5/5 Contact homology and virtual fundamental cycles
Pardon, John
The goal of this course is to give a construction of contact homology in the sense of Eliashberg-Givental-Hofer. I will begin with an introduction to contact geometry, pseudo-holomorphic curves as introduced by Gromov, and some target applications of contact homology. The main focus will then be on extracting enough enumerative information (i.e. "virtual fundamental cycles") out of the relevant moduli spaces of pseudo-holomorphic curves. I will present a general framework for doing this, and then discuss the specific application to contact homology.
 Ah Okay, so today I think we'll talk about gluing. So that is what I want to do is show how to describe the local structure of the modularized space near curves which are split, curves which are broken at ray-bore bits. So let me begin by just recalling the setup. So we have our target y, we have our contact form, and we have our almost complex structure on r cross y and satisfies the following three properties. So it should be r invariant, j of d by ds is equal to the ray vector field and j sends xc to itself and restricted to xc is compatible with d lambda. So then we denote by p the set of closed orbits of our lambda, unparameterized. And now, right, so gluing is simply a statement about one fixed moduli space. So I may as well just fix once and for all gamma plus, gamma minus, and the homology class of curves that were incidentally. So b is the curve, all i lambda generated closed orbits? Ah, yes, I'll know the general. So we fix this data and then the moduli space m bar is, so that's the set of curves like this. Sorry, t minus is a set of n. Oh, gamma minus. And it's a set. Actually, this is gamma plus and these gamma minus. And it's, so this is, each of these is a holomorphic curve mapping to r cross y and it's considered up to translation. Each of these individual pieces is considered up to translation in r direction and in homology plus beta. So from now on, I'll just forget about gamma plus, gamma minus, and beta. That is, I'll leave them out of the notation because we're just interested in this fixed moduli space. And then denote by m bar reg of the locus where the linearized operator is subjective. So then the theorem is that, so first, this is an open set and near any point in it. There exists a local holomorphism from rk cross r to the n minus k to m bar reg by sending 0, sending 0 to the base point. And where n is the virtual dimension of m bar, k is number of nodes at this point p. And it's compatible with the stratification. That is, the left hand side is stratified by which of these first k coordinates are 0 and the right hand side is stratified by which places did we resolve the node. And it's compatible with that stratification. And the compatibility with stratifications is, okay, it comes sort of trivially from the construction, but of course it's crucial to include in the actual statement because the homomorphism type of this only depends on whether k is 0 or positive. Stratification is important data to keep track of. Yeah? So when you say no, you mean you get a little bit of it? Yeah. Yeah. You see them as models. Right. So the first remark is that in the actual, like to establish contact homology, we need something slightly stronger than this. What we need is, actually, need to establish the same statement for these thickened moduli spaces, m bar, set i. The addition of these sort of thickening terms in this moduli space, though, doesn't really change the argument. And so for sake of simplicity, I just want to explain the proof of this. And moreover, not only do we have to deal with these thickened guys, we also have to prove this submersion property. So we have to basically show that this m bar reg inside, m bar j reg, is cut out transversially. Like this. We need this to be a need to be a topological submersion. OK. So I'll let me not say more about this right now. That's exactly the submersion axiom in the implicit Atlas definition which I gave earlier. The point is just that both of these are really only slight extensions of the proof of this. And the other remark is that we don't need to establish any sort of compatibility results, particularly properties between these local gluing charts. They're just local homomorphisms. Probably if one works hard enough using gluing profiles, one can show that we get a smooth structure on this space as a transition mass between these charts are smooth, but this is not needed for what I want to do. Yeah. When you want to do the same thing for thickening the space, do you use something like seeing these elements of thickening the space as homomorphic curves in something bigger? When we do in-originous question and question, how is it simply that you've translated the analysis? You can think of it like that. I mean, I think in the end it's the same. You can write something as the inhomogeneous equation or you can write something as the true homomorphic curve equation on this bigger thing. I think when it comes down to the other things, you actually have to check are the same no matter which perspective you use. The closest result in the literature to this theorem is apparently, as far as I know, this result gluing theorem of Jinx-Tobbs, I believe, in 2009 in their construction of embedded contact tomology. So what they do is they prove a gluing theorem for embedded curves. Embedded isn't very important. They do gluing for SFT curves basically in R cross a three-manifold. In fact, their result is somehow more, much more involved in this statement because they have this extra thing that they do. They stick on branched covers of trivial cylinders in between the curves that they glue. They also have this obstruction bundle gluing that happens. So the techniques which go into this come from the same body of work that this originates from the gauge three literature with Tobbs, Donaldson, Kronheimer, and eventually applied to holomorphic curves by Fleur and FL cubed and Hoverer-Saskis-Ender. So I just want to show how these techniques are applied to prove this theorem in the case of contact tomology. All right, so we begin with some setup. So first we have to fix our base point, which instead of calling P, I'm not going to call x0. So we fix our base point x0 in m bar reg. And what that really is is a map u0 from some curve c0 to x0 hat. So this is just the domain of the map. This is the target, which is a disjoint union of a bunch of copies of r cross y. But I'm just going to call it x hat not. Let mu denote the index, the index of this moduli space that is the virtual dimension. T0 will denote the dual graph of c0. So maybe I'll draw an example. So let's even take this example. So maybe this is c0, then x0 hat is the disjoint union of four copies of r cross y, a separate copy for each of the components of the curve. The map is considered only up to translation. And then T0 is the dual graph. It looks like this. I'll let v of t0. This is just the set of irreducible components of c0 and e. This is the interior edges. Here I really mean the interior edges, so this is the intermediate ray of orbits. In this simpletization, r cross y is exact, so there won't be any bubbling. But one could also add bubbles. In that case, I'm not going to keep track of the bubbles in the tree just because we don't need that stratification. Yeah, yeah. So for the purposes of this argument, it really doesn't matter the genus. But yeah, assuming genus 0 for now. So now our goal is to construct a map of local homeomorphism. We'll get the polynomial from this to here, respecting stratifications and landing in the regular locus. So this, right, since it's a germ of homeomorphism, that means it covers an open neighborhood of x0. And since it lands in m bar reg, that means m bar reg contains an open neighborhood of x0. In particular, m bar reg is open since this is true for every x0. So this is what we have to show. Right, so first, set up, what we want to do is domain stabilization. So we really want to, what's the purpose of this? Set up, over a neighborhood of this point x0 in the moduli space, we want to sort of fix canonical coordinates on the domain. The problem is for this and these two components, they are by holomorphic with a cylinder on the plane respectively, but not canonically so. And we have to get rid of this extra freedom or do some gauge fixing in order to be able to produce local parameters for the moduli space like this. So what we do is we fix some points qi and c0 and our invariant divisors, di inside r cross y such that u0 is transverse to di at qi. So I don't care whether it's transverse to di anywhere else just at qi. Such that this holds and c0 equipped with these points is stable. So here's really, c0 is already equipped with, it's marking by the positive and negative punctures. So yeah, so we just have to stabilize these two guys. So it looks something like this. So this is di at qi. So now since u0 is transverse to di at qi, this property is preserved under small deformations of the holomorphic curve. That is over an open neighborhood of x0, you get corresponding points qi in the domains of all of those other holomorphic curves. And therefore you've done the gauge fixing over in neighborhood. So second, introduce the glued cobertism target. So for a gluing parameter, I'll call an element of this space a gluing parameter, we define this glued target to be obtained from x0 by truncating the ends. Over to the ends, the ends look like 0 infinitely cross y. You truncate them to 0, the gluing parameter cross y. Do the same thing for a negative end and then you glue via this identification. So as I've described, it's a little bit. Yeah, yeah, yes, I forgot about this. So the way I've described it, it's not entirely accurate. So suppose we're taking this curve. So we have x0 is like four copies of r cross y, so let me just draw them as just draw the r factor, it looks something like this. And suppose I glue all of the parameters using exactly this recipe. I get this, this is like x0 hat, xg hat, then looks like, so I have this first piece, second piece, I glue both of these here, I get something sort of non-Hausdorff like this. Obviously, I don't want a non-Hausdorff target, but there's sort of an obvious canonical way of just closing it back up to r cross y. So xg again, so xg of course is again just a big distant union of copies of r cross y, but somehow this particular description is the one we care about. Yeah, yeah, so if we had something more general, then we'd have to, then there would be some constraint on the space of allowable gluing parameters. So now it's important to keep track of the zero sections here. So for v in v, that is a very irreducible component of the original curve, we're going to let zero, so v inside x bar zero, x hat not denote corresponding zero section. And we let the corresponding, and we also denote by that the corresponding zero section. It's descent. So there's some canonical, yeah, zero cross y here, and we get some points here. For each component of the original curve. Okay, now finally, this is, again an important point, I want to fix sections, qv prime of this universal curve over, well, the linear monologous space of curves of genus zero, and equipped with points marked by the input output edges, and these extra marked points, ui. So I already commented earlier that by choosing these divisors di, we identify the domains of all the holomorphic curves, all the points of the modular space of our neighborhood of x zero with fibers of this map. That was somehow the purpose of adding these divisors to gauge fix the target, the domain, rather, gauge fix the domain. What I want to do now is gauge fix the target, and gauge fixing on the target uses these sections. So sections qv prime, such that when you evaluate them at this, at c zero, you land, qv lands in the corresponding irreducible component of c zero. So there's one point, qv prime in every irreducible component of c zero. It is on a, it is on a, it is on a. And gauge fix it, the truth, but are you? Yeah, yeah, yeah, yeah. So, right, so, in, in the, this is sections defined in neighborhood of c zero. I mean, obviously, maybe this whole vibration does not have a section. We can only curve up locally. Yes, and yes, so the whole point of fixing these sections, qv prime is to gauge fix the target. So we're going to now only look at maps which send qv prime to the corresponding zero section zero v. So for group maps, we gauge, we gauge fix the domain via these qi's, which are essentially the inverse image of di, and the gauge fix the target via requiring the u of qv prime lie in this zero section. So really what that means is that at a point close to x zero, I get a unique gluing parameter. So if you have a map close to x zero, your, your domain is canonically identified with, with one of the fibers of this universal curve because you have these points qi from intersecting with di. And then that gives you points qv prime via this arbitrary choice of sections. And then this requirement that qv prime be mapped to the, the zero, the zero section tells you what the gluing parameter has to be. Because this, this length here is exactly the inverse. So I think this is all the gauge fixing we need to do, which is good. Yeah, I guess you could intersect di several times, but you just follow the intersection that p is equal to 0. So we're going to furthermore fix the trivialization of the universal curve over this. Except it's not exactly the universal curve here that I want to fix the trivialization. It's something slightly different. So I'm going to denote by m bar 0. So here I'm looking at a compactified nodulite space of curves of genus 0 with m plus n marked points. But the first n marked points are in addition going to be equipped with a marking of a tangent space by c. So this is, of course, a forgetful map. And this is a principle gl1 to the n nodulite. So this is classifying. Xx classifies curves c with p1, pn, q1, qm, and sigma i from c to the tangent space of pd. So n lower parentheses 2. So why do I care about this curves with an extra marking here? So when you have an end, if you have a map, why do we care? So if you have a holomorphic map, say u from standard half cylinder to r cross y asymptotic to a ray of orbit, it means that u of s comma t is equal to the length of the ray of orbit times s plus some constant gamma tilde of t plus the low of 1 as s goes to infinity, where gamma prime is the parameterization of the ray of orbit that it's asymptotic to. But the point here now is that the coordinates on r cross y give rise to a marking of the tangent space of the puncture by c, namely by requiring that gamma tilde be not rotated at all and b be 0. That is somehow, and this is the way I want to think of these curves. I want to think of c0 as equipped with a marking of the tangent space at each of the punctures. And with respect to those coordinates, I want sort of b to be 0 and gamma tilde to be not rotated. Right. So now we fix coordinate local trivializations of this modular space. I'll fix local diffeomorphisms from J, so Jv. I'll denote c2 times. Oh, this is just the dimension of this modular space. So for each irreducible component, so that's for each vertex, we fix a local diffeomorphism of this modular space, where you have the points qi and you have a tangent space marking at each of the punctures. So yes. Right. So now, OK, so I also want to fix cylindrical coordinates on x0 and c0. That is, so we fix coordinates like this for, oh, these are valid for s sufficiently large near all the punctures. And near all punctures, such that with respect to these coordinates, we have u0 st is asymptotic to the length of the corresponding rate of times s and gamma t, where s goes to infinity. And of course, I mean that, right, the point here is that this coordinate b is 0 and that this is not twisted. OK, so finally, I think the last piece of setup is we have to glue the curves, glue the domain. So really, these gluing parameters g, g lying in r greater than or equal to 0 to the e, this should be thought of as the gluing parameters for the target, not the gluing parameters for the source. So we would glue the curves c sub g. So we basically do exactly the same operation. So we truncate c0 from on each end to 0 g inverse. And we need a factor of the length of the rape orbit also. Right, this is one reason why. Right, because the trivial cylinder has this factor of l in it. So if we glue the cobertism by truncating and translating by g inverse, we better glue the curve by g inverse times l inverse. Right, so it's always the and identify as before. So the picture looks like this. We have an end of c. This is c0. I'm not sure. We have another end of c0. We truncate the ends as we cut off here. And then we glue them together. So this is cg. Now, I want to write them. The rotational coordinate is chosen by the asymptotic matching of the two ends, since they're both asymptotic for the same rape orbit. So it's crucial to fix a point qe double prime in the center of each glued edge. And so this qe double prime in cg exists for every e which has a positive gluing parameter. And this is needed for some gauge fixing. The problem somehow is you have, right, so what happens? So can you tell me what is this qe double prime? So it's the point in the middle of, in the exact middle of the neck. Just fix such a point. I mean, the s1 coordinate doesn't matter. But just as long as you fix some way of defining it. So why do we need this? So the point is that there's a map from. So we have these trivializations of the modular space, each of these components. But that somehow only tells us about curves on this stratum. We also need curves on the other strata, too. So what happens? If we fix a g, we get a map from this product of j's to what? Well, it's a product over all vertices of the glued tree of m bar 0. So again, we look at the qi's, which lie in the vertices which are mapped there, plus qe double primes in the edges which got collapsed, plus the punctures. And they have a double marking, not just the points, but the tens of spaces also. So this is a local defemorphism. And this would not be true if we did not include the data of qe double prime. I mean, the dimension wouldn't even be correct. There. Yes, somehow, if we just do this gluing construction, look at all possible, almost complex structures, j, we get two extra degrees of freedom, because we don't know what the marking of the tangent space at these ends was for c0. This is some sort of tricky point. A local defemorphism. Yeah, yeah, no, the f is this. Oh, yeah, homeomorphism would be enough. It's a smooth map. I think that's all I need. A smooth map, which is a local homeomorphism. I think that's all that's actually needed. And in all these constructions, do you ever need to have the full universal curve of full modulus in the human port space at any point? Or do you only need those local charts of modulus spaces of curves around? Yeah, I only need the local part near the particular curve we're gluing. You never need to know anything about the lowest of the element for an arc in the circular arc. Yeah, I mean, gluing is a completely local statement. So it should come out as this can work more generally with the full story. Oh. Yeah, so you mean for the construction of the apples and stuff. Yeah, I don't need it. It's not really used there either. Um, yeah, I mean, there's some right there. Like, in certain setups, you might sort of be interested in the question of, look at the modulite space. No, you don't look at this modulite space. You ask, does this have a section? This is basically saying, oh, is there a sort of global way of choosing asymptotic markers? I think this is not true. I think it doesn't have a section. But yes, for us in the setup, we don't need a section. Did I? Oh, I already flipped the boards. Great, so we've now defined gauge fixed. Everything near our point. We've defined this glued culprits, the glued curves. So the final thing is we have to define the glued maps. So of course, first we do the pre-glued maps, ug. So we fix cutoff function, chi. chi. And OK, so these are the two steps. So we first define a, I'll call a flattened map, u0 bar g from c0 to x hat 0. So what is the picture? So u, the map u approaches a trivial cylinder. So the trivial cylinder here. So this map is u0. So this is a trivial cylinder. Trivial. And I'm just going to cut u, cut off u0 to the trivial cylinder. So this is that. Oh, and in symbols, it's this. So we take u0 for, right? So capital S will denote g minus 1. So maybe I'll just write that instead. Ls. I'll denote that. Now I remember why I decided to call it s. Anyway, so this is for. And then here we interpolate. That is, we take an exponential trivialization, the sort of diffeomorphism of, right? We take exponential coordinates near this rate of orbit and just cut off. So this is for s in 1, 6. 6 is the same as 5 in all the calculations. Yeah, something like that. Yeah, that doesn't, I mean, yeah. Yeah, the things will happen over this neck. And this is the first thing that happens. And a bunch of things will happen. The other things that happen will happen only on the part of u0, g, which is a trivial cylinder, which is. Sorry. Is it a gamma 2, then? Yeah, it's just a 2 gauge fixed, then to the gamma 2, then. Yeah, I guess, oh, I've usually been thinking of gamma as the unparameterized orbit. So whenever it's something parameterized, I put a tilde. Yeah, I do it. Right, so then we get a, so this preglue map, u, g from c, g to x hat, g is just the descent of u0, g. There's a map of grease with this description of c0, g. So now we have a map. So this gives a map for every, so we get a map from every gluing parameter, we get something, this map. So I don't know what to write here. Of course, I can't write m bar because this map isn't holomorphic. But what we want to do is somehow correct this map to be holomorphic so that it lands in m bar. And we also want to somehow add this extra set of parameters on the left, namely r to the mu minus e, corresponding to the kernel of the linearized operator at this point. So now we're going to get into some actual analysis here. So gluing estimates. So first, I need to tell you what weighted Sobolev norms we're using. So we need some weighted Sobolev spaces of functions and maps of sections over c0. So we're going to be looking at spaces like w, k, p sections of the pullback of a tangent bundle of xg to cg. So we define weighted w, k, p delta norms given in the ends of the necks by the following. So in an end, something like this, p. So basically in the end, we just add a factor of e to the delta s to the function, or multiply the function by a factor of e to the delta s. That is the functions here, dk, like e to the delta s. So I'm just writing the contribution to the pth power of the norm over this end. And away from the ends, it's just the usual norm. And then in the necks, something like 0 to capital S cross s1. We, of course, do something similar. And now the weight is p delta times the minimum of s and capital S minus s. So in an end, the weight looks like this, e to the delta s. And in the neck, the weight looks like this. It's e to the delta times the minimum of s and s minus s. So these are somehow the norms which are relevant for this problem. So here delta s is a small positive number. And by small, you can say precisely what you mean by small. So specifically, delta should be less than the smallest eigenvalue of the asymptotic linearized operator of the Ray-Borbit in each of the ends. And yeah, in this Frett-Hellman analysis, this is where we really need the ends, where we see the need to have ends be non-degenerate. It's one of the places. The other place is in the a priori estimate. A priori estimate, if you have a trivial cylinder over a non-degenerate Ray-Borbit, then if you take a holomorphic curve which is c0 close to a trivial cylinder over a non-degenerate Ray-Borbit, decays exponentially. For that trivial cylinder, exponentially, meaning at least as fast as e to the minus delta s, where for any delta less than the first positive eigenvalue. This is the fundamental a priori estimates of Hofer-Rasovsky's ender. So this defines for us weighted so-bluff spaces like this over each of the glued curves, c sub g. And it suffices to take p equals 2 for this argument. We don't need anything else. But of course, we could, I mean, everything I'm going to be saying is true for any p in this range. But there's no way I need p equal to 1 to impose p equal to 1. Not for this. Chris Wendell has this blog post about this, whether you need, whether at any point you need to use this p equals 1. Yeah. So for this, it's really not needed. I think the, what is the point? Yeah, we somehow, given all the a priori estimates, which, which sometimes require p equals, which, or maybe they don't never require p equals 1. At least some ways you write them, you can, p equals 1 seems necessary. Yeah, I think the conclusion of the discussion with Chris was that maybe gluing was the last remaining place where p equals 1 was needed. And he said gluing doesn't need. Yeah, I think it really doesn't need p equals 1. Yeah, p equals 2 should be really fine. I mean, the point is you're already, you're only dealing with curves, which you've already proved your a priori estimates for. I mean, we take k equal, we can take k to be large. Yeah, I think the difference between p equals. Yeah. But wait, the issue with. Oh, sorry, I wanted to ask k equals 1. Oh, you meant k equals 1. Do you need sometimes p to be bigger or two times k equals 1? Yeah, yeah. Yes, okay. So k to k will be large. Right, okay, so maybe I should summarize since we were saying the wrong thing. So for this gluing argument, all of our curves are, we have, we're sort of doing this argument after we've proved all the a priori estimates. And we know our curves are very highly differentiable, so we're going to just fix a large k. And since k is large, we can take p equals 2. If k were 1, yeah, maybe we need p larger than 2. I have a question. We must think of how this flow theory would be able to find for less and less simple, like the A-symplec. Yeah, I don't think any of these, I mean, I don't know anything about it. Like for p-symplec, if you consider less and less smooth conditions, like p-A-symplec manifolds. I don't think anyone ever considered doing more of the curves in such situations. And I guess I just, again, I'm thinking about p-A-symplec manifolds, and I think that everything is completely different, even the local questions. I think that's just the way it is. So, nonlinear. So we have this, right, so we want to understand the moduli space of holomorphic curves near this pre-glued map. And so what we do is any such map is given by some exponential of some vector field and an almost complex structure on the domain. But actually, we're not going to use all vector fields, so we have to preserve this condition of intersecting di at qi. So we only consider the subspace where c of qi is in the tangent space to di. And we also only consider the subspace where this sort of the same condition at qe double prime, 0. So we have this point qe double prime. And in order to gauge fix the domain again, we have to pick, we have to specify qe double prime. On our priori specify what qe double prime is on the domains of all the curves in the moduli space near x0. So what I'm going to say is that qe double prime is the inverse image, sorry, the image of this space under the exponential map. I'm just declared to be that. That gives a local divisor. The inverse image of that under the map will be qe double prime. So we have a vector field and the most complex structure. And from the vector field and almost complex structure, we can then ask what is del bar and we get a section. Sorry? What is this? J. It's on the target. On the source, on the c. We're varying. J was, I think I've erased it by now. J was this set of almost complex structures on the source. Yeah. Yeah, the almost complex structure on the target is always fixed here. I'll make a 0, 1. Right, so what is this? The map, fg. So it takes a vector field and an almost complex structure. I'll denote by j sub y on there. So what do we do? So x, u, g. See, that's the map. We take differential. Take the anti-holomorphic part with respect to this jy, almost complex structure, and j on the target. This is a section of over cg, not of this bundle, but rather this is in sections, w, k minus 1, p delta sections of this part of omega 0, 1, cg with the jy almost complex structure and with x, u, g, c pullback of this tangent bundle. So to put it here, we use parallel transport on this part. We use some fixed identification map. It doesn't really matter, just some complex linear map. So parallel transport for a j linear connection, certainly need it. Otherwise, a tensor product won't make sense. So j linear connection on x hat, r cross y, say r invariance, j linear connection just to be safe. And then I tensor with this just fixed identification map. It doesn't really matter what we do here, just any map which is c linear and smooth and depends smoothly on y, and is the identity for y equals 0. Now we're not done with the definition because we also have this other component of the target. So we add, so what is this component telling us? This tells us the vertical positions of all of the images of all of these qv primes. So remember, since we did this gauge fixing, we're now interested in maps which map these qv primes to the corresponding 0 sections. So we take x u g c of qv prime of y. So I write it qv prime of y because at this point in c0 and cg really depends on this almost complex structure y. Yeah, because this qv prime was a section of the universal curve. So we have this, and then we just project to r. This is an element of r cross y. We project to r, and we do this for all of the qv primes. So now with this complicated definition, our modular space near x0, our modular space is union over g in, OK, I'm writing it like this, but of course I really just mean small, small gluing parameters, g, sort of a germ of identification of fg inverse of 0. Now this local, this gives a local map from m bar to the space of gluing parameters, but this is not at all a canonical map. It's dependent on our choices of divisors and sections qv prime. Right, and so I, as far as I know all of this should, yeah, works for, right, so another comment about k equals 1 which is larger k, I should apologize for screwing up this discussion the first time. So, yeah, this operator, well it makes sense. Seems to make sense for any k at least 1 and any p between 1 and infinity. If you have k equals 1 and p 2 or smaller, or you have k equals 2 and p equal to 1, I guess we don't care about that, we just want to make sure that the dimensions here are not continuous, which is a little bit disconcerting. But, yeah, right, so of course, yes, no, of course yes, no, you need this, right, because we can't do this exponential. So this only makes sense for either k at least 3, k equals 2 and p greater than 1 or k equals 1 and p greater than 2, right. The differential of this map makes sense for any kp, but in k equals 1, but, yeah. And we actually, so for the gluing argument we're doing right now with just the holomorphic curve equation, we could really do it with any of these that we want to. It's somehow the Fred-Holm analysis is a bit easier with p equals 2. And, but when we do the same thing for the thickened modulized spaces, it's really important to be able to choose large k because there's an extra term in this part of the equation, the thickening term, and it depends on the places where this exponentiated curve intersects certain divisors. And this map from the space of c to cg, given by, take the inverse image under this exponentiated map of a fixed divisor, this map is a regularity class, something depending on k, something like ck minus 2. So if we want this to be a nice smooth map that we can apply the inverse function theorem to and understand it's zero set, we need k to be large. I guess that. Yeah, so I mean, yeah. Great. So now there are a few fundamental estimates which I could spend a whole lecture doing the calculations for, but I'm not. I mean, just write down some things which one can prove about this map. So first, estimate the norm of fg of 0. So lemma this norm fg of 0, that is, so if this measures how far the pre-glued map is from being holomorphic, of course using this norm, this goes to 0 as g goes to 0. Maybe I will sort of prove it. OK, first one comment about the norms. You can define norms with actual explicit formulas including trivializations of these bundles in the ends. But all that really matters is the commensurability class, the commensurability class of these norms, uniform in g. So you could pick another norm which is commensurable with a commensurability constant which doesn't depend on the, that would be fine too. The point is just to have estimates which are true for all g. And this is somehow why this, for any fixed g, this norm defined by this weight is commensurable with a norm defined by not doing any weight, weight 1. But as g goes to 0, of course, they're not uniformly commensurable. OK, so the one form part, so this guy has two parts, the one form and the zero one form and the vertical coordinate part. So the one form part of fg of 0 is supported in this region if you recall the definition, right, that this is just saying where is ug not j holomorphic? It was a splicing together of this holomorphic guy of trivial cylinders, so the only place it could, it might not be j holomorphic is in this green part, the part where we spliced. And OK, now we just use the exponential convergence of ug to a trivial cylinder and the fact that delta is less than this delta gamma. So we have a weighting function which is e to the s delta, but this difference that we cut off decays by even faster exponential since delta is less than delta gamma. And then the second part for this part is just the statement that qv prime of cg j0 approaches the descent of this qv prime in c0. j0, just because as g goes to 0 because these sections are continuous. So there's some calculation to do, but this is, yeah. So the other thing we really, the other key part of understanding the 0 set of fg of 0 near c comma y equals 0 is the following quadratic estimates. And this should be really thought of as a uniform bound on this modulus of smoothness of fg. It's something basically saying f is like a c3 map from this ballast space to this ballast space. And moreover, we're giving an effective bound on its c3 norm independent of g. OK, now we actually don't need it to be c3. We need it to be, we need something considerably weaker. And what we need is this. So proposition for, we have, so you take the derivative at 0 and the derivative at 0 is close to the norm of the derivative at 0 minus the derivative at zeta is bounded by a constant times zeta. So it's a statement which is very similar to saying the derivative is of uniformly elliptics. This is 4k. So you, at least 4c. OK, this k equals at least 4 condition is really only needed when you have the thickening terms in intersection with divisors for this guy. So this should be true for any k. So I won't prove this here. So now we come to the part which has somehow most, a lot of the content. So this is to construct a bounded right inverse. So let dg denote this derivative of fg at 0. fg prime is the direction of the, of the constant. In the direction of the second argument. Yes. Oh, sorry. This is a terrible notation. Zeta and over here, zeta and c are both elements of this direct sum. Oh, there used to be elements. So on the left side is on the, yes, so this notation here and the definition of this notation are different. 0 is an element of, is really 0, 0, c is really c, y, zeta is, zeta, y prime. So yes, so this, yes. So this is the derivative of f at 0, 0, this operator. So it's a linear map from, from the source to the target. Then we give them by the derivative of fg at 0. So that's dg. So by assumption, d0 is, this is a, a, we want to show that dg is also subjective. Okay, we actually want to show a bit more. We want to show that we want to, out of a right inverse for d0, dG. And it'll have two key properties. One is that, right, so in fact, given this right inverse for d0, we will construct a QG with, which is a right inverse for G, such that QG is uniformly bounded, independent of G. And the image of QG, I'll just put this in quotation marks, varies continuously in G. So I'll say precisely, in what sense it varies continuously in G. But I mean, these spaces don't vary nicely in terms of G. So there's not a well-defined, there's not an obvious notion of varies continuously. But I'll say what I mean by that precisely later. So this construction is really a hands-on construction. Given a right inverse d0, we simply sort of write down a right inverse for dG. So we have the key diagram as the following. So if you look at this, C0, CQI and P di plus j. So this gets parallel transported with WK to delta C0. A flattened guy, again with this, there'll be a map which is called a gluing map, which basically takes a trace down to CG. And finally, will be a calibration map, which takes this and ensures that it satisfies this extra condition at Q E double prime, so pi Rds plus R is 0. So now we have all these linearized operators.. This again goes by parallel transport. There's a map in the other direction called break. And this is the same. I won't write it again. So what's written in the last article on the left? Calibrate, calibration map. So this is something extra which has to happen for contact homology, which doesn't need to happen when you're gluing just nodes and say it grow off in theory or even Hamiltonian for theory. Yeah, the point is we need to ensure this condition. OK, so let me just sketch the definitions of these maps. So defining gluing break. So here we have C0 and here we have CG. So the gluing map is supposed to take a section over C0 and give a section over CG. So what's it going to do? Going to cut off here, cut off, and we cut off here. So these are these other series of complicated operations. So this is going to happen at something like 1 third s. So in the region where we've already made it a trivial cylinder. And then we sum, sum of cut off. So this is glue. It's the sum of the, under this map of the cut off guys. And now the map break is we pull back and we cut off at the center. Cut off at 1 half s. Right, so this is. Maybe I should. So the way we cut off for the glue map is not really relevant. We just use whatever cut off function we want. The way we cut off for the break map is crucial. These cut off, so first of all, has to happen at the center. And secondly, these cut off functions have to be symmetric. That is, if you break and then take the sum of fibers back down, you should get the same thing. So this cut off function is 1 minus this cut off function. OK, so defining, finally, defining calibration map. So here we have a picture of Cg. And there's this vector field, which is just constant, but cut off. So we look at this cut off function. So this is cut off. Chi times ds ug. So here's qv double prime. And this is the vector field. So calibrate simply adds the correct multiple of C linear multiple of this so that this condition projection applied to your vector field becomes 0. Now I want to observe that both of these maps, or all three of these maps, actually all the maps in this diagram, are uniformly bounded. For glue and break, you can check. For the calibration map, it's important here that delta is positive for this to be uniformly bounded. And it's important that qe double prime is in the exact center of the neck, whereas it's the point with the maximum weight. So there's this constant vector field here, which is in the kernel of the d operator. Now this constant vector field, in your usual norm, just has constant norm 1. But that's somehow not the right norm to think of it in. We should think of it in the weighted norm. And in the weighted norm, the norm of this constant vector field looks something like this. It reaches a maximum at, or say normalizes with a maximum equals 1 in the center of this guy at qe double prime. And then it decays exponentially as we go towards the boundary of the neck. And this is somehow why this is an OK thing to do, cut it off here. Because whatever xe is at qe double prime, it's very small. It's like e to the minus 1 half capital S delta times the norm of xe. So this thing that we added is actually very small. OK. So this key diagram does not commute, but it almost commutes. And this is basically the property we need to use it to construct this bounded right inverse. So key diagram almost commutes. OK, so the statement is the following. So first, for parallel transport, the statement is very simple, simply the commutator of the maps. Approaches 0 in norm. So these are all, will be three statements, and they're all as g goes to 0. The second statement concerns the bottom. This goes to 0. And finally, the statement for the middle square is a little bit complicated, because glue and break go in opposite directions. And the statement that you can is both sufficient and can actually be proven as the following. So this norm is little o of 1 times the norm of eta, where break applied to eta is equal to d0gx. OK, I should comment that lots of treatments of gluing the two upper squares here are combined into 1. It's very common to not consider this flattening u0 bar g. But I think it's a good sort of intermediate part and sort of divides your estimate into two constituent pieces, which are sort of different. It's also very important that all this cutting off is happening in the region where u0g is a trivial cylinder, which makes the estimates much, doesn't make no all that much easier, but there's less to worry about. So what is the quantifier of eta? You have to solve all eta, such that break is of all sort eta. Yeah, for all. Yeah, so it's, how does it work? So c lies here, and eta lies here. And they get mapped to the same thing here. And the statement is that when you take eta this way, sorry, when you take c this way, so they're mapped to the same thing here. And then the statement is that when you take c around this way, it's very close. Sort of a strange sounding statement, but I don't know any other formulation which is both correct and simpler to state. I mean, you could ask that like going this way and going this way are really close, but that's not true. Or either it's not true or it's not enough to show what we want, one of the two. I forget which. Oh, I saw, do I have time? Oh, I saw that I mean, maybe I'll say in words why this is true, that I mean, one can do some explicit calculation. For this, the point is, again, just u0 decays exponentially to a trivial cylinder. And these only really differ over the part where you cut off u0 to a trivial cylinder. And over that part, we have this estimate that u0 approaches a trivial cylinder, at least exponentially, with any delta strictly less than delta gamma. And our chosen delta is less than delta gamma, so we have this estimate. For the calibration one, the point is just that this vector field dsug is in the kernel of the linearized operator. It just corresponds to translating the trivial cylinder or rotating the trivial cylinder. And we cut it off, but the point is that cutoff happened really far away from qg. And so we use the product rule to expand this. This term is very small. Again, like e to the minus something times delta. So it's crucial that delta is positive for this. And finally, this one. Sorry, finally, this one. Yeah, so maybe I'll just not say why this is true. I mean, you can explicitly analyze what's going on. The point is somehow that the weight, the key point is somehow to compare weights. On C0, you have weights, which look like this. And on CG, you have weights, which look like this. And what you turn out to, the error term in this difference, turns out to be something supported here in this type of region. But the norm you care about is this norm. And you have a bound on its norm in this norm. And so the fact, the ratio between these two norms is very small. And so it goes to 0. OK, so now let's just define these bounded. Right, it versus. So what do I mean very incontinuously in G? This is very crucial for the gluing map to be continuous in G. This image really has to vary continuously in G. And this is somehow a non-trivial statement because the space of gluing parameters could look something like this. And continuity for this gluing parameter and this gluing parameter is some non-trivial statement. So what do I mean continuous in G? So we'll just fix some linear map from this source WK delta to some vector space given by evaluating at some points. So for any subspace B, such that the L0 restricted to the kernel of D0 is an isomorphism. So you pick some points and look at this evaluation map. So first you pick some points. So this evaluation map is injective on the kernel. The kernel is fine dimensional. So you can certainly do this. Just pick some points on the domain, C0 such that this is, yeah, and J is fine dimensional. You can do this. And then quotient out by a suitable subspace B, so it's an isomorphism. This, of course, gives us a corresponding map LG on this sort of domain of FG. Let me not write it again. It's just this. OK. Now we're going to fix, or write, x q0, which is the right inverse. So d0 q0 is the identity. And the image of q0 is the kernel of L0. So since L0 restricted to the kernel of D0 is an isomorphism, the kernel of L0 is a complement for the kernel of D0. And given the choice of complement, the choice of complement determines uniquely a choice of bounded randomness. Closed complement. This is continuous map. So it's a closed subspace. All right. So then this lemma implies, this lemma on the key diagram implies that this guy, so Tg, which by definition is calibration composed of gluing, composed of parallel transport, composed of q0, parallel transport. And breaking is an approximate right inverse for Dg. Meaning that this difference, Dg, Tg minus the identity, is less than 1. In fact, it goes to 0 as g goes to 0. Right. So I mean, what happens? We fix our right inverse here, q0. And then we define our approximate right inverse by just going around. That's Tg. And this, of course, if a diagram commuted on the nose, this would be a true right inverse. This approximate commutativity gives us that it's an approximate right inverse. And now we can use the standard geometric series to invert this. Thus, we get true right inverse qg with the same image. Now, by construction, the image of Tg is contained in the kernel of lg. This basically just follows by definition. So Tg was, well, we passed through some map. We apply q0, and we land in the kernel of l0. And if we look at the definition of parallel transport glue and calibration, they're somehow only happening on the necks. And these evaluation points, at least for sufficiently small g, are not in the necks. So this map, this vertical composition, commutes with the l. So this is our conclusion. Right now, an index calculation shows that this is an isomorphism. So one can calculate the index of dg, and it will be the same as d0. So one concludes that this is an equality. Calculation. I guess really sort of combined with this. So really, we also get a kernel pre-gluing map, a kernel gluing isomorphism from kernel d0 to the kernel of dg, just given by simply applying these maps and then taking the projection onto the kernel induced by this particular choice of write inverse. So this is just projection off of kernel lg. So now finally, we can define the gluing map, which gives us what we want. So the bounded inverse qg determines a local diffeomorphism between the kernel of dg and the inverse image of 0. So this is basically the implicit function theorem. OK, implicitly, I'm also, of course, using, I mean, crucially in here goes the estimate on the norm of fg of 0 and the boundedness of this write inverse and the quadratic estimates. So the picture is like this. We have kernel dg, the image of qg. This is the complement. And there's some manifold here, which is fg inverse of 0. And the map here is just projection off of image qg. So this is the map. And this gives a local diffeomorphism projection. So moreover, this is valid in a ball of some fixed size independent of g. And that's where the boundedness is important. So combining these, so combining, we get for different values of g, we get this map from r e cross kernel of d0 to m bar x. Using our chosen isomorphism of the kernels. So now, to claim this map is one continuous to injective, meaning its restriction to some neighborhood of 0 is injective. And surjective meaning, so surjective means that, for germ, means that for every neighborhood of 0 in the source, there exists a neighborhood of x0 in the target, which is covered by this map restricted to the neighborhood in the source. OK, so these three properties taken together imply that it's a local homeomorphism, since the source is locally compact, how storage space. So what is the content here? Surjectivity is the content here is using these key up-priority estimates of, hopefully, Sasuke-Zender about necks, which are C0 close to trivial cylinders. Injectivity essentially follows from the definition of this stuff I was talking about gauge fixing at the beginning. Continuity, the content is that, well, somehow, this image of QG varies continuously. And what I mean by that is just the image of QG is given by the kernel of this map, LG, which, however you define various continuously, it certainly should be satisfied by this evaluation of points map. The content here basically is that we have this map from kernel DG to kernel. We have this kernel gluing map from kernel D0 to kernel DG, and this image is the same. And image QG equals kernel LG. So this finishes the proof. Thanks for coming. Thank you.
 Ah  Okay, so today I think we'll talk about gluing.  So that is what I want to do is show how to describe the local structure of the modularized  space near curves which are split, curves which are broken at ray-bore bits.  So let me begin by just recalling the setup.  So we have our target y, we have our contact form, and we have our almost complex structure  on r cross y and satisfies the following three properties.  So it should be r invariant, j of d by ds is equal to the ray vector field and j sends  xc to itself and restricted to xc is compatible with d lambda.  So then we denote by p the set of closed orbits of our lambda, unparameterized.  And now, right, so gluing is simply a statement about one fixed moduli space.  So I may as well just fix once and for all gamma plus, gamma minus, and the homology  class of curves that were incidentally.  So b is the curve, all i lambda generated closed orbits?  Ah, yes, I'll know the general.  So we fix this data and then the moduli space m bar is, so that's the set of curves like  this.  Sorry, t minus is a set of n.  Oh, gamma minus.  And it's a set.  Actually, this is gamma plus and these gamma minus.  And it's, so this is, each of these is a holomorphic curve mapping to r cross y and it's considered  up to translation.  Each of these individual pieces is considered up to translation in r direction and in homology  plus beta.  So from now on, I'll just forget about gamma plus, gamma minus, and beta.  That is, I'll leave them out of the notation because we're just interested in this fixed  moduli space.  And then denote by m bar reg of the locus where the linearized operator is subjective.  So then the theorem is that, so first, this is an open set and near any point in it.  There exists a local holomorphism from rk cross r to the n minus k to m bar reg by sending  0, sending 0 to the base point.  And where n is the virtual dimension of m bar, k is number of nodes at this point p.  And it's compatible with the stratification.  That is, the left hand side is stratified by which of these first k coordinates are 0  and the right hand side is stratified by which places did we resolve the node.  And it's compatible with that stratification.  And the compatibility with stratifications is, okay, it comes sort of trivially from the  construction, but of course it's crucial to include in the actual statement because the  homomorphism type of this only depends on whether k is 0 or positive.  Stratification is important data to keep track of.  Yeah?  So when you say no, you mean you get a little bit of it?  Yeah.  Yeah.  You see them as models.  Right.  So the first remark is that in the actual, like to establish contact homology, we need  something slightly stronger than this.  What we need is, actually, need to establish the same statement for these thickened moduli  spaces, m bar, set i.  The addition of these sort of thickening terms in this moduli space, though, doesn't really  change the argument.  And so for sake of simplicity, I just want to explain the proof of this.  And moreover, not only do we have to deal with these thickened guys, we also have to  prove this submersion property.  So we have to basically show that this m bar reg inside, m bar j reg, is cut out transversially.  Like this.  We need this to be a need to be a topological submersion.  OK.  So I'll let me not say more about this right now.  That's exactly the submersion axiom in the implicit Atlas definition which I gave earlier.  The point is just that both of these are really only slight extensions of the proof of this.  And the other remark is that we don't need to establish any sort of compatibility results,  particularly properties between these local gluing charts.  They're just local homomorphisms.  Probably if one works hard enough using gluing profiles, one can show that we get a smooth  structure on this space as a transition mass between these charts are smooth, but this  is not needed for what I want to do.  Yeah.  When you want to do the same thing for thickening the space, do you use something like seeing  these elements of thickening the space as homomorphic curves in something bigger?  When we do in-originous question and question, how is it simply that you've translated the  analysis?  You can think of it like that.  I mean, I think in the end it's the same.  You can write something as the inhomogeneous equation or you can write something as the  true homomorphic curve equation on this bigger thing.  I think when it comes down to the other things, you actually have to check are the same no  matter which perspective you use.  The closest result in the literature to this theorem is apparently, as far as I know, this  result gluing theorem of Jinx-Tobbs, I believe, in 2009 in their construction of embedded  contact tomology.  So what they do is they prove a gluing theorem for embedded curves.  Embedded isn't very important.  They do gluing for SFT curves basically in R cross a three-manifold.  In fact, their result is somehow more, much more involved in this statement because they  have this extra thing that they do.  They stick on branched covers of trivial cylinders in between the curves that they glue.  They also have this obstruction bundle gluing that happens.  So the techniques which go into this come from the same body of work that this originates  from the gauge three literature with Tobbs, Donaldson, Kronheimer, and eventually applied  to holomorphic curves by Fleur and FL cubed and Hoverer-Saskis-Ender.  So I just want to show how these techniques are applied to prove this theorem in the case  of contact tomology.  All right, so we begin with some setup.  So first we have to fix our base point, which instead of calling P, I'm not going to call  x0.  So we fix our base point x0 in m bar reg.  And what that really is is a map u0 from some curve c0 to x0 hat.  So this is just the domain of the map.  This is the target, which is a disjoint union of a bunch of copies of r cross y.  But I'm just going to call it x hat not.  Let mu denote the index, the index of this moduli space that is the virtual dimension.  T0 will denote the dual graph of c0.  So maybe I'll draw an example.  So let's even take this example.  So maybe this is c0, then x0 hat is the disjoint union of four copies of r cross y, a separate  copy for each of the components of the curve.  The map is considered only up to translation.  And then T0 is the dual graph.  It looks like this.  I'll let v of t0.  This is just the set of irreducible components of c0 and e.  This is the interior edges.  Here I really mean the interior edges, so this is the intermediate ray of orbits.  In this simpletization, r cross y is exact, so there won't be any bubbling.  But one could also add bubbles.  In that case, I'm not going to keep track of the bubbles in the tree just because we  don't need that stratification.  Yeah, yeah.  So for the purposes of this argument, it really doesn't matter the genus.  But yeah, assuming genus 0 for now.  So now our goal is to construct a map of local homeomorphism.  We'll get the polynomial from this to here, respecting stratifications and landing in  the regular locus.  So this, right, since it's a germ of homeomorphism, that means it covers an open neighborhood  of x0.  And since it lands in m bar reg, that means m bar reg contains an open neighborhood of  x0.  In particular, m bar reg is open since this is true for every x0.  So this is what we have to show.  Right, so first, set up, what we want to do is domain stabilization.  So we really want to, what's the purpose of this?  Set up, over a neighborhood of this point x0 in the moduli space, we want to sort of  fix canonical coordinates on the domain.  The problem is for this and these two components, they are by holomorphic with a cylinder on  the plane respectively, but not canonically so.  And we have to get rid of this extra freedom or do some gauge fixing in order to be able  to produce local parameters for the moduli space like this.  So what we do is we fix some points qi and c0 and our invariant divisors, di inside r  cross y such that u0 is transverse to di at qi.  So I don't care whether it's transverse to di anywhere else just at qi.  Such that this holds and c0 equipped with these points is stable.  So here's really, c0 is already equipped with, it's marking by the positive and negative  punctures.  So yeah, so we just have to stabilize these two guys.  So it looks something like this.  So this is di at qi.  So now since u0 is transverse to di at qi, this property is preserved under small deformations  of the holomorphic curve.  That is over an open neighborhood of x0, you get corresponding points qi in the domains  of all of those other holomorphic curves.  And therefore you've done the gauge fixing over in neighborhood.  So second, introduce the glued cobertism target.  So for a gluing parameter, I'll call an element of this space a gluing parameter, we define  this glued target to be obtained from x0 by truncating the ends.  Over to the ends, the ends look like 0 infinitely cross y.  You truncate them to 0, the gluing parameter cross y.  Do the same thing for a negative end and then you glue via this identification.  So as I've described, it's a little bit.  Yeah, yeah, yes, I forgot about this.  So the way I've described it, it's not entirely accurate.  So suppose we're taking this curve.  So we have x0 is like four copies of r cross y, so let me just draw them as just draw the  r factor, it looks something like this.  And suppose I glue all of the parameters using exactly this recipe.  I get this, this is like x0 hat, xg hat, then looks like, so I have this first piece, second  piece, I glue both of these here, I get something sort of non-Hausdorff like this.  Obviously, I don't want a non-Hausdorff target, but there's sort of an obvious canonical  way of just closing it back up to r cross y.  So xg again, so xg of course is again just a big distant union of copies of r cross y,  but somehow this particular description is the one we care about.  Yeah, yeah, so if we had something more general, then we'd have to, then there would be some  constraint on the space of allowable gluing parameters.  So now it's important to keep track of the zero sections here.  So for v in v, that is a very irreducible component of the original curve, we're going  to let zero, so v inside x bar zero, x hat not denote corresponding zero section.  And we let the corresponding, and we also denote by that the corresponding zero section.  It's descent.  So there's some canonical, yeah, zero cross y here, and we get some points here.  For each component of the original curve.  Okay, now finally, this is, again an important point, I want to fix sections, qv prime of  this universal curve over, well, the linear monologous space of curves of genus zero,  and equipped with points marked by the input output edges, and these extra marked points,  ui.  So I already commented earlier that by choosing these divisors di, we identify the domains  of all the holomorphic curves, all the points of the modular space of our neighborhood of  x zero with fibers of this map.  That was somehow the purpose of adding these divisors to gauge fix the target, the domain,  rather, gauge fix the domain.  What I want to do now is gauge fix the target, and gauge fixing on the target uses these  sections.  So sections qv prime, such that when you evaluate them at this, at c zero, you land, qv lands  in the corresponding irreducible component of c zero.  So there's one point, qv prime in every irreducible component of c zero.  It is on a, it is on a, it is on a.  And gauge fix it, the truth, but are you?  Yeah, yeah, yeah, yeah.  So, right, so, in, in the, this is sections defined in neighborhood of c zero.  I mean, obviously, maybe this whole vibration does not have a section.  We can only curve up locally.  Yes, and yes, so the whole point of fixing these sections, qv prime is to gauge fix the  target.  So we're going to now only look at maps which send qv prime to the corresponding zero section  zero v.  So for group maps, we gauge, we gauge fix the domain via these qi's, which are essentially  the inverse image of di, and the gauge fix the target via requiring the u of qv prime  lie in this zero section.  So really what that means is that at a point close to x zero, I get a unique gluing parameter.  So if you have a map close to x zero, your, your domain is canonically identified with,  with one of the fibers of this universal curve because you have these points qi from intersecting  with di.  And then that gives you points qv prime via this arbitrary choice of sections.  And then this requirement that qv prime be mapped to the, the zero, the zero section  tells you what the gluing parameter has to be.  Because this, this length here is exactly the inverse.  So I think this is all the gauge fixing we need to do, which is good.  Yeah, I guess you could intersect di several times, but you just follow the intersection  that p is equal to 0.  So we're going to furthermore fix  the trivialization of the universal curve over this.  Except it's not exactly the universal curve here  that I want to fix the trivialization.  It's something slightly different.  So I'm going to denote by m bar 0.  So here I'm looking at a compactified nodulite  space of curves of genus 0 with m plus n marked points.  But the first n marked points are in addition  going to be equipped with a marking of a tangent space by c.  So this is, of course, a forgetful map.  And this is a principle gl1 to the n nodulite.  So this is classifying.  Xx classifies curves c with p1, pn, q1, qm, and sigma i from c  to the tangent space of pd.  So n lower parentheses 2.  So why do I care about this curves with an extra marking  here?  So when you have an end, if you have a map,  why do we care?  So if you have a holomorphic map, say u from standard half  cylinder to r cross y asymptotic to a ray of orbit,  it means that u of s comma t is equal to the length of the ray  of orbit times s plus some constant gamma tilde of t  plus the low of 1 as s goes to infinity, where gamma prime is  the parameterization of the ray of orbit that it's asymptotic  to.  But the point here now is that the coordinates on r cross y  give rise to a marking of the tangent space of the puncture  by c, namely by requiring that gamma tilde be not rotated at all  and b be 0.  That is somehow, and this is the way  I want to think of these curves.  I want to think of c0 as equipped  with a marking of the tangent space at each of the punctures.  And with respect to those coordinates,  I want sort of b to be 0 and gamma tilde to be not rotated.  Right.  So now we fix coordinate local trivializations  of this modular space.  I'll fix local diffeomorphisms from J, so Jv.  I'll denote c2 times.  Oh, this is just the dimension of this modular space.  So for each irreducible component,  so that's for each vertex, we fix a local diffeomorphism  of this modular space, where you have the points qi  and you have a tangent space marking at each of the punctures.  So yes.  Right.  So now, OK, so I also want to fix cylindrical coordinates  on x0 and c0.  That is, so we fix coordinates like this for,  oh, these are valid for s sufficiently large near  all the punctures.  And near all punctures, such that with respect  to these coordinates, we have u0 st  is asymptotic to the length of the corresponding rate  of times s and gamma t, where s goes to infinity.  And of course, I mean that, right, the point here  is that this coordinate b is 0 and that this is not twisted.  OK, so finally, I think the last piece of setup  is we have to glue the curves, glue the domain.  So really, these gluing parameters  g, g lying in r greater than or equal to 0 to the e,  this should be thought of as the gluing parameters for the target,  not the gluing parameters for the source.  So we would glue the curves c sub g.  So we basically do exactly the same operation.  So we truncate c0 from on each end to 0 g inverse.  And we need a factor of the length of the rape orbit also.  Right, this is one reason why.  Right, because the trivial cylinder  has this factor of l in it.  So if we glue the cobertism by truncating and translating  by g inverse, we better glue the curve by g inverse times l  inverse.  Right, so it's always the and identify as before.  So the picture looks like this.  We have an end of c.  This is c0.  I'm not sure.  We have another end of c0.  We truncate the ends as we cut off here.  And then we glue them together.  So this is cg.  Now, I want to write them.  The rotational coordinate is chosen  by the asymptotic matching of the two ends,  since they're both asymptotic for the same rape orbit.  So it's crucial to fix a point qe double prime  in the center of each glued edge.  And so this qe double prime in cg  exists for every e which has a positive gluing parameter.  And this is needed for some gauge fixing.  The problem somehow is you have, right, so what happens?  So can you tell me what is this qe double prime?  So it's the point in the middle of,  in the exact middle of the neck.  Just fix such a point.  I mean, the s1 coordinate doesn't matter.  But just as long as you fix some way of defining it.  So why do we need this?  So the point is that there's a map from.  So we have these trivializations of the modular space,  each of these components.  But that somehow only tells us about curves on this stratum.  We also need curves on the other strata, too.  So what happens?  If we fix a g, we get a map from this product of j's to what?  Well, it's a product over all vertices  of the glued tree of m bar 0.  So again, we look at the qi's, which lie in the vertices which  are mapped there, plus qe double primes  in the edges which got collapsed,  plus the punctures.  And they have a double marking, not just the points,  but the tens of spaces also.  So this is a local defemorphism.  And this would not be true if we did not include the data  of qe double prime.  I mean, the dimension wouldn't even be correct.  There.  Yes, somehow, if we just do this gluing construction,  look at all possible, almost complex structures, j,  we get two extra degrees of freedom, because we don't know  what the marking of the tangent space at these ends  was for c0.  This is some sort of tricky point.  A local defemorphism.  Yeah, yeah, no, the f is this.  Oh, yeah, homeomorphism would be enough.  It's a smooth map.  I think that's all I need.  A smooth map, which is a local homeomorphism.  I think that's all that's actually needed.  And in all these constructions, do you  ever need to have the full universal curve of full  modulus in the human port space at any point?  Or do you only need those local charts of modulus spaces  of curves around?  Yeah, I only need the local part near  the particular curve we're gluing.  You never need to know anything about the lowest  of the element for an arc in the circular arc.  Yeah, I mean, gluing is a completely local statement.  So it should come out as this can work more generally  with the full story.  Oh.  Yeah, so you mean for the construction of the apples  and stuff.  Yeah, I don't need it.  It's not really used there either.  Um, yeah, I mean, there's some right there.  Like, in certain setups, you might sort of  be interested in the question of, look at the modulite space.  No, you don't look at this modulite space.  You ask, does this have a section?  This is basically saying, oh, is there  a sort of global way of choosing asymptotic markers?  I think this is not true.  I think it doesn't have a section.  But yes, for us in the setup, we don't need a section.  Did I?  Oh, I already flipped the boards.  Great, so we've now defined gauge fixed.  Everything near our point.  We've defined this glued culprits, the glued curves.  So the final thing is we have to define the glued maps.  So of course, first we do the pre-glued maps, ug.  So we fix cutoff function, chi.  chi.  And OK, so these are the two steps.  So we first define a, I'll call a flattened map, u0 bar g  from c0 to x hat 0.  So what is the picture?  So u, the map u approaches a trivial cylinder.  So the trivial cylinder here.  So this map is u0.  So this is a trivial cylinder.  Trivial.  And I'm just going to cut u, cut off u0 to the trivial cylinder.  So this is that.  Oh, and in symbols, it's this.  So we take u0 for, right?  So capital S will denote g minus 1.  So maybe I'll just write that instead.  Ls.  I'll denote that.  Now I remember why I decided to call it s.  Anyway, so this is for.  And then here we interpolate.  That is, we take an exponential trivialization,  the sort of diffeomorphism of, right?  We take exponential coordinates near this rate of orbit  and just cut off.  So this is for s in 1, 6.  6 is the same as 5 in all the  calculations.  Yeah, something like that.  Yeah, that doesn't, I mean, yeah.  Yeah, the things will happen over this neck.  And this is the first thing that happens.  And a bunch of things will happen.  The other things that happen will happen only  on the part of u0, g, which is a trivial cylinder, which is.  Sorry.  Is it a gamma 2, then?  Yeah, it's just a 2 gauge fixed, then to the gamma 2, then.  Yeah, I guess, oh, I've usually been thinking of gamma  as the unparameterized orbit.  So whenever it's something parameterized, I put a tilde.  Yeah, I do it.  Right, so then we get a, so this preglue map,  u, g from c, g to x hat, g is just the descent of u0, g.  There's a map of grease with this description of c0, g.  So now we have a map.  So this gives a map for every, so we get a map from every  gluing parameter, we get something, this map.  So I don't know what to write here.  Of course, I can't write m bar because this map isn't  holomorphic.  But what we want to do is somehow correct this map  to be holomorphic so that it lands in m bar.  And we also want to somehow add this extra set of parameters  on the left, namely r to the mu minus e,  corresponding to the kernel of the linearized operator  at this point.  So now we're going to get into some actual analysis here.  So gluing estimates.  So first, I need to tell you what weighted Sobolev norms  we're using.  So we need some weighted Sobolev spaces of functions  and maps of sections over c0.  So we're going to be looking at spaces like w, k, p sections  of the pullback of a tangent bundle of xg to cg.  So we define weighted w, k, p delta norms given in the ends  of the necks by the following.  So in an end, something like this, p.  So basically in the end, we just add a factor of e  to the delta s to the function, or multiply the function  by a factor of e to the delta s.  That is the functions here, dk, like e to the delta s.  So I'm just writing the contribution  to the pth power of the norm over this end.  And away from the ends, it's just the usual norm.  And then in the necks, something like 0 to capital S cross s1.  We, of course, do something similar.  And now the weight is p delta times the minimum of s  and capital S minus s.  So in an end, the weight looks like this, e to the delta s.  And in the neck, the weight looks like this.  It's e to the delta times the minimum of s and s minus s.  So these are somehow the norms which are relevant for this problem.  So here delta s is a small positive number.  And by small, you can say precisely what you mean by small.  So specifically, delta should be less than the smallest eigenvalue  of the asymptotic linearized operator of the Ray-Borbit  in each of the ends.  And yeah, in this Frett-Hellman analysis,  this is where we really need the ends, where we see the need  to have ends be non-degenerate.  It's one of the places.  The other place is in the a priori estimate.  A priori estimate, if you have a trivial cylinder over a non-degenerate Ray-Borbit,  then if you take a holomorphic curve which is c0 close to a trivial cylinder  over a non-degenerate Ray-Borbit, decays exponentially.  For that trivial cylinder, exponentially, meaning at least as fast as e to the minus delta s,  where for any delta less than the first positive eigenvalue.  This is the fundamental a priori estimates of Hofer-Rasovsky's ender.  So this defines for us weighted so-bluff spaces like this over each  of the glued curves, c sub g.  And it suffices to take p equals 2 for this argument.  We don't need anything else.  But of course, we could, I mean, everything I'm going to be saying  is true for any p in this range.  But there's no way I need p equal to 1 to impose p equal to 1.  Not for this.  Chris Wendell has this blog post about this, whether you need, whether at any point you  need to use this p equals 1.  Yeah.  So for this, it's really not needed.  I think the, what is the point?  Yeah, we somehow, given all the a priori estimates, which,  which sometimes require p equals, which, or maybe they don't  never require p equals 1.  At least some ways you write them, you can, p equals 1 seems necessary.  Yeah, I think the conclusion of the discussion with Chris was that maybe  gluing was the last remaining place where p equals 1 was needed.  And he said gluing doesn't need.  Yeah, I think it really doesn't need p equals 1.  Yeah, p equals 2 should be really fine.  I mean, the point is you're already, you're only dealing with curves,  which you've already proved your a priori estimates for.  I mean, we take k equal, we can take k to be large.  Yeah, I think the difference between p equals.  Yeah.  But wait, the issue with.  Oh, sorry, I wanted to ask k equals 1.  Oh, you meant k equals 1.  Do you need sometimes p to be bigger or two times k equals 1?  Yeah, yeah.  Yes, okay.  So k to k will be large.  Right, okay, so maybe I should summarize since we were saying the wrong thing.  So for this gluing argument, all of our curves are, we have,  we're sort of doing this argument after we've proved all the a priori estimates.  And we know our curves are very highly differentiable,  so we're going to just fix a large k.  And since k is large, we can take p equals 2.  If k were 1, yeah, maybe we need p larger than 2.  I have a question.  We must think of how this flow theory would be able to find for less and less  simple, like the A-symplec.  Yeah, I don't think any of these, I mean, I don't know anything about it.  Like for p-symplec, if you consider less and less smooth  conditions, like p-A-symplec manifolds.  I don't think anyone ever considered doing more of the curves in such situations.  And I guess I just, again, I'm thinking about p-A-symplec manifolds,  and I think that everything is completely different, even the local questions.  I think that's just the way it is.  So, nonlinear.  So we have this, right, so we want to understand the moduli space of holomorphic  curves near this pre-glued map.  And so what we do is any such map is given by some  exponential of some vector field and an almost complex structure on the domain.  But actually, we're not going to use all vector fields, so we have to preserve this  condition of intersecting di at qi.  So we only consider the subspace where c of qi is in the tangent space to di.  And we also only consider the subspace where this sort of the same condition at qe  double prime, 0.  So we have this point qe double prime.  And in order to gauge fix the domain again, we have to pick,  we have to specify qe double prime.  On our priori specify what qe double prime is on the domains of all the curves in the  moduli space near x0.  So what I'm going to say is that qe double prime is the inverse image,  sorry, the image of this space under the exponential map.  I'm just declared to be that.  That gives a local divisor.  The inverse image of that under the map will be qe double prime.  So we have a vector field and the most complex structure.  And from the vector field and almost complex structure,  we can then ask what is del bar and we get a section.  Sorry?  What is this?  J.  It's on the target.  On the source, on the c.  We're varying.  J was, I think I've erased it by now.  J was this set of almost complex structures on the source.  Yeah.  Yeah, the almost complex structure on the target is always fixed here.  I'll make a 0, 1.  Right, so what is this?  The map, fg.  So it takes a vector field and an almost complex structure.  I'll denote by j sub y on there.  So what do we do?  So x, u, g.  See, that's the map.  We take differential.  Take the anti-holomorphic part with respect to this jy,  almost complex structure, and j on the target.  This is a section of over cg, not of this bundle,  but rather this is in sections, w, k minus 1, p delta sections  of this part of omega 0, 1, cg with the jy almost complex  structure and with x, u, g, c pullback of this tangent bundle.  So to put it here, we use parallel transport on this part.  We use some fixed identification map.  It doesn't really matter, just some complex linear map.  So parallel transport for a j linear connection,  certainly need it.  Otherwise, a tensor product won't make sense.  So j linear connection on x hat, r cross y,  say r invariance, j linear connection just to be safe.  And then I tensor with this just fixed identification map.  It doesn't really matter what we do here, just any map which  is c linear and smooth and depends smoothly on y,  and is the identity for y equals 0.  Now we're not done with the definition because we also  have this other component of the target.  So we add, so what is this component telling us?  This tells us the vertical positions of all of the images  of all of these qv primes.  So remember, since we did this gauge fixing,  we're now interested in maps which map these qv primes  to the corresponding 0 sections.  So we take x u g c of qv prime of y.  So I write it qv prime of y because at this point  in c0 and cg really depends on this almost complex structure y.  Yeah, because this qv prime was a section of the universal curve.  So we have this, and then we just project to r.  This is an element of r cross y.  We project to r, and we do this for all of the qv primes.  So now with this complicated definition,  our modular space near x0, our modular space  is union over g in, OK, I'm writing it like this,  but of course I really just mean small, small gluing parameters,  g, sort of a germ of identification of fg inverse of 0.  Now this local, this gives a local map from m bar  to the space of gluing parameters,  but this is not at all a canonical map.  It's dependent on our choices of divisors and sections qv prime.  Right, and so I, as far as I know all of this should,  yeah, works for, right, so another comment about k equals 1  which is larger k, I should apologize for screwing up  this discussion the first time.  So, yeah, this operator, well it makes sense.  Seems to make sense for any k at least 1  and any p between 1 and infinity.  If you have k equals 1 and p 2 or smaller,  or you have k equals 2 and p equal to 1,  I guess we don't care about that,  we just want to make sure that the dimensions here  are not continuous, which is a little bit disconcerting.  But, yeah, right, so of course, yes, no, of course yes,  no, you need this, right, because we can't do this exponential.  So this only makes sense for either k at least 3,  k equals 2 and p greater than 1 or k equals 1 and p greater  than 2, right.  The differential of this map makes sense for any kp,  but in k equals 1, but, yeah.  And we actually, so for the gluing argument we're doing  right now with just the holomorphic curve equation,  we could really do it with any of these that we want to.  It's somehow the Fred-Holm analysis is a bit easier  with p equals 2.  And, but when we do the same thing for the thickened  modulized spaces, it's really important to be able  to choose large k because there's an extra term in this  part of the equation, the thickening term,  and it depends on the places where this  exponentiated curve intersects certain divisors.  And this map from the space of c to cg,  given by, take the inverse image under this  exponentiated map of a fixed divisor, this map is a  regularity class, something depending on k,  something like ck minus 2.  So if we want this to be a nice smooth map that we can  apply the inverse function theorem to and understand it's  zero set, we need k to be large.  I guess that.  Yeah, so I mean, yeah.  Great.  So now there are a few fundamental estimates which I could  spend a whole lecture doing the calculations for,  but I'm not.  I mean, just write down some things which one can  prove about this map.  So first, estimate the norm of fg of 0.  So lemma this norm fg of 0, that is,  so if this measures how far the pre-glued map is from  being holomorphic, of course using this norm,  this goes to 0 as g goes to 0.  Maybe I will sort of prove it.  OK, first one comment about the norms.  You can define norms with actual explicit formulas  including trivializations of these bundles in the ends.  But all that really matters is the commensurability class,  the commensurability class of these norms,  uniform in g.  So you could pick another norm which is commensurable  with a commensurability constant which doesn't depend on  the, that would be fine too.  The point is just to have estimates which are true for all  g.  And this is somehow why this, for any fixed g,  this norm defined by this weight is commensurable with a  norm defined by not doing any weight, weight 1.  But as g goes to 0, of course, they're not uniformly  commensurable.  OK, so the one form part, so this guy has two parts,  the one form and the zero one form and the vertical  coordinate part.  So the one form part of fg of 0 is supported in this region  if you recall the definition, right,  that this is just saying where is ug not j holomorphic?  It was a splicing together of this holomorphic guy  of trivial cylinders, so the only place it could,  it might not be j holomorphic is in this green part,  the part where we spliced.  And OK, now we just use the exponential convergence of ug to  a trivial cylinder and the fact that delta is less than this  delta gamma.  So we have a weighting function which is e to the s delta,  but this difference that we cut off decays by even faster  exponential since delta is less than delta gamma.  And then the second part for this part is just the statement  that qv prime of cg j0 approaches the descent of this qv prime  in c0.  j0, just because as g goes to 0 because these sections are  continuous.  So there's some calculation to do, but this is, yeah.  So the other thing we really, the other key part of understanding  the 0 set of fg of 0 near c comma y equals 0 is the following  quadratic estimates.  And this should be really thought of as a uniform bound  on this modulus of smoothness of fg.  It's something basically saying f is like a c3 map from this  ballast space to this ballast space.  And moreover, we're giving an effective bound on its c3 norm  independent of g.  OK, now we actually don't need it to be c3.  We need it to be, we need something considerably weaker.  And what we need is this.  So proposition for, we have, so you take the derivative  at 0 and the derivative at 0 is close to the norm of the  derivative at 0 minus the derivative at zeta is bounded  by a constant times zeta.  So it's a statement which is very similar to saying  the derivative is of uniformly elliptics.  This is 4k.  So you, at least 4c.  OK, this k equals at least 4 condition is really only needed  when you have the thickening terms in intersection  with divisors for this guy.  So this should be true for any k.  So I won't prove this here.  So now we come to the part which has somehow most,  a lot of the content.  So this is to construct a bounded right inverse.  So let dg denote this derivative of fg at 0.  fg prime is the direction of the, of the constant.  In the direction of the second argument.  Yes.  Oh, sorry.  This is a terrible notation.  Zeta and over here, zeta and c are both elements of this  direct sum.  Oh, there used to be elements.  So on the left side is on the,  yes, so this notation here and the definition of this notation  are different.  0 is an element of, is really 0, 0, c is really c, y,  zeta is, zeta, y prime.  So yes, so this, yes.  So this is the derivative of f at 0, 0, this operator.  So it's a linear map from, from the source to the target.  Then we give them by the derivative of fg at 0.  So that's dg.  So by assumption, d0 is,  this is a,  a,  we want to show that dg is also  subjective.  Okay, we actually want to show a bit more.  We want to show that we want to,  out of a right inverse for d0,  dG. And it'll have two key properties.  One is that, right, so in fact, given this right inverse for d0,  we will construct a QG with, which is a right inverse for G,  such that QG is uniformly bounded, independent of G.  And the image of QG, I'll just put this in quotation marks,  varies continuously in G. So I'll say precisely,  in what sense it varies continuously in G.  But I mean, these spaces don't vary nicely  in terms of G. So there's not a well-defined,  there's not an obvious notion of varies continuously.  But I'll say what I mean by that precisely later.  So this construction is really a hands-on construction.  Given a right inverse d0, we simply sort of write down  a right inverse for dG.  So we have the key diagram as the following.  So if you look at this, C0, CQI and P di plus j.  So this gets parallel transported with WK to delta C0.  A flattened guy, again with this,  there'll be a map which is called a gluing map, which  basically takes a trace down to CG.  And finally, will be a calibration map,  which takes this and ensures that it satisfies  this extra condition at Q E double prime, so pi Rds  plus R is 0.  So now we have all these linearized operators.  This again goes by parallel transport.  There's a map in the other direction called break.  And this is the same.  I won't write it again.  So what's written in the last article on the left?  Calibrate, calibration map.  So this is something extra which  has to happen for contact homology, which  doesn't need to happen when you're gluing just nodes  and say it grow off in theory or even  Hamiltonian for theory.  Yeah, the point is we need to ensure this condition.  OK, so let me just sketch the definitions of these maps.  So defining gluing break.  So here we have C0 and here we have CG.  So the gluing map is supposed to take a section over C0  and give a section over CG.  So what's it going to do?  Going to cut off here, cut off, and we cut off here.  So these are these other series of complicated operations.  So this is going to happen at something like 1 third s.  So in the region where we've already  made it a trivial cylinder.  And then we sum, sum of cut off.  So this is glue.  It's the sum of the, under this map of the cut off guys.  And now the map break is we pull back  and we cut off at the center.  Cut off at 1 half s.  Right, so this is.  Maybe I should.  So the way we cut off for the glue map  is not really relevant.  We just use whatever cut off function we want.  The way we cut off for the break map is crucial.  These cut off, so first of all, has to happen at the center.  And secondly, these cut off functions  have to be symmetric.  That is, if you break and then take the sum of fibers back  down, you should get the same thing.  So this cut off function is 1 minus this cut off function.  OK, so defining, finally, defining calibration map.  So here we have a picture of Cg.  And there's this vector field, which is just constant,  but cut off.  So we look at this cut off function.  So this is cut off.  Chi times ds ug.  So here's qv double prime.  And this is the vector field.  So calibrate simply adds the correct multiple of C linear  multiple of this so that this condition projection applied  to your vector field becomes 0.  Now I want to observe that both of these maps,  or all three of these maps, actually all the maps  in this diagram, are uniformly bounded.  For glue and break, you can check.  For the calibration map, it's important here  that delta is positive for this to be uniformly bounded.  And it's important that qe double prime  is in the exact center of the neck,  whereas it's the point with the maximum weight.  So there's this constant vector field here, which is  in the kernel of the d operator.  Now this constant vector field, in your usual norm,  just has constant norm 1.  But that's somehow not the right norm to think of it in.  We should think of it in the weighted norm.  And in the weighted norm, the norm  of this constant vector field looks something like this.  It reaches a maximum at, or say normalizes with a maximum  equals 1 in the center of this guy at qe double prime.  And then it decays exponentially as we  go towards the boundary of the neck.  And this is somehow why this is an OK thing to do,  cut it off here.  Because whatever xe is at qe double prime, it's very small.  It's like e to the minus 1 half capital S delta  times the norm of xe.  So this thing that we added is actually very small.  OK.  So this key diagram does not commute, but it almost commutes.  And this is basically the property  we need to use it to construct this bounded right inverse.  So key diagram almost commutes.  OK, so the statement is the following.  So first, for parallel transport,  the statement is very simple, simply the commutator of the maps.  Approaches 0 in norm.  So these are all, will be three statements,  and they're all as g goes to 0.  The second statement concerns the bottom.  This goes to 0.  And finally, the statement for the middle square  is a little bit complicated, because glue and break  go in opposite directions.  And the statement that you can is both sufficient  and can actually be proven as the following.  So this norm is little o of 1 times the norm of eta,  where break applied to eta is equal to d0gx.  OK, I should comment that lots of treatments  of gluing the two upper squares here are combined into 1.  It's very common to not consider this flattening u0 bar g.  But I think it's a good sort of intermediate part  and sort of divides your estimate into two constituent  pieces, which are sort of different.  It's also very important that all this cutting off  is happening in the region where u0g is a trivial cylinder,  which makes the estimates much, doesn't make  no all that much easier, but there's less to worry about.  So what is the quantifier of eta?  You have to solve all eta, such that break is of all  sort eta.  Yeah, for all.  Yeah, so it's, how does it work?  So c lies here, and eta lies here.  And they get mapped to the same thing here.  And the statement is that when you take eta this way,  sorry, when you take c this way, so they're mapped  to the same thing here.  And then the statement is that when you take c around this way,  it's very close.  Sort of a strange sounding statement,  but I don't know any other formulation which  is both correct and simpler to state.  I mean, you could ask that like going this way  and going this way are really close, but that's not true.  Or either it's not true or it's not enough  to show what we want, one of the two.  I forget which.  Oh, I saw, do I have time?  Oh, I saw that I mean, maybe I'll  say in words why this is true, that I mean,  one can do some explicit calculation.  For this, the point is, again, just u0 decays exponentially  to a trivial cylinder.  And these only really differ over the part  where you cut off u0 to a trivial cylinder.  And over that part, we have this estimate  that u0 approaches a trivial cylinder,  at least exponentially, with any delta strictly less  than delta gamma.  And our chosen delta is less than delta gamma,  so we have this estimate.  For the calibration one, the point  is just that this vector field dsug  is in the kernel of the linearized operator.  It just corresponds to translating the trivial cylinder  or rotating the trivial cylinder.  And we cut it off, but the point is that cutoff happened  really far away from qg.  And so we use the product rule to expand this.  This term is very small.  Again, like e to the minus something times delta.  So it's crucial that delta is positive for this.  And finally, this one.  Sorry, finally, this one.  Yeah, so maybe I'll just not say why this is true.  I mean, you can explicitly analyze what's going on.  The point is somehow that the weight, the key point  is somehow to compare weights.  On C0, you have weights, which look like this.  And on CG, you have weights, which look like this.  And what you turn out to, the error term in this difference,  turns out to be something supported here  in this type of region.  But the norm you care about is this norm.  And you have a bound on its norm in this norm.  And so the fact, the ratio between these two norms  is very small.  And so it goes to 0.  OK, so now let's just define these bounded.  Right, it versus.  So what do I mean very incontinuously in G?  This is very crucial for the gluing map to be continuous in G.  This image really has to vary continuously in G.  And this is somehow a non-trivial statement  because the space of gluing parameters  could look something like this.  And continuity for this gluing parameter  and this gluing parameter is some non-trivial statement.  So what do I mean continuous in G?  So we'll just fix some linear map from this source WK delta  to some vector space given by evaluating at some points.  So for any subspace B, such that the L0 restricted  to the kernel of D0 is an isomorphism.  So you pick some points and look at this evaluation map.  So first you pick some points.  So this evaluation map is injective on the kernel.  The kernel is fine dimensional.  So you can certainly do this.  Just pick some points on the domain, C0 such that this is,  yeah, and J is fine dimensional.  You can do this.  And then quotient out by a suitable subspace B,  so it's an isomorphism.  This, of course, gives us a corresponding map LG  on this sort of domain of FG.  Let me not write it again.  It's just this.  OK.  Now we're going to fix, or write,  x q0, which is the right inverse.  So d0 q0 is the identity.  And the image of q0 is the kernel of L0.  So since L0 restricted to the kernel of D0 is an isomorphism,  the kernel of L0 is a complement for the kernel of D0.  And given the choice of complement,  the choice of complement determines uniquely  a choice of bounded randomness.  Closed complement.  This is continuous map.  So it's a closed subspace.  All right.  So then this lemma implies, this lemma on the key diagram  implies that this guy, so Tg, which by definition  is calibration composed of gluing,  composed of parallel transport, composed of q0,  parallel transport.  And breaking is an approximate right  inverse for Dg.  Meaning that this difference, Dg, Tg minus the identity,  is less than 1.  In fact, it goes to 0 as g goes to 0.  Right.  So I mean, what happens?  We fix our right inverse here, q0.  And then we define our approximate right  inverse by just going around.  That's Tg.  And this, of course, if a diagram commuted on the nose,  this would be a true right inverse.  This approximate commutativity gives us  that it's an approximate right inverse.  And now we can use the standard geometric series  to invert this.  Thus, we get true right inverse qg with the same image.  Now, by construction, the image of Tg  is contained in the kernel of lg.  This basically just follows by definition.  So Tg was, well, we passed through some map.  We apply q0, and we land in the kernel of l0.  And if we look at the definition of parallel transport  glue and calibration, they're somehow only  happening on the necks.  And these evaluation points, at least for sufficiently small  g, are not in the necks.  So this map, this vertical composition,  commutes with the l.  So this is our conclusion.  Right now, an index calculation shows  that this is an isomorphism.  So one can calculate the index of dg,  and it will be the same as d0.  So one concludes that this is an equality.  Calculation.  I guess really sort of combined with this.  So really, we also get a kernel pre-gluing map,  a kernel gluing isomorphism from kernel d0 to the kernel  of dg, just given by simply applying these maps  and then taking the projection onto the kernel induced  by this particular choice of write inverse.  So this is just projection off of kernel lg.  So now finally, we can define the gluing map,  which gives us what we want.  So the bounded inverse qg determines  a local diffeomorphism between the kernel of dg  and the inverse image of 0.  So this is basically the implicit function theorem.  OK, implicitly, I'm also, of course,  using, I mean, crucially in here goes the estimate  on the norm of fg of 0 and the boundedness  of this write inverse and the quadratic estimates.  So the picture is like this.  We have kernel dg, the image of qg.  This is the complement.  And there's some manifold here, which is fg inverse of 0.  And the map here is just projection off of image qg.  So this is the map.  And this gives a local diffeomorphism projection.  So moreover, this is valid in a ball  of some fixed size independent of g.  And that's where the boundedness is important.  So combining these, so combining,  we get for different values of g,  we get this map from r e cross kernel of d0 to m bar x.  Using our chosen isomorphism of the kernels.  So now, to claim this map is one continuous to injective,  meaning its restriction to some neighborhood of 0  is injective.  And surjective meaning, so surjective means that,  for germ, means that for every neighborhood of 0 in the source,  there exists a neighborhood of x0 in the target, which  is covered by this map restricted  to the neighborhood in the source.  OK, so these three properties taken together  imply that it's a local homeomorphism,  since the source is locally compact, how storage space.  So what is the content here?  Surjectivity is the content here is  using these key up-priority estimates of,  hopefully, Sasuke-Zender about necks, which are C0 close  to trivial cylinders.  Injectivity essentially follows from the definition  of this stuff I was talking about gauge fixing  at the beginning.  Continuity, the content is that, well, somehow,  this image of QG varies continuously.  And what I mean by that is just the image of QG  is given by the kernel of this map, LG, which, however you  define various continuously, it certainly  should be satisfied by this evaluation of points map.  The content here basically is that we have this map from kernel  DG to kernel.  We have this kernel gluing map from kernel D0 to kernel DG,  and this image is the same.  And image QG equals kernel LG.  So this finishes the proof.  Thanks for coming.  Thank you.
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Ah Okay, so today I think we'll talk about gluing. So that is what I want to do is show how to describe the local structure of the modularized space near curves which are split, curves which are broken at ray-bore bits. So let me begin by just recalling the setup. So we have our target y, we have our contact form, and we have our almost complex structure on r cross y and satisfies the following three properties. So it should be r invariant, j of d by ds is equal to the ray vector field and j sends xc to itself and restricted to xc is compatible with d lambda. So then we denote by p the set of closed orbits of our lambda, unparameterized. And now, right, so gluing is simply a statement about one fixed moduli space. So I may as well just fix once and for all gamma plus, gamma minus, and the homology class of curves that were incidentally. So b is the curve, all i lambda generated closed orbits? Ah, yes, I'll know the general. So we fix this data and then the moduli space m bar is, so that's the set of curves like this. Sorry, t minus is a set of n. Oh, gamma minus. And it's a set. Actually, this is gamma plus and these gamma minus. And it's, so this is, each of these is a holomorphic curve mapping to r cross y and it's considered up to translation. Each of these individual pieces is considered up to translation in r direction and in homology plus beta. So from now on, I'll just forget about gamma plus, gamma minus, and beta. That is, I'll leave them out of the notation because we're just interested in this fixed moduli space. And then denote by m bar reg of the locus where the linearized operator is subjective. So then the theorem is that, so first, this is an open set and near any point in it. There exists a local holomorphism from rk cross r to the n minus k to m bar reg by sending 0, sending 0 to the base point. And where n is the virtual dimension of m bar, k is number of nodes at this point p. And it's compatible with the stratification. That is, the left hand side is stratified by which of these first k coordinates are 0 and the right hand side is stratified by which places did we resolve the node. And it's compatible with that stratification. And the compatibility with stratifications is, okay, it comes sort of trivially from the construction, but of course it's crucial to include in the actual statement because the homomorphism type of this only depends on whether k is 0 or positive. Stratification is important data to keep track of. Yeah? So when you say no, you mean you get a little bit of it? Yeah. Yeah. You see them as models. Right. So the first remark is that in the actual, like to establish contact homology, we need something slightly stronger than this. What we need is, actually, need to establish the same statement for these thickened moduli spaces, m bar, set i. The addition of these sort of thickening terms in this moduli space, though, doesn't really change the argument. And so for sake of simplicity, I just want to explain the proof of this. And moreover, not only do we have to deal with these thickened guys, we also have to prove this submersion property. So we have to basically show that this m bar reg inside, m bar j reg, is cut out transversially. Like this. We need this to be a need to be a topological submersion. OK. So I'll let me not say more about this right now. That's exactly the submersion axiom in the implicit Atlas definition which I gave earlier. The point is just that both of these are really only slight extensions of the proof of this. And the other remark is that we don't need to establish any sort of compatibility results, particularly properties between these local gluing charts. They're just local homomorphisms. Probably if one works hard enough using gluing profiles, one can show that we get a smooth structure on this space as a transition mass between these charts are smooth, but this is not needed for what I want to do. Yeah. When you want to do the same thing for thickening the space, do you use something like seeing these elements of thickening the space as homomorphic curves in something bigger? When we do in-originous question and question, how is it simply that you've translated the analysis? You can think of it like that. I mean, I think in the end it's the same. You can write something as the inhomogeneous equation or you can write something as the true homomorphic curve equation on this bigger thing. I think when it comes down to the other things, you actually have to check are the same no matter which perspective you use. The closest result in the literature to this theorem is apparently, as far as I know, this result gluing theorem of Jinx-Tobbs, I believe, in 2009 in their construction of embedded contact tomology. So what they do is they prove a gluing theorem for embedded curves. Embedded isn't very important. They do gluing for SFT curves basically in R cross a three-manifold. In fact, their result is somehow more, much more involved in this statement because they have this extra thing that they do. They stick on branched covers of trivial cylinders in between the curves that they glue. They also have this obstruction bundle gluing that happens. So the techniques which go into this come from the same body of work that this originates from the gauge three literature with Tobbs, Donaldson, Kronheimer, and eventually applied to holomorphic curves by Fleur and FL cubed and Hoverer-Saskis-Ender. So I just want to show how these techniques are applied to prove this theorem in the case of contact tomology. All right, so we begin with some setup. So first we have to fix our base point, which instead of calling P, I'm not going to call x0. So we fix our base point x0 in m bar reg. And what that really is is a map u0 from some curve c0 to x0 hat. So this is just the domain of the map. This is the target, which is a disjoint union of a bunch of copies of r cross y. But I'm just going to call it x hat not. Let mu denote the index, the index of this moduli space that is the virtual dimension. T0 will denote the dual graph of c0. So maybe I'll draw an example. So let's even take this example. So maybe this is c0, then x0 hat is the disjoint union of four copies of r cross y, a separate copy for each of the components of the curve. The map is considered only up to translation. And then T0 is the dual graph. It looks like this. I'll let v of t0. This is just the set of irreducible components of c0 and e. This is the interior edges. Here I really mean the interior edges, so this is the intermediate ray of orbits. In this simpletization, r cross y is exact, so there won't be any bubbling. But one could also add bubbles. In that case, I'm not going to keep track of the bubbles in the tree just because we don't need that stratification. Yeah, yeah. So for the purposes of this argument, it really doesn't matter the genus. But yeah, assuming genus 0 for now. So now our goal is to construct a map of local homeomorphism. We'll get the polynomial from this to here, respecting stratifications and landing in the regular locus. So this, right, since it's a germ of homeomorphism, that means it covers an open neighborhood of x0. And since it lands in m bar reg, that means m bar reg contains an open neighborhood of x0. In particular, m bar reg is open since this is true for every x0. So this is what we have to show. Right, so first, set up, what we want to do is domain stabilization. So we really want to, what's the purpose of this? Set up, over a neighborhood of this point x0 in the moduli space, we want to sort of fix canonical coordinates on the domain. The problem is for this and these two components, they are by holomorphic with a cylinder on the plane respectively, but not canonically so. And we have to get rid of this extra freedom or do some gauge fixing in order to be able to produce local parameters for the moduli space like this. So what we do is we fix some points qi and c0 and our invariant divisors, di inside r cross y such that u0 is transverse to di at qi. So I don't care whether it's transverse to di anywhere else just at qi. Such that this holds and c0 equipped with these points is stable. So here's really, c0 is already equipped with, it's marking by the positive and negative punctures. So yeah, so we just have to stabilize these two guys. So it looks something like this. So this is di at qi. So now since u0 is transverse to di at qi, this property is preserved under small deformations of the holomorphic curve. That is over an open neighborhood of x0, you get corresponding points qi in the domains of all of those other holomorphic curves. And therefore you've done the gauge fixing over in neighborhood. So second, introduce the glued cobertism target. So for a gluing parameter, I'll call an element of this space a gluing parameter, we define this glued target to be obtained from x0 by truncating the ends. Over to the ends, the ends look like 0 infinitely cross y. You truncate them to 0, the gluing parameter cross y. Do the same thing for a negative end and then you glue via this identification. So as I've described, it's a little bit. Yeah, yeah, yes, I forgot about this. So the way I've described it, it's not entirely accurate. So suppose we're taking this curve. So we have x0 is like four copies of r cross y, so let me just draw them as just draw the r factor, it looks something like this. And suppose I glue all of the parameters using exactly this recipe. I get this, this is like x0 hat, xg hat, then looks like, so I have this first piece, second piece, I glue both of these here, I get something sort of non-Hausdorff like this. Obviously, I don't want a non-Hausdorff target, but there's sort of an obvious canonical way of just closing it back up to r cross y. So xg again, so xg of course is again just a big distant union of copies of r cross y, but somehow this particular description is the one we care about. Yeah, yeah, so if we had something more general, then we'd have to, then there would be some constraint on the space of allowable gluing parameters. So now it's important to keep track of the zero sections here. So for v in v, that is a very irreducible component of the original curve, we're going to let zero, so v inside x bar zero, x hat not denote corresponding zero section. And we let the corresponding, and we also denote by that the corresponding zero section. It's descent. So there's some canonical, yeah, zero cross y here, and we get some points here. For each component of the original curve. Okay, now finally, this is, again an important point, I want to fix sections, qv prime of this universal curve over, well, the linear monologous space of curves of genus zero, and equipped with points marked by the input output edges, and these extra marked points, ui. So I already commented earlier that by choosing these divisors di, we identify the domains of all the holomorphic curves, all the points of the modular space of our neighborhood of x zero with fibers of this map. That was somehow the purpose of adding these divisors to gauge fix the target, the domain, rather, gauge fix the domain. What I want to do now is gauge fix the target, and gauge fixing on the target uses these sections. So sections qv prime, such that when you evaluate them at this, at c zero, you land, qv lands in the corresponding irreducible component of c zero. So there's one point, qv prime in every irreducible component of c zero. It is on a, it is on a, it is on a. And gauge fix it, the truth, but are you? Yeah, yeah, yeah, yeah. So, right, so, in, in the, this is sections defined in neighborhood of c zero. I mean, obviously, maybe this whole vibration does not have a section. We can only curve up locally. Yes, and yes, so the whole point of fixing these sections, qv prime is to gauge fix the target. So we're going to now only look at maps which send qv prime to the corresponding zero section zero v. So for group maps, we gauge, we gauge fix the domain via these qi's, which are essentially the inverse image of di, and the gauge fix the target via requiring the u of qv prime lie in this zero section. So really what that means is that at a point close to x zero, I get a unique gluing parameter. So if you have a map close to x zero, your, your domain is canonically identified with, with one of the fibers of this universal curve because you have these points qi from intersecting with di. And then that gives you points qv prime via this arbitrary choice of sections. And then this requirement that qv prime be mapped to the, the zero, the zero section tells you what the gluing parameter has to be. Because this, this length here is exactly the inverse. So I think this is all the gauge fixing we need to do, which is good. Yeah, I guess you could intersect di several times, but you just follow the intersection that p is equal to 0. So we're going to furthermore fix the trivialization of the universal curve over this. Except it's not exactly the universal curve here that I want to fix the trivialization. It's something slightly different. So I'm going to denote by m bar 0. So here I'm looking at a compactified nodulite space of curves of genus 0 with m plus n marked points. But the first n marked points are in addition going to be equipped with a marking of a tangent space by c. So this is, of course, a forgetful map. And this is a principle gl1 to the n nodulite. So this is classifying. Xx classifies curves c with p1, pn, q1, qm, and sigma i from c to the tangent space of pd. So n lower parentheses 2. So why do I care about this curves with an extra marking here? So when you have an end, if you have a map, why do we care? So if you have a holomorphic map, say u from standard half cylinder to r cross y asymptotic to a ray of orbit, it means that u of s comma t is equal to the length of the ray of orbit times s plus some constant gamma tilde of t plus the low of 1 as s goes to infinity, where gamma prime is the parameterization of the ray of orbit that it's asymptotic to. But the point here now is that the coordinates on r cross y give rise to a marking of the tangent space of the puncture by c, namely by requiring that gamma tilde be not rotated at all and b be 0. That is somehow, and this is the way I want to think of these curves. I want to think of c0 as equipped with a marking of the tangent space at each of the punctures. And with respect to those coordinates, I want sort of b to be 0 and gamma tilde to be not rotated. Right. So now we fix coordinate local trivializations of this modular space. I'll fix local diffeomorphisms from J, so Jv. I'll denote c2 times. Oh, this is just the dimension of this modular space. So for each irreducible component, so that's for each vertex, we fix a local diffeomorphism of this modular space, where you have the points qi and you have a tangent space marking at each of the punctures. So yes. Right. So now, OK, so I also want to fix cylindrical coordinates on x0 and c0. That is, so we fix coordinates like this for, oh, these are valid for s sufficiently large near all the punctures. And near all punctures, such that with respect to these coordinates, we have u0 st is asymptotic to the length of the corresponding rate of times s and gamma t, where s goes to infinity. And of course, I mean that, right, the point here is that this coordinate b is 0 and that this is not twisted. OK, so finally, I think the last piece of setup is we have to glue the curves, glue the domain. So really, these gluing parameters g, g lying in r greater than or equal to 0 to the e, this should be thought of as the gluing parameters for the target, not the gluing parameters for the source. So we would glue the curves c sub g. So we basically do exactly the same operation. So we truncate c0 from on each end to 0 g inverse. And we need a factor of the length of the rape orbit also. Right, this is one reason why. Right, because the trivial cylinder has this factor of l in it. So if we glue the cobertism by truncating and translating by g inverse, we better glue the curve by g inverse times l inverse. Right, so it's always the and identify as before. So the picture looks like this. We have an end of c. This is c0. I'm not sure. We have another end of c0. We truncate the ends as we cut off here. And then we glue them together. So this is cg. Now, I want to write them. The rotational coordinate is chosen by the asymptotic matching of the two ends, since they're both asymptotic for the same rape orbit. So it's crucial to fix a point qe double prime in the center of each glued edge. And so this qe double prime in cg exists for every e which has a positive gluing parameter. And this is needed for some gauge fixing. The problem somehow is you have, right, so what happens? So can you tell me what is this qe double prime? So it's the point in the middle of, in the exact middle of the neck. Just fix such a point. I mean, the s1 coordinate doesn't matter. But just as long as you fix some way of defining it. So why do we need this? So the point is that there's a map from. So we have these trivializations of the modular space, each of these components. But that somehow only tells us about curves on this stratum. We also need curves on the other strata, too. So what happens? If we fix a g, we get a map from this product of j's to what? Well, it's a product over all vertices of the glued tree of m bar 0. So again, we look at the qi's, which lie in the vertices which are mapped there, plus qe double primes in the edges which got collapsed, plus the punctures. And they have a double marking, not just the points, but the tens of spaces also. So this is a local defemorphism. And this would not be true if we did not include the data of qe double prime. I mean, the dimension wouldn't even be correct. There. Yes, somehow, if we just do this gluing construction, look at all possible, almost complex structures, j, we get two extra degrees of freedom, because we don't know what the marking of the tangent space at these ends was for c0. This is some sort of tricky point. A local defemorphism. Yeah, yeah, no, the f is this. Oh, yeah, homeomorphism would be enough. It's a smooth map. I think that's all I need. A smooth map, which is a local homeomorphism. I think that's all that's actually needed. And in all these constructions, do you ever need to have the full universal curve of full modulus in the human port space at any point? Or do you only need those local charts of modulus spaces of curves around? Yeah, I only need the local part near the particular curve we're gluing. You never need to know anything about the lowest of the element for an arc in the circular arc. Yeah, I mean, gluing is a completely local statement. So it should come out as this can work more generally with the full story. Oh. Yeah, so you mean for the construction of the apples and stuff. Yeah, I don't need it. It's not really used there either. Um, yeah, I mean, there's some right there. Like, in certain setups, you might sort of be interested in the question of, look at the modulite space. No, you don't look at this modulite space. You ask, does this have a section? This is basically saying, oh, is there a sort of global way of choosing asymptotic markers? I think this is not true. I think it doesn't have a section. But yes, for us in the setup, we don't need a section. Did I? Oh, I already flipped the boards. Great, so we've now defined gauge fixed. Everything near our point. We've defined this glued culprits, the glued curves. So the final thing is we have to define the glued maps. So of course, first we do the pre-glued maps, ug. So we fix cutoff function, chi. chi. And OK, so these are the two steps. So we first define a, I'll call a flattened map, u0 bar g from c0 to x hat 0. So what is the picture? So u, the map u approaches a trivial cylinder. So the trivial cylinder here. So this map is u0. So this is a trivial cylinder. Trivial. And I'm just going to cut u, cut off u0 to the trivial cylinder. So this is that. Oh, and in symbols, it's this. So we take u0 for, right? So capital S will denote g minus 1. So maybe I'll just write that instead. Ls. I'll denote that. Now I remember why I decided to call it s. Anyway, so this is for. And then here we interpolate. That is, we take an exponential trivialization, the sort of diffeomorphism of, right? We take exponential coordinates near this rate of orbit and just cut off. So this is for s in 1, 6. 6 is the same as 5 in all the calculations. Yeah, something like that. Yeah, that doesn't, I mean, yeah. Yeah, the things will happen over this neck. And this is the first thing that happens. And a bunch of things will happen. The other things that happen will happen only on the part of u0, g, which is a trivial cylinder, which is. Sorry. Is it a gamma 2, then? Yeah, it's just a 2 gauge fixed, then to the gamma 2, then. Yeah, I guess, oh, I've usually been thinking of gamma as the unparameterized orbit. So whenever it's something parameterized, I put a tilde. Yeah, I do it. Right, so then we get a, so this preglue map, u, g from c, g to x hat, g is just the descent of u0, g. There's a map of grease with this description of c0, g. So now we have a map. So this gives a map for every, so we get a map from every gluing parameter, we get something, this map. So I don't know what to write here. Of course, I can't write m bar because this map isn't holomorphic. But what we want to do is somehow correct this map to be holomorphic so that it lands in m bar. And we also want to somehow add this extra set of parameters on the left, namely r to the mu minus e, corresponding to the kernel of the linearized operator at this point. So now we're going to get into some actual analysis here. So gluing estimates. So first, I need to tell you what weighted Sobolev norms we're using. So we need some weighted Sobolev spaces of functions and maps of sections over c0. So we're going to be looking at spaces like w, k, p sections of the pullback of a tangent bundle of xg to cg. So we define weighted w, k, p delta norms given in the ends of the necks by the following. So in an end, something like this, p. So basically in the end, we just add a factor of e to the delta s to the function, or multiply the function by a factor of e to the delta s. That is the functions here, dk, like e to the delta s. So I'm just writing the contribution to the pth power of the norm over this end. And away from the ends, it's just the usual norm. And then in the necks, something like 0 to capital S cross s1. We, of course, do something similar. And now the weight is p delta times the minimum of s and capital S minus s. So in an end, the weight looks like this, e to the delta s. And in the neck, the weight looks like this. It's e to the delta times the minimum of s and s minus s. So these are somehow the norms which are relevant for this problem. So here delta s is a small positive number. And by small, you can say precisely what you mean by small. So specifically, delta should be less than the smallest eigenvalue of the asymptotic linearized operator of the Ray-Borbit in each of the ends. And yeah, in this Frett-Hellman analysis, this is where we really need the ends, where we see the need to have ends be non-degenerate. It's one of the places. The other place is in the a priori estimate. A priori estimate, if you have a trivial cylinder over a non-degenerate Ray-Borbit, then if you take a holomorphic curve which is c0 close to a trivial cylinder over a non-degenerate Ray-Borbit, decays exponentially. For that trivial cylinder, exponentially, meaning at least as fast as e to the minus delta s, where for any delta less than the first positive eigenvalue. This is the fundamental a priori estimates of Hofer-Rasovsky's ender. So this defines for us weighted so-bluff spaces like this over each of the glued curves, c sub g. And it suffices to take p equals 2 for this argument. We don't need anything else. But of course, we could, I mean, everything I'm going to be saying is true for any p in this range. But there's no way I need p equal to 1 to impose p equal to 1. Not for this. Chris Wendell has this blog post about this, whether you need, whether at any point you need to use this p equals 1. Yeah. So for this, it's really not needed. I think the, what is the point? Yeah, we somehow, given all the a priori estimates, which, which sometimes require p equals, which, or maybe they don't never require p equals 1. At least some ways you write them, you can, p equals 1 seems necessary. Yeah, I think the conclusion of the discussion with Chris was that maybe gluing was the last remaining place where p equals 1 was needed. And he said gluing doesn't need. Yeah, I think it really doesn't need p equals 1. Yeah, p equals 2 should be really fine. I mean, the point is you're already, you're only dealing with curves, which you've already proved your a priori estimates for. I mean, we take k equal, we can take k to be large. Yeah, I think the difference between p equals. Yeah. But wait, the issue with. Oh, sorry, I wanted to ask k equals 1. Oh, you meant k equals 1. Do you need sometimes p to be bigger or two times k equals 1? Yeah, yeah. Yes, okay. So k to k will be large. Right, okay, so maybe I should summarize since we were saying the wrong thing. So for this gluing argument, all of our curves are, we have, we're sort of doing this argument after we've proved all the a priori estimates. And we know our curves are very highly differentiable, so we're going to just fix a large k. And since k is large, we can take p equals 2. If k were 1, yeah, maybe we need p larger than 2. I have a question. We must think of how this flow theory would be able to find for less and less simple, like the A-symplec. Yeah, I don't think any of these, I mean, I don't know anything about it. Like for p-symplec, if you consider less and less smooth conditions, like p-A-symplec manifolds. I don't think anyone ever considered doing more of the curves in such situations. And I guess I just, again, I'm thinking about p-A-symplec manifolds, and I think that everything is completely different, even the local questions. I think that's just the way it is. So, nonlinear. So we have this, right, so we want to understand the moduli space of holomorphic curves near this pre-glued map. And so what we do is any such map is given by some exponential of some vector field and an almost complex structure on the domain. But actually, we're not going to use all vector fields, so we have to preserve this condition of intersecting di at qi. So we only consider the subspace where c of qi is in the tangent space to di. And we also only consider the subspace where this sort of the same condition at qe double prime, 0. So we have this point qe double prime. And in order to gauge fix the domain again, we have to pick, we have to specify qe double prime. On our priori specify what qe double prime is on the domains of all the curves in the moduli space near x0. So what I'm going to say is that qe double prime is the inverse image, sorry, the image of this space under the exponential map. I'm just declared to be that. That gives a local divisor. The inverse image of that under the map will be qe double prime. So we have a vector field and the most complex structure. And from the vector field and almost complex structure, we can then ask what is del bar and we get a section. Sorry? What is this? J. It's on the target. On the source, on the c. We're varying. J was, I think I've erased it by now. J was this set of almost complex structures on the source. Yeah. Yeah, the almost complex structure on the target is always fixed here. I'll make a 0, 1. Right, so what is this? The map, fg. So it takes a vector field and an almost complex structure. I'll denote by j sub y on there. So what do we do? So x, u, g. See, that's the map. We take differential. Take the anti-holomorphic part with respect to this jy, almost complex structure, and j on the target. This is a section of over cg, not of this bundle, but rather this is in sections, w, k minus 1, p delta sections of this part of omega 0, 1, cg with the jy almost complex structure and with x, u, g, c pullback of this tangent bundle. So to put it here, we use parallel transport on this part. We use some fixed identification map. It doesn't really matter, just some complex linear map. So parallel transport for a j linear connection, certainly need it. Otherwise, a tensor product won't make sense. So j linear connection on x hat, r cross y, say r invariance, j linear connection just to be safe. And then I tensor with this just fixed identification map. It doesn't really matter what we do here, just any map which is c linear and smooth and depends smoothly on y, and is the identity for y equals 0. Now we're not done with the definition because we also have this other component of the target. So we add, so what is this component telling us? This tells us the vertical positions of all of the images of all of these qv primes. So remember, since we did this gauge fixing, we're now interested in maps which map these qv primes to the corresponding 0 sections. So we take x u g c of qv prime of y. So I write it qv prime of y because at this point in c0 and cg really depends on this almost complex structure y. Yeah, because this qv prime was a section of the universal curve. So we have this, and then we just project to r. This is an element of r cross y. We project to r, and we do this for all of the qv primes. So now with this complicated definition, our modular space near x0, our modular space is union over g in, OK, I'm writing it like this, but of course I really just mean small, small gluing parameters, g, sort of a germ of identification of fg inverse of 0. Now this local, this gives a local map from m bar to the space of gluing parameters, but this is not at all a canonical map. It's dependent on our choices of divisors and sections qv prime. Right, and so I, as far as I know all of this should, yeah, works for, right, so another comment about k equals 1 which is larger k, I should apologize for screwing up this discussion the first time. So, yeah, this operator, well it makes sense. Seems to make sense for any k at least 1 and any p between 1 and infinity. If you have k equals 1 and p 2 or smaller, or you have k equals 2 and p equal to 1, I guess we don't care about that, we just want to make sure that the dimensions here are not continuous, which is a little bit disconcerting. But, yeah, right, so of course, yes, no, of course yes, no, you need this, right, because we can't do this exponential. So this only makes sense for either k at least 3, k equals 2 and p greater than 1 or k equals 1 and p greater than 2, right. The differential of this map makes sense for any kp, but in k equals 1, but, yeah. And we actually, so for the gluing argument we're doing right now with just the holomorphic curve equation, we could really do it with any of these that we want to. It's somehow the Fred-Holm analysis is a bit easier with p equals 2. And, but when we do the same thing for the thickened modulized spaces, it's really important to be able to choose large k because there's an extra term in this part of the equation, the thickening term, and it depends on the places where this exponentiated curve intersects certain divisors. And this map from the space of c to cg, given by, take the inverse image under this exponentiated map of a fixed divisor, this map is a regularity class, something depending on k, something like ck minus 2. So if we want this to be a nice smooth map that we can apply the inverse function theorem to and understand it's zero set, we need k to be large. I guess that. Yeah, so I mean, yeah. Great. So now there are a few fundamental estimates which I could spend a whole lecture doing the calculations for, but I'm not. I mean, just write down some things which one can prove about this map. So first, estimate the norm of fg of 0. So lemma this norm fg of 0, that is, so if this measures how far the pre-glued map is from being holomorphic, of course using this norm, this goes to 0 as g goes to 0. Maybe I will sort of prove it. OK, first one comment about the norms. You can define norms with actual explicit formulas including trivializations of these bundles in the ends. But all that really matters is the commensurability class, the commensurability class of these norms, uniform in g. So you could pick another norm which is commensurable with a commensurability constant which doesn't depend on the, that would be fine too. The point is just to have estimates which are true for all g. And this is somehow why this, for any fixed g, this norm defined by this weight is commensurable with a norm defined by not doing any weight, weight 1. But as g goes to 0, of course, they're not uniformly commensurable. OK, so the one form part, so this guy has two parts, the one form and the zero one form and the vertical coordinate part. So the one form part of fg of 0 is supported in this region if you recall the definition, right, that this is just saying where is ug not j holomorphic? It was a splicing together of this holomorphic guy of trivial cylinders, so the only place it could, it might not be j holomorphic is in this green part, the part where we spliced. And OK, now we just use the exponential convergence of ug to a trivial cylinder and the fact that delta is less than this delta gamma. So we have a weighting function which is e to the s delta, but this difference that we cut off decays by even faster exponential since delta is less than delta gamma. And then the second part for this part is just the statement that qv prime of cg j0 approaches the descent of this qv prime in c0. j0, just because as g goes to 0 because these sections are continuous. So there's some calculation to do, but this is, yeah. So the other thing we really, the other key part of understanding the 0 set of fg of 0 near c comma y equals 0 is the following quadratic estimates. And this should be really thought of as a uniform bound on this modulus of smoothness of fg. It's something basically saying f is like a c3 map from this ballast space to this ballast space. And moreover, we're giving an effective bound on its c3 norm independent of g. OK, now we actually don't need it to be c3. We need it to be, we need something considerably weaker. And what we need is this. So proposition for, we have, so you take the derivative at 0 and the derivative at 0 is close to the norm of the derivative at 0 minus the derivative at zeta is bounded by a constant times zeta. So it's a statement which is very similar to saying the derivative is of uniformly elliptics. This is 4k. So you, at least 4c. OK, this k equals at least 4 condition is really only needed when you have the thickening terms in intersection with divisors for this guy. So this should be true for any k. So I won't prove this here. So now we come to the part which has somehow most, a lot of the content. So this is to construct a bounded right inverse. So let dg denote this derivative of fg at 0. fg prime is the direction of the, of the constant. In the direction of the second argument. Yes. Oh, sorry. This is a terrible notation. Zeta and over here, zeta and c are both elements of this direct sum. Oh, there used to be elements. So on the left side is on the, yes, so this notation here and the definition of this notation are different. 0 is an element of, is really 0, 0, c is really c, y, zeta is, zeta, y prime. So yes, so this, yes. So this is the derivative of f at 0, 0, this operator. So it's a linear map from, from the source to the target. Then we give them by the derivative of fg at 0. So that's dg. So by assumption, d0 is, this is a, a, we want to show that dg is also subjective. Okay, we actually want to show a bit more. We want to show that we want to, out of a right inverse for d0, dG. And it'll have two key properties. One is that, right, so in fact, given this right inverse for d0, we will construct a QG with, which is a right inverse for G, such that QG is uniformly bounded, independent of G. And the image of QG, I'll just put this in quotation marks, varies continuously in G. So I'll say precisely, in what sense it varies continuously in G. But I mean, these spaces don't vary nicely in terms of G. So there's not a well-defined, there's not an obvious notion of varies continuously. But I'll say what I mean by that precisely later. So this construction is really a hands-on construction. Given a right inverse d0, we simply sort of write down a right inverse for dG. So we have the key diagram as the following. So if you look at this, C0, CQI and P di plus j. So this gets parallel transported with WK to delta C0. A flattened guy, again with this, there'll be a map which is called a gluing map, which basically takes a trace down to CG. And finally, will be a calibration map, which takes this and ensures that it satisfies this extra condition at Q E double prime, so pi Rds plus R is 0. So now we have all these linearized operators.. This again goes by parallel transport. There's a map in the other direction called break. And this is the same. I won't write it again. So what's written in the last article on the left? Calibrate, calibration map. So this is something extra which has to happen for contact homology, which doesn't need to happen when you're gluing just nodes and say it grow off in theory or even Hamiltonian for theory. Yeah, the point is we need to ensure this condition. OK, so let me just sketch the definitions of these maps. So defining gluing break. So here we have C0 and here we have CG. So the gluing map is supposed to take a section over C0 and give a section over CG. So what's it going to do? Going to cut off here, cut off, and we cut off here. So these are these other series of complicated operations. So this is going to happen at something like 1 third s. So in the region where we've already made it a trivial cylinder. And then we sum, sum of cut off. So this is glue. It's the sum of the, under this map of the cut off guys. And now the map break is we pull back and we cut off at the center. Cut off at 1 half s. Right, so this is. Maybe I should. So the way we cut off for the glue map is not really relevant. We just use whatever cut off function we want. The way we cut off for the break map is crucial. These cut off, so first of all, has to happen at the center. And secondly, these cut off functions have to be symmetric. That is, if you break and then take the sum of fibers back down, you should get the same thing. So this cut off function is 1 minus this cut off function. OK, so defining, finally, defining calibration map. So here we have a picture of Cg. And there's this vector field, which is just constant, but cut off. So we look at this cut off function. So this is cut off. Chi times ds ug. So here's qv double prime. And this is the vector field. So calibrate simply adds the correct multiple of C linear multiple of this so that this condition projection applied to your vector field becomes 0. Now I want to observe that both of these maps, or all three of these maps, actually all the maps in this diagram, are uniformly bounded. For glue and break, you can check. For the calibration map, it's important here that delta is positive for this to be uniformly bounded. And it's important that qe double prime is in the exact center of the neck, whereas it's the point with the maximum weight. So there's this constant vector field here, which is in the kernel of the d operator. Now this constant vector field, in your usual norm, just has constant norm 1. But that's somehow not the right norm to think of it in. We should think of it in the weighted norm. And in the weighted norm, the norm of this constant vector field looks something like this. It reaches a maximum at, or say normalizes with a maximum equals 1 in the center of this guy at qe double prime. And then it decays exponentially as we go towards the boundary of the neck. And this is somehow why this is an OK thing to do, cut it off here. Because whatever xe is at qe double prime, it's very small. It's like e to the minus 1 half capital S delta times the norm of xe. So this thing that we added is actually very small. OK. So this key diagram does not commute, but it almost commutes. And this is basically the property we need to use it to construct this bounded right inverse. So key diagram almost commutes. OK, so the statement is the following. So first, for parallel transport, the statement is very simple, simply the commutator of the maps. Approaches 0 in norm. So these are all, will be three statements, and they're all as g goes to 0. The second statement concerns the bottom. This goes to 0. And finally, the statement for the middle square is a little bit complicated, because glue and break go in opposite directions. And the statement that you can is both sufficient and can actually be proven as the following. So this norm is little o of 1 times the norm of eta, where break applied to eta is equal to d0gx. OK, I should comment that lots of treatments of gluing the two upper squares here are combined into 1. It's very common to not consider this flattening u0 bar g. But I think it's a good sort of intermediate part and sort of divides your estimate into two constituent pieces, which are sort of different. It's also very important that all this cutting off is happening in the region where u0g is a trivial cylinder, which makes the estimates much, doesn't make no all that much easier, but there's less to worry about. So what is the quantifier of eta? You have to solve all eta, such that break is of all sort eta. Yeah, for all. Yeah, so it's, how does it work? So c lies here, and eta lies here. And they get mapped to the same thing here. And the statement is that when you take eta this way, sorry, when you take c this way, so they're mapped to the same thing here. And then the statement is that when you take c around this way, it's very close. Sort of a strange sounding statement, but I don't know any other formulation which is both correct and simpler to state. I mean, you could ask that like going this way and going this way are really close, but that's not true. Or either it's not true or it's not enough to show what we want, one of the two. I forget which. Oh, I saw, do I have time? Oh, I saw that I mean, maybe I'll say in words why this is true, that I mean, one can do some explicit calculation. For this, the point is, again, just u0 decays exponentially to a trivial cylinder. And these only really differ over the part where you cut off u0 to a trivial cylinder. And over that part, we have this estimate that u0 approaches a trivial cylinder, at least exponentially, with any delta strictly less than delta gamma. And our chosen delta is less than delta gamma, so we have this estimate. For the calibration one, the point is just that this vector field dsug is in the kernel of the linearized operator. It just corresponds to translating the trivial cylinder or rotating the trivial cylinder. And we cut it off, but the point is that cutoff happened really far away from qg. And so we use the product rule to expand this. This term is very small. Again, like e to the minus something times delta. So it's crucial that delta is positive for this. And finally, this one. Sorry, finally, this one. Yeah, so maybe I'll just not say why this is true. I mean, you can explicitly analyze what's going on. The point is somehow that the weight, the key point is somehow to compare weights. On C0, you have weights, which look like this. And on CG, you have weights, which look like this. And what you turn out to, the error term in this difference, turns out to be something supported here in this type of region. But the norm you care about is this norm. And you have a bound on its norm in this norm. And so the fact, the ratio between these two norms is very small. And so it goes to 0. OK, so now let's just define these bounded. Right, it versus. So what do I mean very incontinuously in G? This is very crucial for the gluing map to be continuous in G. This image really has to vary continuously in G. And this is somehow a non-trivial statement because the space of gluing parameters could look something like this. And continuity for this gluing parameter and this gluing parameter is some non-trivial statement. So what do I mean continuous in G? So we'll just fix some linear map from this source WK delta to some vector space given by evaluating at some points. So for any subspace B, such that the L0 restricted to the kernel of D0 is an isomorphism. So you pick some points and look at this evaluation map. So first you pick some points. So this evaluation map is injective on the kernel. The kernel is fine dimensional. So you can certainly do this. Just pick some points on the domain, C0 such that this is, yeah, and J is fine dimensional. You can do this. And then quotient out by a suitable subspace B, so it's an isomorphism. This, of course, gives us a corresponding map LG on this sort of domain of FG. Let me not write it again. It's just this. OK. Now we're going to fix, or write, x q0, which is the right inverse. So d0 q0 is the identity. And the image of q0 is the kernel of L0. So since L0 restricted to the kernel of D0 is an isomorphism, the kernel of L0 is a complement for the kernel of D0. And given the choice of complement, the choice of complement determines uniquely a choice of bounded randomness. Closed complement. This is continuous map. So it's a closed subspace. All right. So then this lemma implies, this lemma on the key diagram implies that this guy, so Tg, which by definition is calibration composed of gluing, composed of parallel transport, composed of q0, parallel transport. And breaking is an approximate right inverse for Dg. Meaning that this difference, Dg, Tg minus the identity, is less than 1. In fact, it goes to 0 as g goes to 0. Right. So I mean, what happens? We fix our right inverse here, q0. And then we define our approximate right inverse by just going around. That's Tg. And this, of course, if a diagram commuted on the nose, this would be a true right inverse. This approximate commutativity gives us that it's an approximate right inverse. And now we can use the standard geometric series to invert this. Thus, we get true right inverse qg with the same image. Now, by construction, the image of Tg is contained in the kernel of lg. This basically just follows by definition. So Tg was, well, we passed through some map. We apply q0, and we land in the kernel of l0. And if we look at the definition of parallel transport glue and calibration, they're somehow only happening on the necks. And these evaluation points, at least for sufficiently small g, are not in the necks. So this map, this vertical composition, commutes with the l. So this is our conclusion. Right now, an index calculation shows that this is an isomorphism. So one can calculate the index of dg, and it will be the same as d0. So one concludes that this is an equality. Calculation. I guess really sort of combined with this. So really, we also get a kernel pre-gluing map, a kernel gluing isomorphism from kernel d0 to the kernel of dg, just given by simply applying these maps and then taking the projection onto the kernel induced by this particular choice of write inverse. So this is just projection off of kernel lg. So now finally, we can define the gluing map, which gives us what we want. So the bounded inverse qg determines a local diffeomorphism between the kernel of dg and the inverse image of 0. So this is basically the implicit function theorem. OK, implicitly, I'm also, of course, using, I mean, crucially in here goes the estimate on the norm of fg of 0 and the boundedness of this write inverse and the quadratic estimates. So the picture is like this. We have kernel dg, the image of qg. This is the complement. And there's some manifold here, which is fg inverse of 0. And the map here is just projection off of image qg. So this is the map. And this gives a local diffeomorphism projection. So moreover, this is valid in a ball of some fixed size independent of g. And that's where the boundedness is important. So combining these, so combining, we get for different values of g, we get this map from r e cross kernel of d0 to m bar x. Using our chosen isomorphism of the kernels. So now, to claim this map is one continuous to injective, meaning its restriction to some neighborhood of 0 is injective. And surjective meaning, so surjective means that, for germ, means that for every neighborhood of 0 in the source, there exists a neighborhood of x0 in the target, which is covered by this map restricted to the neighborhood in the source. OK, so these three properties taken together imply that it's a local homeomorphism, since the source is locally compact, how storage space. So what is the content here? Surjectivity is the content here is using these key up-priority estimates of, hopefully, Sasuke-Zender about necks, which are C0 close to trivial cylinders. Injectivity essentially follows from the definition of this stuff I was talking about gauge fixing at the beginning. Continuity, the content is that, well, somehow, this image of QG varies continuously. And what I mean by that is just the image of QG is given by the kernel of this map, LG, which, however you define various continuously, it certainly should be satisfied by this evaluation of points map. The content here basically is that we have this map from kernel DG to kernel. We have this kernel gluing map from kernel D0 to kernel DG, and this image is the same. And image QG equals kernel LG. So this finishes the proof. Thanks for coming. Thank you.
data and then the moduli space m bar is, so actual, like to establish contact homology, we statement for these thickened moduli spaces, m in their construction of embedded contact right, so we begin with some setup. So first we the index of this moduli space that is the coordinates on the domain. The problem is for target to be obtained from x0 by truncating the about the lowest of the element for an arc in the fundamental a priori estimates of curves are, we have, we're sort of doing this this goes to 0 as g goes to 0. Maybe I will sort want to show that we want to, out of a right