 I wanted to mention that, so I don't have actual lecture notes for my mini course, but on the other hand, most of the contents of what I'm talking about today are in a blog post that I wrote several months ago, and this is the address. So if you're curious about that, in particular, theorem three, this was the theorem that I have to talk about the proof now, about transversality for holomorphic curves and simpletizations for generic R invariant J. This is a theorem that, so the original proof was by Dragnev, about 2004 or 2005, 2004, I guess. There's another proof of that in a paper by Bourgeois, in the appendix of one of his early papers. But both of those prove something slightly less general than what I stated. They don't deal with stable Hamiltonian structures. They deal with context structures and use that assumption in an essential way. So the proof I'm talking about is very different, and the only place where it's really written down anywhere is in that blog post. So let me say what is the main thing that I need to prove. The general outline of the proof is the same as before, but the main difference is that if J belongs to this space of R invariant, almost complex structures, then what is the tangent space to that space look like? Well, if Y is an element in that, I imagine that as really a section. It's equivalent to a section of the bundle psi. Rather, the endomorphism bundle psi, let's even say complex anti-linear endomorphism bundle psi. So in any case, that's what kind of object it is. And then I would define some fluricy epsilon type of space and talk about why having some small c epsilon norm. But the point is the tangent bundle of the manifold I'm working with is this is r cross m. This has a natural splitting into one complex sub-bundle that's generated by r and the red vector field, and another complex sub-bundle that's generated by psi. And in this form, Y is some kind of block matrix, which is 0 in the upper left, 0 everywhere in fact, except in the lower right. So when I think about the universal modulized space at some point, so the linearization of the operator that defines that space locally looks the same as usual. So that's exactly the same formula I wrote down before. But Y is much more constrained than it was before. And Y was really what made the theorem work the first time around because I was able to make quite general choices of Y and contradict the possibility that there exists this section in the dual that annihilates everything in the image of this. So last time you emphasized that you had an infinite dimensional space. It's still an infinite dimensional space. It's like one dimension less, but it's still infinite. It's still, well, no, it's, yeah, no. I mean, I would say it's dimension is something like n minus 1 times infinity compared with n times infinity. I mean, I've still gotten rid of infinity. So in order to actually take care of this, I need to think more about splittings. I have Y written in this block form here, but it turns out to be helpful to think in terms of a slightly different splitting, which one can even do in completely general situations. And that's to think in terms of the normal bundle with holomorphic curve. And now you're going to ask me, are you assuming that the curve is immersed and therefore has a normal bundle? And I'm going to say no. But it has a normal bundle anyway. So here's what I mean by that. In general, let's go back to the general situation for the moment. We can be mapping to any symplectic or boredism completion. This is a holomorphic curve. Now, this gives rise to an operator, which I'll denote by dUc, what's that? It is the complex linear part of the usual linearized Cauchy-Riemann operator, dU. So you can take always a complex linear part of any real linear operator. And now it's not too hard to check that this one also is a Cauchy-Riemann type operator. So since it's complex linear, this defines a holomorphic bundle structure on the pulled-back tangent bundle, where you say a section is holomorphic if and only if it's annihilated by this complex linear Cauchy-Riemann type operator. The advantage of doing that is that we then have already a holomorphic section in the picture, not of precisely this bundle, but of a bundle associated to it proposition, is that dU, what's that? That is a section of the bundle of complex linear homomorphisms from t sigma dot to u star tw. And now, since u star tw has a holomorphic bundle structure and the tangent bundle of Riemann surface obviously does, this thing also has a holomorphic structure inherited. And the claim is that this thing is a holomorphic section of that bundle. So this is an observation which I first learned about from a paper by Ivashkovic and Shavtushin. And it's not so hard to check that this is true. It really just follows from the Cauchy-Riemann equation for you. But it has some non-trivial consequences. For instance, this implies already, if you didn't know this, that critical points of a J-holomorphic curve have to be isolated. Because those are simply the zeros of this section, which is a holomorphic section with respect to this holomorphic structure we defined. And it implies a lot more than that. It implies, well, think about what zeros of holomorphic sections look like in general. Locally, you can think of them as, let's say, this is a cn-valued function on neighborhoods in c. Now, if it has a 0 at a point, that's going to be some power of z times some other non-zero holomorphic function. And that holomorphic function has a non-zero value at that point. That points in some direction. It defines some complex line in cn. So what this is doing is defining a tangent space to our curve, even at a point where it's critical. So there exists always this well-defined tangent space. One way to say that is there exists a smooth sub-bundle, I'll call t sub u, subscript u, in the pulled back tangent bundle such that t u is equal to the image of du at non-critical points. So the image of du has a natural extension to a smooth sub-bundle. And then I can define what I mean by normal bundle. It's just going to be some complex complement of that sub-bundle. OK? If I have a family of these things, I can say one parameter family of these things, do I get a section of the holomorphic section of the pull back of this u up to star kw? You know what? Just like you know, could I spend some time? So one parameter family of what? I mean, you're saying that if I understood this pullback of tw, that's a perfectly good holomorphic bundle in sigma. And if it was an L-thread geometry and I was trying to get it, and I had a map, and I had a family of maps, one parameter family of maps, then I get the tangent space to the space of maps is the sections of the pullback of the tangent bundle of u. Sure. And I just wondered, do you get such a thing here, too? I mean, you're asking whether the tangent space to the family then is going to be a holomorphic section of this bundle. Is that the question? Basically, yeah. Not quite, I think. The tangent space to that section is going to be annihilated by du, but I'm not sure about its complex linear part. That's something slightly different. So I wouldn't make a claim about that. What I have claimed is that, for reasons similar to what you're talking about, the complex linear part, well, both du and its complex linear part, define a, well, they induce a holomorphic structure on this homomorphism bundle such that the derivative of u is a holomorphic section. I can't claim more than that. I think if you know the answer in algebraic geometry, holomorphic curves in the values are entirely algebraic. So I mean, I suppose in algebraic geometry, you probably assume the complex structure is integral anyway, in which case du is already its own complex linear part. And then the answer to your question is yes. Oh, have you said this is a holomorphic bundle? So by holomorphic bundle, I mean the standard definition. I mean local trivializations whose transition maps are holomorphic. I had to do something slightly artificial to create that here. Because the linearized Cauchy-Riemann operator I have for you is not a complex linear operator because j is not integrable, it's only almost complex. The smooth structure coming from the complex structure is not smoothly compatible with the original one. I think it actually is. It is. And it's sort of non-obvious, but I think it's true. I did think about that at some point. So let's just move. I don't think. I think it is smooth. So why does du seem by holomorphic bundle structure in this circuit? It is a Cauchy-Riemann type operator. That's something to be checked. You have to check that it satisfies the Leibniz rule for Cauchy-Riemann operators. And then it is complex linear by definition. So then any complex linear Cauchy-Riemann operator defines a holomorphic bundle structure because there's local existence of holomorphic sections. But I think it's only double. Now, I think I did something with Sikaral symbol. I think the two differentials of structure, the overlaps, are only double and one p. Well, were your maps smooth in what you did? Yeah. Because, yeah, so I think the Cauchy-Riemann operator is something which is L-infinity. No, no, you're talking about proving the similarity principle or something. Yeah, this is different. Nothing here is L-infinity. Everything's smooth. I'm not so sure that you don't have to do something. I mean, the operator has smooth coefficients because all I did is take the original operator and remove its complex anti-linear part. That's some smooth 0th order term. So the local solutions to that linear equation are smooth. And we can debate it during office hours, if you like. But just understand, if W had a complex structure in the honest sense, there'd be no need for any of this. Then I wouldn't have had to take the complex linear part. Yeah, I could have just taken du itself. So this is some fix for that. So in that sense, the holomorphic structure I'm defining on the pulled-back tension model is the one you think it is. All right. So du is a bundle only on the complement of critical points? No, it's defined everywhere. Image du is only defined on the complement of critical points. du is an extension of it that's also defined at the critical points. This is the point of this discussion. So it's due to the fact that if you have a cn-valued holomorphic map on c that is 0 somewhere, even at the point where it's 0, it has in some sense a well-defined direction, right? Because it's some vanishing holomorphic function times another one that's not 0. That determines the direction. I would say, can't you also say the dross mining of the direction is proper? Is there any limit along the curve in this? I don't know, maybe. I haven't tried to say it that way. There could be several spaces there. No, because you take no limit along the complex curve. All right, so there are other ways to see this also that this bundle is well-defined, seeing that it smooths us a little bit harder. But to see that there's some continuous bundle is easy. If you look, for instance, at the Mikalov white formula that tells you what you can look like in the neighborhood of a critical point, then you do see very clearly that there's a well-defined tangent space at the critical point. In any case, the complementary bundle is what I'm calling generalized normal bundle. Now that I have this splitting, du was not looking at the complex linear part, but just the original du. That's a Cauchy-Riemann type operator on u star T w, which is now split into tangent and normal parts. So I can write this operator in block form. I'll write it as du t is the tangent to tangent part. There's a normal to normal part. There's a normal to tangent part. And the other part is 0, which is a claim that can easily be checked. So I claim this is what it looks like, and du and dn, du t and du n, are Cauchy-Riemann type operators on the bundles Tu and nu respectively. So the reason for this vanishing part, what I'm saying here is the Cauchy-Riemann operator du takes sections that are in tangent direction to other sections in the tangent direction. And that's basically an expression of the fact that given a holomorphic curve, I have lots of other holomorphic curves that I immediately signed by just reparameterizing the domain and perhaps changing the complex structure. The word Cauchy-Riemann type. Sorry? What is the word Cauchy-Riemann type? Maybe this is a good time for that. So say e is a complex vector bundle over sigma. I have a first-order differential operator takes sections of e to 0, 1 forms valued in e, and it satisfies real linear. Yeah, real linear. Well, that'll follow from what I'm about to write. d of f times eta equals d bar times eta plus f times d eta for all functions f in c infinity on the base, real valued, not complex value. So that's the definition of a real linear Cauchy-Riemann type operator. And d sub u is 1 in general. So now I've got these two new Cauchy-Riemann type operators in the picture. dut is defined on a line bundle. d urn is defined on a vector bundle of rank 1 less than u star tw. So that's a slight simplification. So the main step in the proof of theorem 3, of course, we need to prove that this linearized d bar operator is subjective. Well, what does that look like with respect to this splitting? Now, let me write eta as a section of u star tw. So that has a splitting of its own into a tangent part and a normal part. And I have the general formula up there still. Jtu of y is the first term. What is that? Well, y is acting on the domain. Tu is sending you from the domain into the tangent part of u star tw. And j is preserving that because the curve is jhalomorphic. So that only is a tangent part. There's no normal part there. And since actually, tu commutes with j, I can rewrite that as tu of little jy. The next term, du eta, I need to write that as so there's the tangent part, dut, acting on eta t. And then there's also going to be dun acting on eta n from that term for the normal part. There's going to be more stuff in the tangent part of the target. I'm going to ignore that stuff because it won't matter. The only thing I have left is there's a normal part of the last term, which looks something like, let's write pi n for the projection to the normal subbundle of y of tu of j. Why didn't you have a y in the first term? That's what it's behind the dot, dot, dot. Because what I'm about to tell you says you don't have to care about it. So main lemma, following lemma is surprisingly useful for a whole lot of things. The operator that sends the pair y on tangent part of eta to what I have up here in the top of the tangent part, tu of gy plus tangent part du eta tangent part. That's subjective. So let me postpone the proof of that for the moment. It is, in particular, let's say it's what's it mapping onto. Mapping onto the space of Lp delta sections of bundle of anti-linear homomorphisms of t sigma dot to tu tangent part. OK? Can you, I've never been lost in why we're doing what we're doing. The order at the very top, the one that's solving this problem, how is that related to the tangent direction and the normal direction? I haven't told you yet how it's related to the tangent and the normal direction. Presumably that was behind the line, I mean, the problem that we're dealing with is that we have a different splitting of the tangent bundle to this inflectation. And the y is non-trivial in only one block with respect to that splitting. You have to be patient. I want to know, fundamentally, I just want to know if this operator is subjective. I've just shown you that the top part of it is. Well, I haven't shown you what I've claimed it is. What this means is in order to show this whole thing is subjective, actually, all I have to do is equivalent, this is subjective if and only if the map that takes the normal part of eta and y to this stuff. That needs to be subjective, which would mean that's mapping onto sections of class Lp delta of hum bar from t sigma dot to the normal bundle. So I simplified my task a little bit because I can now, in fact, I no longer have to pay any attention to Teichner space. All I have in the picture is this normal Cauchy V-mount operator that's on a bundle that's a little bit smaller than my original one, plus this extra term, which looks very similar to a term that I already dealt with in this morning's proof. So my task is simpler. There is also a relationship between the tangent normal splitting and the splitting of psi versus the other stuff up there. Because remember, one of the conditions in the theorem said, I need my curve U to be not everywhere tangent to psi. I'm going to focus on a point where that's true. So you are using here that the dots kind of don't depend on eta, right? The dots do depend on eta, but I don't care because the rest of it's subjective, right? I mean, whatever the dots are, I can minus what you want to call them. No, but whatever the dots are, oh, sorry. The dots don't depend on eta t in particular. They depend on eta n. Yeah. Right. I mean, there's a cross term in the splitting of the Cauchy V-mount operator, which that's part of the dots. Can you say again, what was the condition that something was not everywhere, what? You can ask someone who's taken notes. I'm sorry. So U is not everywhere tangent to psi. That was the main condition. All right, so let's go ahead and prove that this last thing is true. This operator, this normal Cauchy V-mount operator plus the extra term is always going to be surjective. And again, y is varying in an infinite dimensional space, but it no longer seems like I've lost so much information by only having one block of y vary because I'm also only looking at a block of similar size when I'm looking at the normal bundle. So if it's not surjective, then again, there exists some object theta, which is not 0, that's orthogonal to the image of that operator. And that means that'll be a section of class LQ with exponential weight minus delta of that bundle, such that du normal acting on eta n paired with theta vanishes for all eta n, that's condition 1, and condition 2, the second term, pi n. So normal projection of y of tu of j paired with theta equals 0 for all y. Only two conditions. I had three conditions before, one of which I didn't use. The thing is I've already sort of used the condition involving Taichmuller space. The main lemma that I stated here depends on that. This first term here is what depends on Taichmuller space, and the lemma wouldn't be true without that term. So I'll come back to that in a little bit. Now, again, condition 1 tells me I have a weak solution of a formal adjoint equation. So that's going to mean theta has isolated zeros. It's annihilated by some Cauchy-Riemann type operator, and it was non-trivial by assumption. Sorry, maybe. That's better. So the game is, again, choose an injective point, choose y in the neighborhood of the image of that point so that we make this term point-wise positive, and then we get our contradiction. So can we do that? Well, let's pick z as an injective point, by which I actually mean I want it to be an injective point for the map u m, which takes my domain into m, not r cross m. And I can do that because there is this lemma that I stated saying that injective points of that map are also dense. And also, d u m at z is then non-zero. Another way of saying that, which is convenient, is this implies that u at z at z is not tangent to the sub-bundle generated by r in the rate vector field. Because by the holomorphic curve equation, that's something that's true if and only if the derivative of this map u m vanishes. So that's usually not true. I'm free to assume at this point it's not true. But also, I assume u is not everywhere tangent to psi. So let's also assume u at z is also not tangent to psi, which means it's necessarily transverse to psi as the tangent space to u is complex, and so is psi. So they're transverse. That means I have quite a lot of freedom in how I define this generalized normal bundle. I could define it so that it actually matches psi at this point and nearby. So let's say without loss of generality, the normal bundle equals psi near that point. Do you have some sort of density for when your curve is not tangent to psi? Yeah, that's a great point. I don't have that. That's not true. And is there something at the intersection of injective points and not tangent? So at least I know that the injective points are dense. And the points where this is not tangent to this are dense. So I just have to find any point where this condition is true and perturb it a little bit so that these are also true. I didn't specify that I'm only perturbing in some perturbation domain. I can have this be anywhere in M. Of course, one can prove theorems that say you just perturbed J in some particular region, but then you have to be careful on precisely this detail. So having chosen this, I can now pick y in a neighborhood of the image of z so that, well, t u of j is some non-trivial map in the tangent directions. And y is going to act on that also non-trivially because of the fact that what would be problematic is if my curve were tangent to the other bundle at that point. Then y would not do anything. But as long as that's not true, as long as those two spaces are not identical, y acts non-trivially on something in the tangent direction. And now I can use the usual linear algebraic lemma to make y do whatever I want. And this normal projection no longer does anything. Why is sending you to somewhere in psi? And I've arranged it so that psi and the normal bundle are the same. I didn't understand the usual linear algebraic lemma in the last lecture on this. Yeah, I didn't state it. So the linear algebraic lemma basically says in the class of matrices that arise in the tangent space to the space of complex structures compatible with omega, you can always, so given any non-trivial vector and any other vector, you can choose one of these matrices that maps that first one to the second vector. That's what you need to know. So I can do this now such that point-wise, these guys match. So their point-wise inner product is 0, or sorry, is positive, but 0 outside a neighborhood. And then I have my contradiction again, same as before. So that proves the theorem up to this detail about this functional analytic lemma. So reducing it to the normal bundle again, this has the advantage that I reduce it to a problem involving a bundle that has the same rank as the bundle in which y is doing perturbations. That at least makes it plausible that I should be able to perturb enough to make the operator surjective. I just have to be careful about the issue whether u is tangent to psi or not. If u is everywhere tangent to psi, then this argument can't work. As long as I can find a point where that's not true, I'm fine. So what about this main lemma? Let's move over here. So I don't think I'll do the general case, but let's consider the case where u is immersed, which means the normal bundle of u is what you think it is. You don't have to do anything fancy. So the lemma says I have an operator that takes y tangent to a Taichmuller slice and eta t, a section in the tangent part of u star tw, sends it onto this bundle. So one of the nice things about the immersed case is that my bundle t sub u has a very simple definition as simply the image of du everywhere. So actually, du now defines a bundle isomorphism from t sigma dot to t sub u. And one can then show that the lemma is then equivalent to showing that the following operators surjective send something in the tangent space to the Taichmuller slice plus, let's say, w1p sections of t sigma dot. I'm going to put a subscript in here, which I'll explain in a moment. Well, let's just say these are the sections. So vector fields of class w1p, let's actually pretend we're on a closed surface for the moment, and talk about vector fields of class w1p, which vanish at the punctures and at the marked points. So I'm going to send that to lp sections of bundle of anti-linear maps from t sigma to itself, specifically sending the pair of yx to jy plus d sigma x. What's d sigma? That's the standard Cauchy-Riemann operator on t sigma that comes from just the holomorphic structure of t sigma. So one way to think about that is it's the linearized Cauchy-Riemann operator at the identity when you're talking about holomorphic maps from sigma to itself. So my claim, which is not terribly deep, it's just a matter of staring at it, is that under this isomorphism you get between the tangent bundle and t sub u, the operator I'm claiming is surjective, that's equivalent to this operator here being surjective. Take yx to jy, standard Cauchy-Riemann operator acting on x. The place where I'm really being a bit sloppy is that I changed my punctured surface into a closed surface to make my life easier. So technically, I skipped a step. I should have said something about these operators acting on a punctured surface, and this d sigma would be the linearized Cauchy-Riemann operator for the identity as an asymptotically cylindrical map from sigma dot to itself. OK? That would have been more complicated to write down, but one can compute without too much trouble that that map is equivalent to this map in the sense one is surjective if and only if the other is, and they're both fred home with the same index. So to finish this up, I really just need to take a closer look at this standard Cauchy-Riemann operator on d sigma. It has its kernel and co-kernel have meanings, which are significant. So in particular, the kernel of d sigma acting on the space of sections that vanish at the punctures and the mark points, that's just the tangent space to the automorphism group at the identity. As the holomorphic vector fields that vanish at those points, the tangent space at the identity to odd sigma gamma union theta. That's nice. What is the co-kernel? Well, let's think for a moment about what the image is. Well? So where did sigma learn about the punctures and mark points? Where did d signal learn about the punctures and mark points? So yeah, that's the step that I skipped. So what's the question exactly? Is it the case that someone's definition of d sigma is some term having to do with punctures? So right now, it's not the case. Right now, d sigma is acting on vector fields on a closed dream on surface. I put the punctures back. But I required the vector fields to vanish at the punctures. OK. That vanishing condition is what makes this operator equivalent to the one for the punctured surface. Yeah, the punctures disappeared. So the exponential weights don't make a sense anymore either. So here's what I really want to say. What would be the tangent space at j to the following object? Take the action of the diffeomorphism group of sigma fixing mark points and punctures acting on j. So the orbit of j under that group action. I'm asking, what would this tangent space be? Well, I'm just going to call this an exercise so that I don't bore you too much right now. That's the image of this operator. That's just a computation one can do basically with a lead derivative of diffeomorphisms acting on j. You can get everything in the image of this operator in that way. Now, remember the definition of Teichmühler space? It's the quotient of all the complex structures modulo this diffeomorphism action. So that tells you actually there's a natural correspondence between the co-kernel of this standard operator. And Teichmühler space, I think I called it curly t. So this Teichmühler space, remember, is a space of all complex structures divided by the action of this group. Co-kernel similarly is everything divided by the image of that operator. So that's the meaning of the co-kernel. And in particular, I'm sorry, I really mean to say a tangent space to Teichmühler space, which is the same thing by definition as a tangent space to j in my Teichmühler slice. So the right way actually to design a Teichmühler slice would have been just pick any smooth family of complex structures that at j is tangent to some complement of this Cauchy-Riemann operator, the standard Cauchy-Riemann operator. That's analytically the cleanest definition. And now it actually becomes more or less a tautology that this operator is surjective. Because you look at it, what does it do? It takes y, which is in a tangent space to the Teichmühler slice, just multiplies it by j, which is nothing, takes some vector field x and acts on it with the Cauchy-Riemann operator. Well, this jy is still going to sit in this subspace, which by definition is a complement of the image of the Cauchy-Riemann operator. So that's the reason for that being surjective. It's just that almost by definition this tangent space of the Teichmühler slice is complementary to the image of the Cauchy-Riemann operator. So that's at least a sketch of the proof of the immersed case. For the non-immersed case, I'm just going to tell you it becomes more likely for this operator to be surjective rather than less. It becomes an operator of larger index. Do you formally reduce to the immersed case by blowing up the target? Blowing up would mean what in this case? Complex or real? I think the smooth structure puts you on that. If it's not there, I mean, if j, if it's not into the ball, it's going to behave nicer than the ball. If it doesn't on the ball, it's not into the ball. I mean, it's not going to seem like that. Um, I don't have very much time left. So I'm going to take a vote about what to talk about in the remaining time. Uh, one possibility would be to talk about automatic transversality, which was in my abstract. And that is, follows on to this discussion very nicely because it's another application of the normal Cauchy-Riemann operator. Another possibility would be to say a bit about why the exponential weights are there. I'm not sure if I can do both. I can try to do both. Automatic transversality? OK. So let me say a brief word verbally about the exponential weights while I'm erasing. OK? The key, ah, unfortunately, I erased the splitting. Remember, before I erased it a moment ago, there was this splitting. So on this inflectation, the tangent bundle looks like this line bundle term generated by r in the rate vector field plus psi. So the relevance of that is there's a lemma that's fundamental to fluorohemology. If you read, for instance, Solomon's lecture notes on fluorohemology in chapter 2, you come across this lemma that tells you under what conditions a linearized Cauchy-Riemann operator on a cylinder will be Fred Holm. And that operator has to have a certain asymptotic form. At infinity, it decays to this so-called asymptotic operator. So the asymptotic operator, if you think about by analogy with Morse homology, that plays the role of the Hessian of the Morse functional. A function is Morse if and only if that Hessian is an invertible operator. Similarly, the asymptotic operator, that's something that, in theory, is supposed to be invertible if and only if your rape orbit is non-degenerate. And that's precisely the situation where the Fred Holm property holds. But what you actually have here is because of this splitting, the asymptotic operator also is written in block form. So it ends up looking something like minus i times derivative of t, so acting on functions on the circle, and then some other thing down here with a 0th order term. This part is going to be invertible if and only if the orbit is non-degenerate. But this part is not invertible no matter what. It's a nice Fred Holm operator of index 0, but it has kernel. And that's sort of automatically there because there is some kind of degeneracy built into the problem because you can reparameterize rape orbits in an S1 direction. And there's this additional artificial R direction in this emplactization. So there's two directions of degeneracy there. The exponential weight is a crutch that lets us get around that because if you work in an exponentially weighted space, your Cauchy Riemann operator is equivalent. So the exponentially weighted space has a not quite canonical but kind of natural isomorphism to the usual Sobolev spaces. But under that isomorphism, the Cauchy Riemann operator changes. It changes precisely by plus delta or possibly minus delta. So for any non-zero delta sufficiently small, this operator becomes invertible. Can I vote for this topic instead? Anyone else? I can also talk about this more in office hours. Sorry. So that's really that's just a technical crutch. And actually, I will also say that once you've applied this lemma and reduced everything to a problem about the normal bundle, now you can forget about the exponential weights. They no longer play any role if your ray orbits are non-degenerate because reducing to the normal bundle, well, asymptotically, the normal bundle is the same as psi. Tangent directions are the same as this. So you lose this part, and you're just left with the part that actually is invertible. And now the exponential weight is still there, but it doesn't matter if it's there. If you adjust it a little bit, your operator remains invertible. So now you can forget about it, and you still have a perfectly reasonable Fretholm operator for the linearization. All right. So what's the other nice consequence of this normal splitting I talked about? Well, again, just talking about a general holomorphic curve in completed symplectic cubordism, one of the things we learned is, I didn't say it explicitly, but it is implicit in this lemma I stated. This thing is Fretholm regular if and only if the normal Cauchy Riemann operator is surjective. So now you can forget entirely about Taichmuller slices. You can forget about the tangent direction, just restrict to the normal direction. This is the operator you need to be surjective. And at least if the curve is immersed, it turns out that the index of this operator is the same as the dimension of the modularized space. If you have critical points, that's not quite true. Let me actually write that down. So remember, du is a holomorphic section of a certain bundle. It's not just homomorphisms from t sigma dot to u star tw, but actually it has its image by definition in the tangent part. Now, this is a line bundle here. I can count its sections. Let me make my life easier and remove the dots. Let's just talk about a closed Riemann surface for the moment. And now I can compute c1 of this bundle. That's going to be the number of critical points, the algebraic number of critical points. Each of them counts positively, possibly with some extra positive order. And that c1 of this bundle, which is c1 of tu minus c1 of t sigma, that's the Euler characteristic of sigma. So the difference between c1 of tu and the Euler characteristic is exactly this number of critical points. And that means the generalized normal bundle has c1 equal to what I would usually call c1 of u. That's by definition c1 of the pullback tangent bundle, minus c1 of the tangent part, which now is Euler characteristic plus number of critical points. So I have c1 minus Euler characteristic minus number of critical points. So having more critical points makes c1 of my normal bundle smaller. Now the point is, I can say a lot more of c if the normal bundle happens to be a line bundle. So question, if E is a line bundle, when must Akoshi Riemann operator on E be subjective? Well, why should you believe there is any such condition? Well, on a line bundle, the index formula involves c1, which is also related to some count of zeros of sections. Let's try and use that fact. D, Akoshi Riemann operator D, let's call it that, is going to be surjective if and only if its formal adjoint is injective, call that D star. So what's the formal adjoint? I don't really need to know much about this, but just in pure generalities, the formal adjoint is something that's conjugate to Akoshi Riemann type operator on some vector bundle. And its index is minus the index of D. So this is Akoshi Riemann type operator on some bundle, let's call it E hat, such that the index of D star equals minus the index of D. This is a very unusual sort of mini course for me, because I haven't, until this point, needed to claim that I know what the index formula is. But now I really need it. So if we know the index of D, if we know c1 of E, we can deduce c1 of E hat from all this. Index of D star by Riemann rock. We're talking about a line bundle, so it's going to be for algebraic geometers. This is the real index, not the complex index. Euler characteristic of the domain plus 2c1 of E hat. That's minus index of the original D, which means minus Euler characteristic of the domain plus 2c1 of the original bundle. And this tells me that 2c1 of E hat equals minus Euler characteristic times 2 minus 2c1. Let's erase the 2s and just say this is minus Euler characteristic minus c1 of E. Now observe, if c1 of E hat is negative, then D star is necessarily injective. Why is that? It's a count of zeros. So if there's something in the kernel of D star, that's a holomorphic section for some definition of holomorphic. Its zeros are necessarily positive. So it cannot have any since, well, it can't exist, in fact, because a negative count of zeros would not be possible. This is where what I'm saying is absolutely unique to a complex line bundle. You can't do it with higher rank. So that, of course, is equivalent to saying the original operator was surjective. So I get this criterion that says, if c1 of E hat is negative, now what does that mean? That means this Euler characteristic plus c1 of E is positive. That implies D is surjective. Now, let's put that back into what I had in the nonlinear setting. Let's say, suppose the dimension of my ambient symplectic manifold is 4 so that my normal bundle is a line bundle. And the criterion now says c1 of the generalized normal bundle has to satisfy this relation. Euler characteristic plus c1 of the normal bundle. I've got a formula for it up there somewhere. It says it's c1 of u minus Euler characteristic minus number of critical points. So Euler characteristic gets canceled out. I have c1 of u. And I have the number of critical points. This is what has to be positive. If this is positive, then it says u must be regular. And that's completely general. So there was a version of this proof for immersed holomorphic curves in the paper by Hofer-Lizant-Sikharov. I think it was already at least sketched by Gromov in his 1985 paper. This version that allows for critical points appeared first in a paper by Vatskiewicz and Chevtysian. And then there is also a punctured version, which is in a paper of mine, which I think I don't have time to talk about. But you're free to ask me questions about it. The only other thing I want to say, so this does, it does not make any assumption at all about you being somewhere injective. This sometimes can be made to work for multiply covered holomorphic curves. And let me just mention one case where that's actually useful. So let me rewrite this formula a little bit because I have the index of u equals, so in this four-dimensional case, minus the Euler characteristic plus 2c1. So this criterion is equivalent to this number being greater than minus the Euler characteristic plus twice the number of critical points. So let's write that as 2g minus 2 plus twice the number of critical points. That's a sufficient condition for transversality. Now, the punctured version says, the following, the only difference in this whole discussion with the punctured version is I need to work a little bit harder to get my criterion on a line bundle for what guarantees a Cauchy Riemann type operator being surjective. I talked about it in terms of counting zeroes of holomorphic sections. Now, c1 of a line bundle over a punctured surface is zero. It's a trivial bundle, so that's somehow not the right question to ask. But these holomorphic sections also satisfy asymptotic conditions that have to do with the asymptotic operator determined by the rape orbits. So there are winding numbers associated to those rape orbits that are somehow predetermined, and those are related to conlethander indices. Those apply the kind of constraints you need, so what you end up with after you explore all of that is a criterion that says, if the index is strictly greater than twice the genus minus 2 plus the number of punctures where the conlethander index is even plus twice the number of critical points, then u is regular. So my last comment for today is going to be following exercise. Prove that for generic J in a four-dimensional symplectic co-bordism, all index 1 J holomorphic cylinders are regular. That's part of the exercise. But isn't the whole point of solitude that you don't need? So kind of. But the problem with this formula is this term that counts the number of critical points. That's not really something you have any topological control over in general. However, if you're looking at curves of low index, there are more things you can say, because being immersed is a generic condition in some sense. It's also something I'd be happy to talk about in office hours if anyone wants to talk about that. There are results that say, looking at somewhere-injective curves, the subspace, the space of somewhere-injective curves that are not immersed lives inside a manifold of a certain co-dimension, co-dimension at least two compared with everything else. So if you're looking at curves of low index, you can exclude curves with critical points entirely. And then you have a much better chance of being able to apply this criterion. So I believe another place where you could look up the answer to this exercise is Jo's thesis, because she made considerable use of this. Or at least in simpletizations, she definitely made use of this, because it means you can honestly count index one holomorphic cylinders, even if they're multiply-covered. There is actually a better version on the archive. There's a better version on the archive. We forgive you. Yeah? So when you talk about voodoo-ex-index, it was stated in the exercise, that voodoo-ex-index is an unparameterized problem, right? The index of the unparameterized problem, yeah. So literally the virtual dimension of the modular space of unparameterized curves. That's what I mean by index. So in the immersed case, that happens to be identical to the Fredholm index of the normal Cauchy Riemann operator. And that's something that's easy to show. OK, so I'm done.