Sorry about that.  And the betas also have R. Is it changed the R to 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 understood the two gluing from yesterday. Is this like the same kind of thing  in one dimension down?  Yeah, it's the same kind of thing. So I mean the picture, if you want a picture, then I  had 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, or 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 Cotto functions kick in and  we start interpolating to U plus and then we follow U plus like that.  And we have some coordinate chart here. So all the Cotto function stuff is taking place  inside this coordinate chart. So it makes sense.  Can you do the understanding on that side?  Yes. So as S increases, so we're following U minus until we're very close to Q. And then  these, these Cotto 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 beta minus and beta plus like beta  minus equals 1 minus.  I don't, I don't need that.  It could be, well, that you have two times. So what is the, you assume that the image  of it could be 0?  Right. Yes. So, 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.  Okay. So that's the pre-glue curve and the usual, the usual way that's presented is  to say you would now try to, you'd argue that this can be perturbed to an actual flowline.  Then I'm going to do something a little different, which is I'm going to perturb the things before,  before pre-glueing them and then pre-glue them. So it's going to look like this. So we're  going to, 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, 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, it's perturbed by  some, some combination of psi minus and psi plus. Okay. Yes. So U plus and U minus are  always shifted. Okay. And I want to say what, I want to solve for this to be a flowline.  So we're going to solve, so, so when is this a flowline? So I'll just write the equation.  And if you've done this before, the perturbation answer would have been, yeah. If the size  is zero, then it won't be a flowline. So I want to choose the size to make it a flowline.  So I want to solve for that. I'm going to write down the equation for this to be a flowline  and then solve it. Okay. So this is a little messy, but this is really the very simplest  case I could think of. And also, let's, to make my life even easier, let's assume that,  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, well, I mean, so I wrote these gluing papers  with TOVs and so this is like my analysis training. So what I've learned is basically  you sort of write down an equation, then there's 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.  That's, I tried to make, but I'm trying to make this so there's as little crap as possible.  Okay. So let's just write the equation. So I have ds minus V of this thing, beta minus  times u minus plus i minus plus beta plus times u plus plus i plus. So what is this?  So there's a ds beta minus times u minus plus i 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 side 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 one. So I can, I can, I'm allowed  to do that. It 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 side minus.  And then in general, there's going to be some additional error term. I'll write this as  q minus of side minus. So this is, 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 is all, this only comes up away for, away from outside of  my coordinate chart.  Outside of your coordinate chart, you just take exponentials instead of sums.  So you put them by sums and it only makes sense in this coordinate chart?  Yeah, yeah. So this is to be interpreted as the exponential map of u minus evaluated  on side minus. So I'm sorry for my sloppy notation. I just tried to make it simple.  Okay. 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 zero. So I can cross this out.  And 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, so I get beta minus times, so, 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 nav s of psi minus, minus nav of 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, it 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 zero. In the more general, the more general situations like in the papers with  the tobs, there's a bunch of extra crap in here, which you then have to estimate. Okay.  How much time did I start? Or how much time do I have left? Okay. 22 minutes? Okay, 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 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, 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 rejected. And its 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 zero. And then now if we have  this lemma, then we've glued because if both of the theaters are zero, then certainly their  sum is zero 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 this stuff from this paper which was written  eight years ago. It's one of the reason 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, well let me first  rewrite the equation a little bit. So let, what?  You were ravens that were tying. Uh-oh. So what's my notation? Right. 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. Okay. Sorry, never mind.  Which term is guaranteeing that after the term, the end of one goes to the start of the other  one. I mean these are decaying as you go to the ends. So these go to zero. So you don't,  you're not going to escape this coordinate chart and it's going to be okay.  I think the initial set of guarantees will tell us the zero and it's zero. So, I think  so there's a, so there's an inverse. So d plus or minus inverse, which is going to go  to the target. Use your favorite, favorite bonnet space completion. So we'll give you  an isomorphism to the orthogonal complement of the kernel of d plus or minus. Okay. So  I'm going to rewrite this equation. So the equation, so what is theta? Let's look at  the equation theta minus equals zero. 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 zero 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, these are the equations we need to solve. And then you want to use,  use the contraction mapping theorem in your favorite bonnet space completion. So you pick,  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, this, this term of 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 one 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, 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 c zero small then this, then this part  will also be a contraction. So you get a unique solution. So tada, we've  got to include. And then you can, the 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,  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, 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 the, 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 it is linear or not.  Well 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 there's just some extra error terms and it's not a big deal. Question?  That's right. 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. Okay so now let's do an example where there's  an obstruction.  But are you saying that it would be possible maybe to do using different theta minus and  theta plus to get that equation to be zero even though individually they're not zero?  I mean you can change psi minus and psi plus. So in the region where they were both of the  cota 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 no longer satisfy the equations theta  minus and theta plus equals zero.  Okay now here's warmup number two and it's going to get a little more interesting. So  warmup number two I have a one parameter family of functions and at time t equals zero there's  a flow line, an index zero 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 one  and the flow line u plus. Okay and then I want to glue these to an actual flow line from  p to r for t not equal to zero. Now the thing is you actually 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. The  analysis is actually going to tell us which. Okay so how does this work? So what's the  construction? So again we choose r very large we translate as before. And then what's the  data for the inputs? So there's no assumption that we have t, g, t for t to zero it's generic.  Yeah I do need generic. I do need generic. So this I'm going to assume that this. Okay  so the co-index of this would be the co-index of the co-index of the co-index of this would  be one that you would prefer to go for. Where is the fact that you want in this situation  appear in the previous. What? In four walls that you could align when you were gluing them.  So what's different in this situation is that the linearized operator for d zero is no longer  subjective. 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. Okay and then from these inputs we get.  And what's the larger one? All of this discussion. We're trying to show that something about  the homotopy invariance and this construction in the start or?  That is one thing you could do with this. Right now I'm just doing it as an example of  the gluing construction. But yeah if you want to understand how Morse complex has changed  under bifurcations and this is part of what you have to do. Okay so then this you get  a flow line if and only if again beta minus of 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 zero.  Okay 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 is zero of psi minus psi plus  t is equal to d zero, ah, zero. So it's equal to d zero of psi zero 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  zero u zero plus psi zero plus t times the derivative of v plus and other stuff. You  said we used the index differences being one for surjectivity but we don't actually need  surjectivity. We don't need the v plus minus two. 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 will just be defined on the image  of the operator. No, well it will work up to an error which I'm about 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 zero  psi plus such that so as before psi zero is perpendicular to the kernel of d zero.  Psi plus is perpendicular to the kernel of d plus. Theta plus of psi zero psi plus t  is equal to zero. And as for theta zero I can't necessarily get it to equal zero. All  I can get is that this is perpendicular to the image of d zero. 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, well maybe I should write it down. So let's let pi zero be the projection  to the image of d zero. So then we have a d zero inverse goes from the image of d zero  to the orthogonal complement of the kernel of d zero. Now we're going to solve an equation  that looks like well what I want is that pi zero of theta zero is equal to zero. That's  the equation I want to solve. So I can write this as pi zero of d zero psi zero equals  blah blah blah. And then I can write this as so now I can apply the operator d zero inverse  to it. So I can write this as psi zero equals d zero inverse of blah blah blah. I guess  there's a pi zero here also. So d zero inverse pi zero of blah blah blah. And then you can  do the contraction mapping theorems before. 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 zero to both equal zero. And I can't get that. Well I can get theta plus to equal  zero but theta zero it lives in a one dimensional vector space. It lives in the complement in  the co-kernel of d zero. It's a one dimensional vector space. I can't get that to equal zero.  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. This is a trivial bundle. The fiber over  any point is the co-kernel of d zero. And I have a section s. And what is the section  s? So s of t comma r is equal to the state of zero. So the gluing construction says I  can almost get the gluing to work except the state of zero might not equal to zero. The  state of zero is the obstruction to gluing. So when this is equal to zero I can actually  glue. Okay? So this is some function of t and r. When this function is zero 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, you know, the Titanic if it has compartments. 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.  Are you and Chris having 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.  All at office hours. Any questions? In this example the D0 doesn't depend on where you  are in the space. It's a very trivial adjustment.  Other examples will be less trivial. But this one is, as I said I tried to make the easiest  one I could think of. Any questions? Let's thank Michael again.
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Okay, thanks to the organizers. So I'm supposed to talk about obstruction bundle gluing and the references are, so these two unreadable papers by myself and Taubs, 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. And 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 and 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. And obstruction bundle gluing is relevant to this question, as we'll see. Okay, so here's the Morse theory example I want to start with, because it's easier. And then we'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. And then, so Morse means, I mean, if you have a circle-valued function, then locally it's 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. Okay. And then you can define a Morse complex where you do the usual thing. You count gradient flow lines between critical points. The only issue is 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 defined 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 one. And then we have the sum from K equals zero to infinity. 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 across the level set sigma K times. So the flow line like this is counted with T to the zero. If it goes across 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. Yeah. 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. So you want to define an explicit chain map. So you want to know sort of what, so you want to sort of 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. Okay, good. Okay. So you have a... So about this circle value, why not just unwrap the circle? Yeah, you can do that. Although you still have to account with the nova covering because you'll have a non-compact manifold. And then if I know how usual most functions change. Ah, this is... You're about to see why it's not that simple. Okay. So now I have a homotopy. And let's say it's not generic at time t equals zero. 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 zero. Or actually, you'll call it U zero. Okay, so U zero is a flow line from Q to itself. And so this... The reason why I'm doing circle-valued Morse theories for this to be possible, you can't... This can't happen in real-valued Morse theory. In circle-valued Morse 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 one. 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 one, this fault line. And then sort of unwrapping the circle, as Vivek suggested, to draw the picture, we could have a bunch of copies of this flow line U zero 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 zero, 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, how many ways to glue this to a flow line for T not equal to zero? And that's going to tell us how the chain complex changes. Like, if there's 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 my, 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 equals one plus or minus T and then I don't know what. Such that, so the differential after slight, where T is slightly positive, so D plus and it's a differential for T slightly positive. And the differential for D 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 one plus or minus T to the plus or minus one 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 parts that goes through this and then each such part will be one of those matrices which is part. 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 Q plus you could reverse T across in the other direction. Opposite power series, yeah. 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 will, 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. I just wanted to make sure that that was correct. Yeah. Right. Let's call this Q. No, use Q already. Okay, well, we're just going to have to learn to tolerate some ambiguity. We're never going to, but let's make this a capital T. How about that? Is that better? Yeah. All right. Any more? Okay. Sorry about that. Also this lemma, this is a fact I know of in the usual moist 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. Okay. Other questions about this? Yeah. Is there a straightforward criterion for which version A is applied? There's a non-straightforward. So we're going to, there are like 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 count, 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. Okay. So how do we glue this stuff? Well, so I'm going to, let's do a warm up before we get to this. Let's see. Where is the hook? Ah. I'm really impressed with what it says audience numbers. Not yet. Wait. How do I get that one? Oh gosh. Here we go. I'm sort of afraid that I'm going to be like hanging onto the boards and then like lift it up. Okay. Right. 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. Okay. So this is something which is supposed to be easy. I'm going to make it difficult. Okay. 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. Okay. 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 moduli space of curves from P to R. So these, let's write V for the gradient of F. I think V is the upward gradient and then these, these flow line U will satisfy the equation DSU plus V of U equals 0. So I'm thinking of, I really should draw the arrows going the other way. So they're parameterized like this. So my flow lines actually go up. Okay. So U is a map from R to X. And I want to glue these things. Also there's a, there's a, sorry. These signs are going to kill me. Okay. All right. And there's also a, there's 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 monochromatic 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 sum 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. Okay. So that's, 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. Okay. 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. Okay. What does that mean? Okay. Mark, 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, right, and I'm going to, 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. Okay. And then what? So now, this translation by R is making them overlap more or less? So basically, you're pulling them apart so that sort of at time, at S equals 0, they're sort of close to the critic, they're both close to the critical point Q. Oh, no. And then we choose a coordinate chart on X in a neighborhood of Q. Okay. And then we can preglue 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. Okay. 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 though. Shouldn't you put the shift in by R? The U plus were shifted. I guess I'm using the same notation. Well, U minus is the, we're supposed to be shifted rather than be close, isn't it? 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 have, these are the shifted, the translated U minus and U plus. Sorry about that. And the betas also have R. Is it changed the R to 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 understood the two gluing from yesterday. Is this like the same kind of thing in one dimension down? Yeah, it's the same kind of thing. So I mean the picture, if you want a picture, then I had 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, or 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 Cotto functions kick in and we start interpolating to U plus and then we follow U plus like that. And we have some coordinate chart here. So all the Cotto function stuff is taking place inside this coordinate chart. So it makes sense. Can you do the understanding on that side? Yes. So as S increases, so we're following U minus until we're very close to Q. And then these, these Cotto 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 beta minus and beta plus like beta minus equals 1 minus. I don't, I don't need that. It could be, well, that you have two times. So what is the, you assume that the image of it could be 0? Right. Yes. So, 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. Okay. So that's the pre-glue curve and the usual, the usual way that's presented is to say you would now try to, you'd argue that this can be perturbed to an actual flowline. Then I'm going to do something a little different, which is I'm going to perturb the things before, before pre-glueing them and then pre-glue them. So it's going to look like this. So we're going to, 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, 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, it's perturbed by some, some combination of psi minus and psi plus. Okay. Yes. So U plus and U minus are always shifted. Okay. And I want to say what, I want to solve for this to be a flowline. So we're going to solve, so, so when is this a flowline? So I'll just write the equation. And if you've done this before, the perturbation answer would have been, yeah. If the size is zero, then it won't be a flowline. So I want to choose the size to make it a flowline. So I want to solve for that. I'm going to write down the equation for this to be a flowline and then solve it. Okay. So this is a little messy, but this is really the very simplest case I could think of. And also, let's, to make my life even easier, let's assume that, 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, well, I mean, so I wrote these gluing papers with TOVs and so this is like my analysis training. So what I've learned is basically you sort of write down an equation, then there's 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. That's, I tried to make, but I'm trying to make this so there's as little crap as possible. Okay. So let's just write the equation. So I have ds minus V of this thing, beta minus times u minus plus i minus plus beta plus times u plus plus i plus. So what is this? So there's a ds beta minus times u minus plus i 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 side 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 one. So I can, I can, I'm allowed to do that. It 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 side minus. And then in general, there's going to be some additional error term. I'll write this as q minus of side minus. So this is, 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 is all, this only comes up away for, away from outside of my coordinate chart. Outside of your coordinate chart, you just take exponentials instead of sums. So you put them by sums and it only makes sense in this coordinate chart? Yeah, yeah. So this is to be interpreted as the exponential map of u minus evaluated on side minus. So I'm sorry for my sloppy notation. I just tried to make it simple. Okay. 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 zero. So I can cross this out. And 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, so I get beta minus times, so, 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 nav s of psi minus, minus nav of 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, it 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 zero. In the more general, the more general situations like in the papers with the tobs, there's a bunch of extra crap in here, which you then have to estimate. Okay. How much time did I start? Or how much time do I have left? Okay. 22 minutes? Okay, 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 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, 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 rejected. And its 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 zero. And then now if we have this lemma, then we've glued because if both of the theaters are zero, then certainly their sum is zero 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 this stuff from this paper which was written eight years ago. It's one of the reason 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, well let me first rewrite the equation a little bit. So let, what? You were ravens that were tying. Uh-oh. So what's my notation? Right. 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. Okay. Sorry, never mind. Which term is guaranteeing that after the term, the end of one goes to the start of the other one. I mean these are decaying as you go to the ends. So these go to zero. So you don't, you're not going to escape this coordinate chart and it's going to be okay. I think the initial set of guarantees will tell us the zero and it's zero. So, I think so there's a, so there's an inverse. So d plus or minus inverse, which is going to go to the target. Use your favorite, favorite bonnet space completion. So we'll give you an isomorphism to the orthogonal complement of the kernel of d plus or minus. Okay. So I'm going to rewrite this equation. So the equation, so what is theta? Let's look at the equation theta minus equals zero. 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 zero 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, these are the equations we need to solve. And then you want to use, use the contraction mapping theorem in your favorite bonnet space completion. So you pick, 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, this, this term of 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 one 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, 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 c zero small then this, then this part will also be a contraction. So you get a unique solution. So tada, we've got to include. And then you can, the 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, 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, 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 the, 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 it is linear or not. Well 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 there's just some extra error terms and it's not a big deal. Question? That's right. 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. Okay so now let's do an example where there's an obstruction. But are you saying that it would be possible maybe to do using different theta minus and theta plus to get that equation to be zero even though individually they're not zero? I mean you can change psi minus and psi plus. So in the region where they were both of the cota 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 no longer satisfy the equations theta minus and theta plus equals zero. Okay now here's warmup number two and it's going to get a little more interesting. So warmup number two I have a one parameter family of functions and at time t equals zero there's a flow line, an index zero 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 one and the flow line u plus. Okay and then I want to glue these to an actual flow line from p to r for t not equal to zero. Now the thing is you actually 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. The analysis is actually going to tell us which. Okay so how does this work? So what's the construction? So again we choose r very large we translate as before. And then what's the data for the inputs? So there's no assumption that we have t, g, t for t to zero it's generic. Yeah I do need generic. I do need generic. So this I'm going to assume that this. Okay so the co-index of this would be the co-index of the co-index of the co-index of this would be one that you would prefer to go for. Where is the fact that you want in this situation appear in the previous. What? In four walls that you could align when you were gluing them. So what's different in this situation is that the linearized operator for d zero is no longer subjective. 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. Okay and then from these inputs we get. And what's the larger one? All of this discussion. We're trying to show that something about the homotopy invariance and this construction in the start or? That is one thing you could do with this. Right now I'm just doing it as an example of the gluing construction. But yeah if you want to understand how Morse complex has changed under bifurcations and this is part of what you have to do. Okay so then this you get a flow line if and only if again beta minus of 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 zero. Okay 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 is zero of psi minus psi plus t is equal to d zero, ah, zero. So it's equal to d zero of psi zero 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 zero u zero plus psi zero plus t times the derivative of v plus and other stuff. You said we used the index differences being one for surjectivity but we don't actually need surjectivity. We don't need the v plus minus two. 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 will just be defined on the image of the operator. No, well it will work up to an error which I'm about 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 zero psi plus such that so as before psi zero is perpendicular to the kernel of d zero. Psi plus is perpendicular to the kernel of d plus. Theta plus of psi zero psi plus t is equal to zero. And as for theta zero I can't necessarily get it to equal zero. All I can get is that this is perpendicular to the image of d zero. 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, well maybe I should write it down. So let's let pi zero be the projection to the image of d zero. So then we have a d zero inverse goes from the image of d zero to the orthogonal complement of the kernel of d zero. Now we're going to solve an equation that looks like well what I want is that pi zero of theta zero is equal to zero. That's the equation I want to solve. So I can write this as pi zero of d zero psi zero equals blah blah blah. And then I can write this as so now I can apply the operator d zero inverse to it. So I can write this as psi zero equals d zero inverse of blah blah blah. I guess there's a pi zero here also. So d zero inverse pi zero of blah blah blah. And then you can do the contraction mapping theorems before. 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 zero to both equal zero. And I can't get that. Well I can get theta plus to equal zero but theta zero it lives in a one dimensional vector space. It lives in the complement in the co-kernel of d zero. It's a one dimensional vector space. I can't get that to equal zero. 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. This is a trivial bundle. The fiber over any point is the co-kernel of d zero. And I have a section s. And what is the section s? So s of t comma r is equal to the state of zero. So the gluing construction says I can almost get the gluing to work except the state of zero might not equal to zero. The state of zero is the obstruction to gluing. So when this is equal to zero I can actually glue. Okay? So this is some function of t and r. When this function is zero 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, you know, the Titanic if it has compartments. 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. Are you and Chris having 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. All at office hours. Any questions? In this example the D0 doesn't depend on where you are in the space. It's a very trivial adjustment. Other examples will be less trivial. But this one is, as I said I tried to make the easiest one I could think of. Any questions? Let's thank Michael again.
think of, which is for Morse theory. And that flurhomology.wordpress.com, in a posting from Bau and Kohanda on contact homology, 1412.0276. bundle gluing is a way of gluing things when transversality fails, but it doesn't fail co-index of the co-index of the co-index of this