 Thus far in this mini-course, I think it's fair to say all the results I've been talking about are pretty well considered standard, and the proofs I've described are also using more or less standard methods. The title of the mini-course has the word classical, classical transversality results. What I mean by that is any result that you can express in the form for generic domain independent, almost complex structures, J, we have transversality for some particular class of J-halomorphic curves. So that's allowed to include some methods which cannot be called standard, and that's what I'm going to talk about today in particular, because everything I talked about so far with one minor exception when I talked about automatic transversality, everything else makes the restriction that we're only talking about somewhere injective curves, which of course is a big problem if you want to define big invariance such as Gromov-Witton-Thierier or SFT, and to a large extent you cannot really hope for classical transversality results to be true in the generality that you would need to define those theories, and I do already illustrated this briefly and I'll reiterate that a bit in a moment, but there are situations when those results can hold, when you don't need to go to much more general frameworks such as the polyfold theory in order to define everything. If you do get transversality for your honest holomorphic curves, it can make your life easier because the Cauchy-Riemann equation is much easier to handle and carry some sort of natural geometric information that you don't necessarily have in whatever perturbed equation you're going to solve in a more general framework, so sometimes it just requires much more originality in your way of thinking to get what you need. Let me give you an illustration, so for today I'm going to depart a little bit from the title of the course and not really talk about SFT, so I'm not going to talk about punctured holomorphic curves, but just closed holomorphic curves, let's say sigma and sigma prime will be closed surfaces, almost all of what I'm going to say can very likely be generalized to punctured holomorphic curves, but that's work in progress, so I'm not really going to touch upon it besides a few general ideas. I will just tell you about things that I know. So a multiple cover looks like this, I have a k to one holomorphic branched cover from one Riemann surface to another, I assume k is greater than one, sigma prime is the domain of some j-holomorphic curve into an almost complex manifold Wj, capital J, and then the composition of these two holomorphic maps gives me the multiple cover, so let's assume V is somewhere injective, U is now a k-fold covered holomorphic curve, and it'll be helpful to note that there's a relationship between the Euler characteristics of these two domains as they both figure into the index formula, so there's the Riemann-Herwitz formula that says minus Euler characteristic of sigma plus the degree of the cover times Euler characteristic of sigma prime equals this quantity that I like to call Z of d phi because that's literally an algebraic count of the zeros of d phi, or in other words it's the number of branch points counted with the orders of branching, number of critical points of phi, and since phi is holomorphic all of those count positively, so this is an integer that's always greater than or equal to zero. So we have this as a constraint relating the domains, if you've never seen the formula before you can just think of it this way, d phi is a section of a certain complex line bundle which you can easily write down, and then you can compute C1 of that line bundle, the answer is the left-hand side of this formula. That's so hard to understand, but about me, please. You can also draw some picture with, you can triangulate things and draw a picture of what branch points looks like, maybe you prefer that, but it's very quick thinking of it as C1, if I ever forget the formula that's how I remember. So now let's write down the index formulas that we get from Riemann-Roch. So the index of V, so W I assume is 2n dimensional, real 2n dimensional, the index of V by which I mean the virtual dimension of the modularized space of unparameterized holomorphic curves that V lives in. So that's not literally the front home index of the linearized Cauchy-Riemann operator, but it is by basically what I explained before, it is that plus the dimension of the relevant type mirror space. So it's the actual dimension of the modularized space, if transversality holds. And the formula is n minus 3 times Euler characteristic of the domain plus 2 first turn class evaluated on the homology class of V. How does that relate to the index of U? So same formula with a different domain. Now Riemann-Hurwitz tells me I can rewrite Euler characteristic of sigma as k times Euler characteristic of sigma prime minus this count of branch points. And of course C1 of U is just k times C1 of V, so I have 2 times that. And now I see sitting inside this formula, k times the index of V, because there's n minus 3 times Euler characteristic of sigma prime plus C1 of V over here. I just have an extra term. So I get k times the index of V minus n minus 3 times this count of branch points. So in particular, to be a little bit more concrete on what can go wrong with transversality for multiple covers, suppose that the underlying somewhere injective curve had its index 0. So that's generically a rigid isolated object in its modular space. And for generic J, it's also going to be stable under small perturbations of J, which means there's no way you can get rid of the multiply covered curve since there will always be a perturbed, multiple cover of your perturbed somewhere injective curve. But the index of U is then going to be minus n minus 3 times this non-negative count of branch points, which can easily be negative. At least if we're in dimension 8 or upward, unless the cover is unbranched, that's going to be a negative number. Dimension 8 and upwards, that's precisely where we stop being able to assume that our symplectic manifold is semi-positive. So there we run into trouble. This is likely to be a negative number in general. But we see that we cannot perturb that curve away. The situation is actually even worse. If you think about what kind of curves we already know must exist in a neighborhood of U, there's not just U itself. There's the other nearby branched covers of the same V. And those come in a non-trivial modular space in general. So the actual dimension of the modularized space of holomorphic curves near U, and when I say actual dimension, I'm not making the assumption that it's a smooth manifold or anything. It might not be. It might not even be a smooth orbifold. But it does contain a smooth orbifold, which I can identify very clearly, namely the space of branched covers of sigma over sigma prime of the same degree. So we get at least the dimension of the space of k to 1 branched covers modulo reparameterization. And that's a very well understood modularized space. That's just another modularized space of holomorphic curves living in dimension 2, if you like. And you can compute the dimension from the Riemann-Roch formula as usual. The answer is 2 times the number of branch points. And there's sort of a geometric interpretation. You might want to follow up on that. But you can just see that. This is my next sentence that you just interrupted. There's a geometric interpretation of this, because given a branched cover, you can find other branched covers nearby by moving around the positions of the branch points in the image. And those are not equivalent up to reparameterization. They give you different branched covers. And this is precisely the number of parameters you see by doing that. So this is basically classical. So we see that number is never going to match the actual index of U, unless possibly the cover is unbranched, or what, or if we're in dimension 2, but so what. We don't really care about dimension 2. So here it's fair to say transversality is generally not plausible for the multiple covers. I do want to make another observation about this, though. So again, the index of V is 0. So generically, by which I mean after possibly perturbing J, I'm allowed to assume that somewhere injective index 0 curves are immersed. Why is that? So I've discussed this with a couple of people since my last talk or discussion session or whatever it was. There was an exercise I posed about using automatic transversality to prove that in a four-dimensional symplectic co-boardism, generically an index 1 holomorphic cylinder will always be regular. And part of what you have to do to prove that is observe that whatever kind of holomorphic cylinder that one covers, if it's a multiple cover, is also going to be a sufficiently low index so that you can assume it's immersed. And then you can apply the automatic transversality criterion after you know that. So there's this general fact, which I don't have time to talk about in earnest. But one can show that if you take your usual moduli space, add a marked point to it. So that increases the dimension of the moduli space by 2. But now constrain that marked point by asking for the first derivative of your map to vanish at that marked point. So in general, that decreases the dimension of the moduli space by 2n. So the upshot of that is for generic j, the space of a holomorphic somewhere injective curves that are not immersed may be considered to be of co-dimension 2 minus 2n compared to the larger moduli space. So that means if my index is 0, I can assume that that space of non-immersed curves is empty. Therefore, this one's immersed. So I'm not going to say more about that. Let's just accept it for now. It's a j-holomorphic fact analogous to the standard differential topological fact that you can perturb smooth maps to be immersed, given the right dimensional conditions at least. So we have this. Now that means let's look at the normal bundle. I've got. Can you at least say why there's more to say than what you just said about why this statement is true? Yeah, because you have to work out the details. Go ahead and try it. Let me know how it goes. So the normal bundle is not the generalized, but the usual definition of normal bundle here. That's what it's C1 is. The generalized normal bundle of U is very easy to describe. U is not immersed because there can be branch points. There are critical points of U. But of course, the generalized normal bundle of U is just going to be the pullback of this normal bundle in V. So that's 2k times that number over there, C1 of V minus or the characteristic of sigma prime. So remember, I talked last time about restricting the linearized Cauchy-Riemann operator to the generalized normal bundle. That also gives you a Cauchy-Riemann-type operator on a bundle of rank 1 lower. Its threshold index is given directly by the Riemann-Roch formula. Let's see what it is. So d urn is the restriction to the normal bundle. And Riemann-Roch tells me, well, the rank of this bundle is n minus 1 as a complex bundle times the early characteristic of the domain plus 2c1, the generalized normal bundle. So Riemann-Hurwitz tells me what that first term is. We have this extra term counting the number of critical points. And then 2c1 and U, I've just written up here. It's am I missing something? People are saying there's a k missing. But where? No, there shouldn't be. It shouldn't be a k. Yeah. The last one. And the a is missing. Well. From the first line, you have two times the restriction. That's V, and this is U, right? That's why there's the k. No, I think this is OK. I'm doing this a slightly more roundabout way than I planned, accidentally. I've got k times n minus 1 or the characteristic minus another 2 or the characteristic plus twice c1 of V, then minus n minus 1 times count of Bradford. This is fine, because this number in the brackets here is 0, because I assume that V has index 0. That's the index formula. So I'm just left with minus n minus 1 Z of d phi. So what I notice about this is unlike the index of U that I wrote down, just looking at the normal operator, there's some predictable pattern. This number is always non-positive, which means it's conceivable that this normal operator might actually always be injective. And that's something geometrically meaningful, if it's true. So I'm going to actually state this as a conjecture. About a year ago at this time, I was calling it a theorem, but then an error was discovered in that proof. So this is a conjecture that says, so for generic J, all multiple covers U of somewhere injective index 0 curves V have normal Cauchy-Riemann operator injective, which has a nice consequence if it's true. What that actually implies is all of the other curves in the neighborhood of U are precisely the ones you already know about. They're just the other multiple covers of V. So I'm saying this result would give you a precise description of the modularized space of curves in near U as having exactly this dimension of the space of branched colors. Is that strong enough to get PPS integrality from a number of curves? That would be the intention, right? So the intention of a result like this would be to prove something like the Gopakumar-Vaffa conjecture, right? So the reason why this kind of thing is supposed to be interesting is that it means, in certain settings, if you want to compute Gromov-Witton invariance, you really only have to understand the somewhere injective curves. And the rest of it, of course, the multiple covers are not regular in the usual sense. So you have to do some kind of perturbation if you want to actually count them. But there's a standard way of doing this within homogeneous perturbations to the Cauchy-Riemann equation. And you can predict the count that you'll get because you see the entire modularized space in terms of the space of branched covers. It has an obstruction bundle. And you can compute the Euler class of that obstruction bundle. That will give you the answer, right? So this has nice corollaries in Gromov-Witton theory. What was the answer to that question? I think the answer to your question was yes. I'm not really sure how to interpret the index of this kind of view. Right. The main thing I want to say about it right now is that since it is generally negative, the operator can be injective. And what I really want to explain is how you interpret the fact that that operator is injective. So if the kernel of this operator is trivial, think about it like this. The following is a scenario you don't want. You don't want to have a sequence of curves that have different images from you converging to you. My claim is that all the sequences of curves that can converge to you are of the form v composed with some other branched cover. So they all have the same image. Now, if you have a sequence of curves with different images converging towards you, you can do this trick where you look at that as curves living in the normal bundle of you and rescale that normal bundle. So that as the curves approach, you rescale so that you see them not actually approaching but staying in some bounded subset of the normal bundle, apply Gromov compactness to that. That sequence is going to converge to some generally nodal holomorphic curve, which will have some component that you can interpret as something either in the kernel of this operator or in the kernel of some related operator corresponding to a branched cover of lesser degree. So if you know that these kernels are all trivial, that precludes this scenario. So it says really that the space of branched covers or the space of use that are branched covers of v is an open subset of the whole modular space. This is all speculation. I mean, the argument that I just described can be made fully rigorous. But of course, the conjecture is only a conjecture. We don't know if this is true. So I do want to talk about a special case of it, which we do know. Yes? Can you give an example when the hypotheses of the conjecture aren't satisfied where the desired result isn't true? You know, hopefully covered curves are approximated by curves with different images? I'd have to think about that a little bit. I mean, I certainly could come up with examples given enough time where, I mean, in particular, there's hardly any conditions here. The main condition is just that the simple curve has indexed zero. So in situations where that's not true, the possibility of a sequence of simple curves converging to something multiply covered is always something you have to worry about. You usually have to make some effort to avoid it. Doesn't Tyvers have some examples when he attacks Tori, and he has multiply-covered Tori that's a generically branched off double-covered in most kind of times? And I'm going to talk about that a little bit. But I don't think it's an example of this phenomenon. It's much nicer than that. Richard Heim has some nice example where you need generalistic analysis. OK, I'm not aware of that. OK. So here's an actual theorem, which is one of the cases of this conjecture. And this is in a joint paper that I wrote with Chris Garrick last year. I'm going to make the statement a little bit unnecessarily more complicated than I need just to illustrate how different it is from the results we talked about so far. So let's say fix an open subset u in our closed manifold m, closed symplectic manifold of dimension 2n, and fix also a tame, almost complex structure, J fix. Then I will say there exists a commieger subset J reg living, as you'd expect, in the space of all j's that are omega tame and match J fix outside of this subset, such that for all j's of this class, all unbranched covers, u, of somewhere injective index 0 curves, considered, call it v, contained in the perturbation domain are regular. So remember, if I'm talking about unbranched covers, then the disaster scenario I described with this index relation doesn't happen. That's the case where index v equals u implies index v equals 0 implies index u is also 0. So in that case, regularity in the usual sense is plausible numerically. And I'm saying for generic j, it actually happens, at least if we're looking at curves contained entirely inside this perturbation domain. That's one major difference with the theorems I explained earlier. I'm not saying we get regularity for all the curves that intersect or have any point mapping into the perturbation domain. But they have to be contained in it entirely. So we'll see why that seems to be necessary. I mean, I don't know if that condition can be dropped. I certainly don't know a way of dropping it. Is there anything you can't just say to you to be the whole manifold? No, no, you can certainly take u to be the whole manifold, right? But if you wanted to just do perturbations in some subset, then you have to restrict yourself a bit. Do you ask that u has to have compact closure or something like that? Well, u is closed. Yeah. u is closed. Or sorry, u, m is closed. Therefore, u has compact closure. Yes. I mean, I could also allow m to be not compact, and then I would indeed have to require that u has compact closure. That's a good question, in fact. OK? So I'm not going to explain the proof in quite this level of generality, but I'm going to explain a case of it, which is somewhat older than our result, but somehow very badly known or badly understood. So in the case n equals 2, so it's just dimension 4, and where both domains are the torus, and the underlying simple curve is actually embedded. So this is what Dusa alluded to with her question a moment ago. This was done by Taubes in 1996, a paper in Journal of the AMS. The proof there is kind of hidden. In fact, when I went back to it last night to figure out where it was, it took me a while. And it's very sketchy, but somehow the ideas in it are extremely potent. So of course, this means we're getting regularity for a multiply-covered holomorphic tori that are covering embedded tori. Taubes needed this because his definition of the Gromov invariant actually counted those things, and it did it without doing abstract perturbations. It did it for generic j. So in this situation on the torus, there are some convenient things. The fact that v has indexed 0 means that its normal bundle has to be trivial. So I can write normal bundle of v is going to be identified with trivial bundle over t2, complex line bundle, and use similarly. And a Cauchy-Riemann operator, if I write it in those terms, just looks like the usual d bar operator plus some 0th order term. So here, by d bar operator, I mean literally just partial by s plus i times partial by t acting on complex valued functions. And a is, say, a c infinity map from the torus to the space of real linear n-demorphisms of c. That's what a 0th order term looks like. And then the pulled-back normal operator is literally the pulled-back of that. So the normal operator for u is going to be the standard d bar plus this 0th order term a composed with the cover of phi. So just a few general points before I really get into the argument. It suffices to show the following. Suffices to show that for all j, of course, of the tame class I'm thinking about, v is a j-holomorphic embedding of the torus into this perturbation domain. And u equals v composed with some cover phi. Phi is necessarily unbranched, since it's a torus covering a torus. Then I want to show that I can perturb j to j prime such that v is still j prime holomorphic. But the normal operator for u becomes an isomorphism, defined with respect to the perturbed almost complex structure, j, j prime. Can you explain, like, one more time how to think about the normal operator geometry? Definition is very simple. You take the ordinary Cauchy-Riemann operator, which is defined in the pulled-back tangent bundle. You restrict it to sections of the normal bundle. And now you get some section of a larger bundle that you don't want, but also has a projection to a corresponding normal part. It takes you to some section of the bundle hom bar of t sigma to u star tw, tm, whatever it's called. There's a normal projection in that as well. It takes you to hom bar of t sigma to normal bundle. So you just compose the Cauchy-Riemann operator with that normal projection. This is the terms of deformation, sort of, you know, match the tangent bundle, and I think arbitrary definitions by now. Ah, yeah. OK, yeah, there is another nice way to think about this. So there's an alternative way of describing a neighborhood of a curve in its modularized space that I didn't talk about. If the curve is immersed, specifically. So if the curve is immersed, one way of describing all the other curves nearby is actually write this a little bit. So u is immersed. All the other curves nearby can be assumed to be of the form x u h for some h section of the normal bundle. OK, now, in fact, that's going to hit a unique parameterization of every nearby curve in the modularized space, but you don't get to choose what complex structure you have in the domain at all. So what you have to do then is not actually look for nearby maps of this form that solve the Cauchy-Riemann equation with respect to some specific j, but these are all going to be immersed also. So just look for nearby maps like this whose tangent spaces are j invariant, so that automatically you can pull that back to some complex structure in the domain. You don't get to prescribe it. So that's another way of seeing all the holomorphic curves in the vicinity of this one immersed curve. And well, you can write down some nonlinear operator that does that for you. Its linearization is essentially this normal Cauchy-Riemann operator. That also tells you why it happens that the front home index of the normal operator is the same as the dimension of the modularized space in the immersed case. That's basically because if you preserve in the tangent direction, then you just re-parameterize in the case of this. Right, perturbations in the tangent direction are sort of not meaningful for studying this modularized space because this is giving you re-parameterizations of the same curve. OK, so I don't know if everyone had a chance in the meantime to think about why what I just said here is true. Suffice is to show given a curve and a multiple cover, you can perturb J to one that makes that specific multiple cover regular in the sense of the normal operator being an isomorphism. So this is an exercise using the Taub strict that I explained last time. You can exhaust the space you're interested in with a countable union of compact subsets. In this case, finite subsets, even. So as long as you're able to achieve transversality for each of those subsets, then you can find some set of Js that's a countable intersection of open dense sets that does everything you want. So that's all I'm going to say. This is a version of the Taub strict to get you from there to the result we really want. The other thing I'm going to say is we can reduce this problem to something that's really only involving linear Cauchy-Riemann type operators on an abstract vector bundle. Because I can say for all 0th order terms a prime, so 0th order term is just some c infinity function valued in the space of real linear maps on c, we can find the J prime such that J prime equals J in the tangent directions on the curve v. So T sub v is what I was calling the generalized tangent bundle before. Literally that just means the image of the differential of v. And we can also say J prime equals J outside some neighborhood of the image of v. But the normal operator for v expressed in this trivialization and expressed with respect to the perturbed almost complex structure is d bar plus a prime. So I'm only saying here, give me any perturbed Cauchy-Riemann operator you want in the space of all real linear Cauchy-Riemann operators, I can find a perturbed J that realizes that operator for you. And this is not terribly deep. Actually in the embedded case, when v is embedded, this is fairly easy to prove. It's just a matter of choosing the normal first derivative of your perturbed J in the right way to produce the right 0th order term. One can also do it in the immersed case, and that's a bit more painful. I'm not going to talk about that. OK. So let's also take this as given. And just look at a problem involving Cauchy-Riemann operators on line bundles. Sorry, I have a question about this. I don't understand in the middle of the first line, such that j prime is equal to j on t sub v is equal to t. The image of dv, in other words, the tangent space is to the curve. Oh, I thought that was the index of v, I see. Yeah. Any other questions? All right, so here's a claim. And I'm even going to label this one an improbable claim. Because when I first saw this in Chaubes' 96th paper, I had no idea why I should believe this is true. And I'm still not sure I can explain to you why you should believe this is true, but I can prove it. So here we go. Let's suppose d of the form d bar plus a is a Cauchy-Riemann operator on the trivial bundle over t2, trivial line bundle. And b is a bundle map on that trivial bundle, which I'm going to assume is a complex anti-linear bundle isomorphism. So the two key properties are it's complex anti-linear and it's a bundle isomorphism. So given that the bundle is trivial, it's obvious that you can do this. This is more or less equivalent to saying b of z acts on a vector eta in a certain fiber by some complex valued function beta of z times complex conjugate of eta, where this complex valued function beta is assumed to be nowhere 0. So it's mapping into c star. So that's the assumption. I want to just mention quickly, if I were not working on the torus, but with an unbranched cover with more general domains, sigma and sigma prime, the assumption that the index of the simple curve is 0 allows me again to do this, then I would be able to find a complex anti-linear bundle isomorphism between the relevant bundles, even though they're not trivial. That's actually equivalent to the fact that the index of the simple curve is 0. So that's something also that is easy to check. Now, the statement will be that I can define a perturbed operator d tau as tau plus a real parameter times this extra bundle map b treated as a 0th order term. That is an isomorphism for all tau real numbers outside some discrete subset. So this is one way of perturbing a Cauchy-Riemann operator that might not be an isomorphism and making it into an isomorphism. In particular, what you need to notice about this perturbation is it doesn't care at all about symmetry. I require this extra term b to be a complex anti-linear isomorphism. I don't require it to be anything else. If my bundle is a pullback of some bundle that's defined by a simple curve, but I have been looking actually at the normal operator on a multiply-covered curve, I can do a perturbation like this just by changing j along the simple curve. Now, I pull that back to the multiple cover. My perturbation of my operator is going to be invariant now under deck transformations of the cover. This claim does not care, where symmetry messes up the usual argument and this is the reason why I need somewhere injective in the usual arguments. Symmetry ruins this sard-smale argument. This is impervious to that. So I need to convince you this is true. Sorry, like before, an isomorphism, would it still predict the index of the original b? Well, it's a compact perturbation of the operator. So yeah, this doesn't change the index. So basically, all the Fredholm operators in this talk have index zero. So they're either isomorphisms or they have both kernel and co-crow. Yeah? Can you remind me or tell me how we're going to let's say uproar? So would we have this or are we going to use it? Right, so if we have this, that means I can find, sorry if I have my given j and my given cover, which is maybe not regular, I can find a perturbed j that perturbs the normal operator along the cover in this way and therefore makes that curve regular. So first step, this one parameter family of operators, d tau, is injective, which of course, since its index zero, implies it's an isomorphism for all tau sufficiently large. So this step is interesting because I'm quite convinced that this argument couldn't have come from somebody who was mainly a symplectic topologist. It had to come from a gauge theorist. In particular, if you're familiar with Tauves's work relating Gromov invariance and Cybergwitten invariance, you'll notice a parallel here. There's something that Tauves does in Cybergwitten theory where you write down the Cybergwitten equations with this perturbation term depending on a real parameter. And you can prove that for topological reasons, if you make that parameter very large, you don't have any solutions. Or you do have solutions, but they're converging in the sense of currents to holomorphic curves, something like that. So this is going to be a much easier version of that. And it's an argument that I've never seen anywhere else in symplectic topology. So let's say, well, think about, well, I haven't really specified at all what Bonnock space is I'm working with. But just to make my life easy, let's choose Hilbert spaces and say the operators are going from h1 to l2. So let's suppose I have some non-zero element eta and h1 on the torus. And the idea is to operate on that with dTau and look at the l2 norm squared. So dTau is a sum of two terms, which means I can expand this l2 pairing. And I'll get three terms. I have d eta l2 norm squared plus tau squared times l2 norm of the perturbation, B, which I'm writing here as beta times eta bar. So let's write it beta eta bar l2 norm squared. And then there's a cross term. I'm assuming that my inner product is real valued because the linear operator is real linear, not complex linear. So I can use a Hermitian inner product on the trivial complex line bundle, but then I have to take the real part. So this is going to look like two tau times the real part of the l2 inner product of beta eta prime with d of eta. Let's write that out as d bar eta plus a eta. So a few observations about this. Of course, this first term is non-negative, obviously. The second term, since I assumed that B is a bundle isomorphism, which means this map beta is nowhere zero, I can bound this term from below by the l2 norm of eta. So this is greater than or equal to some constant c1 times the l2 norm of eta squared. And that's all I need to care about that term. Over here, I really have two terms in the cross term. So let's look at the more harmless one first. The pairing of beta eta bar with eta, that also is clearly its absolute value is bounded above by another constant times the l2 norm of eta squared. And I have to worry a little bit about the other part of the cross term. So I need to work a little bit to estimate that properly. So I'd be much happier with that term if it didn't involve a derivative of eta. So you have an l2 pairing of something with something else that's a derivative. What do you do? You can integrate by parts. Let's see. Real part of inner product beta eta bar with d bar eta is literally the real part of the integral of the conjugate beta bar times eta times d bar eta integrated over t2. Now use the Leibnitz rule and say that's integral of d bar of the whole thing beta bar eta times eta minus the integral of d bar beta bar eta times eta. Integrating d bar of something over a closed manifold would give me 0 due to Stokes theorem, or probably even reduce that to the fundamental theorem of calculus. That's 0. Over here, I can expand a little bit further and say minus real part d bar beta bar times eta times eta, still minus. This is one of those arguments where if you get one sign wrong, you're really dead, beta bar d bar eta. OK, hopefully I got the signs right. And I am missing an eta. Yeah, here. Thank you. That's important because I want this term to be the same as that term that I had on the left-hand side. So I can put this last term on the left-hand side. So I have twice that equals this. So the thing I'm trying to estimate actually equals minus 1 half real part integral of d bar beta bar eta times eta. And I don't really have to care about the details of this anymore either. I just want to say the absolute value of all this is now less than or equal to some other constant times the L2 norm of eta squared. So I put that all together. Now this whole thing. Sorry, Chris, are you somehow saying that d bar operated as self-projector? No. I am saying that if you take a complex valued function on the torus and integrate d bar of it over the whole torus, you will always get 0. And that you can prove that using Stokes theorem. But that means you're just throwing d bar from one term onto the other. Yeah. Isn't that some L2 in her product? No, no, no. I mean, it was an L2 product to start with, but I wrote it as a product of complex numbers over here. I took conjugates of the terms on the left. Yeah? So this is an integral of a complex valued function that's expressed as a product of complex valued functions. It equals that L2 in a product. It's just the real part of it. Right. So I'm defining my L2 product to be the real part of this integral. So when you write eta times eta, what do you mean? You've got eta's a complex number. Eta's a complex number. So it's just a square of that. Yes. It's a function. Yes. And you've got some measure on T2 that you're integrating. The usual one. Yes. So these are good questions, and they kind of allude to the fact that one can do all this in a much more general setting, but it causes a bit more of a headache. So this is not a uniquely low-dimensional phenomenon I'm describing. Even though I'm describing a low-dimensional proof, you can do it in higher dimensions. You can also do it on more general domains. It just requires several extra steps that I don't have time for. So summarize this. This d tau of eta L2 norm squared is now greater than or equal to some constant, which I'm going to change the name of the constant, c1 prime times t squared minus another constant, c2 prime times tau, oops, tau squared tau, times L2 norm of eta. And that's the proof. If eta is non-trivial, then this cannot be zero. As long as tau is sufficiently large. Why did you do it? You felt like there's less than 1, but you immediately say let's take tau to the big, rich stuff? That the first one goes with tau squared and the second one? First one, yes, goes like tau squared. I mean, the main difficulty in this argument was that I had to get control over this term that has the derivative of eta in it and relate that to just the plain L2 norm of eta. That's what the integration by parts is for. And that's also where I used. I mean, you may not have noticed it so explicitly, but that's where I used the assumption that my perturbation term is a complex anti-linear perturbation. So it's specifically because it appears in there as beta times the conjugate of eta. If I didn't have eta bar on the upper left, this argument wouldn't have worked. As close as I can come to giving you an intuitive reason to believe this. I'm sorry. The constants there, like C3 and C2, but C3 is like the norm of del bar beta bar. It's whatever is convenient. There is one. Do you believe that there is some constant that makes that true? You stare at it a little bit longer and you'll be fine. All right, so I'm almost done. I have one more bit of magic to pull off to finish this proof. And that's a bit of analytic perturbation theory. So I won't have too much time to. How much time do I actually have? That's always the answer, isn't it? Analytic perturbation theory. So what I'm about to say can be done in the real analytic category, but I don't want to because I'm a little bit allergic to the real analytic category. So instead, I'm going to work in the complex analytic category and complexify my operator first. My operator is only real linear, even though it's acting in a complex vector bundle. So what I can do at the expense of having two complex structures in the picture instead of one is I can complexify the domain and the target of the operator and consider the canonical complex linear operator, the extension of that, to the complexification. So, oops, more precisely. So we've got d tau is a real linear operator from some space of complex-valued h1 functions to complex-valued l2 functions. Let's take, actually to be more precise, we can say this is h1 functions from s1 to r2. And calling it r2 instead of c will help avoid some confusion coming up, as I'm about to complexify it. Sorry, yeah, t2, not s1. I'm a nasty feeling it even says s1 in my notes. That's not so good. So if we complexify, this becomes an operator I'll call d tau c from, call it h1c, which you can think of as, well, the tensor product of this Hilbert space with c, which is equivalent to the space of h1 functions from t2 to c2. And it's going to map that to corresponding space l2c. And it's complex linear. And what I can do now is allow my parameter tau to be complex instead of just real. So now the map taking tau to the operator d tau is a holomorphic map from c to the space of Fredholm index 0 operators, complex linear, from h1 complexified to l2 complexified, which, of course, is an open subset of the space of all bounded linear operators, complex linear between those spaces. So this is just a complex Bonnock space. I have a map from c into that Bonnock space. It is easily seen to be differentiable with respect to the complex variable tau. It's, in fact, it's an affine map. So that's a holomorphic map into this open subset of Fredholm index 0 operators. So the last step is then to observe the following. The space of non-invertible operators, so operators that are not isomorphisms, sitting inside this space of Fredholm index 0 operators, is what we call an analytic sub-variety, a complex analytic sub-variety, which means locally you can express it as the zero set of some holomorphic function on that infinite dimensional, so open subset of an infinite dimensional complex Bonnock space. But that function is valued in something finite dimensions, valued in c, in fact. So what we end up with is the set of all parameters tau in c with the property that the complexified operator d tau c is not an isomorphism looks locally like the zero set of a holomorphic function from c to c. That's why that set is discrete. We already showed that it's not everything, because for very large tau, the operator is an isomorphism. So once you know then that the zero set must be discrete, you're done. And then, of course, you have to think a little bit about relating this statement about the complexification to the original real linear operator. That's fine. Basically, you can convince yourself that if tau is real, then the complexified operator being an isomorphism implies that the real linear operator is an isomorphism. So I was going to prove this lemma. In fact, I was going to say other stuff after that, but there's no time. So if you want to know the proof of the lemma, I can tell you whenever, if my voice holds out. But not now. But not now. So let's have lunch. Are there any other questions? Why don't you say it's a discrete set? Do you really mean it's a finite set? I mean, it can't accumulate on zero on count. So I don't know what happens in, for other, large, complex parameters that are not necessarily in positive real line, that's probably how it works. I mean, you were talking of, I mean, your main interest was in the zeroes when tau was real, and near zero, because we were making a little count of it. This was why I haven't thought about the answer to your question very much. So it's probably true. But if you're saying a zero is a homomorphic function, they're isolated, aren't they? They're isolated. They're isolated. So in other words, if it's zero, it's zero, and it's a whole neighborhood where it's not zero. Correct. There's a whole neighborhood where it's discrete. Well, oh, I see, a discrete including zero. OK. Please repeat what I mean up for just the picture itself. It's always d tau, I mean, if this is what you do with this, d is d, and d can't be tau. How are we going to kind of show you the picture? So, OK, my goal, I had a specific normal Cauchy V1 operator for this multiply-covered curve. So this is what in this abstract discussion at the end, I'm calling just plain d. And then d tau becomes just some perturbation of this normal operator, which corresponds to some specific choice of perturbation of the almost complex structure in the neighborhood of the curve. I'm going to reset the clock and restart at 2. If you're new, let's thank Chris again.