Introduction to Regularization Problems
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McDuff, Dusa
Moduli spaces of pseudoholomorphic curves arise as the zero set of a Fredholm section of a suitable bundle, and one expects and hopes that they can be regularized in order to define invariants that are stable under perturbations. This lecture provides an overview of some of the analytic difficulties that must be solved in order to construct such a regularization, and briefly explains some traditional approaches to their solution, namely via geometric regularizations and finite dimensional reductions.
 Well, hello everybody. I'm going to give what I hope is a very elementary talk, so you'll probably be very bored. But for those of you who don't know this or have forgotten a little bit, I'm just going to go over the problem of regularization and explain the traditional methods for dealing with it a little bit. So the question is this, we have a symplectic manifold and it's got an almost complex structure J, which is omega-tame. And we're looking at J-halomorphic curves. Now, to be simple, I'm just going to think of the domain being a sphere, so I'm going to write, say, m tilde top 0k of, well, decorated by all these things, J, A, say, which is going to be the set of maps. Now, I'm just going to take the sphere as my domain into m, which are J-halomorphic, which represent the class A, so that's a class in the second homology of m. And, oh yeah, I'm going to have some points, z1 through zk, which is just underline z, and these points are different. And that's just a space of J-halomorphic maps. So these elements are maps and the tilde means that the elements are maps. Now, does everybody know what all these words mean? Because I'm just assuming you do. And if so, if you don't, please ask. Okay? It's up to you. And then we're interested in this space, but also we're interested in, if you've got two of these elements, u, z, we can say one element is equivalent to another element, u prime, z prime, if there's a diagram of this kind. So we've got phi from S2 to S2, which is phi is an amherbius group. So that's a bi-halomorphic map. We've got u going there to m. We've got u prime going there, and the diagram's meant to commute, and say we've got z prime there, and that goes to phi z. Right? So this would be something of the form u prime would be phi composed with u, and this would be phi inverse of z. Okay? So we have this equivalence relation. And we're not interested really in the space of actual maps. We're more interested in the space of equivalence classes of maps. So that's going to be written as m top, 0k. So that's going to be m tilde top, 0k, divided out by this equivalence relation. So these are equivalence classes. So we're not really interested so much in the way they're parameterized, just that they have a parameterization. We're interested in the images, the geometric images in them. And I put top here because this is the top strata of some completed space. Okay? Now, for example, well, so in the best case, this m tilde top, 0k of mja is a smooth manifold, and it's vastly cut out. Oh, I forgot, I said it was j holomorphic. Well, I forgot to say what that means is that the Cauchy-Riemann operator is 0. So del bar j of u is a half of du composed with plus j, du composed with j. So that's the Cauchy-Riemann operator. And so this is 0 set on operator, and if you're saying it's transversely cut out, you mean that the derivative of the operator is subjective. I'm going to say more about that later on. But anyway, you want it to be transversely cut out, and the dimension of this thing then would be equal to the index of the operator is 2n plus 2c1 of a plus 2k, which is the number of mark points because they're varying in a two-dimensional space. So that's a dimension. And why you want it to be transversely cut out is that there's a Fredholm theory, and that means that if you vary j, the space is not going to change very much. So that's the sort of basic space of j holomorphic maps. And so in the best case, the first situation is that this is a nice manifold. And then, as I say, what we're really interested in is the Cauchy space. So this is m top. That this is almost never compact, but it's compact enough. And why this space is almost never compact? You can see that it's almost never compact. Because one thing I'm insisting, I've got these points z1 through zk, and they're meant to be all different from each other. So two of them could come together, and of course, that would not be in this space. And then another problem about the compactness is that this group of equivalences, this group here, psotc, is not a compact group. So that means you're quotienting out by something which is not compact, and that can cause problems. But what do I mean by compact enough? Well, sort of for the purposes, it's compact. So compact, what are the purposes? Well, we're interested in the evaluation map from m tilde, well really from m top, no. And let me, well, m a, so just put an a in there. I'm mapping to m to the k. So we've got a pair uz, which is an equivalence class, so I can put brackets like that, going to u evaluated at z1 up to u evaluated at zk. And that, of course, is defined on the space of maps, but it goes, it maps through the quotient. So there's a well defined map here. And what you want for this is, you want the image, image so that it represent a homology class. And so that, you see, I mean, if this were a compact manifold, then of course it would have a fundamental group, fundamental class. So you could just look at the push forward of that fundamental class. But in fact, for it to represent homology class, it doesn't have to have a compact image. What it needs is that the IE, if you look at the closure of the image, so this is the image of the evaluation map minus the image itself, so this is the boundary of the image. This should have co-dimension at least two. So that means what we say is it represents a pseudo cycle. And to give you an example, well let's take the very simplest example, I think it's a really simple example, take the projective plane with a standard J and the class to be the class of a line. And then C1, the first shown class of the line is 3. And so the dimension of M, if we take two mark points, the dimension here is going to be, it's 2n plus 2c1 of A plus 2k, which is 4, and then minus 6, which is the dimension of the reprametrization group that I've quotiented out by. Okay, so this is the dimension of Psl2c, and there will be a group. And this is 6, and this is 4, and so we get 8. And that of course is the same as the dimension of if we take Cp2 times Cp2. So if we look at the evaluation map from M top 0, 2 in the class of a line going to Cp2 times Cp2, we have uz goes to uz1, uz2. And what this is, remember these things represent a line. And so this is the elements here represent, you've got an s2, you've got two points on it, which in fact we can normalize, we can take the first one to be at 0, and the second one to be at infinity say, because the group is triply transitive, so we can fix these two points, and then we've still got an action of C star acting on these things. So this is M tilde top with a fixed point set to 0 and infinity and quotiented out by C star. So it's still something, a space of maps quotiented out by a non-compact group. And anyway, you've got this space of maps, you're looking at where they go, so you've got two points in Cp2, and you sort of take the line through them. Well, there's a unique line through any two points. Therefore, this map is injective. If we take the standard complex structure, we know that that's injective, but of course it's not surjective. And you can see it's not surjective because the two points it's going through are different. They have to be different because I've assumed that these two points are different and allowed to come together. So the image is equal to Cp2 times Cp2 minus the diagonal. Right? Very clearly. So it's not a compact image. You can see the space is not compact. On the other hand, the set of maps is somehow compact. That's by Gromov's theorem. And you can compactify this. I mean, this diagonal has codimension two. It's real codimension four, I suppose, because it's where these two points are the diagonal here. And so it does represent a pseudo-cycle. It does represent a fundamental class because the image of the boundary is so small. I just understand that the words codimension of these two means inside the closure of the image or inside the ambient? Inside the m to the k. So we're not talking about the structure of the domain here, which is to, I mean, in good cases, you can assume it's a manifold, but it's not going to be compact, you see. But you're saying, so this, it's a, well, it's a non-compact manifold. But if you look at its image in here, it has a fundamental class because the boundary is so small. Is there really a key codimension condition? What? The way that case is. No, no, it's a huge thing. Yeah. What? Codimension in the closure. Well, it's a codimension. It's a real codimension of at least two in m to the k. So whatever dimension that has. So in this case, I mean, this is a, you try to, in this case, what's wrong with it? What if m to the k was an enormous thing and this was like one curve? We really care about the closure. Oh, I see. I see what you're saying. Yeah, okay. So what do I mean by, yeah, so I'm saying this incorrectly. What I mean is that the boundary of the image minus the image should be a union of maps from some manifolds v. So v is some, this is some finite union, say. And the dimension of v should be less than or equal to the dimension of our original space m top minus two. That's what I mean. I'm sorry. You're right. Right. That makes sense. So it's codimension. It means it should have at least two dimensions less than the dimension of the image. And this is assuming m top has the expected dimension? That's assuming that m top has the expected dimension, which it's going to have if it's a, that was my first assumption that it's a transversely cut out manifold of the right dimension. So it means that it has the correct dimension. And then, right, that you're assuming that this thing is going to, then this is meant to represent a homology class of the right dimension, the dimension of the index. That's okay now. Okay. So now, let's say, so there are several things here. I mean, the regularity problem is the problem of arranging the first thing that it's transversely cut out to find if you could find a J to make it a nice manifold. That's the regularity problem. But then there's also the compactness problem that you want to make sure that this image is compact enough. Well, how can you make sure about that? Now, the compactness is actually much easier to deal with than the regularity. So let me just say a few words about compactness. So there's Grumhoff's compactness theorem, which tells you that if you have a, you know, you look at these are sequences of spheres, right? And you look at these are spheres in the same way. So we have these. So we've got some curves UK, say, from, say, S2 into M. We've got a sequence of curves. And they're in a fixed homology class. And so that means that the pullback of a mega on the S2 is fixed. It's a mega of A, the integral. And this is actually, analytically, it's the energy. It's equal to, I hope you got the formula wrong, the pullback of du squared over S2. That's the energy of u. So it's basically the L2 norm of u, measured with respect to the correct, I mean, you have to choose a metric on there and stuff, measure it with the correct things. And you get, this is the energy. So this is a sequence of curves. So these are C infinity maps, the holomorphic curves, and they have bounded L2 norm. Well, they're actually W12 norm. This is the L2 norm of the first derivative. So this is what's called the, it's a sort of boundary case. If we had, because these are holomorphic maps on a two dimensional space, if it were a bounded W1P norm for P bigger than 2, then the standard sort of compactness arguments that tell you there would be a compact subsequence or a convergent subsequence. That's just standard elliptic theory. And why the W1P norm, why P is bigger than 2 is because this norm, if you have something bounded in W1P norm, that controls the C0 norm. So that means it's an equicontinuous family in the C0 norm. So you use sort of standard results about equicontinuity to extract a convergent subsequence. But what we have is this borderline case where it's a 2, not bigger than 2. So in fact, it's possible for this thing not to have a convergent subsequence. But if it doesn't have a convergent subsequence, what's happening is that the energy is somehow concentrating near a point and you can rescale it and get a convergent subsequence. So, and that comes down to the fact that, for example, in this case here, we're quotienting out even here by a non-compact group. That's a non-compact group there. And so that sort of affects the convergence. But anyway, this is really well understood. And what Gromov proved was that there exists a subsequence. You also called UK say that converges to what Gromov called the Cusp curve or what we now call stable map. And what a stable map is, you see bubbles can develop. I'll explain a little bit in this case later on. But what it's going to be is the domain now, sigma, is a nodal sphere. So it's going to be a union of spheres joined at points. So it could look like this. It could be a bubble, you know, a set of spheres like that. That would be the domain. There'd be some marked points on here. Because your marked points would converge somewhere, we've got three marked points, they converge somewhere like that. And the map, you would then map each component into n. So you'd have a sequence of, you'd have a set of maps. And that's what, so for example, just suppose we were looking at a map from S2 into CP2, which had degree two. So that would be a quadric. But a quadric can degenerate into two lines. So that's in class two up. That's one degeneration that could happen. You can have, so that's a geometric degeneration where you have a sequence of things representing a quadric which degenerates to two lines. But you could also have a degeneration that happens because, for example, suppose you're in this case and the marked points come together. Well, the marked points are not allowed to come together here. If they do come together, then you could have a situation like this where you actually have a bubble. And that bubble contains both marked points. And here, this is what's called a ghost component in the sense that the map you restricted to this component is constant. So what you have here is a situation where you have this goes, this is a line. So you've got a map of a line and your two marked points are sitting on the domain in this sphere which is called a ghost component. So I should give you a proper definition of a stable map so you can see you can put these all together into the space of stable maps. Now, is there any way of retrieving that board? Oh, wait a minute. What about this? No. It's got a very tall person. There's a rod. Ah, this thing. Doesn't have a... What's... Oh, what? This is it. This is it. This one's got a hook. Okay. Let's see. All right. Now... Oh, God. Well... That's good enough. Oh, look, it's going up by itself. Very obedient. Okay. Any of the most points you integrate, like the thing you integrate with the energy, did you develop poles there or is it just the same? Well, no, the energy just comes from the map. So it doesn't see where the mark points are. So I want to give you a definition of a stable map. And of course, these can occur in any genus. They don't have to be genus zero. I was just restricting to genus zero for simplicity. But... So let's just do genus zero because the point about genus zero is that then you can think of your domain as a union of these two spheres. And each two sphere, you can parameterize in a nice way and give it the standard complex structure. So you don't have to worry about variations of the complex structure on the domain. Typically, you'd have a nodal surface which you'd have to allow the complex structure on the surface to vary in type of a space. You can put that in if you want, but that's just make it simple. So it's a nodal Riemann surface. So you have something and it's all... It's meant to... You know, you can have something like that. But it's meant to be connected in a tree diagram. So you're not allowed to have something like this. That would have genus. That would be a genus one thing. So it's a tree here, a table tree. And you have a certain set of mark points. You sprinkle your points around. Once you've got five mark points, you could sprinkle them around like that. The mark points are all distinct. And so they're on particular spheres. So you have to say which sphere they're on. Then you have some maps, U, which is a one map for each component. And so we have... So that's... So we're looking at two poles, which is a form of sigma, complex structure is understood here, Z, U. And then you put an equivalence relation. Well, the obvious equivalence relation that just like I had before, you just have... You have some holomorphic. This is a holomorphic map here. So you're interested in these modulo reparameterization. And then there's an all-important stability condition. Now, you don't need to... When you're talking about a stable map, you doesn't have to be holomorphic. I mean, eventually we're interested in the holomorphic ones. But in general, you can just look at arbitrary maps with nodal domains like this. But it's very important to have a stability condition. And what that is telling you, this is equivalent to saying that for each element, if you have sigma, J, U, and you look at the group of automorphisms, so a map of that to itself, this is finite. So you never... You don't allow yourself to have an infinite number of automorphisms. And what that means is... Now, if, say, on a component like this component, the map is non-constant, then the map is either injective or it's a multiple covering, and there could only be a finite number of automorphisms of that component. So if you have a component where the map is non-constant, there's no condition. But if you have a component, if the map U were constant here, then you see, what would an automorphism as this would mean? Well, it would have to fix Z0, Z1. It would have to fix this nodal point. But you see, if we've only got two... What are called special points? We've only got two special points on this component. There's a mark point and there's a nodal point. And so they're, in principle, as a one complex parameter family of automorphisms of this thing. So that would not be allowed. So the claim is that if U alpha restricted to the alpha's component is constant, S alpha to alpha has at least three special points. And special points are nodal points or mark points. That's the important condition. So that means, for example, in this particular case, I could allow that to be... That could be a ghost component. So that one could be a ghost component. This one could be a ghost component. This one's not allowed to because there's only got two mark points. That one couldn't be. So this would be a possible degeneration if you were looking at maps, say, from U from S2 to CP2 of degree two. So that it's in the class twice a line. Then this one could represent a line. That one could represent a line. And we've got five mark points. That would be a perfectly good stable map of this kind. And the claim is that if you look... So we define M bar 0k MjA to be the equivalence classes of stable maps in genus 0. So this is a genus thing. And K mark points. And representing the class A. And if you've got a J, they're J-halomorphic. So that's the space of J-halomorphic stable maps. And it could be... It's a stratified space. I mean, why I wrote top there is that top was its top component, if you like, where its domain just had a smooth sphere. But then it has all these other components where the domains are nodal curves. And some of them, if you like, are sort of artificial. Like, this is an artificial one because the two mark points have come together. And so, you know, nearby things to this would be something where you had a... And then you had this would still be constant, say, this one could have a Z2 on it. But now we'd have three mark points on that component. So... And then when these two mark points come together, we'd have this bubble sort of artificially created in the domain. This is... This was Konsevich's idea of the way of describing what's happening, way of describing a compactification. So the claim is that we have this space here. This is compact. It's a compact space. And if everything is regular... So in other words, if all the underlying maps, you see, a regular thing would be... For this, would be that this map was a j-halomorphic map transversely cut out, which I'm going to say more about later. So is that. And then all these intersections here, you've got a space of spheres here, a space of spheres here, these are kind of a concept map. All these intersections are meant to be transverse so that it all sort of comes out and it's all in the right dimension. And the strata, the formal dimension, if you like, of the strata is equal to the dimension of the original, of the top space minus twice the number of nodes. So each node gives you, contributes sort of a co-dimension two to the thing. And that's a formal dimension. That's the index. So that means that if you can arrange everything to be trans... You know, all the things that transverse, that will be the dimension that it has. But of course, it doesn't have to be like that. But if it were, then you see, if you're in a situation like this where you can arrange that all the lower strata have basically have the right dimension, then the top thing of this will represent a pseudo-cycle because this is compact. So you can compact by the image of the map by just looking at the image of these things. And then the boundary of it is going to be covered by the images of the strata, you know, the higher co-dimension strata here. And they're all going to have at least co-dimension two. So that's a sort of standard regularization picture that you compactify the image by putting in the images of the evaluation maps from the strata and this... In this space of stable maps. If you can arrange, if you can control our dimensions reasonably, then they all have co-dimension at least two and then you get a pseudo-cycle. Now, of course, the kicker is that you can't always control a dimension. That's the whole problem. But if you could, then you'd be... You'd solve this problem. You'd have got... I mean, because our aim, of course, is to try and use these spaces of maps to get some invariance. We want to get these homology classes out that we can then intersect and get the grammar for an invariance or something. And this is just for closed curves, but if you're doing open curves, you have boundaries and you put boundary conditions on. The formal structure is exactly the same. The equation looks a little bit more complicated because you have to deal with what happens at the near the non-compact parts of the domain, but formally it's exactly this idea that you can always compactify the space you're looking at by making some space of stable maps or stable buildings or something. And then you just need to control the dimensions. In principle, all the boundary components you've added should have co-dimension at least two. You need to control that and make sure it does. Okay? So that's telling you all I was going to say about compactness. But you see, I must have erased... Here, I erased the first condition which was that everything was transversely cut out. That's what I need to talk about because that's the hard thing. The compactness is sort of understood. So, so... Why don't you have to worry about the unstable maps? Like, why aren't you there somewhere in the computer? Well, the unstable maps would be... Well, the thing is that they don't appear in the compactification because if you have an unstable component, suppose that were an unstable component, so that map was constant. Then you could just forget about that bubble and put that point there. You see, the unstable components are things which don't actually contribute to the image because the map on them is zero. It's just a constant map. And so you can just collapse any unstable components. And the thing is that it's... I mean, that's a good question. I do not understand. Can you say again? Yeah, I'm saying it. Suppose we had this situation, but this sphere was a constant map there. And that would be an unstable component. But the thing is that you... So, I'm saying that that's not going to be in the compactification because we don't allow that in the compactification. What we do is, in that particular case, we just get rid of that completely and just put the mark point there and there wouldn't be a bubble there at all. Because the image would be the same. If you have like a sequence, I mean, like in your top stratum, like you can just define it so that it converges to a stable map with the stability condition. When you look at the compactness argument, what's happening when a bubble forms, typically, I mean, in this particular case, if this is what we had, all that would have happened here is that some of the mark points would have come together. And the underlying map here, you just have a sequence of maps which was basically converging to a map. And all that would have happened to create this set of bubbles is that the mark points would have come together in a particular way, sort of these two points coming together first and then these two points coming together second and then they would form this bubble. And you can actually see that completely clearly just by sort of rescaling and blowing up and looking at, you know, nearby this. You'd have... Here would be the point where you'd be rescaling. This point is going to go into these bubbles. You'd have a Z4 and Z5 very, very close and then sort of further away, you'd have a Z2 and Z3. So there'd be two scales of bubbling here. This would be the... These ones would be the furthest ones and these ones would be the later ones. But the map itself would be constant. So the map to M would be constant, but you're talking about how the mark points are coming together. And it turns out, I mean, we've proved this in exhaustively in our book on J-Homorff-Curse, a deep mark. If you just look at the way you prove the compactness, you can get... I mean, you could describe... I mean, of course, you could put in something like that and allow that as a limit. But then you wouldn't have a sort of unique limit. And it doesn't give you a good structure. I mean, this was Komsovic's realization that the good structure is... Because there would be an infinite automorphism group of those, which you wouldn't want. You wouldn't want. You'd have to divide out, you see, by a C star here. You want the set of automorphisms of all these things to be finite. That's very important. We are already doing that now. The other component that still has a C star automorphism... No, because you see, here, the map is non-constant. So you've got some map U, and you're saying U is equal to U composed with phi, if you had an automorphism. So phi would have to be something that fixed these two points. And so you'd have to have phi Z1 is Z1. Phi of the nodal point, let's say infinity is infinity. And it would have to satisfy this. Well, if U is non-constant, you wouldn't have... I mean, you could... It's a condition, right? I mean, it could happen. Well, you would only have a finite number. You see, what you could have... You could be a map, say you could have U from... You could think of C, U and infinity to M. And it's going to take Z to say Z cubed. And then you could have an automorphism which took Z to E to the 2 pi i over 3 Z. That would be an automorphism, so U of phi would equal phi. But it would be a finite group. You're not saying there aren't any automorphisms. You're saying it's a finite group of automorphisms. So if it was a multiply covered curve, it could have an automorphism. But if it's an injective curve, somewhere injective curve. So that's another thing. You have this idea of a somewhere injective curve, U from S2 to M. What that means is that there is some point Z naught such that if you look at its image and then its pre-image, that's just Z naught. So that it's a map that... Here is the domain as 2. Here's your Z naught. On this little neighborhood, it's injective. And then you see if it's injective there and you have something that behaves like this, U of phi. Phi would have to be the identity in that little neighborhood where it was injected. And that doesn't happen if you have a Mobius transformation that fixes three points. It's the identity. So really, the stability condition, the sort of geometric way of understanding it is telling you that the group of automorphisms of this stable map is finite. That's what the stability condition is saying. And that's a very important condition that allows you to... I mean, it's the fact that you could have automorphisms means that you're actually... The space here, the strata are typically overfalls rather than manifolds because you could have automorphisms. So you're modeled on an overfold. But it's very important that the finite... It's overfold with finite isotropic groups. Yeah. It just occurred to me that I think you didn't say that stable maps have to have finite energy, which is a guarantee that the stability condition actually implies finite automorphisms. Ah, okay. Well, I was saying that we're in some class A. Right. So then omega of A is a number. Right. So they do have finite energy. But you're right. I mean, if you're doing more general things, the finiteness of the energy is actually what makes this whole compactness serum work. You've got to have a bound on the energy, bound on the W12 norm to get started with the analysis. Excuse me. Yeah. How do you put the topology on and bar the location? Oh, something called... How awkward of you. Something called the Gromov topology, which I don't particularly want to go into. I mean, each stratum, it's clear enough how you to apologize it. And then nearby, well, there's a structure you see. When you have a stable map, I mean, I have to mention this at some point if I'm doing sort of the elementary beginning anyway. If you want to know what's a neighborhood of something like this, you have to do gluing. So you have to say gluing is a way of, if you've got a nodal domain, then sort of nearby, there are things which nearly have the bubbles, nearly have the nodes, but don't quite. So nearby, so here we've got two nodes. So we've got one gluing parameter here and another gluing parameter there. And for each, if you have... Let me get a slightly bigger board. So suppose we've got a single node. So here's a... We've got some map. So this... Let's have no mark points just for simplicity. This is going to some line and that's going to some line. So the non-trivial map. So this is going to a pair of lines. So the image would be a pair of lines. Well, nearby, there's pair of lines in CP2. You actually have, of course, a quadric. Impossible to draw. But how analytically do you see that? Well, you have this sphere. You take this sphere and you take... With minus a disc, you take that sphere minus a disc. So you've got S2 minus a disc of size epsilon. You've got another one, S2 minus a disc of size epsilon. And then you sort of glue the boundaries together. You identify the boundaries so that you have something like this, right? I mean, that's what we'd smooth it out a little bit so it's actually a sphere. Now, when you glue the boundaries, you have a twist here. You have a theta twist. So the parameters you have here, you have an epsilon which measures how big the disc is here. You have a theta. So that you can think of as A epsilon times e to the 2 pi i theta. That is in C. That's a little complex gluing parameter. So when epsilon is zero, it's when you have the bubble. And when epsilon's a little bit bigger than zero, you've joined the domains together. Now, of course, in practice, what you do is you use cylindrical coordinates here. So it sort of looks like an infinite end. So you think of something looking like this and something looking like this. So you take away a point here and you have this infinite cylindrical end and then you chop off here at some point and join them together. And then because a map, you see, how do you get a holomorphic map here? Well, the map you've got u1 here and u2 there, say, well, it's near this point. It's essentially constant. So you can just on this glued together thing here, you just make your new map you've got u tilde, say, which is u1, u2 glued together with some gluing A. It's basically equal to u1 here. It's equal to u2 here. And then here you just patch together with some kind of bump function. You just interpolate. So this is almost homomorphic, not quite homomorphic. And then you use a Newton process to make it homomorphic. So that's an analytic thing called gluing. And again, Dietmar and I do the very, very simplest case of that in our J-Holm-Hoffit-Kerf books. Now gluing, you know, you can, the analysis of this is pretty well understood. It's horrible. I mean, it's real analysis, right? So if you're a topologist like me and you don't like analysis, I mean, if you're an analyst, you love it. But otherwise, it's a sort of pain that has to be survived, right? But it does exist. It is well understood. But it can be done in different contexts. And of course, in the Fredholm, in the polyphol context, they have a very beautiful way of doing the gluing. But, and then that sort of gives you the topology. I mean, there is something called the Gromov topology that you can measure, which I really don't want to go into. But basically, you can take, if you have a map that's, you know, sort of, the domain sort of looks like that, you can, you can, you're basically, well, I'm just not, you have to rescale, you have to restrict the bits of the domain and rescale it and sort of measure it. I don't particularly want to go into that. All I want to say is, is a topology. Which, so you have these strata and for each node, so if you have a strata with one node, like consisting of elements like that, then there's a, over that there's a line bundle consisting of the gluing parameters, and then there's a map which takes you that line bundle into, into sort of resolves it. So you get, you can get a topology. Of course, Gromov, I mean, the Gromov topology, you don't need the gluing. You can describe it without the gluing. Anyway, I did want to say a little bit about the regularity question. What is the dimension of the strata? Right. So you have this two times the number of nodes. So this two is just for each of you, you have this complete parameter. Exactly. Yeah. So that, that, that's exactly, that's, that's exactly right where those parameters come from. Okay, so let me, let's, but I think I, you know, I should say something better than to justify the type of my topic. Okay, so, okay we might just kind of forgot OK, so I had the notion about some more injective curve. It's actually right up here. Now, what does transversely cut out? Well, so we have the Cauchy-Riemann operator. Del by j of u is a half of du plus j composed with du composed with little j. And this thing, well, du is a differential. So that we're thinking of as a one form. This is a one form on the domain with coefficients of the pullback of the tangent model. So it's an anti-holomorphic one form. And du itself is a one form. The fact that we add that to make it anti-j homomorphic, so anti-commute with j. And so we can think in general, we have x is some space of maps. Perhaps I should write w1p maps from s2 into m. We've got j as a space of tame, almost complex structures. And we have a pair of elements in it. We have uj in there. So we have a new does it. And so this is just a map. And that's a j. And then we have this operator. Now this operator lives in a one form. But you see its image depends on u. So that means that we really have a bundle over here, e. And we're thinking of del bar j as a section of this bundle here. And the fiber of this bundle, e uj. So for each pair uj here, we get a map in there. That's precisely the space of. And actually, if I'm completing this to w1p maps, I mean, you can do this in the c infinity case. But if you want to do Fredholm theory or something, you better put some balance spaces here. So then this is better be the instead of the, these are c infinity things. Let's have the LP. Let's take the LP sections because we've differentiated. So you get LP. So we have this bundle here. You can prove this is a nice bundle over this space. It's a balance bundle. You've got this operator. What we'd like to say transversely cut out is we'd like that we're interested in the zero set. And what we want is for this, we want del bar j to be transverse to the zero section. So we want that the solutions. So to say transverse, that means we have this operator. We can look at the linearization L of uj, which is the linearization of this operator. I mean, this is from a Banff manifold to a Banff manifold. So you can differentiate it. C1 should be C1. So you can differentiate it. And then you can project onto the fiber. And using a connection in general, but along the zero set, it's well defined. And then the linearization is sort of the linearization of the. So if we evaluate that on a psi and a y. So psi is in the tangent space to the mapping space at u. And y is in the tangent space at j to the space of almost complex structures. This linearized operator takes the form while it's the ordinary linearization of the Cauchy-Riemann operator, where you fix j and just linearize it. And then you add to it the sort of linear term coming from varying j. Well, j actually appears here only linearly. So this is a half of y composed with gu composed with j. So you have the linearization because it's these two terms. Now, if you fix j, then of course this term goes out. And the transversely cut out means that du, j is subjective for all u in the solution space. And then when it is, you see this operator here, if you just look at du, you fix j and look at that operator, this is a Fredholm operator. So that means it has finite dimensional kernel, has a closed range in a finite dimensional co-carnal. So to say it's subjective that it just has a kernel, the kernel has a fixed dimension equal to the index of the operator. And then it's an open condition to have that subjective linearization. So there's an implicit function theorem in this Danuk space context for Fredholm operators. So if it's subjective for all u here and you vary j a little bit, it'll remain subjective. So that's some kind of nice stable situation. So that's what you're looking for. And so if it's subjective, that means the solution space is a manifold of the right dimension just using the implicit function theorem. But now, what's the basic theorem in this setting? The basic theorem in this setting is that you can't unfortunately prove. I mean, we'd like to say that if we allow variations in j, we'd like to say that it's always a manifold. But that's not quite true. So the theorem is that if you look at m star a jc. So this is all somewhere injective. J holomorphic curves ua. I'm out of time, so I'm just going to. You have five minutes. I have five minutes. OK. Well, that would be helpful. Thank you. So you look at all somewhere injective curves. And you allow yourself to vary j. And this is a Van Aak's manifold. And I suppose I should actually be a little bit more careful. Jl would consist of Cl, almost complex structures, because we'd better put ourselves in a situation of a Van Aak manifold. So instead of lying C infinity j, we'd better complete them to be, they could be W1p, but you can, or Wkp, but you can make them just Cl, almost complex structures. So that's a Van Aak's space. So that's a Van Aak manifold. And then if you look at the map, just taking a pair uj to u, so j, sorry, that this is Fredholm for L large enough. Strictly, you're actually, perhaps it's always Fredholm, but really what I want to say is if L is large enough, does a set sitting inside there of regular values. So that would be where the thing is transversely cut out. It's called regular if the thing is transversely cut out. And the point here is that this is a residual in the sense of bear. Or I suppose you can say commiga. Or if you're in the first edition of the book with DeepMal, we said of second category, but that's actually a wrong thing. That doesn't mean what we meant it to mean. But anyway, this is a large set. In other words, it's a countable intersection of dense open sets. So it's a large set, that commiga thing. It's not open and dense itself, but it's a countable intersection of open dense things. So it means it's always non-empty, lots of stuff in it. And in fact, there's a trick of tags that allows you, I mean, in principle, you do this first when you have a CL thing, but you can take L equal to infinity here by some trick of tabs. So what that's telling you, and then so that means then that there's a large set in here of regular j where there are all the solutions of regular. And so that, but the kicker here is that the curves have to be somewhere injective. You can't do this. You see, if it's a multiply covered curve, you may not be able to make it regular. The difficulty is that if you have, suppose you have u from S2 to M, and this could be regular, that one could be fine. So du could be onto, and it could be somewhere injective. But then we could take a map which is a composite of, we could take phi from S2 to itself of degree k, and then look at phi composed with u. So perhaps I should call this psi, because it's not a holomorphism. So this is now, this is a multiple covering, because you're doing your sort of, this, for example, could be u at z to the k or something. Now, if you have a regular curve that exists, and you perturb j, it's still exist, they persist. So you may not be able to, this may be a perfectly good regular curve, you perturb j, you always have u. Then you always have these multiple covers. But what's wrong with these multiple covers? The fact is that the index of this thing, well it's 2n plus 2c1 now of k times a plus whatever. Well, perhaps we don't have any mark points, so that's what it is. Now, the difficulty here is that if c1a is negative, which it suddenly could be. I mean, there's no reason. In order to have the index, if you have the index of u, is 2n plus 2c1 of a. And because if we're just looking at a map from a sphere into m, it always, if it exists, there's always a sixth parameter family, because we have all the reparameterizations. So the index of this has to be bigger than or equal to 6 in order for this curve to be there. But suppose you're in a dimension, 2n is 10. You could suddenly have this being 10, and this being minus 4. And so you have a curve in that class with a negative shown class. But then you see, if you look at something like this, this thing could be so negative that this actual index could be negative. And yet, this curve would always be there. You can't perturb it away, because the underlying curve is always there, and therefore it's multiple curves are there. So if you have a strata in the stable map, which actually contains something like this, I mean, you could have a curve in class, suppose you have a curve in class B, and then it could decompose into something of class B minus ke plus ke, where c1 of e is negative. It could decompose like that. Well, then you see, you're not going to necessarily be able to get rid of the curves in class ke. They may always exist. There could be multiple curves of curves in class e, which are allowed to exist. But then the index of this thing would be big, because you see the first chunk class here of c minus ke, well, the first chunk class of e is negative. And so the first chunk class of B minus ke is bigger than the first chunk class of B. So that means you'd have a strata, and the curves in class B minus ke, would, in the best situation, even if you could make them all generic, they would live in a space which has got too high a dimension. So that means you lose control of the boundary. And therefore, you can't necessarily, you can't have simple methods of getting rid of this. So that's the problem, why you can't use regular standard thing, and why you need this to be injective. Well, I don't have time. If people don't know the argument I can explain to you, you can ask questions about, I don't have time to explain why it has to be somewhere injective here. But basically, the reason is that you've got, here's your manifold, and here's where you're varying j. So when you're varying, you see you're allowed to vary j here to get regularity. But you're varying j on m. But the image of your curve, here's s2, where the domain of your map is, your map is going to some one form on s2. So you have to take your variation in j, which lives in m, and pull it back to s2. Now if it's somewhere injective, then there's going to be a little neighborhood here. You can control it, you can control j here. If it's somewhere injective, you can pull it back and control it there. But if it's not somewhere injective, when you pull back something from here, it would have several pre-images here. And so it doesn't give you so much control over what the variations in j do for you. So really, as I explained, if it's not somewhere injective of the curve, you can really have these bad components of the compactification. And it is too big that can't be dealt with. Now I was going to say more about the analytic difficulties in dealing with this in the traditional way. You see what you have to do, I'll be explaining sort of geometric regularization. You have to have more complicated variations. You have to allow yourself more complicated perturbations, which don't just depend. I mean, the perturbations I'm allowing myself here are just to vary j, which is say, it's sort of a big variation. That's just on the manifold. You vary j on the manifold. Well, you can have much finer variations of j to correct. But then there are all kinds of problems which I luckily don't have time to get into. So there we are. OK. So since we have office hours this evening, I propose that long questions get asked then. But do people have any quick, shorter questions, producer? Does pretty much everyone say that the regular locus is connected? Yes, it does. One parameter family is because everything sort of happens in code dimension 2. Well, you don't actually have to know it's connected, the regular locus. What you need to know is that there's a regular family, a regular homotopies. And regular homotopies, you see, if you have one parameter family in here, that sort of gives you an extra dimension because you've got a variation of j in one dimensional families. So you can have a one parameter family of things here which doesn't actually consist of everywhere regular things, but consists of things where the code dimension of the image is just one dimension. Right. So yeah, the part of this theorem tells you if you have two regular values, you can find a regular homotopy in between where the inverse image would be a manifold of the right dimension. But you can't necessarily pass between them through regular j's. No, you can't necessarily pass between two regular j's. Because if you could, it would mean that the space of j homomorphic curves for j0 and for j1 were isomorphic, basically. And basically, all you can say is that they're co-important. For spheres in certain situations, you can. But in general, you can't. There are more questions for Dusa. All right. Let's thank Dusa for her nice introduction. Thank you.
 Well, hello everybody. I'm going to give what I hope is a very elementary talk, so you'll  probably be very bored. But for those of you who don't know this or have forgotten a little  bit, I'm just going to go over the problem of regularization and explain the traditional  methods for dealing with it a little bit. So the question is this, we have a symplectic  manifold and it's got an almost complex structure J, which is omega-tame. And we're looking at  J-halomorphic curves. Now, to be simple, I'm just going to think of the domain being a  sphere, so I'm going to write, say, m tilde top 0k of, well, decorated by all these things,  J, A, say, which is going to be the set of maps. Now, I'm just going to take the sphere  as my domain into m, which are J-halomorphic, which represent the class A, so that's a class  in the second homology of m. And, oh yeah, I'm going to have some points, z1 through zk,  which is just underline z, and these points are different. And that's just a space of  J-halomorphic maps. So these elements are maps and the tilde means that the elements  are maps. Now, does everybody know what all these words mean? Because I'm just assuming  you do. And if so, if you don't, please ask. Okay? It's up to you. And then we're interested  in this space, but also we're interested in, if you've got two of these elements, u, z,  we can say one element is equivalent to another element, u prime, z prime, if there's a diagram  of this kind. So we've got phi from S2 to S2, which is phi is an amherbius group. So  that's a bi-halomorphic map. We've got u going there to m. We've got u prime going there,  and the diagram's meant to commute, and say we've got z prime there, and that goes to  phi z. Right? So this would be something of the form u prime would be phi composed with  u, and this would be phi inverse of z. Okay? So we have this equivalence relation. And  we're not interested really in the space of actual maps. We're more interested in the  space of equivalence classes of maps. So that's going to be written as m top, 0k. So that's  going to be m tilde top, 0k, divided out by this equivalence relation. So these are equivalence  classes. So we're not really interested so much in the way they're parameterized, just  that they have a parameterization. We're interested in the images, the geometric images in them.  And I put top here because this is the top strata of some completed space. Okay? Now,  for example, well, so in the best case, this m tilde top, 0k of mja is a smooth manifold,  and it's vastly cut out. Oh, I forgot, I said it was j holomorphic. Well, I forgot to say  what that means is that the Cauchy-Riemann operator is 0. So del bar j of u is a half  of du composed with plus j, du composed with j. So that's the Cauchy-Riemann operator.  And so this is 0 set on operator, and if you're saying it's transversely cut out, you mean  that the derivative of the operator is subjective. I'm going to say more about that later on.  But anyway, you want it to be transversely cut out, and the dimension of this thing  then would be equal to the index of the operator is 2n plus 2c1 of a plus 2k, which is the  number of mark points because they're varying in a two-dimensional space. So that's a dimension.  And why you want it to be transversely cut out is that there's a Fredholm theory, and  that means that if you vary j, the space is not going to change very much. So that's the  sort of basic space of j holomorphic maps. And so in the best case, the first situation  is that this is a nice manifold. And then, as I say, what we're really interested in  is the Cauchy space. So this is m top. That this is almost never compact, but it's compact  enough. And why this space is almost never compact? You can see that it's almost never  compact. Because one thing I'm insisting, I've got these points z1 through zk, and they're  meant to be all different from each other. So two of them could come together, and of  course, that would not be in this space. And then another problem about the compactness  is that this group of equivalences, this group here, psotc, is not a compact group. So that  means you're quotienting out by something which is not compact, and that can cause problems.  But what do I mean by compact enough? Well, sort of for the purposes, it's compact. So  compact, what are the purposes? Well, we're interested in the evaluation map from m tilde,  well really from m top, no. And let me, well, m a, so just put an a in there. I'm mapping  to m to the k. So we've got a pair uz, which is an equivalence class, so I can put brackets  like that, going to u evaluated at z1 up to u evaluated at zk. And that, of course, is  defined on the space of maps, but it goes, it maps through the quotient. So there's a  well defined map here. And what you want for this is, you want the image, image so that  it represent a homology class. And so that, you see, I mean, if this were a compact manifold,  then of course it would have a fundamental group, fundamental class. So you could just  look at the push forward of that fundamental class. But in fact, for it to represent homology  class, it doesn't have to have a compact image. What it needs is that the IE, if you look  at the closure of the image, so this is the image of the evaluation map minus the image  itself, so this is the boundary of the image. This should have co-dimension at least two.  So that means what we say is it represents a pseudo cycle. And to give you an example,  well let's take the very simplest example, I think it's a really simple example, take  the projective plane with a standard J and the class to be the class of a line. And then  C1, the first shown class of the line is 3. And so the dimension of M, if we take two  mark points, the dimension here is going to be, it's 2n plus 2c1 of A plus 2k, which  is 4, and then minus 6, which is the dimension of the reprametrization group that I've quotiented  out by. Okay, so this is the dimension of Psl2c, and there will be a group. And this  is 6, and this is 4, and so we get 8. And that of course is the same as the dimension  of if we take Cp2 times Cp2.  So if we look at the evaluation map from M top 0, 2 in the class of a line going to  Cp2 times Cp2, we have uz goes to uz1, uz2. And what this is, remember these things represent  a line. And so this is the elements here represent, you've got an s2, you've got two points on  it, which in fact we can normalize, we can take the first one to be at 0, and the second  one to be at infinity say, because the group is triply transitive, so we can fix these  two points, and then we've still got an action of C star acting on these things. So this  is M tilde top with a fixed point set to 0 and infinity and quotiented out by C star.  So it's still something, a space of maps quotiented out by a non-compact group. And anyway, you've  got this space of maps, you're looking at where they go, so you've got two points in  Cp2, and you sort of take the line through them. Well, there's a unique line through  any two points. Therefore, this map is injective. If we take the standard complex structure,  we know that that's injective, but of course it's not surjective. And you can see it's  not surjective because the two points it's going through are different. They have to  be different because I've assumed that these two points are different and allowed to come  together. So the image is equal to Cp2 times Cp2 minus the diagonal. Right? Very clearly.  So it's not a compact image. You can see the space is not compact. On the other hand, the  set of maps is somehow compact. That's by Gromov's theorem. And you can compactify this.  I mean, this diagonal has codimension two. It's real codimension four, I suppose, because  it's where these two points are the diagonal here. And so it does represent a pseudo-cycle.  It does represent a fundamental class because the image of the boundary is so small.  I just understand that the words codimension of these two means inside the closure of the  image or inside the ambient? Inside the m to the k. So we're not talking about the structure  of the domain here, which is to, I mean, in good cases, you can assume it's a manifold,  but it's not going to be compact, you see. But you're saying, so this, it's a, well,  it's a non-compact manifold. But if you look at its image in here, it has a fundamental  class because the boundary is so small.  Is there really a key codimension condition?  What?  The way that case is.  No, no, it's a huge thing.  Yeah.  What?  Codimension in the closure.  Well, it's a codimension. It's a real codimension of at least two in m to the k. So whatever  dimension that has. So in this case, I mean, this is a, you try to, in this case, what's  wrong with it?  What if m to the k was an enormous thing and this was like one curve? We really care about  the closure.  Oh, I see. I see what you're saying. Yeah, okay. So what do I mean by, yeah, so I'm saying  this incorrectly. What I mean is that the boundary of the image minus the image should  be a union of maps from some manifolds v. So v is some, this is some finite union, say.  And the dimension of v should be less than or equal to the dimension of our original  space m top minus two. That's what I mean. I'm sorry. You're right. Right. That makes  sense. So it's codimension. It means it should have at least two dimensions less than the  dimension of the image.  And this is assuming m top has the expected dimension?  That's assuming that m top has the expected dimension, which it's going to have if it's  a, that was my first assumption that it's a transversely cut out manifold of the right  dimension. So it means that it has the correct dimension. And then, right, that you're assuming  that this thing is going to, then this is meant to represent a homology class of the  right dimension, the dimension of the index. That's okay now.  Okay. So now, let's say, so there are several things here. I mean, the regularity problem  is the problem of arranging the first thing that it's transversely cut out to find if  you could find a J to make it a nice manifold. That's the regularity problem. But then there's  also the compactness problem that you want to make sure that this image is compact enough.  Well, how can you make sure about that? Now, the compactness is actually much easier to  deal with than the regularity. So let me just say a few words about compactness.  So there's Grumhoff's compactness theorem, which tells you that if you have a, you know,  you look at these are sequences of spheres, right? And you look at these are spheres in  the same way. So we have these. So we've got some curves UK, say, from, say, S2 into M.  We've got a sequence of curves. And they're in a fixed homology class. And so that means  that the pullback of a mega on the S2 is fixed. It's a mega of A, the integral. And this is  actually, analytically, it's the energy. It's equal to, I hope you got the formula wrong,  the pullback of du squared over S2. That's the energy of u. So it's basically the L2  norm of u, measured with respect to the correct, I mean, you have to choose a metric on there  and stuff, measure it with the correct things. And you get, this is the energy. So this is  a sequence of curves. So these are C infinity maps, the holomorphic curves, and they have  bounded L2 norm. Well, they're actually W12 norm. This is the L2 norm of the first derivative.  So this is what's called the, it's a sort of boundary case. If we had, because these  are holomorphic maps on a two dimensional space, if it were a bounded W1P norm for P  bigger than 2, then the standard sort of compactness arguments that tell you there would be a compact  subsequence or a convergent subsequence. That's just standard elliptic theory. And  why the W1P norm, why P is bigger than 2 is because this norm, if you have something  bounded in W1P norm, that controls the C0 norm. So that means it's an equicontinuous  family in the C0 norm. So you use sort of standard results about equicontinuity to extract  a convergent subsequence. But what we have is this borderline case where it's a 2, not  bigger than 2. So in fact, it's possible for this thing not to have a convergent subsequence.  But if it doesn't have a convergent subsequence, what's happening is that the energy is somehow  concentrating near a point and you can rescale it and get a convergent subsequence. So, and  that comes down to the fact that, for example, in this case here, we're quotienting out even  here by a non-compact group. That's a non-compact group there. And so that sort of affects the  convergence. But anyway, this is really well understood. And what Gromov proved was that  there exists a subsequence. You also called UK say that converges to what Gromov called  the Cusp curve or what we now call stable map. And what a stable map is, you see bubbles  can develop. I'll explain a little bit in this case later on. But what it's going to  be is the domain now, sigma, is a nodal sphere. So it's going to be a union of spheres joined  at points. So it could look like this. It could be a bubble, you know, a set of spheres  like that. That would be the domain. There'd be some marked points on here. Because your  marked points would converge somewhere, we've got three marked points, they converge somewhere  like that. And the map, you would then map each component into n. So you'd have a sequence  of, you'd have a set of maps. And that's what, so for example, just suppose we were looking  at a map from S2 into CP2, which had degree two. So that would be a quadric. But a quadric  can degenerate into two lines. So that's in class two up. That's one degeneration that  could happen. You can have, so that's a geometric degeneration where you have a sequence of things  representing a quadric which degenerates to two lines. But you could also have a degeneration  that happens because, for example, suppose you're in this case and the marked points  come together. Well, the marked points are not allowed to come together here. If they  do come together, then you could have a situation like this where you actually have a bubble.  And that bubble contains both marked points. And here, this is what's called a ghost component  in the sense that the map you restricted to this component is constant. So what you have  here is a situation where you have this goes, this is a line. So you've got a map of a line  and your two marked points are sitting on the domain in this sphere which is called a ghost  component. So I should give you a proper definition of a stable map so you can see you can put  these all together into the space of stable maps. Now, is there any way of retrieving  that board? Oh, wait a minute. What about this? No. It's got a very tall person. There's  a rod. Ah, this thing. Doesn't have a... What's... Oh, what? This is it. This is it. This one's  got a hook. Okay. Let's see. All right. Now... Oh, God. Well...  That's good enough. Oh, look, it's going up by itself. Very obedient. Okay.  Any of the most points you integrate, like the thing you integrate with the energy, did  you develop poles there or is it just the same? Well, no, the energy just comes from  the map. So it doesn't see where the mark points are. So I want to give you a definition  of a stable map. And of course, these can occur in any genus. They don't have to be  genus zero. I was just restricting to genus zero for simplicity. But... So let's just  do genus zero because the point about genus zero is that then you can think of your domain  as a union of these two spheres. And each two sphere, you can parameterize in a nice  way and give it the standard complex structure. So you don't have to worry about variations  of the complex structure on the domain. Typically, you'd have a nodal surface which you'd have  to allow the complex structure on the surface to vary in type of a space. You can put that  in if you want, but that's just make it simple. So it's a nodal Riemann surface. So you have  something and it's all... It's meant to... You know, you can have something like that.  But it's meant to be connected in a tree diagram. So you're not allowed to have something like  this. That would have genus. That would be a genus one thing. So it's a tree here, a  table tree. And you have a certain set of mark points. You sprinkle your points around.  Once you've got five mark points, you could sprinkle them around like that. The mark points  are all distinct. And so they're on particular spheres. So you have to say which sphere they're  on. Then you have some maps, U, which is a one map for each component. And so we have...  So that's... So we're looking at two poles, which is a form of sigma, complex structure  is understood here, Z, U. And then you put an equivalence relation. Well, the obvious  equivalence relation that just like I had before, you just have... You have some holomorphic.  This is a holomorphic map here. So you're interested in these modulo reparameterization.  And then there's an all-important stability condition. Now, you don't need to... When  you're talking about a stable map, you doesn't have to be holomorphic. I mean, eventually  we're interested in the holomorphic ones. But in general, you can just look at arbitrary  maps with nodal domains like this. But it's very important to have a stability condition.  And what that is telling you, this is equivalent to saying that for each element, if you have  sigma, J, U, and you look at the group of automorphisms, so a map of that to itself,  this is finite. So you never... You don't allow yourself to have an infinite number  of automorphisms. And what that means is... Now, if, say, on a component like this component,  the map is non-constant, then the map is either injective or it's a multiple covering, and  there could only be a finite number of automorphisms of that component. So if you have a component  where the map is non-constant, there's no condition. But if you have a component, if  the map U were constant here, then you see, what would an automorphism as this would mean?  Well, it would have to fix Z0, Z1. It would have to fix this nodal point. But you see,  if we've only got two... What are called special points? We've only got two special points  on this component. There's a mark point and there's a nodal point. And so they're, in  principle, as a one complex parameter family of automorphisms of this thing. So that would  not be allowed. So the claim is that if U alpha restricted to the alpha's component is  constant, S alpha to alpha has at least three special points. And special points are nodal  points or mark points. That's the important condition. So that means, for example, in  this particular case, I could allow that to be... That could be a ghost component. So  that one could be a ghost component. This one could be a ghost component. This one's  not allowed to because there's only got two mark points. That one couldn't be. So this  would be a possible degeneration if you were looking at maps, say, from U from S2 to CP2  of degree two. So that it's in the class twice a line. Then this one could represent a line.  That one could represent a line. And we've got five mark points. That would be a perfectly  good stable map of this kind. And the claim is that if you look... So we define M bar  0k MjA to be the equivalence classes of stable maps in genus 0. So this is a genus thing.  And K mark points. And representing the class A. And if you've got a J, they're J-halomorphic.  So that's the space of J-halomorphic stable maps. And it could be... It's a stratified  space. I mean, why I wrote top there is that top was its top component, if you like, where  its domain just had a smooth sphere. But then it has all these other components where the  domains are nodal curves. And some of them, if you like, are sort of artificial. Like,  this is an artificial one because the two mark points have come together. And so, you  know, nearby things to this would be something where you had a... And then you had this would  still be constant, say, this one could have a Z2 on it. But now we'd have three mark points  on that component. So... And then when these two mark points come together, we'd have this  bubble sort of artificially created in the domain. This is... This was Konsevich's idea  of the way of describing what's happening, way of describing a compactification. So the  claim is that we have this space here. This is compact. It's a compact space. And if everything  is regular... So in other words, if all the underlying maps, you see, a regular thing  would be... For this, would be that this map was a j-halomorphic map transversely cut out,  which I'm going to say more about later. So is that. And then all these intersections  here, you've got a space of spheres here, a space of spheres here, these are kind of  a concept map. All these intersections are meant to be transverse so that it all sort  of comes out and it's all in the right dimension. And the strata, the formal dimension, if you  like, of the strata is equal to the dimension of the original, of the top space minus twice  the number of nodes. So each node gives you, contributes sort of a co-dimension two to  the thing. And that's a formal dimension. That's the index. So that means that if you  can arrange everything to be trans... You know, all the things that transverse, that  will be the dimension that it has. But of course, it doesn't have to be like that. But  if it were, then you see, if you're in a situation like this where you can arrange that all the  lower strata have basically have the right dimension, then the top thing of this will  represent a pseudo-cycle because this is compact. So you can compact by the image of the map  by just looking at the image of these things. And then the boundary of it is going to be  covered by the images of the strata, you know, the higher co-dimension strata here. And they're  all going to have at least co-dimension two. So that's a sort of standard regularization  picture that you compactify the image by putting in the images of the evaluation maps from  the strata and this... In this space of stable maps. If you can arrange, if you can control  our dimensions reasonably, then they all have co-dimension at least two and then you get  a pseudo-cycle. Now, of course, the kicker is that you can't always control a dimension.  That's the whole problem. But if you could, then you'd be... You'd solve this problem.  You'd have got... I mean, because our aim, of course, is to try and use these spaces  of maps to get some invariance. We want to get these homology classes out that we can  then intersect and get the grammar for an invariance or something. And this is just  for closed curves, but if you're doing open curves, you have boundaries and you put boundary  conditions on. The formal structure is exactly the same. The equation looks a little bit more  complicated because you have to deal with what happens at the near the non-compact parts  of the domain, but formally it's exactly this idea that you can always compactify the space  you're looking at by making some space of stable maps or stable buildings or something.  And then you just need to control the dimensions. In principle, all the boundary components you've  added should have co-dimension at least two. You need to control that and make sure it  does. Okay?  So that's telling you all I was going to say about compactness. But you see, I must have  erased... Here, I erased the first condition which was that everything was transversely  cut out. That's what I need to talk about because that's the hard thing. The compactness  is sort of understood. So, so...  Why don't you have to worry about the unstable maps? Like, why aren't you there somewhere  in the computer?  Well, the unstable maps would be... Well, the thing is that they don't appear in the  compactification because if you have an unstable component, suppose that were an unstable component,  so that map was constant. Then you could just forget about that bubble and put that point  there. You see, the unstable components are things which don't actually contribute to  the image because the map on them is zero. It's just a constant map. And so you can just  collapse any unstable components. And the thing is that it's... I mean, that's a good  question.  I do not understand. Can you say again?  Yeah, I'm saying it. Suppose we had this situation, but this sphere was a constant map there.  And that would be an unstable component. But the thing is that you... So, I'm saying that  that's not going to be in the compactification because we don't allow that in the compactification.  What we do is, in that particular case, we just get rid of that completely and just put  the mark point there and there wouldn't be a bubble there at all. Because the image would  be the same.  If you have like a sequence, I mean, like in your top stratum, like you can just define  it so that it converges to a stable map with the stability condition.  When you look at the compactness argument, what's happening when a bubble forms, typically,  I mean, in this particular case, if this is what we had, all that would have happened  here is that some of the mark points would have come together. And the underlying map  here, you just have a sequence of maps which was basically converging to a map. And all  that would have happened to create this set of bubbles is that the mark points would have  come together in a particular way, sort of these two points coming together first and  then these two points coming together second and then they would form this bubble. And  you can actually see that completely clearly just by sort of rescaling and blowing up and  looking at, you know, nearby this. You'd have... Here would be the point where you'd be rescaling.  This point is going to go into these bubbles. You'd have a Z4 and Z5 very, very close and  then sort of further away, you'd have a Z2 and Z3. So there'd be two scales of bubbling  here. This would be the... These ones would be the furthest ones and these ones would  be the later ones. But the map itself would be constant. So the map to M would be constant,  but you're talking about how the mark points are coming together. And it turns out, I mean,  we've proved this in exhaustively in our book on J-Homorff-Curse, a deep mark. If you just  look at the way you prove the compactness, you can get... I mean, you could describe...  I mean, of course, you could put in something like that and allow that as a limit. But then  you wouldn't have a sort of unique limit. And it doesn't give you a good structure. I mean,  this was Komsovic's realization that the good structure is... Because there would be an  infinite automorphism group of those, which you wouldn't want. You wouldn't want. You'd  have to divide out, you see, by a C star here. You want the set of automorphisms of all these  things to be finite. That's very important. We are already doing that now. The other component  that still has a C star automorphism... No, because you see, here, the map is non-constant.  So you've got some map U, and you're saying U is equal to U composed with phi, if you  had an automorphism. So phi would have to be something that fixed these two points.  And so you'd have to have phi Z1 is Z1. Phi of the nodal point, let's say infinity is  infinity. And it would have to satisfy this. Well, if U is non-constant, you wouldn't have...  I mean, you could...  It's a condition, right? I mean, it could happen.  Well, you would only have a finite number. You see, what you could have... You could be  a map, say you could have U from... You could think of C, U and infinity to M. And it's  going to take Z to say Z cubed. And then you could have an automorphism which took Z to  E to the 2 pi i over 3 Z. That would be an automorphism, so U of phi would equal phi.  But it would be a finite group. You're not saying there aren't any automorphisms. You're  saying it's a finite group of automorphisms. So if it was a multiply covered curve, it  could have an automorphism. But if it's an injective curve, somewhere injective curve.  So that's another thing. You have this idea of a somewhere injective curve, U from S2  to M. What that means is that there is some point Z naught such that if you look at its  image and then its pre-image, that's just Z naught. So that it's a map that... Here is  the domain as 2. Here's your Z naught. On this little neighborhood, it's injective. And  then you see if it's injective there and you have something that behaves like this, U of  phi. Phi would have to be the identity in that little neighborhood where it was injected.  And that doesn't happen if you have a Mobius transformation that fixes three points. It's  the identity. So really, the stability condition, the sort of geometric way of understanding  it is telling you that the group of automorphisms of this stable map is finite. That's what  the stability condition is saying. And that's a very important condition that allows you  to... I mean, it's the fact that you could have automorphisms means that you're actually...  The space here, the strata are typically overfalls rather than manifolds because you could have  automorphisms. So you're modeled on an overfold. But it's very important that the finite...  It's overfold with finite isotropic groups. Yeah.  It just occurred to me that I think you didn't say that stable maps have to have finite energy,  which is a guarantee that the stability condition actually implies finite automorphisms.  Ah, okay. Well, I was saying that we're in some class A. Right. So then omega of A is  a number. Right. So they do have finite energy. But you're right. I mean, if you're doing  more general things, the finiteness of the energy is actually what makes this whole compactness  serum work. You've got to have a bound on the energy, bound on the W12 norm to get started  with the analysis. Excuse me. Yeah. How do you put the topology on and bar the location?  Oh, something called... How awkward of you. Something called the Gromov topology, which  I don't particularly want to go into. I mean, each stratum, it's clear enough how you  to apologize it. And then nearby, well, there's a structure you see. When you have a stable  map, I mean, I have to mention this at some point if I'm doing sort of the elementary beginning  anyway. If you want to know what's a neighborhood of something like this, you have to do gluing.  So you have to say gluing is a way of, if you've got a nodal domain, then sort of nearby, there  are things which nearly have the bubbles, nearly have the nodes, but don't quite. So  nearby, so here we've got two nodes. So we've got one gluing parameter here and another gluing  parameter there. And for each, if you have... Let me get a slightly bigger board. So suppose  we've got a single node. So here's a... We've got some map. So this... Let's have no mark  points just for simplicity. This is going to some line and that's going to some line.  So the non-trivial map. So this is going to a pair of lines. So the image would be a pair  of lines. Well, nearby, there's pair of lines in CP2. You actually have, of course, a quadric.  Impossible to draw. But how analytically do you see that? Well, you have this sphere.  You take this sphere and you take... With minus a disc, you take that sphere minus a disc.  So you've got S2 minus a disc of size epsilon. You've got another one, S2 minus a disc of  size epsilon. And then you sort of glue the boundaries together. You identify the boundaries  so that you have something like this, right? I mean, that's what we'd smooth it out a little  bit so it's actually a sphere. Now, when you glue the boundaries, you have a twist here.  You have a theta twist. So the parameters you have here, you have an epsilon which measures  how big the disc is here. You have a theta. So that you can think of as A epsilon times  e to the 2 pi i theta. That is in C. That's a little complex gluing parameter. So when  epsilon is zero, it's when you have the bubble. And when epsilon's a little bit bigger than  zero, you've joined the domains together. Now, of course, in practice, what you do is you  use cylindrical coordinates here. So it sort of looks like an infinite end. So you think  of something looking like this and something looking like this. So you take away a point  here and you have this infinite cylindrical end and then you chop off here at some point  and join them together. And then because a map, you see, how do you get a holomorphic  map here? Well, the map you've got u1 here and u2 there, say, well, it's near this point.  It's essentially constant. So you can just on this glued together thing here, you just  make your new map you've got u tilde, say, which is u1, u2 glued together with some gluing  A. It's basically equal to u1 here. It's equal to u2 here. And then here you just patch together  with some kind of bump function. You just interpolate. So this is almost homomorphic,  not quite homomorphic. And then you use a Newton process to make it homomorphic. So that's  an analytic thing called gluing. And again, Dietmar and I do the very, very simplest case  of that in our J-Holm-Hoffit-Kerf books. Now gluing, you know, you can, the analysis of  this is pretty well understood. It's horrible. I mean, it's real analysis, right? So if you're  a topologist like me and you don't like analysis, I mean, if you're an analyst, you love it.  But otherwise, it's a sort of pain that has to be survived, right? But it does exist.  It is well understood. But it can be done in different contexts. And of course, in the  Fredholm, in the polyphol context, they have a very beautiful way of doing the gluing. But,  and then that sort of gives you the topology. I mean, there is something called the Gromov  topology that you can measure, which I really don't want to go into. But basically, you  can take, if you have a map that's, you know, sort of, the domain sort of looks like that,  you can, you can, you're basically, well, I'm just not, you have to rescale, you have  to restrict the bits of the domain and rescale it and sort of measure it. I don't particularly  want to go into that. All I want to say is, is a topology. Which, so you have these strata  and for each node, so if you have a strata with one node, like consisting of elements  like that, then there's a, over that there's a line bundle consisting of the gluing parameters,  and then there's a map which takes you that line bundle into, into sort of resolves it.  So you get, you can get a topology. Of course, Gromov, I mean, the Gromov topology, you don't  need the gluing. You can describe it without the gluing. Anyway, I did want to say a little  bit about the regularity question.  What is the dimension of the strata? Right. So you have this two times the number of nodes.  So this two is just for each of you, you have this complete parameter.  Exactly. Yeah. So that, that, that's exactly, that's, that's exactly right where those  parameters come from. Okay, so let me, let's, but I think I, you know, I should say something  better than to justify the type of my topic. Okay, so, okay we might just kind of forgot  OK, so I had the notion about some more injective curve.  It's actually right up here.  Now, what does transversely cut out?  Well, so we have the Cauchy-Riemann operator.  Del by j of u is a half of du plus j  composed with du composed with little j.  And this thing, well, du is a differential.  So that we're thinking of as a one form.  This is a one form on the domain with coefficients  of the pullback of the tangent model.  So it's an anti-holomorphic one form.  And du itself is a one form.  The fact that we add that to make it anti-j  homomorphic, so anti-commute with j.  And so we can think in general, we  have x is some space of maps.  Perhaps I should write w1p maps from s2 into m.  We've got j as a space of tame, almost complex structures.  And we have a pair of elements in it.  We have uj in there.  So we have a new does it.  And so this is just a map.  And that's a j.  And then we have this operator.  Now this operator lives in a one form.  But you see its image depends on u.  So that means that we really have a bundle over here, e.  And we're thinking of del bar j as a section of this bundle  here.  And the fiber of this bundle, e uj.  So for each pair uj here, we get a map in there.  That's precisely the space of.  And actually, if I'm completing this to w1p maps,  I mean, you can do this in the c infinity case.  But if you want to do Fredholm theory or something,  you better put some balance spaces here.  So then this is better be the instead of the,  these are c infinity things.  Let's have the LP.  Let's take the LP sections because we've differentiated.  So you get LP.  So we have this bundle here.  You can prove this is a nice bundle over this space.  It's a balance bundle.  You've got this operator.  What we'd like to say transversely cut out  is we'd like that we're interested in the zero set.  And what we want is for this, we want del bar j to be transverse  to the zero section.  So we want that the solutions.  So to say transverse, that means we have this operator.  We can look at the linearization L of uj, which  is the linearization of this operator.  I mean, this is from a Banff manifold to a Banff manifold.  So you can differentiate it.  C1 should be C1.  So you can differentiate it.  And then you can project onto the fiber.  And using a connection in general,  but along the zero set, it's well defined.  And then the linearization is sort of the linearization of the.  So if we evaluate that on a psi and a y.  So psi is in the tangent space to the mapping space at u.  And y is in the tangent space at j to the space  of almost complex structures.  This linearized operator takes the form  while it's the ordinary linearization of the Cauchy-Riemann  operator, where you fix j and just linearize it.  And then you add to it the sort of linear term  coming from varying j.  Well, j actually appears here only linearly.  So this is a half of y composed with gu composed with j.  So you have the linearization because it's these two terms.  Now, if you fix j, then of course this term goes out.  And the transversely cut out means  that du, j is subjective for all u in the solution space.  And then when it is, you see this operator here,  if you just look at du, you fix j and look at that operator,  this is a Fredholm operator.  So that means it has finite dimensional kernel,  has a closed range in a finite dimensional co-carnal.  So to say it's subjective that it just has a kernel,  the kernel has a fixed dimension equal to the index  of the operator.  And then it's an open condition to have  that subjective linearization.  So there's an implicit function theorem  in this Danuk space context for Fredholm operators.  So if it's subjective for all u here and you vary j a little bit,  it'll remain subjective.  So that's some kind of nice stable situation.  So that's what you're looking for.  And so if it's subjective, that means the solution space  is a manifold of the right dimension just  using the implicit function theorem.  But now, what's the basic theorem in this setting?  The basic theorem in this setting  is that you can't unfortunately prove.  I mean, we'd like to say that if we allow variations in j,  we'd like to say that it's always a manifold.  But that's not quite true.  So the theorem is that if you look at m star a jc.  So this is all somewhere injective.  J holomorphic curves ua.  I'm out of time, so I'm just going to.  You have five minutes.  I have five minutes.  OK.  Well, that would be helpful.  Thank you.  So you look at all somewhere injective curves.  And you allow yourself to vary j.  And this is a Van Aak's manifold.  And I suppose I should actually be a little bit more careful.  Jl would consist of Cl, almost complex structures,  because we'd better put ourselves in a situation  of a Van Aak manifold.  So instead of lying C infinity j,  we'd better complete them to be, they could be W1p,  but you can, or Wkp, but you can make them just Cl,  almost complex structures.  So that's a Van Aak's space.  So that's a Van Aak manifold.  And then if you look at the map, just taking a pair uj to u,  so j, sorry, that this is Fredholm for L large enough.  Strictly, you're actually, perhaps it's always Fredholm,  but really what I want to say is if L is large enough,  does a set sitting inside there of regular values.  So that would be where the thing is transversely cut out.  It's called regular if the thing is transversely cut out.  And the point here is that this is a residual in the sense of bear.  Or I suppose you can say commiga.  Or if you're in the first edition of the book with DeepMal,  we said of second category, but that's actually a wrong thing.  That doesn't mean what we meant it to mean.  But anyway, this is a large set.  In other words, it's a countable intersection  of dense open sets.  So it's a large set, that commiga thing.  It's not open and dense itself, but it's  a countable intersection of open dense things.  So it means it's always non-empty, lots of stuff in it.  And in fact, there's a trick of tags that allows you,  I mean, in principle, you do this first when you have a CL thing,  but you can take L equal to infinity here by some trick of tabs.  So what that's telling you, and then so that means then  that there's a large set in here of regular j  where there are all the solutions of regular.  And so that, but the kicker here is  that the curves have to be somewhere injective.  You can't do this.  You see, if it's a multiply covered curve,  you may not be able to make it regular.  The difficulty is that if you have,  suppose you have u from S2 to M, and this could be regular,  that one could be fine.  So du could be onto, and it could be somewhere injective.  But then we could take a map which is a composite of,  we could take phi from S2 to itself of degree k,  and then look at phi composed with u.  So perhaps I should call this psi,  because it's not a holomorphism.  So this is now, this is a multiple covering,  because you're doing your sort of, this, for example,  could be u at z to the k or something.  Now, if you have a regular curve that exists,  and you perturb j, it's still exist, they persist.  So you may not be able to, this may be a perfectly good  regular curve, you perturb j, you always have u.  Then you always have these multiple covers.  But what's wrong with these multiple covers?  The fact is that the index of this thing,  well it's 2n plus 2c1 now of k times a plus whatever.  Well, perhaps we don't have any mark points,  so that's what it is.  Now, the difficulty here is that if c1a is negative, which  it suddenly could be.  I mean, there's no reason.  In order to have the index, if you have the index of u,  is 2n plus 2c1 of a.  And because if we're just looking at a map from a sphere  into m, it always, if it exists, there's always  a sixth parameter family, because we have all  the reparameterizations.  So the index of this has to be bigger than or equal to 6  in order for this curve to be there.  But suppose you're in a dimension, 2n is 10.  You could suddenly have this being 10,  and this being minus 4.  And so you have a curve in that class  with a negative shown class.  But then you see, if you look at something like this,  this thing could be so negative that this actual index  could be negative.  And yet, this curve would always be there.  You can't perturb it away, because the underlying curve  is always there, and therefore it's multiple curves are there.  So if you have a strata in the stable map,  which actually contains something like this,  I mean, you could have a curve in class,  suppose you have a curve in class B,  and then it could decompose into something of class  B minus ke plus ke, where c1 of e is negative.  It could decompose like that.  Well, then you see, you're not going  to necessarily be able to get rid of the curves in class ke.  They may always exist.  There could be multiple curves of curves in class e,  which are allowed to exist.  But then the index of this thing would be big,  because you see the first chunk class here of c minus ke,  well, the first chunk class of e is negative.  And so the first chunk class of B minus ke is bigger  than the first chunk class of B. So that  means you'd have a strata, and the curves in class B minus ke,  would, in the best situation, even if you could make them  all generic, they would live in a space which has got  too high a dimension.  So that means you lose control of the boundary.  And therefore, you can't necessarily, you can't have  simple methods of getting rid of this.  So that's the problem, why you can't use regular standard  thing, and why you need this to be injective.  Well, I don't have time.  If people don't know the argument I can explain to you,  you can ask questions about, I don't have time to explain why  it has to be somewhere injective here.  But basically, the reason is that you've got, here's your  manifold, and here's where you're varying j.  So when you're varying, you see you're allowed to vary j here  to get regularity.  But you're varying j on m.  But the image of your curve, here's s2, where the domain of  your map is, your map is going to some one form on s2.  So you have to take your variation in j, which lives in  m, and pull it back to s2.  Now if it's somewhere injective, then there's going to be a  little neighborhood here.  You can control it, you can control j here.  If it's somewhere injective, you can pull it back and  control it there.  But if it's not somewhere injective, when you pull back  something from here, it would have several pre-images here.  And so it doesn't give you so much control over what the  variations in j do for you.  So really, as I explained, if it's not somewhere  injective of the curve, you can really have these bad  components of the compactification.  And it is too big that can't be dealt with.  Now I was going to say more about the analytic difficulties  in dealing with this in the traditional way.  You see what you have to do, I'll be explaining sort of  geometric regularization.  You have to have more complicated variations.  You have to allow yourself more complicated perturbations,  which don't just depend.  I mean, the perturbations I'm allowing myself here are just  to vary j, which is say, it's sort of a big variation.  That's just on the manifold.  You vary j on the manifold.  Well, you can have much finer variations of j to correct.  But then there are all kinds of problems which I luckily  don't have time to get into.  So there we are.  OK.  So since we have office hours this evening, I propose that  long questions get asked then.  But do people have any quick, shorter questions, producer?  Does pretty much everyone say that the regular locus is  connected?  Yes, it does.  One parameter family is because everything sort of  happens in code dimension 2.  Well, you don't actually have to know it's connected, the  regular locus.  What you need to know is that there's a regular family, a  regular homotopies.  And regular homotopies, you see, if you have one parameter  family in here, that sort of gives you an extra dimension  because you've got a variation of j in one  dimensional families.  So you can have a one parameter family of things here  which doesn't actually consist of everywhere regular  things, but consists of things where the code dimension of  the image is just one dimension.  Right.  So yeah, the part of this theorem tells you if you have  two regular values, you can find a regular homotopy in  between where the inverse image would be a manifold of the  right dimension.  But you can't necessarily pass between them through  regular j's.  No, you can't necessarily pass between two regular j's.  Because if you could, it would mean that the space of j  homomorphic curves for j0 and for j1 were isomorphic,  basically.  And basically, all you can say is that they're  co-important.  For spheres in certain situations, you can.  But in general, you can't.  There are more questions for Dusa.  All right.  Let's thank Dusa for her nice introduction.  Thank you.
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