 At the end last time, I left myself the task of proving a theorem that you can achieve transversality for holomorphic curves and symplectic cubordisms for generic J. Let me repeat the statement quickly. So the setting is the completion of a symplectic cubordism which has stable Hamiltonian structures at the positive and negative ends. And I fixed some open subset, which I said is pre-compact source, an open and bounded subset. And let me introduce some new notation slightly better than what I did last time. Let's say J fix is just some choice of almost complex structures, some choice of almost complex structure in this special class I defined, compatible with the symplectic form and with the stable Hamiltonian structures. And now the statement of theorem two says that there exists, I guess I'll say the word commieger since everyone else is, a commieger subset J u reg sitting in the space of all almost complex structures compatible with the data which also match our choice J fix outside of this subset u such that all, let's say for all J in this special class, all holomorphic curves in the moduli space defined by J with injective points in the subset u are regular. So somebody asked the question last time whether I really need to say that the curve is not a multiple cover or somewhere injective or just that there exists an injective point in the perturbation domain u. The answer is all those things are equivalent in this situation but I'm stating it this way because in the proof that's precisely what we need. We need the existence of an injective point that gets mapped into the perturbation domain. So in other situations where those conditions are not all equivalent, still this is what you need. All right, so I've got a few choices of the technicalities when I go about something like this. I first need to decide in what space exactly of almost complex structures do I want to consider perturbations? And well, there's a natural space in the picture which is the wrong one because what I'm really interested of course is this space of almost complex structures that I assume are all smooth with the C infinity topology on it and you can call that a fresh A manifold at least because I'm saying M when I mean W hat, apologies. So W hat of course is not compact and it's not so easy to define either Banach norms or even semi-norms on a non-compact domain but of course I'm only allowing these perturbations to be non-trivial outside of this compact subset or inside this compact subset. So it's easy enough to define semi-norms and I can call this object here a fresh A manifold but fresh A manifolds are not very useful to me because the implicit function theorem doesn't hold, the Sartz-Mill theorem doesn't hold. There's a whole lot of analytical results I need which I can't use. So really I need some Banach manifold. What do you mean by withinjective points in U? So injective point is right, you understand that part. So I mean an injective point whose image is contained in the subset U. That's the assumption. I mean I'm not saying all curves have that property. I'm saying I'm only considering curves that have that property. So when you apply this theorem of course it's going to miss some curves. The typical way you'd apply it is you'd let U be the compact part of the co-boardism not including the cylindrical ends and then it doesn't say anything about curves that are contained entirely in the cylindrical ends but we'll have another theorem for that in a little bit. So I have to make a choice of some Banach manifold of almost complex structures. And Duse explained one possible choice yesterday which is instead of looking at smooth almost complex structures ask them to be of class CK for some finite K. That's a good choice but I'm a little bit allergic to it myself just because if I then write down the nonlinear Cauchy-Riemann operator for a J which is not smooth then the operator on the so the section of the Banach space bundle will also not be smooth. It will have finitely many derivatives. Sartz's male theorem requires my maps to have at least some number of derivatives. I'd have to look it up to find out how many. And these are details I don't really want to worry about if I have a choice. I'd rather have everything in the picture be smooth except for of course the maps themselves are going to be of sub level class. But I'd like J to be smooth so that my nonlinear Fredholm section is smooth. So I'm going to adopt Fleur's solution to that problem which is to define the following object. So proof. Let's fix another almost complex structure that I would call the reference J. J ref that's in this space as well and matching J fix outside of you. And I'm just going to consider for the moment almost complex structures that are perturbations of J ref in some precise sense. So the other piece of data I need to fix is a set of positive numbers. I'll talk to you collectively by epsilon. It's a sequence and the sequence is converging to zero. And then define J epsilon to be the space of all J's of the form X along J ref of some Y. So I'm taking X, imagine X to be the exponential map on the manifold of complex structures on each tangent space. That's something you can define however you like. It is a manifold so that's fine. Why? I can consider to be in the tangent space to that fresh a manifold and also vanishing outside of you. So literally this just means it's a smooth section of the bundle whose fibers are the tangent spaces to the manifolds of complex structures on the tangent spaces. And I'm saying a smooth section but I'm also going to require that it's finite in this so-called Fleur C epsilon norm that's designed as the sum from zero to infinity is epsilon m times the Cm norm. So I want that to be finite. In fact I'm going to require that it's small for some constant C. So there's an exercise one needs to do and it's a standard lemma that you can find in various places originally in some of Fleur's original papers. That if you choose this sequence of epsilons converging to zero fast enough then this norm not only defines a Banach space but it defines one which is large enough to contain smooth bump functions supported and arbitrarily small neighborhoods of points. That's the key thing we need to know about this. So that defines a Banach space of y's and just exponentiating that from j ref we get a Banach manifold of j's. It's trivially a Banach manifold. I've just defined one chart on this Banach manifold so it's a small one and this is in some sense not any kind of global construction. I cannot get anywhere near a dense set of j's with this but I can get j's at least arbitrarily close to j ref and j ref to start with was arbitrary. So now the universal moduli space m star of j epsilon is going to be the set of pairs u and j where j belongs to the j epsilon space and u is a holomorphic curve for j and also I wanted to satisfy the extra conditions that u is somewhere injective or namely u has an injective point mapped to the perturbation domain u. So that's traditionally called universal moduli space. I think given the way we're doing it it's not a terribly good name because it is not universal in the sense of being able to consider all possible j's that we might be interested in. We're only looking at j's that are in some neighborhood of j ref and in fact it's it's not just a smooth neighborhood it's a neighborhood of j ref in a very peculiar topology on the space of smooth functions. At the end of the day I'm just going to care about the fact that I can get arbitrarily see infinity close to j ref by objects in this space. What is this j, why is roughly a complex structure, what's a tangent space to complex structure? So literally it's a section of the endomorphism bundle which anticommutes with j. And how do you suggest we do this? Pick a Riemannian metric and measure it. No, not integrate, just the sup norm. So important detail there is of course the thing is only nontrivial in this compact subset. Any other questions? So if this is a different topology on the space of functions for all epsilon's? It's going to be yes. And how am I supposed to think about this topology? Am I not supposed to think about this topology? Yeah, that's a difficult question. I also would not say that I really know how to think about this topology. The main things we need to know are again the space is large enough to contain bump functions of small support. That's a that's a non-obvious lemma but it has a pretty quick proof due to Fleur. Otherwise you just need to know that this space includes continuously into C infinity with the ordinary C infinity topology. Yeah, just to make sure I understand it. So if I weren't so thinking about only considering C k j's and I could just work with C k j's and the usual. You can work with C k j's is also a personally fine way to do this. Yes? I have just a question with the notation. So like about the epsilon, it means that you have like the limit of epsilon's like k going to infinity is zero. Do we also need to bound the j's being proved in j epsilon to ensure omega compatibility? Or is that taken care of by? Yeah, so omega compatibility is is implicit here when I said that y belongs to the tangent space of this space. Okay. So yeah, that is a that's a linear algebraic condition on y point wise. Yes? Why do we have to choose the j ref? I mean we have to pay for this. Right. The reason I didn't just use j fix here is that if I did, what I will have proved in the end is that a small perturbation of j fix achieves transversality. I would like to actually know that given any j in my admissible set, I can perturb that one to achieve transversality. So j ref is arbitrary except for the conditions I stated. Okay. So hopefully we're all on the same page. Now, we have this functional analytics setup that we introduced last time. We know how to understand the modular space of holomorphic curves locally as the zero set of a Cauchy-Riemann operator possibly divided by some symmetry action. Let's just write that down in this case. So the result comparable to what I ended with last time is we can say a neighborhood of the element, let's say u naught is the j holomorphic curve with complex structure, j naught, and the target almost complex structure is called capital J naught. So that defines an element of the universal modular space. And the neighborhood of that element is now going to be in a bijective correspondence with a neighborhood of the equivalence class, sorry, yes, neighborhood of little j naught, u naught, capital J naught in some zero set of a section divided by the automorphisms of u naught, where this section I'm talking about maps. So again, I have my Teichmuller slice. It's a finite dimensional space manifold of complex structures on the domain parametrizing a neighborhood in Teichmuller space. B is my Banach manifold of maps of some Sobolev class from sigma dot into w hat. And then I have j epsilon. That's the one element that I didn't have in this picture before, mapping to some Banach space bundle. You can infer from this picture what the fibers of this Banach space bundle are supposed to be able to explain to write down the map, of course, since little j u big j to Tu plus big j of Tu little j. So that again is a smooth section of some Banach space bundle. It is smooth because all the capital J's are smooth. If the capital J's were not smooth, then I would have a bit more of a headache to figure out exactly how differentiable this section is. That's why I introduced the C epsilon space. On the right hand side were the parentheses. The parentheses. As in, what do you divide by art? You divide the neighborhood by art or you divide the art and then you divide it? I divide the zero set of this section by art. So this is some diffeomorphisms of the domain which act on all three of these. Well, it doesn't do anything to big j, but it acts on these two. The same way that we had yesterday. So I can actually eliminate that from the picture here because actually u naught is somewhere injective. So I can get rid of the automorphism group. It's trivial. So I'm in the world of manifolds at this point if I'm talking about simple codes. Yeah. So the neighborhood on the right hand side includes a little variation in the little j naught, right? Yes. So does that mean that m star includes varying the complex structure? Sure. Absolutely. I mean, m star is, so this is just shorthand notation. It's not just a map, u. It's an element of this modular space which includes the data of the domain. So the domain complex structure is absolutely allowed to vary throughout this picture. Okay. So there's my section. And of course, I would like to know whether the universal modular space is smooth. It's going to be a smooth barnock manifold if the linearization of that section is always surjective. And that's the main thing one has to prove in any argument of this sort. And we already heard a little bit from Dusa about this yesterday on why you have to prove it and why it goes wrong sometimes, specifically if you don't have the existence of an injective point. So let's talk about that. Let's write down the linearization first. It's going to take a triple y eta, capital y 2. Well, differentiating that thing with respect to lowercase j gives us j of t u of y. Differentiating with respect to u gives us the usual Cauchy Riemann linearized Cauchy Riemann operators associated to u, which I'm just going to denote by d u of eta. And then we additionally have differentiations with respect to capital J gives y of t u of j. So note the first two terms in this expression together are precisely the operator that I wanted to be surjective when I talked about regularity yesterday. So these two form a surjective operator if and only if u is Fredholm regular. That was a definition, which I didn't really write down, but I implied it. And I have this additional term, the advantage of which is that y is now varying in an infinite dimensional vector space. So it makes it a lot more plausible that you could force this operator to be always surjective. So how do we see if it is? That's the claim. This thing is surjective. And let me be a little bit more specific about my functional analytic setting at this point. So last time I said that the maps u are in some bonnock manifold of maps that are class w, k, p. And I didn't specify k and p, except to say that k times p is greater than 2. So the Sobolev embedding tells you this is contained in C0. For convenience right now, just to make my life easier, I'm going to assume k equals 1. So let's say this operator is really going to take the tangent space to a Teichmuller slice plus some space of w, 1, p class sections of u star t, w plus the tangent space to big j. I'm going to take those to sections of class l, p. I need the exponential weights, as Helmut would point out to me if I don't say it myself. The subject of why I need the exponential weights should be discussed, but I don't have time for it in this talk. So I would be happy to do that in the discussion. What is the next time? So remember the definition of this space, right? We had this asymptotic condition on the behavior of sections that said it's not just that they are of class w, k, p on the cylinder, but e to the delta s times the section is of class w, k, p. So it forces exponential decay. The delta itself is just some small positive number that's, we said, as long as it's small enough, you're not losing any holomorphic curves by imposing this condition. So I'm very fond of the philosophy, by the way, that nothing you do in this whole story should ever really depend on your choice of Sobolev space as long as you satisfy the condition for inclusion to C0 or even some larger differentiability if you want. So I don't like making this choice, but it's only the first step in the proof. If you want to then prove that the same thing is true for w, k, p to w, k minus 1p, that becomes easy if you have already, you can prove it inductively, basically. So I'm just going to do this case. Now, so if it's not surjective, then we can say there exists some element of the dual of Lp delta here that annihilates the whole image. So the image is killed by some non-zero element, I'll call theta, in the dual of Lp w. Quick digression. I don't really like this distinction people like to make between analysts and topologists and whatever else people want to call themselves. Partly because I don't personally feel like I am either an analyst or a topologist. I have no idea what I am, but enough people think of me as an analyst, I guess, that they often ask me the following question. Why do you need to worry about Sobolev spaces in particular in this whole picture? Why can't you just say let's consider maps of class C, K, isn't that, doesn't give you a reasonable Bonach manifold? It does give you a reasonable Bonach manifold. Of course, reason number one why we might prefer the Sobolev spaces is that we have certain standard estimates for Cauchy Riemann operators that work in Sobolev spaces, and this is what tells you that the linearized operator is Fred Holm. Okay, those estimates also do exist, for instance, in Holder spaces. So you can not talk about CK, but CK alpha for some positive alpha. I don't know how to prove the estimates there, but they do exist. You get the Fred Holm property, so why don't you use that? Isn't that still somehow simpler than Sobolev spaces? Well, the answer to that is what I'm about to do. Of course, we know what the dual space of LP is in a very concrete way, and that's not true for Holder spaces. The answer is known, but it's not pleasant. The dual of LP is a perfectly pleasant object. I can write it down concretely. So this is the step where it really pays off. Actually, even I'm aware of one of the mid-90s papers by Hofer, Wroclaw and Sander where they do use Holder spaces for most of the way, and precisely at this step, they switch to Sobolev spaces. So I've got this element of the dual. Now, the dual is just LQ with exponential weight minus delta for 1 over Q plus 1 over P equals 1. And saying that that annihilates everything in the image means, well, I can write that as three separate conditions, in fact, depending on whether I operate with this thing on any of the three variables. So it means, first of all, that if I take any j of t u of y and integrate that against theta, which I can abbreviate as an L2 pairing. So that's the pairing of LP with LQ. That's going to be 0 for all y. Same thing will be true for operating with, so, du eta. That's going to be 0. And, whoops, y of t u of j pair with theta will also be 0. So again, this for all eta, this for all y. So that's what comes as saying that theta annihilates the image. Now, the second condition in particular, you can interpret that as saying that theta is a weak solution of another Kauty-Riemann equation, namely for the formal adjoint of du. You can replace du on the right-hand side with its formal adjoint. And if that paired with all eta is 0, that means that the formal adjoint annihilates theta. Now, there's various regularity results that tell you anything annihilated by a Kauty-Riemann operator, even if it's just a weak solution of class LQ, is going to be smooth. And even better, solutions of Kauty-Riemann equations, if they have zeros anyway, those zeros are isolated, unless the whole thing is trivial. That's the similarity principle. So the second equation really says formal adjoint du star of theta equals 0. So by regularity, theta is c infinity and has only isolated zeros. I mean, you first need to show that the adjunct is going to be some Kauty-Riemann type operator. Yeah, absolutely. So in the background of this is a very general fact that for any Kauty-Riemann type operator, there is a formal adjoint which is conjugate to a Kauty-Riemann type operator. Now, what about the second thing? Well, remember, y is living in an infinite dimensional space, which is large enough to contain bump functions with small support. So I can exploit that fact and say, let's choose an injective point whose image is contained in the perturbation domain. And now, I can choose y to be something that is non-zero only on a neighborhood of that point inside the perturbation domain. But I can make fairly arbitrary choices in that small neighborhood. In particular, I can choose y to be supported near u of z, such that point-wise, this pairing is positive, but it vanishes everywhere else. And that's my contradiction, because if, well, I'm able to do this because I know, in fact, I should have also added this without loss of generality. By moving this point, this injective point a little bit, I can also assume theta of z is not zero, right? Because they're only isolated zeros of theta. That's what I got from condition two. Sorry, so is one injective point enough? Well, one injective point implies a neighborhood consisting of injective points. It's an open condition. So having that just in a small neighborhood is certainly enough. Yes? Is that because you can always find a y which takes a non-zero vector to another one? Is that what you're doing? Yes, it is that you can find y is taking any non-zero vector to any other non-zero vector. It's a slightly non-obvious fact, given that I want y to be also tangent to the space of compatible complex structures. So, again, there's a linear algebra lemma that needs to be proof. There is a proof in our book that I do not. There is a proof. There's a proof in my lecture notes as well, which is copied from her book, basically, because it's one of these proofs that one can write it down very short and still get no insight from it. I don't know why it's true, but it's true. Yes, it's a crazy formula. So you don't use condition one? Excellent question. Actually, I was expecting Augustine to ask that question. I didn't use condition one in this proof, which kind of suggests that I didn't use the Teichmüller slice at all. That's where condition one came from. That's the small y, tangent to the Teichmüller slice. But this does not mean that the Teichmüller slice isn't somehow important. I was able to achieve transversality here, well, for the universal modular space, without worrying about that. On the other hand, it's still true that there exist j-holomorphic curves which are stable, which have positive index. So they exist for generic big j, but not for generic little j. They will only exist for certain choices of little j, and you have to allow that j to be arbitrary in order to find those curves. Because if you fix the conformal structure of the domain, you get a problem that may have negative index and those things will then not exist. So that's really, it's not so much a transversality question, but it's an index question. My second remark is I will have occasion to use condition one in some form later in this talk, if I get there. Yeah. Yeah, I'm not done. Let me get there. So I prove the universal modular space is smooth. And why was it so important that we knew the dual? Well, I could write down the dual as some kind of section of a bundle. It's a very concrete thing instead of an element of the abstract dual space of some other space of sections. And then I use this L2 pairing. Really, the crux of the argument involves this L2 pairing of sections. And that depends on having a precise picture of what the dual of that target space is. Because in the end, it wasn't, the argument I made was point-wise. It was not some abstract Hilbert space in a product or something. It was a point-wise argument at the end. All right. This thing is a Bonnack manifold. Now we look at the projection of that to J epsilon, taking a pair of UJ to just J. And the Thart's male theorem tells me some generalized sense of almost every J will be regular values. So there exists a Comeger subset, let's call it J epsilon reg in J epsilon, such that for all J in J epsilon reg and UJ belonging to this universal moduli space, so in particular, U has an injective point in the perturbation domain, then U is regular. And in particular, the moduli space with respect to J itself is a smooth manifold. So that's great. A corollary of that is that the set of regular J's in this larger space of smooth, almost complex structures that match J fix outside of U is dense. That's immediate because I've achieved regularity for J's in J epsilon which are arbitrarily C infinity close to J ref. And J ref was arbitrary. That's why I chose J ref instead of J fix. So now this is good. I have density of the regular, almost complex structures. I think Katrin would tell me I should be satisfied, but I'm not. So you're using sards mail, do you need some sort of press? Yeah, so sards mail depends on the fact that indeed this map here is, this projection is Fredholm. That's equivalent to the fact that the Cauchy Riemann operator is Fredholm. That's an easy exercise. So I disagree with Katrin's philosophy in a certain sense that, first of all, it's definitely not true that I should always be satisfied to get a non-empty set of regular complex structures. Maybe that's good enough if I only care about defining a theory that I want to prove as an invariant. But if I then also want to compute that theory, I need something more, right? Because usually I can make some specific choice of J for which I am able to count all the holomorphic curves I care about. And of course that J is not going to be generic, so I need to make a generic perturbation of it. And I need to be assured that I can do such a perturbation of that specific J where I understand everything and attain something regular. So density is at least what I need. I would like a little bit more, in fact, because it's often true that I want to take some space of almost complex structures that I've proven are regular in a certain sense and now intersect it with another space that's regular by a slightly different conditions. Okay? Sometimes I want to do some countable intersection of those. Well, even an intersection of two dense subsets can be empty. So I'd actually like to know that I get commieger, so countable intersection of open dense subsets. Now, how do I get from here to there? There's a trick for this that was introduced as far as I know by Taub's. And it also appears in the book by Dusa and Dietmar using it for different purposes because they've got finitely differentiable almost complex structures. Instead, I'm trying to get now from C epsilon to C infinity. You can apply it in more or less the same way. I think people often don't realize just how widely applicable this is. And I'm going to do a kind of lazy version of it based on the fact that we know the space of holomorphic curves has a nice compactification. Okay? One can be less lazy and formulate conditions that are in some sense equivalent to that but requiring less knowledge. I'm just going to save the following. Let's let M good of J denote the space of all curves U in the moduli space which have an injective point in the perturbation domain. And now for each natural number N define a subset of the whole moduli space Mn of J to be the space of all curves with the property that the distance in some metric that I'm not going to specify but I know where to look up a proof that it exists. Distance from U to the complement of the good space in the compactified moduli space. So compactified moduli space minus M good. I want that distance to be greater than or equal to one over N. Okay? So in order, I mean one can make this precise. To do it, one has to look at the appendix of the SFT compactness paper where they prove that the topology defined on the SFT compactification has a metric. So distance with respect to that metric. So this obviously is a subset of M good. If I take arbitrarily large M, I can exhaust M good with this countable union of these subsets. But also this subset is compact. It's a closed subset of a compact metric space. And it's even compact in a slightly stronger sense because this metric on the compactified moduli space depends continuously on J. So I can also move J around. And if the sequence of J converges, I can also get convergent sub-sequences of objects in this sequence of spaces. So the consequence of that, the space M, the U must come from the space M good. Yes. Yes. Yes, follows from the definition, yeah. Yeah, because the distance is from there is positive. So let's define Jreg N to be simply the space of all J's in my admissible class such that all curves U in M N with respect to J are regular. So that's some set of almost complex structures. It's a subset of the space of smooth almost complex structures. I have the C infinity topology on it. Now the fact that this subset of the moduli space is compact tells me that this space of regular almost complex structures that only cares about that is open. And it's also dense due to the argument I just did above. I can perturb any almost complex structure to one that achieves regularity for all of M good, not just the subset. So due to the above argument, it's dense. So all I have to do now is say the space I'm looking for called Jreg U is the countable intersection of these Jreg N's. And then I'm done because that's a countable intersection of open dense subsets. So the rant that I will give you on this topic is that the flurcy epsilon space is a very nice object but you shouldn't take it too seriously. My personal opinion is it should never appear in the statement of any theorem. It should appear in lemmas and it should appear in proofs of theorems. But just about any theorem you can prove about the flurcy epsilon space, you can also prove about C infinity using this trick. You also get perspective if you have two elements in Jreg, they'll be defined by a regular homotopy. Yeah indeed. I mean, the argument works in more or less the same way if you want to study regular homotopies. Can you explain this trick? Where do you get rid of that epsilon? Yeah, what is epsilon? Which epsilon? I mean, epsilon is a notation representing a space. So I guess the key point is the claim that this set of J's here is dense. So it follows what I said here is that the set of J's that achieves transversality for me is dense in C infinity as a consequence of the fact that it's also a bare subset of J epsilon. This uses the fact that J epsilon has a continuous inclusion into C infinity. So once I get density in J epsilon, I also get density in C infinity. I mean, a type this trick is sort of expressing finding a compact space so that when it's compact you can say the regular elements of the compact thing are open. So type this trick gives you the open ones. And then you take an intersection of open depths, a comfortable intersection of open depths. That's a good way of putting it. Excuse me, could you explain again why this J-eric M is open? So yeah, why is this open? So first of all, this space is compact because it's a closed subset of a compact space. Now, all right, suppose that this thing were not open. What would that mean? It would mean we have we've got some J that achieves transversality for every curve in this space. But we could also find a sequence of curves or a sequence of J's converging to our J that don't achieve transversality for every curve in the corresponding spaces here. But now compactness tells you as the J's converge, the U's in these spaces also have a convergent subsequence. And there are non-regular curves converging to a regular curve. That's what's not possible because regularity is an open condition, right? That's surjectivity of a Fredholm operator. That's an open condition. And can you explain dense again? Okay, did you believe me up to this point? So we get commieger in J epsilon, which implies dense in J epsilon, and that implies dense in C infinity, just because J epsilon includes continuously into C infinity. Well, you have to take over the union for all J fixes. I don't. J refs. Yeah, rather, yeah. Over all J refs. Yeah. Right, I mean, so, exactly. So what this proves is for your original J ref, you can approximate J ref arbitrarily well in C infinity with a regular J. So, a J ref was arbitrary. So that implies that the regular things are dense in C infinity. And dense in that is actually a stronger condition than what we need down here. It implies it immediately. Yeah. So if I had a sequence in M, M, J, I could converge to something you know here, right? No. A sequence in M, M, J will not converge to anything nodal because it is a positive distance away from the nodal part of the compactified modular space. But what if, I mean, there could be nodal things in M good, right? No, there cannot be. M good consists only of somewhere injective, honest J homomorphic curves. Yeah. I mean, it's, it really seems like cheating, right? You're using the knowledge that this compactification exists and just saying let's stay a finite distance, a positive distance away from the bad part of the compactification. But for M of J, you kind of think it's some sort of. I don't have to, well, I assume non-degenerate Ray-Borbits, but I don't have to fix more than that. Well, to get a uniform finite pull for energy. This is okay. Well, no, okay. What, actually, here's a good reason why I want commieger. What I really should do is do this whole argument just looking at holomorphic curves in a fixed genus and fixed relative homology class. And then say, having proven we have a commieger subset for that genus and that homology class, now take the countable intersection of all those choices as well. If you're just looking at Max's domain as a sphere, it would make it really simple. Then, then this M and J, you can define it simply by giving a point-wise bound to the derivative. Just say the derivative is bounded by some constant number. That's a good point. No bubbles can form. I mean, it's a problem. This is what you actually see when people use this in the literature. I mean, I'm using this formulation as a shortcut. But when you write this down in the literature, usually people write down a list of conditions that say things like the C1 norm of d, the sup norm of du is bounded by 1 over n, or sorry, bounded by n. Also, some condition about distance of u of x minus u of y divided by distance from x to y should be within bounds determined by m so that you stay away from the multiple covered curves. And you need something to keep you with positive distance away from nodal curves as well. So one can also, one can write down conditions like that which do everything for you without having to first prove Gromov compactness or SFT compactness, whichever. The argument, in fact, you can do this in, for closed curves at least, you can do this in almost complex manifolds that are not tamed by a symplectic structure. You don't have to prove Gromov compactness first. You just have these conditions, you have elliptic regularity telling you C1 bounds will give you compactness and so forth. Okay. Can I ask one last time why does injective inclusion imply density for C infinity? So this, the claim is this, if I have a sequence of j's converging to some particular j in the C epsilon topology, then it also converges in the C infinity topology. So, as soon as you know that you can approximate whatever j you want arbitrarily well in J epsilon, you have that that's also true in C infinity. And Chris, the bit of the proof that trails when you have to say multiple covers, is that hidden in the board hidden there, right? When you say that, that thing is positive. Because you can have that if you go around the point twice, say. Right. Yeah, let's look at that again. So why does this fail for a multiple cover? I said choose an injective point and we make this perturbation of, we define y in some fashion on a neighborhood of u of z. Now, I want to do that such that this point wise pairing becomes positive in a neighborhood of z. And I want to also know that it's not going to be positive in any other neighborhood somewhere else, which would happen if you had another point going through u of z. That's why this fails for multiple covers. Well, it could be negative for a multiple point. It could be negative sometimes. I can't conclude anything unless I really know that this is the only place where u is going through this region. How much time do I actually have? Yeah. So I would argue that this has been a sort of hybrid lecture and discussion session and that the afternoon discussion session should be the same thing. You haven't used one still, right? I've not used condition one. The organizers, we approve. Very good. So maybe we shouldn't let everyone else know before. So let me at least write down the statement of what I still need to get to in the afternoon. So hopefully, we're happy now with chance of solitude for simple curves in symplectical boredisms. Actually, can you maybe move the tab strip board down? So in order to actually do SFT, we also need to understand transversality for curves and symplectizations. So that's the translation invariant, J. Here's the theorem. So let's say now M is just going to be a closed, odd-dimensional manifold. We've got a stable Hamiltonian structure. It induces our hyperplane distribution, psi and rate vector field, Rh. And remember, we have from that a special class of translation invariant almost complex structures associated to the stable Hamiltonian structure. So theorem three says there exists another space I'm going to call Jreg. It's a commieger subset of JH such that for all Js and Jreg, all J-holomorphic curves with, how should I say this, that are somewhere injective and not everywhere in the region to psi are regular. So let's unpack this a little bit. First of all, we're working in R cross M. So it's a trivial symplectical boredism, which means our map can be written in terms of a real-valued function and a map from a function of V-mon surface into M. I call those U, R, and U, M. And there's really two points of difficulty in this problem. First, the perturbations I'm making for J are no longer really local because I require J-holists to be translation invariant. So any perturbation I make near a particular point is going to affect what J is everywhere up and down from there. So that problem turns out to be relatively easy to solve. There's a technical lemma which appears in one of the mid-90s HWZ papers again that says if U is somewhere injective, then the injective points of the map U, M from the puncture V-mon surface to the odd-dimensional manifold are also dense. I'm not going to prove this. Actually, I've not yet managed to understand the proof because I haven't sat down with it long enough, but I can tell you where to look it up if you want to look it up. Do you need to ask finite energy or something? Yeah, I always assume finite energy. If you want to hear about infinite energy curves and simpletizations, you have to speak with Joel Fish and not with me. So that's one thing. I'm not going to be too worried about the translation invariance. As long as I know that the map to M itself has an injective point, then I can imagine I could do the same trick I did with a perturbation just supported near that point, and that's not going to mess anything up. But what I'm more worried about is the fact that, remember, what are the conditions that these Js must satisfy? Well, for one thing, J maps the unit vector in the R direction to the Rape vector field, and the Rape vector field is not ever changing in this picture. That's fixed by the stable Hamiltonian structure I started with. The only thing I'm allowed to vary is how J maps the hyperplane distribution to itself. So I'm doing perturbations of J on a sub-bundle of real co-dimension 2, and I need to know that those perturbations are going to be a large enough space to achieve transversality, and you might notice from the statement of my theorem that if you're not careful, this isn't true. There are, in fact, for certain stable Hamiltonian structures, J-holomorphic curves that will not be made regular by any perturbation of this sort. Just quickly write down an example, and then at lunchtime, let's say M is S1 times W, where, again, W omega is a closed symplectic manifold. No, let's make it simpler than this. Let's just say sigma is a surface. So S1 times W, and say T is the coordinate on S1, omega is an area form on sigma, and then this is the fluorohomology example that I gave you yesterday, but just with the two-dimensional case. So to do fluorohomology in a two-dimensional manifold, you would write down this stable Hamiltonian structure of the form big omega equals omega plus some Hamiltonian term if you like, or just take the Hamiltonian to be zero for now. Lambda equals DT. So it means that psi here is the foliation by these surfaces sigma. And if I'm perturbing J in this translation invariant class, I'm really only perturbing what J does on that foliation. So that means I'm looking at T-dependent families of complex structures on sigma. Well, for all such choices, there exist closed J-holomorphic curves that look like map sigma to R times S1 times sigma by taking Z to constant, constant Z. And these will sometimes have negative index. So there's an exercise for lunchtime. Convince yourself that if you were really allowed to take a fully generic perturbation of J, you would kill all these curves if the genus is large because they have negative index. So it means the perturbations in this translation or invariant class are not large enough to actually achieve transversality for everything. They don't for these curves. So that's why I have this extra condition. You have to just pay attention to curves that are not everywhere tangent to psi. Now, for the case that most of us care about most, psi is a contact structure, and it's easy to guarantee that your curve will never be everywhere tangent to psi. So I will continue with this in the afternoon. So any cylindrical thing? Any cylindrical thing. Well, this curve is asymptotically cylindrical. All of its punctures approach ray orbits. All zero of them. Actually, I mean, genus wanted, but it's a tourist. It's good enough because you've only got one variation and you're only allowed to vary in the S-1 direction or the other direction. Yeah, I wasn't going to commit myself to a statement like this, but I think it's true. Genus 1 is already a contradiction to transversality. Yeah. OK, I'm done.