 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 moduli 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, almost complex structure on r cross y. Satisfies the following three properties, so it should be our invariant j of d by ds is equal to the rave vector field and j sends c to itself and restricted to c is compatible with the lambda. So then we denote by p the set of closed orbits of our lambda, unparameterized. 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 considered. So we fix this data and then the moduli space and bar is, what's the set of curves like this? Oh, gamma minus and it's, yeah, it's a set, right, so this is gamma plus and these, and it's, okay, so this is, each of these is a homomorphic curve mapping to r cross y and it's considered up to translation, each of these individual pieces is considered up to translation in our direction and in homology class beta. So I'll, 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 surjective. So then the theorem is that, so first this is an open set and near any point in it, there exists a local homomorphism rk cross r to the n minus k to m bar reg by sending zero, sending zero to the base point and where n is the virtual dimension of m bar, k is number of nodes of this, at this point p and, and, and it's compatible with the stratification, that is the, the left hand side is stratified by which of these first k coordinates are zero and the, the right hand side is stratified by which, at which places did we resolve the node and it's compatible with that stratification and the compatibility with stratifications is, okay, comes sort of trivially from the construction but of course it's, it's crucial to include in the actual statement because the, the homomorphism type of this only depends on whether k is zero or positive. Stratification is important data to keep track of, yeah. But when you say node, you, you mean later we get favorable bits? Yeah. Do I can, do you see them as nodes in the, right, so the first remark is that in the actually, in like to, to establish contact with the technology we need something slightly stronger this, stronger than this. What we need is, actually need to establish the same statement for, for these thickened modulized spaces m bar sub i. The, the addition of these sort of thickening terms in this modulized space though doesn't really change the argument and so for sake of simplicity I just want to explain the difference of this. And moreover we, we, not only do we have to deal with these thickened guys, we also have to prove this submersion property. We have to basically show that this m bar reg inside m bar j reg is cut out transversally like this. We need this to be a, need to be a topological submersion. Okay, so I'll, let me not say more about this right now. It's, I mean, this is exactly the submersion axiom in the implicit atlas definition which I, which I gave earlier. The point is just that, oh, both of these are, are really only slight extensions of the proof of this. And, oh, and the other remark is that we don't need to establish any sort of compatibility results. Compatibility properties between these local gluing charts. They're just local homomorphisms. Probably if, if one works hard enough using gluing profiles, one, one can, one can show that we, we get a smooth structure on this space as a transition between these charts are smooth but, but this is not, not needed for, for what I want to do. Yeah. When, when you want to do the same thing for thickened material space, do you use something like seeing this elements of the thickened material space as homomorphic curves with something bigger like in the, when we do inhomogeneous question. How is it simply that you translate the analysis? You can think of it like that. I, I, I mean, I think in the end it's the same. I mean, you can, you can write something as the inhomogeneous, inhomogeneous equation or you can write something as the true homomorphic curve equation on this bigger thing. And I think the thing, when it comes down to the, the things you actually have to, have to check are the same no matter which perspective you use. So, right, so the, the closest result in the literature to this theorem is apparently as far as I know this result gluing theorem of James Tubbs I believe in 2009 in their construction of embedded contact technology. So, what they do is they, they prove a gluing theorem for embedded curves. Embedded isn't, isn't very important. They, they, they do gluing for SFT curves basically in R cross a three manifold. And in fact their, their result is somehow more, much more involved in this statement because they have this, this extra thing that they do. They, they stick on branched covers of trivial cylinders in between the curves that, that they glue. And they also have this obstruction bundle gluing that happens. So the, I mean, somehow the techniques which go into this are basically the, come from the same body of work that, that this lies in this, or sort of originates from the gauge three literature with Tubbs, Donaldson, Cronheimer. And eventually with, sort of, applied to holomorphic curves by Fleur and FL cubed and Hofer and Saski's under. So I just want to give and show how these techniques are applied to prove this theorem in the case of contact homology. All right, so begin with some setup. So, so first we have to fix our base point, which instead of calling P, I'm not going to call X zero. So we fix our base point X zero in M bar reg. And what that really is, is a map U zero from some curve C zero to X zero hat. So, so this, this is just the domain of the map. This is, this is the target, which is a bunch of, a disjoint union of a bunch of copies of, of R cross Y. But I'm just going to call it X hat, mu denote the index, the index of this modular space, that is the, the virtual dimension. T zero will denote the dual graph, C zero. So, maybe I'll draw an example. So let's even take this example. All right, so maybe this is C zero. Then X, X zero hat is the disjoint union of four copies of R cross Y, a separate copy for each of the components of the curve. And the map is considered only up to translation. And then T zero is the dual graph. So it looks like this. And I'll let V, V of T zero, this is just the set of irreducible components of C zero and E, this is the interior edges. Oh, I really, I really, here I really mean the interior edges. So this is the intermediate ray of orbits. Have you been in your last lecture? So you can auto-bubblings, yeah, in some way? Yeah, so in the simpletization, R cross Y is exact, so there won't be any bubbling. But, but one could, one could also add, add bubbles. In that case, I'm not going to keep track of the bubbles in the, in the tree. Just because, just because we don't need that stratification to, in the, in the, Yeah, yeah, yeah, so I mean, for the purposes of this argument, it really doesn't matter the genus, but, but yeah, assuming genus zero for now. Right, so, so now our, our goal is, is to construct a map, a local homeomorphism, germ of, a meal from here, respecting stratifications and landing the regular locus. So this, right, since it's a germ of homeomorphism, that means it covers an open neighborhood of X zero, and since it lands in M bar reg, that means M bar reg contains an open neighborhood of X zero. In particular, M bar reg is open, since this is true for every X zero. So this is what we have to show. So, so first step we want to do is domain stabilization. So, we really want to, what's the purpose of this step? Over a neighborhood of this point X zero in the moduli space, we want to sort of fix canonical coordinates on the domain. And the problem is, for this, and this, these, these two components, they are, they are by holomorphic with the cylinder and the plane respectively, but, but not canonically so. And we have to get rid of this, this, this extra freedom, or do some gauge fixing in order to, to be able to produce local, 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, I don't care whether it's transverse to di anywhere else just at, at qi. Okay, such that this holds and c0 equipped with these points is stable. Okay, so here's really, c0 is already equipped with, it's marking by like the positive and negative punctures. So, yeah, so we just have to stabilize these two guys. So it looks something like this, this is di and qi. So now, since u0 is transverse to di at qi, this property of is, is, is preserved under small deformations of the holomorphic curve. That is over an open neighborhood of x0, you get corresponding points qi in, in the domain of all of those other holomorphic curves. And therefore, you've done this gauge fixing over a neighborhood. So, second, we introduce the glued cobertism as a target. Okay, so, so for a gluing parameter, I'll call an element of this space a gluing parameter, we define this glued target to be obtained at x0 by truncating, truncating the ends, all the way to the ends. The ends look like zero, infinitely, cross y. You truncate them to zero, the gluing parameter, cross y. You do the same thing for a negative end, and then you glue, you glue via this identification. So, so x, as I've described, it's a little bit. But you're taking this because you want to keep it close to zero. 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, so 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. Okay, but 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, so xg again, so xg of course is again just a big disjoint union of copies of r cross y. But somehow this particular description is the one we care about. Because you don't have trouble. Yeah, yeah, so if we had something more general, then we'd have to, then there'd be some constraint on the space of allowable glue and parameters. Okay, 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, its descent to x. Right, 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, now finally, this is, again an important point. I want to fix sections qv prime of this universal curve over the linear monolid space of curves of genus zero and equipped with points marked by the input output edges and these extra marked points, qi. Okay, 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 monolid 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. So it's been run around the circle. It's not one, yeah. Sorry. Ah, this qv prime can be pointed on a circle, yeah. A little set of vertical coordinates. It is on A. I am. I am. It just defeats it. It defeats the curve by R, yeah? Yeah. Yeah, six, yeah. Yeah, six, yeah. Right, so. Right. Okay. This is sections defined in the neighborhood of C zero. I mean, obviously, maybe this whole vibration doesn't have a section. We only care about locally. Yes. And yeah, 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 glued maps, the 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 domain is canonically identified 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 zero section tells you what the gluing parameter has to be because this length here is exactly the g inverse. Oh, 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. Would you just follow the intersection from that used to be... Yeah, yeah, exactly. Right, 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. I'm going to denote by m bar zero. So here I'm looking at a compactified nodular space of curves of genus zero with m plus n marked points. But the first n marked points are in addition going to be equipped with a marking of the tangent space by c. So this is, of course, a forgetful map, and this is a principle gl1 to the n. It's classifying. This classifies curves c with p1, pn, q1, qm, and sigma i from c to the tangent space of pi. Is this star one? Yeah. Which is written in n subset what? So n lower parentheses two. Yeah, so why do I care about this curves with an extra marking here? So when you have an end, so if you have a map, so 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, 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 one as s goes to infinity, where gamma prime is a 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 sort of requiring that gamma tilde be sort of not rotated at all and b be zero. That is somehow, and this is the way I want to think of these curves. And I want to think of c zero 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 zero and gamma tilde to be not rotated. So now we fix coordinate local trivialization of this modular space. So I fix local diffeomorphisms from J, so Jv, we'll denote c two times 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 now, okay, so I also want to fix the coordinates on x zero and c zero, that is, so we fix coordinates like this for, okay, these are valid for s sufficiently large. Near all the punctures, and near all punctures, such that with respect to these coordinates, we have u zero s t is asymptotic to the length of the corresponding rate of orbit 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 zero and that this is not twisted. Okay, so finally, I think the last piece of setup is we have to glue the curves, glue the domain. So you leave these gluing parameters g, g lying in r greater than or equal to zero to the e. This should be thought of as the gluing parameters for the target, not the gluing parameters for the source, c sub g. Right, so we basically do exactly the same operation, so we truncate c zero from on each end to zero g inverse and we need a factor of the length of the rate of 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. Right, so it's always the, and identify. So the picture looks like this, we have an end of c, this is c zero, this is a puncture. We have another end of c zero, we truncate the ends as we cut off here and then we glue them together. So this is c g. Now, I want to, right, the rotational coordinate is chosen by the, you have an asymptotic matching of these, of the two ends, since they're both asymptotic to the same rate of orbit. So the crucial thing to do is to fix a point q e double prime in the center of each glued edge. So this q e double prime in c g exists for every e which has a positive gluing perimeter. And this is needed for some gauge fixing. The problem somehow is you have, right, so what happens? Sir, can you tell me what is this q e double prime, so the point you fixed? So it's the point in the middle of, in the exact middle of the neck here. Just fix, I mean, 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 families of almost, these trivializations of the modular space of each of these components. But that somehow only tells us about curves on this stratum. We also need curves on the other stratum, 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 zero. So again, we look at the q i's, which lie in the vertices which mapped there, plus q e double primes in the edges which got collapsed, plus the punctures. And they have a double marking, not just the points, just the tensor spaces also. So this is a local defeat morphism. And this would not be true if we did not include the data of q e double prime. I mean, the dimension wouldn't even be correct. Somehow, if we just do this, if we just do this gluing construction and look at all possible almost complex structures, j, we get two extra degrees of freedom. Well, basically, 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 defeat morphism. Yeah, yeah, no. 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. In all these constructions, do you ever need to have the full universal curve of full model space at any point? Or do you only need those local charts of model spaces of curves? Yeah, I only need the local part near the particular curve we're gluing. Do you ever need to know anything about the global subject of the model curve? Yeah, I mean, gluing is a completely local statement. Do you ever need to know anything about the full story of... Yeah, so you mean for the construction of the Atlas and stuff? Yeah. Yeah, I don't need it. It's not really used there either. Yeah, I mean, there's some... In certain set ups, you might sort of be interested in the question of... Look at this moduli space. No, you don't look at this moduli space. You ask, does this have a section? This is basically saying, oh, is there a sort of global way of choosing asymptotic markers? This is... I think this is not true. I think it doesn't have a section. But yes, for us in this setup, we don't need a section. Okay, so we've now defined... We've gauged fixed everything near our point. We've defined these glued cobertisms, 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, and... Okay, 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 the trivial cylinder. So the trivial cylinder here. So this map is U0. This is the trivial cylinder. And I'm just going to cut U... cut off U0 to the trivial cylinder. So this is... oh, and in symbols it's this. So we take U0 for... right, so capital S will denote... 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. So this is for... And then here we interpolate. That is, we take an exponential trivialization... the sort of diffeomorphism of... we take exponential coordinates near this great orbit, and just cut off. So this is for S in 1, 6th. Yeah, something like that. Yeah, that doesn't... I mean, 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 the trivial cylinder, which is... Is this a gamma 2, then? Yeah, it's just a two-gauge set of gamma 2, then it's gamma 2. Yeah, I guess... oh, I... I've usually been thinking of gamma as the unparameterized orbit, so whenever it's something parameterized, I put a tilde. Right, so then we get a... So this pre-glued map UG from CG to X hat G is just the descent of U0 G. The map agrees 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... So I don't know what to write here. It's... 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. Going to get into some actual analysis here. 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 WKP sections of the pullback of a tangent bundle of XG to CG. So we define weighted WKP delta norms given in the ends of the necks by the following, in an end something like this. 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. So I'm just writing the contribution to the pth power of the norm over this end. 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. This is e to the delta times the minimum of s and s1. So these are somehow the norms which are relevant for this problem. So here delta s is a small positive number. By small, you can say precisely what you mean by small. So specifically, delta should be less than the smallest eigenvalue of the asymptotically linearized operator of the rib orbit in each of the ends. And yeah, in this Fredhelm 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. If you have a trivial cylinder over a non-degenerate rib orbit, then if you take a C0, another holomorphic curve, which is C0 close to a trivial cylinder over a non-degenerate rib orbit, decays exponentially toward that trivial cylinder. Exponentially, meaning at least as fast as e to the minus delta s were for any delta less than the first positive eigenvalue. This is the fundamental. A priori estimates of Hofer-Rossovsky's Ender. So this defines for us weighted so-bluff spaces like this. Over each of the glued curves C sub g. And OK, it suffices to take p equals 2 for this argument. We don't need anything else. But of course, we could, I mean, be saying it's true for any p in this range. There's no way you need p equal 1 to impose p equal 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 sometimes require p equals, or maybe they don't never require p equals 1. At least some ways you write them, you can, p equals 1 seems necessary. I think the conclusion of the discussion with Chris was that maybe gluing was the last remaining place where people's 1 was needed. She said gluing 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 cursor, which you've already proved your a priori estimates for. And I mean, we take k equal, we can take k to be large. Yeah, I think the difference between p equals, yeah. But wait, no, the issue was- Oh, sorry, I wanted to ask k equals 1. Oh, you meant k equals 1. Yeah, yeah. Yeah, so 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're sort of doing this argument after we've proved all the a priori estimates. And we know our curves are very highly differentiable, 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 the whole, the flow theory, whatever, and find for less and less symmetric manifold. Like, p is symmetric. Yeah, this is- Yeah, I don't think anybody's- I mean, I don't know anything about- Like for p is symmetric, you know? If you consider less and less smooth symmetric, condition of symmetric structure, like p is symmetric manifold. Ah, p. Is it possible to define such a thing? I'm thinking I might never consider doing more morphic curves in such situations today. And I guess I guess I'm thinking about p is symmetric manifold, but it seems that everything is completely related, even the local questions. Yeah, I think we should talk to that. Okay, 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. Okay, 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. So, we have this point qe double prime. In order to gauge fix the domain again, we have to pick- we have to specify qe double prime. On a priori specify what qe double prime is on the domains of all the curves in the 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 can just declare it to be that. That gives a local divisor. The inverse image of that under the map will be qe double prime. Okay, so we have a vector field and an almost complex structure. And from the vector field and almost complex structure, we can then ask what is del bar and we get a section j on the source, on c. We're varying- j was- I think I've erased it by now. Yeah, j was this set of almost complex structures on the source. Yeah. Yeah, the almost complex structure on the target is always fixed here. So, I'll make a zero one. Right, so what is this? 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 c, that's the map. We take differential, take the anti-holomorphic part with respect to this jy almost complex structure in j on the target. There's a section of over cg, not of this bundle, but rather this is in sort of sections, w k minus one p delta sections of this part of omega zero one 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 parallel transport on this part. Use some fixed identification map. I mean, 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. That is r cross y. Say r invariance, j-linear connection just to be safe. And then I tensor with this just fixed identification map doesn't really matter. What we do here is just some, any map which is c-linear and smooth and depends smoothly on y and is the identity for y equals zero. 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 zero 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 moduli space near x0, our moduli space is the union over g in, okay, I'm writing it like this but of course I really just mean small, small gluing parameters g. So, a germ of identification of fg inverse of zero. Now, this local, this gives like a local map from m bar to the space of gluing parameters but this is not at all a canonical map. It depended 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 one versus larger k. I should apologize for screwing up this discussion the first time. So, yeah, this operator, well, it makes sense. It makes sense for any k at least one and any p between one and infinity. If you have k equals one and p two or smaller or you have k equals two and p equal to one, I guess we don't care about that. Then sections here are not continuous which is a little bit disconcerting. But, yeah, right, so of course, yes, no, of course, yes, no, you need, right, because we can't do this exponential. Yeah, so this only makes sense for either k at least three, k equal two and p greater than one or k equals one and p greater than two. The differential of this map makes sense for any kp and k equal at least one. 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 two. But when we do the same thing for the thickened modulite 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 two. 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 mean, so now there are a few fundamental estimates, which I could spend a whole lecture doing the calculations for, but I'm not, so let me just write down some things which one can prove about this map. So first, estimate norm of fg of zero. So lemma this norm fg of zero, that is, so if this measures how far the pre-glued map is from being holomorphic, of course, using this norm, this goes to zero as g goes to zero. Maybe I will sort of prove it. First one, comment about the norms. I mean, 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 g, and that would be fine too. The point is just to have estimates which are true for all g. And this is somehow why this, or for any fixed g, this norm defined by this weight is commensurable with a norm defined by not doing any weight, like one, but as g goes to zero, of course, they're not uniformly commensurable. So the one form part, so this guy has two parts, the one form and the zero one form and the vertical coordinate part. Also the one form part of fg of zero is supported in this region. If you recall the definition, 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 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 j zero approaches the descent of this qv prime in c zero. j zero, just as g goes to zero because these sections, so there's some calculation to do with this. So the other thing we really, the other key part of understanding the zero set of fg of zero near c comma y equals zero is the following quadratic estimates and what this should be really thought of as a uniform bound on this modulus of smoothness of fg. It's something basically saying f is in, say f is like a c three map from this ballax space to this ballax space and moreover we're giving an effective bound on its c three norm independent of g. Okay, now we actually don't need it to be c three. We need something considerably weaker. And what we need is this. So proposition for, so you take the derivative and the derivative at zero is close to the norm of the derivative at zero minus the derivative at zeta is bounded by constant times zeta. So it's some statement which is very similar to what I'm saying with the derivative is of uniformly elliptics. This is four k. So you, at least four c. Okay, this k equals at least four condition is really only needed when you have the thickening terms intersection with divisors for, for this guy it's, you know, this should be true for any k. Okay, so I won't, I won't prove this here. So now we come to the part which has somehow most, a lot, a lot of the content. So this is to construct a bounded right inverse. So, so let dg denote this derivative of fg at zero. The prime is the derivative in the direction of the, of the process. In the direction of the second argument. Oh, sorry. This is a terrible notation. Zeta and, over here, zeta and c are both elements of this direct cell. There used to be elements. So on the left side is only a... Yeah, so this notation here and the definition of this notation are different. Zero is an element of, is really zero comma zero, c is really c comma y, zeta is, zeta comma y prime. Right, so yeah, so, so this, yes. So this is the derivative of f at zero comma zero, this operator. So it's a linear map from, from the source to the target and we give them by the derivative of fg at zero. So that's dg. So by, by assumption, d zero is surjective. So we want to show that dg is also, okay, we actually want to show a bit more. We want to show that we want to out of a right inverse for d zero, we're going to construct the right inverse for dg. And, and it'll, it'll have two key properties. One is that given this right inverse for d zero, we will construct a q, 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 it's, these spaces don't like vary nicely in terms of g. So it's not really, there's not a well defined, there's not an obvious notion continuously, but I'll say, I need to say what I mean by that precisely later. So, so this construction is, is really a hands on construction, given a, given a right inverse d zero, we simply sort of write down a right inverse for, for, for dg. So the, we have the key diagram as the following. We get this c zero, c, qi, and qi. So this gets parallel transported with wk to delta c zero and a flattened guy again with this fluid map which basically takes a trace down to cg, and finally will be a calibration map which takes this ensures that it satisfies this extra condition at q e double prime. So pi rds plus r. So now we have all these linearized operators. This again goes by parallel transport. There's a map in the other direction called break. This is the same. I won't write it again. Calibrate. Calibration. Calibration. So this is the, 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. Okay, so let me just sketch the definitions of these maps. So defining glue 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? We're 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 one-third s. So in the region where we've already made it a trivial cylinder. And then we sum. This is glue. The sum of the under this map of the cut off guy. And now the map break is we pull back and we cut off at the center. Cut off at. 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 it has to happen at the center. And secondly these cut off functions have to be symmetric. That is when you, if you break and then sort of take the sum of fibers back down you should get the same thing. So this cut off function is one minus this cut off function. 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 dsug. So here's qe double prime. And this is the vector field. So calibrate simply adds the correct multiple c linear multiple of this so that this condition projection applied to your vector field becomes zero. 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. It's the point with the maximum weight. So there's constant vector field here which is in the kernel of d operator. Now this constant vector field you have the 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 of this constant vector field looks something like this. It reaches a maximum at, or say normalizes so the maximum equals 1 in the center of this guy at q double prime. And then it decays exponentially as we go towards the boundary of the neck. And this is somehow why this is an okay thing to do, cut it off here. Because whatever c is at qe double prime it's very small. It's like e to the minus one half capital s delta times the norm of c. So this thing that we added is actually very small. Oh, 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 commutes. Okay, so the statement is the following. So first for parallel transport the statement is very simple, simply the commutator of the maps approaches zero in norm. So these are all would be three statements and they're all as g goes to zero. The second statement concerns the bottom this goes to zero. 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 one times the norm of eta where break applied to eta is equal to d zero g c. Okay, I should comment that lots of treatments of gluing this sort of the two upper squares here are sort of combined into one. It's very common to not consider this flattening u zero 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 coming off is happening in the region where u zero g is a trivial cylinder which makes the estimates much easier. It doesn't make it all that much easier but there's less to worry about. So what is the quantifier of eta? For all eta, for all 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 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 in the other formulation which is both correct and simpler to state. You could ask that going this way and going this way are really close but that's not true. Either it's not true or it's not enough to show what we want. One of the two, I forget which. Okay, so maybe I'll just say in words why this is true that 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 puts to the trivial cylinder at least exponentially with any delta 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 okay we cut it off and 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. Finally this one sorry finally this one yeah so maybe I'll just not say why this is true I've explicitly analyzed 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 sort of 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 that the ratio between these two norms is very small and so it goes to 0 as g goes to 0 okay so now let's just define these bounded right it versus so what do I mean continuously in G this is very crucial for the gluing map to be continuous in G this image really has to very continuously in G and this is somehow a non-trivial statement because the space of gluing parameters could look something like this and continuity sort of for this gluing parameter and this gluing parameter 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 like 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 right so first you pick some points so this evaluation map is injective on the kernel the kernel is finite dimensional so you can certainly do this just pick some points on the domain C0 such that this is yeah and J is finite dimensional you can do this and then quotient out by a suitable subspace B so it's an isomorphism so this of course gives us a corresponding map on this sort of domain let me not write it again it's just this now we're going to fix fix Q0 which is the right inverse so 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 bound trainers closed complement this is a continuous map so it's a closed subspace right so then this this lemma implies lemma on the key diagram implies that this guy so T G which by definition is calibration composed of gluing composed of parallel transport to zero parallel transport and breaking is an approximate right inverse for DG meaning that this difference DG minus the identity is less than one in fact it goes to zero as G goes to zero what happens we fix our right inverse here Q0 and then we define our approximate right inverse by just going around that's T of course if a diagram commuted on the nose this would be a true right inverse this approximate commutativity gives us that's an approximate right inverse and now we can use the standard geometric series and 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 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 this is our conclusion right now an index calculation so one can calculate the index of DG and it will be the same as D0 so and so one concludes that this is an equality calculation I guess really sort of combined with this right so we 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 right inverse so this is just projection off of kernel LG okay so now finally we can define the gluing map which gives us what we want so the bounded right inverse QG determines a local diffeomorphism between the kernel of DG and the inverse image of zero so this is basically the implicit function theorem okay implicitly I'm also of course using I mean crucially in here goes the estimate on the norm of FG of zero and the boundedness of this right inverse and the quadratic estimates so the picture is like this kernel DG the image of QG this is the complement and there's some manifold here which is FG inverse of zero and the map here is just projection off of image QG so this is the map and this gives a local diffeomorphism projection so and 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 for combining we get combining for different values of G we get this map from RE of D0 to M bar using our chosen isomorphism of the kernels okay so now we claim this map is one continuous to injective meaning it's restriction to some neighborhood of zero is and surjective meaning so surjective means for germ means that for every neighborhood of zero 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 okay so these three properties taken together implies that it's a local homeomorphism since the source locally compact so what is the content here surjectivity is the content here is using these key a priori estimates of Hoferasowski-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 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 yeah right we have this kernel gluing map from kernel D0 to kernel DG and this and the image is the same QG equals kernel so this finishes the proof thanks for coming