 We do a holomorphic curves. OK, so if x is a smooth manifold and j is endomorphism of the tangent bundle, which squares to minus the identity, then we call this an almost complex manifold. And j is chosen because it's the next letter in the alphabet after i. And you're supposed to think of j as multiplication by i. So when you think of j as multiplication by i, you realize that j endows the tangent bundle with the structure of a complex vector space. And conversely, a complex vector space structure on the tangent bundle is the same thing as just choosing this endomorphism j. OK, so in particular, this can only happen in even dimensions, all along with complex manifolds are even dimensional. So if we have a map from one of these to another, I'll just do the holomorphic. When it's derivative, that's going from the tangent bundle of x to the pullback of the tangent bundle of x prime, is complex linear. In other words, df composed, I guess, on the side by j prime is the same as df imposed on the other side of the j. OK, so this isn't, in general, such a useful PDE to be studying because it's over-determined. So this is equation over-determined, unless, well, OK, so there's some trivial cases, but the only interesting one is when the dimension of x, the dimension of the domain, is 2. OK, in that case, it's called the pseudo-holomorphic curve because x is a surface, it's a two-dimensional real manifold, but a one-dimensional complex manifold, so it's called a curve. And this interesting case, when the dimension of x is 2, this equation, star, is elliptic. And what that means is that we can expect to find a reasonable five-dimensional space of solutions. And moreover, that in favorable cases, that the space of solutions will be a reasonable five-dimensional space. OK, so there's a fact, which is that all two-dimensional, almost complex, manifolds are Riemann surfaces. That is, they have local, local analytic coordinates, i.e. locally equivalent to C with its standard complex structure. And the way to prove that is to use ellipticity of the equation. So you use ellipticity of the equation for, I guess, maps from x to C, analytic maps, pseudo-holomorphic maps from x to C. If you produce a produce of those, you get these. OK, so the holomorphic is a pseudo-holomorphic map from a Riemann surface. So I hope I'm not, well, I may be swiftly corrected by the audience, but as far as I can see right now, these objects are completely useless as individual or discrete objects. To do anything at all with them, you have to study the modular space of all of them, but put them into a family. Any application requires understanding the space of all pseudo-holomorphic maps. A pseudo-holomorphic map. So a map is pseudo-holomorphic when the differential is complex linear. Does that answer the? Well, they will chat back if it didn't answer. OK, so I will write the pseudo-holomorphic curve equation in this form here. But it's the same equation as before. So this is the modular space. OK, so what structure does this have beyond being a set? So first, we could put a topology on the set of holomorphic maps. And well, I guess there are two topologies you might consider. So first, you could equip it with a compact open topology. What's an open set in the compact open topology? Well, if you take a compact subset of the domain and an open subset of the target, then the space of maps which send that compact set of the domain inside the open subset of the target is supposed to be an open subset of the, yes. Yes, so the question is, what did I write here? What is this equation? Yeah, so du, let me write it over here. du is a map from the tangent bundle of c to the pullback of the tangent bundle of x. These are complex vector spaces, and this is a real linear map. So it has a decomposition into a unique decomposition as a sum of a c-linear map and a c-conjugate linear. And so the assertion at du is holomorphic. The same as saying this one is 0. OK, so I just gave a definition of the compact. Yes. Sorry, you mean, oh, well, so maybe I want some more topology on the surface. Maybe I want a domain other than the sphere. Yeah, this is a roman c, not a blackboard, not the complex plate. That's just the curve, c for curve, not c for complex. OK, so I gave this definition of the compact to open topology, which probably people have forgotten by now. Doesn't matter, because it's perhaps better thought of as the topology of uniform convergence on compact subsets. OK, so usually our curve, c, is compact. I might even say it's always compact. So this is just the topology of uniform convergence. OK, and there's a remark, which is that this topology satisfies a very nice property, which is that for any pair of topological spaces, so you call them a and b, where a is locally compact, map from any topological space to the continuous maps from a to b, continuous in the compact open topology, if and only if the associated map, z times a to b, is continuous. So if you come up with any other topology on any other complicated way of defining a topology on the set of continuous maps with this property, a map to it is continuous if in a leaf. This associated map is continuous. Then that topology you defined is the compact open topology. OK, so another way you might topologize, this is using, well, the topology of the infinity convergence on compact subsets. So that's a much stronger topology priori. So it's a much stronger topology on the space of all maps. On the subspace of holomorphic maps, it turns out it's the same. So proposition of the C0 and C infinity topologies on the space of holomorphic, the set of holomorphic maps coincide. And this is a consequence of the elliptic regularity and the Gromov-Schwarz lemma. As far as I know, I think we actually don't care so much about this result in the sense that this topology number two is the right topology to use. Sort of a coincidence that the same as the compact open topology, but any time we work with the space, we're not likely to care about the compact open topology. I'm erasing the board, so now it's a good chance. Yeah, so right. So what does it mean for the pseudo-holomorphic map equation to be over-determined? Right, so I guess there are probably some technical PDE way of formulating it. But I guess all I really mean is there are more equations than there are variables. So a pseudo-holomorphic map to the target with n is the dimension of the target that many functions. And then there's the equation asserting it's pseudo-holomorphic, and that's more equations than. So let me not try to calculate it right now. Of course, the reason that's not really a good answer is that maybe the equations are a little bit redundant. So you have to be careful if you really want to justify that. Oh, OK, I'm done with this page. OK, so let me try to talk about the local structure, the space of pseudo-holomorphic maps. So let's fix a base point. So a base point for this is just a pseudo-holomorphic map U. And let's assume C is compact. So locally, we can describe the modular space as follows. So let's consider the space of maps from C, pullback of the tangent bundle, map that to the pullback of the tangent bundle, tensor with anti-holomorphic one-forms. And we send a section of a section of the pullback of the tangent bundle gives me another nearby map. I might write that nearby map as the exponential of C. OK, so that gives you a new map nearby. No, C is the Riemann surface. It's a complex curve. So it's maybe the Riemann sphere. Maybe it's the Riemann sphere or a torus. Yes, yes. What is HS? HS. So HS is the space of so-bola for regularity S functions. So HS of, say, of a manifold is the space of, well, OK, let's take smooth functions. And let's equip it with the norm, which is the sum of the L2 norms of the derivatives up to order k. That's a positive definite in a product. And we take the completion. Oh, yeah, I messed this up, didn't I? I mean, I need to take the completion with respect to that norm. So let me write it as the completion. I'm going to say m is compact. OK, so there are fractional versions of this space as you define using the Fourier transform for any real value number in place of an integer k. But it doesn't really matter for now. k can be an integer. Yes, the complex conjugate vector space. So the bar is the complex conjugate vector space. So V is a complex vector space. V bar, well, it's the same as V as sets, but the multiplication is conjugated. So lambda bar is lambda bar. So this is the place where this anti-holographic component of the derivative naturally lies. So if you take a real linear map and decompose it into a complex linear map, plus an anti-linear map, then the c-linear map, u, 1, 0, that's a section of u-upper-star-tx tensor over c with the co-tension bundle. And du 0, 1 is a section of that. OK, so I'm not done writing the map, so maybe I'll finish it. So x with c, that's a smooth map. Well, it's not a smooth map. It's a map nearby u. And I can take the anti-holographic component of its derivatives, d of this 0, 1, and this thing, now it's not a section of this bundle. It's a section of x, c pullback tangent bundle of x tensor. So I have to do a parallel transport from c to u of that. OK, so it doesn't really matter the details of what exactly I wrote there and many other ways to write it. The point is that this modular space of holomorphic maps is just the inverse image of 0 near u. Or maybe I should better say it like this. It's an identification of, I guess, germs of topological spaces. OK, so this map is not a linear map. It has a linearization near 0. So this is derivative of d at 0. Oh, it's another map. And it's an elliptic operator. So as you already know from Francesco's lecture this morning, since d is elliptic and c is compact, these together imply that this is a Fred Holm operator. OK, which means that the kernel is dimensional. The image closed and the co-kernel is not dimensional. OK, now there's a non-s for this statement, any real number s for this one, for this one. Yes, OK, what about the, I repeat the question. What values of s are permissible here? Yes, let's see. That's a good question. What values of s? Certainly if it's sufficiently large, you're fine. But what exactly, why exactly do we need that? Well, OK, so first of all, you definitely need, yeah. So I think we definitely need HS to embed into continuous functions. Because if that's not true, then maybe I exponentiate and my map holds non-compact and non-complete or something, that wouldn't be good. So I definitely want a neighborhood of 0 here to be uniformly small. I think that's all I need. Because I think as long as you have HS inside C0, then HS is closed under various operations, like post-composing with a smooth function. But that's actually not all that easy to see. No, I don't think so. Is it obvious whether D is elliptic? Definitely not. Yeah, so I haven't even defined elliptic. So it would be hard for it to be obvious before I define it. And even then, I doubt it. So this operator basically sends C to, OK, it's derivative and the 0, 1. So I can say a little bit. OK, for it to have any chance at all of being elliptic, the vector space in the domain and the vector space in the target better have the same dimension. In this case, they do because this is a tensor over C with a one-dimensional complex vector space. I mean, ellipticity amounts to saying that for any covector in C, basically the highest order term of this operator in the direction of that covector is an isomorphism between here and here. That's what ellipticity means. And so if you have an operator like this, writing it all out in coordinates is a big mess. But writing down the leading order term is a little bit easier because everything else just dies. So to verify ellipticity, you don't have to write out everything that you would otherwise have to. OK, so the stretch on this statement, basically what it says is that these two vector spaces are infinite dimensional. But this d, other than finitely many dimensions in the domain and the target, this d is just an isomorphism. So it cancels the infant dimensions of the domain with the infant dimensions of the target. Now there's a nonlinear analog or nonlinear consequence, but it's really an analog of this thread homeless of the linearization about the nonlinear operator. So consequence d, the bold d, is that there exists a fine dimensional model of the shape rn to rm. So here's d down here, where this is a transverse fiber product. And so mb, upshot of that assertion is that the inverse of 0 is f inverse of 0. And so this is a sort of fine dimensional presentation of our transverse. You're asking what this means? Yeah, so the derivative of bold d is little d. So transverse means that the derivative of this plus, let's call that q, the derivative of q from hs plus rm h minus 1 is surjective. That's a reasonable statement, since the co-colonial of d is fine dimensional. So I only have to, I only need to find out many more dimensions to fill it out. So of course, when I write hs and hs minus 1, I mean those spaces over there. I'm just abbreviating for a sake of time. Yes? Well, since it's a vector space, right, it's tangent space. Yeah, so why is the domain of the derivative the same as the domain of the map itself? And the reason is that, well, it's a vector space. So it's tangent space if the origin is itself. I guess I mean the, yes. What do I mean by this statement? So there's, I guess, I think a pretty apparent map from in this direction. And that is isomorphism. Right, well, it means it's, so what is this? What is this? Tell us about the whole, the space of holomorphic maps, whole CX. And what does it tell us? It tells us that it's locally the zero set of a smooth function, whatever, I mean, whatever structure that has. So topologically, it doesn't have all that much structure. I mean, you can take any closed set in Rn and express it as the zero set of a smooth function. So you need a much more complicated structure to keep track of what exactly this chart tells you about it. Okay, I have, people are asking lots of questions because I have a lot, a lot of pages left. Okay, that's good. Any other questions about this? Okay, so I can say a little bit about the here. As Francesco already said, the index of the Fredholm operator is the kernel, dimension of the kernel minus dimension of the kernel and you deform the elliptic operator. These dimensions might jump, but the index stays the same. And in this case, the two-singer index theorem will tell you that this is the same as I hope I write. The order of the characters gives C times the dimension X plus twice the first turn class of the pullback of the tangent bundle. Sorry? Yes, yes, yes indeed, integrated over C. Okay, so there are two situations to distinguish here. So U called regular or transverse. You've probably forgotten what U is by now. U was the base point in the spaces of the homework maps which we were talking about. So it's the base point around which we make this chart. So U is called regular or transverse when D or equivalently F is surjective. Okay, so and in this case, if U is regular, that means that this modulized base locally is smooth manifold near U. That's, what is that? When D F is surjective, F inverse of zero is a smooth manifold. F being a smooth map from Rn to Rn. So there's a little bit of problem with this word is. I said here because I haven't, so far I've told you what this is as a set. I've told you what it is as a topological space. So I guess what we've proven here is that it's locally Euclidean. It's homeomorphic to Rn. But in order to say it's locally diffeomorphic to an open subset of Rn, I need already to have a smooth structure here. Question from the chat. Is it easy to compute what F is explicitly? I doubt it. It seems rather difficult to make any sort of explicit, at least not directly. The way I presented it, I don't know how you could directly figure out what F is. I mean, if you somehow have a complete current issue structure on the holomorphic map, the space of holomorphic maps, and you calculate what the dimension of the tangent and the obstruction spaces are, then maybe you could use that to deduce what the local model has to be in some simple case. But okay, so there's a problem with this word is. Really what I should be saying is that this gives us a chart for a holomorphic, for this morphology space. And well that's one chart, depending on the base point U and depending on whatever choices went into the existence of my map F. But maybe there are lots of other points U around which you could find a chart and there are also many more choices of F you could use. And you have to know that these charts are compatible. One thing I will say is that the locus of regular points is an open set and that's not so hard to see. So why is that true? Why is the regular locus open? Well, it's the locus of, regular just means this operator is surjective. So if you have a family of fret home operators, I claim that surjectivity is an open condition. Pithy one sentence justification for that. But it's true for finite dimensional. It's a space of linear maps from between five dimensional vector spaces. And you can reduce to that case. Okay, so this is the case when you was regular and in general all you have is this chart. So it's the fiber over zero of some smooth map and that's called the Kurinishi chart. And as you might expect it's a bit more complicated to say what it means for charts of this form to be compatible with all the possible ones that you might define. Okay, so why do these ask together anyway? The hint is to use universal properties. So universal property is like the characterization I gave of a compact open topology, a map from arbitrary topological space into the set of continuous maps is continuous with respect to the compact open topology on the target if and only if this map or the associated map from Z cross A to B is continuous. So let's declare that for, so this is the category of smooth manifolds. So let's declare that for any smooth manifold there's a natural bijection between smooth maps from Z to this regular locus of holomorphic curves and smooth maps B cross C to X smooth whereas the fibers are super holomorphic and regular. So that doesn't really look like a definition it's like a property. Yeah, so what is this vertical map here? Right, so let's begin over here. So the space of holomorphic curves is the zero set of this map bold D and I just sort of postulated the existence of a five-dimensional model of this map bold D. Namely that there should be a fiber product diagram like this and in that case, the inverse of the zero set of D is the same as the zero set of F. So all I've written here is the zero set of F, this fiber product here is the zero set of F. So I shouldn't really put the entire space of holomorphic maps here. I mean, the open neighborhood of U on which we have this chart, I say. Oh yes, yes, the diagram is commutative. Right, yeah, so this is let's name the map. Name the map might be capital F. So then I can take F of a little z comma, something from c to x. And so that map should be regular, is pseudoholomorphic and regular. The regular slash transverse in this sense. Yes, it's the definition. Yes, the question is, is this the definition of the space? It looks like a property, but I'm gonna claim that it's a definition. Yes, yes, fixing complex structure and C. Questions about? So yes, as Paul pointed out, this doesn't really look like a definition. It looks like a property. But I claim that in fact it is a definition or it's maybe 99% of the way there too. It's part partially definition. Maybe I shouldn't say 99%. Okay, so we need a little bit of a abstract discussion. Okay, so let's C be a category. So let's recall what that means. So C has a set of objects. And for every pair of objects, there's a set, x, y, elements upon x, y are called morphisms. And finally there's the data of how you can compose morphisms. And composition is associative. Okay, so you're familiar with many categories. Smooth manifolds and smooth maps. Belian groups and group homomorphisms. Groups and group homomorphisms. Topological spaces and continuous maps. Morphisms in a category are just sets. They don't have any other structure. Bunkter from the opposite of C to the category of sets does what? It assigns each object of C, that and each map in the object of C, a map which I'll call f up or star from f of y as compatible as composition. Okay, so let me give some examples because otherwise who knows what any of this means. First example, a little bit abstract. This is the functor which assigns to an object. The z, the set of maps from z to x. Okay, but let's look at some more, perhaps more interesting ones. So any topological space, let me perhaps assign the space of continuous, a set of continuous maps, z cross. A to B, and here A and B are fixed. Topological spaces. So top is the category of topological spaces and continuous maps. Okay, so if I have a continuous map from z cross A to B, and a continuous map from z prime to z, then I can pull back and get a continuous map from z prime cross A to B. So that's those of the pullback maps. I can do the same thing for smooth. Map folds, so smooth maps cross A. Another one, z maps to smooth embeddings. Sorry, not smooth embeddings. I'll do that one later. Isomorphism classes of vector bundles on z. That pulls back. If I have a map from z to z prime, a vector bundle on z prime pulls back to one on z. Okay, a functor of this form is called representable. Or in definition, f, f, a functor from C op. That is called representable isomorphic to Hx for some x and c. And now what I claim is that this object x is unique up to unique isomorphism if it exists. There's only one of them. If I give you a functor, it determines this object x uniquely and in as canonical way as you can want. Okay, so let's go back to what we were trying to do. Declaring that this right-hand side is a definition of, well, declaring that this bijection here is a definition of this whole Cx reg as a smooth manifold is exactly representing a functor. So over here on the right-hand side, I have a functor on smooth manifolds. To every smooth manifold, you associate z, you associate the set of maps like this satisfying this condition. And what I'm saying is I want that to be of the form that a smooth map from z to something. Okay, so that's what it means to be representable. Then representing object of f is a pair, x and c and u in f of x, u because it's, this is the universal family. It's a pair such that for all b and c, z, x to f of z given by sending f to f whole back of u is a nice morphosome. Representing functors is perhaps something that you've been before if you've been classifying spaces, which I believe you probably have. Yeah, let's see. So what about these other functors? Are they representable? So this functor here, so if a is locally compact, this is representable represented by the set of continuous maps with the compact open topology. If a is not locally compact, I think I'm pretty sure that in general it's not representable. What about this functor? Is this one representable? The smooth map from a to b, a and b are smooth manifolds. Yeah, so if a is a finite set, then it's represented by b to the a. And if a is not a finite set, then it's not representable. It's some big infinitimensional thing. It's not representable by a smooth map. This one, isomorphism classes of vector bundles. So there's a simple way of seeing it's not representable. There are two ways of seeing it's not representable. I'll tell you the first one. I'll tell you the second one later if I have time. The first one is that, well, okay, if this is representable by something, then I can figure out what its points have to be by taking z to be a point. So if I have a smooth manifold, the set of maps from a point to it is its underlying set. So if this is representable by something, I can figure out the underlying set of that something by plugging in z to be a point. If I plug in z to be a point, I can figure out the isomorphism of classes of vector bundles on a point. Well, okay, I might as well write this functor as a disjoint union of a bunch of things and ask whether each one for the rank and ask whether each one is representable. If I fix the rank, then there's only one vector bundle on a point. So then the only possibility of a representing object is a point. It's pretty clear it doesn't represent that. Okay, so the question is, can one define this holomorphic mapping space in the complex category or work in the complex category? So I guess you'd, right, so z would be a complex manifold and you want whole cx to be a complex manifold such that we have this property where presumably this condition here would be that this map capital F is jointly holomorphic. I believe this is true, but it's sort of hard to track down references for it. Maybe I haven't tried so hard. I think there's some, I think this might actually be the context in which the original work of Korean issue was done to try to define mapping spaces between two complex analytic manifolds, between complex analytic spaces and show that that is itself a complex analytic space. So the question is, what is this map? Yeah, so, right, so F is a functor. So that means to any map in our category from z to x, any elements of hom zx, you get a map from f of x to f of z. And I was sort of, I didn't really say what that was in all these cases. It's sort of implicit, there's an obvious one to take, but if I were really doing this right, I'd write it down, right? So if I have a nice morph of the class of vector bundles on z and I map from z prime to z, I pull back to get something on z prime, for example. All these pull back. Okay, so I'll give examples of representable functors right now, but let me just, I don't think I can, I don't think I'm allowed to make this definition without stating the first result about it, which is that representing objects are unique up to unique isomorphism. So for any of representing objects, x, u, x prime, u prime, there exists a unique map, x prime, pulling back the universal family of x prime to the universal family of x. So that's the sense in which if I give you the functor, then the object is determined, yes. Let's see, that was a long question. So it's a little bit difficult to repeat. I'm going to repeat it by saying is what's the, are all, what's the relation between representable functors and forgetful functors? What about it? Yeah, this is what I, yeah. I mean, I mean, it's definitely not, yeah, forgetful functors are representable, but it's sort of, so we have a little bit of wrong intuition, which tells us that if we have two billion groups, then harm from A to B is an abelian group, or x and y are two vector spaces, then harm x, y is a vector space. And if x and y are topological spaces, then at least if they're sufficiently nice, I have this topological space of maps in the compact open topology that I just said. But that intuition isn't, we're sort of breaking that intuition when we do category three this way. Our harms are always set. They're not, they don't have extra structure a priori. We can try to give them extra structure by doing this sort of thing, but they don't begin with extra structure. And you see for topological spaces, there's not always a topological space, harm x, y for all things. Okay, let me, let me keep going a little bit. Let's see, so I've already done a few of these examples. Yeah, let me just do these two. So if we have the homotopy category of CW complexes and look at the functor sending x to the isomorphism classes of vector bundles. So we're familiar with classifying spaces and algebraic topology and that tells us this functor is representable. So this functor represented by, and when you represent a functor, you really should always think of it not as just an object, but as a pair. A pair of the object with the family. So this is this infinite Grasmani and BO together with its tautological vector bundle, right? So this familiar state from algebraic topology that homotopy class of maps to BO is the same thing as an isomorphism class of vector bundles is exactly the statement that this functor is represented by this. Yeah, so I mentioned this space of smooth maps from M to N. So M and N are smooth manifolds, but I can look at the space C infinity M and space. This is a topological space, the space of the maps M to N with topology of C infinity convergence on compact. Subsets of M. So I'm sort of gonna go in the reverse direction here and ask what functor does this represent? But this space C infinity M and N represents functor of what one might call relatively smooth maps Z cross M to N. So this is Z is just a topological space. And here relatively smooth means, well, if you take sort of local Euclidean charts for M and N and you differentiate arbitrarily many times in the M direction, you still get a jointly continuous map. Yeah, so relatively smooth means that, so let's give this a name, call it capital F. I take arbitrarily many derivatives in the M direction, and this is continuous, all for any derivative you might take. And then there's a, right, it's continuous on Z cross M. Okay, so there's probably an implicit question which you might be thinking but have not asked. We just why would anyone in their right mind prefer to define object category C by first defining a functor and then representing it. So there are two answers to this. Answer number one is that you're defining an object in your category C might actually be pretty complicated. So if C is just a topological, if C is just a category of topological spaces, then, well, okay, the set with a topology is not so bad. But if it's say the category of smooth manifolds, then to find an object of C, well, you need a set with a topology and a smooth Alice. And then, well, maybe, okay, maybe you're trying to define say like a scheme or something and then has a structure sheaf you have to define or maybe a complex variety you have to define has a lot of structure or crew need she space has a lot of structure. A lot of categories can have quite complicated objects. Find an object of C can be quite complex compared to defining a functor. Because what's a functor like this is just a bunch of sets and maps between them. That's all. That's answer number one. The answer number two is that it's sort of related to answer number one. Maybe you can say it's a consequence of answer number one, but it's an important consequence, which is that functor often more obviously canonical. So what I'm defining, for example, if you're trying to study modulite spaces of pseudo-hologomorphic curves, you know, I wrote that map, bold D. And to write it, I had to choose a Sobolev completion. I had to choose that real parameter S. And someone might choose to work with some different Sobolev space or some different bottlenecks space of maps, not necessarily Sobolev space. And maybe you have different ways of choosing those charts. But this definition here of this functor here is sort of obviously canonical. I mean, there's not much else I could say. I mean, there's one other thing I could say, which I could sort of lower the regularity, to say it's only CK or something. Yeah, yeah. So this is, yes, the question is how would one ever prove a functor is representable? And why is that any easier than just defining the object in the first place? Yeah, so that indeed is a disadvantage to this approach. The answers are the advantage and the disadvantage is you have to prove representability. One reason the functor approach helps you is that you can prove representability locally. So what I wrote here to give a different smooth chart for the modular space. If I prove that in a neighborhood of that point view that that's a local chart, that that chart I gave it, if I prove that it represents this functor, then I've proven representability locally. The fact that it represents this functor, which didn't depend on any of the data going into the chart, tells me that any other chart gives the same smooth structure. And so the smooth structures locally patch together. Yes, so, okay, the question was just to repeat. And so this is even what I was about to write down next. So let me write out a sketch and then it will be more easy to throw darts at it once it's up here on the board. Okay, so this functor is representable. Okay, so we already have a topological space. Okay, and it suffices to prove representability locally. So that is it suffices to define the open cover of the topological space. And each of the elements of that open cover with the structure of a smooth manifold. So that those smooth manifolds represent the functors of smooth maps landing in those open subsets. That's just because if I wanted to find a smooth structure on a topological manifold, I can define it locally and patch together as long as they agree on overlaps. And since they represent the same functor, they have to agree on overlaps. Okay, so how are we gonna prove representability locally? So, it's our base point. And so we get this map d. And, okay, now, in some sense, I'm gonna cheat a little bit because I'm really hiding a lot of the content is contained simply in the Banach space inverse function theorem or implicit function theorem, which tells me that since the differential is surjective, that this is a smooth manifold. Okay, so I'd rather not spend a lot of time explaining the Banach space implicit function theorem. But that's what it tells you. And that's a non-trivial statement. So, given that, I wanna try to argue that this zero, I think it's zero, sorry, oh, here, yes. Well, right, so smooth manifold near, well, so it's zero in here, which corresponds to u in here. So, I wanna argue, okay. So, to show that this smooth manifold represents this functor, amounts to, amounts to the following claim. It's the claim that, so for z with manifold, a map from z to this inverse image of zero is smooth if and only if the associated map from c to x, that's what we're trying. And there's not, there's some content to it, but it's mostly just elliptic regularity. So, this direction is, I claim, trivial. It's trivial in the sense that you have to, you know what the smooth structure on this HS space is. And this direction, elliptic regularity. So, a priori, a smooth map from z to d inverse of zero okay, it's just a smooth map from z to this HS space. So, let's mark the smooth map is what, say HS of, yeah, HS of c, is a smooth, is a map from z cross r to c where the derivatives in the z direction all exist and are in HS of c. Yes, I really should call that either HS of c is continuous. So, it's sort of smooth in the horizontal direction, in the z direction, and the HS regularity in the c direction. And the elliptic regularity allows you to upgrade that to smoothness in the c-der or joint smoothness. Yeah, my point here is mostly, this is chasing definitions here, but it's a nice way of showing that the charts the local smooth charts patch together nicely. And if you want some sort of more exotic structure on the modular space like a Korean issue structure, then you're essentially obliged to do it this way unless you want to have some nightmare of patching things. I had more stuff on my notes, but this is a nice stopping point. So, repeat the question. So, the question, the statement, the question is, is it the case that the reason we do this is that we had this argument to find a local smooth structure and then we say, well, that local smooth structure represents this functor, and the functor was just this canonical thing and therefore the local smooth structures are all the same, they patch together. And the answer is yes. Does what work in any other category? So, the question is, can we axiomatize locality in any category and can we make a representability a local notion? Yeah, so there's this really general notion of a growth in dectopology on a category and it allows you to make sense of sheaves and stacks on a category. And, yeah, representability is locality. So, yeah, representability being a local statement amounts to some being able to glue objects together. And topological spaces are easy to glue together, but if you go to algebraic varieties and try to start to glue them in bad ways, then you get more exotic things like algebraic spaces. And so, in that context, representability is not local. Yeah, but the question is, is everything just tracing definitions? So, certainly not everything that you care about, but maybe a lot of the foundation, certainly. It's in the eye of the beholder in some sense. But the question is, if the modular space is not transverse, then can you write down this, still write down modularity functor and try to represent it by something? Yes, I believe the answer is yes. You can just write down, I mean, you can write down basically the same thing. So, it's represented by a current issue space. And so you need to allow Z to be a current issue space and say what it means for map from Z cross C to X to be fiber-wise holomorphic when Z is a current issue space. So, if you can make sense of that, then you can make sense of the functor and you can chase some more definitions and prove it's representable. So, the question is we've proved local representability. We get the local representing objects, what's the global representing object? Right, so the global representing object is a smooth manifold. So, how do you define a smooth manifold? Well, it's a topological space with an atlas of charts. So, here's the topological space. And what are the charts? Well, the charts are the same charts that I defined much earlier. The content is that the transition maps between the charts are smooth and that's somehow some fact we proved here. And then if you're really pedantic and you defined your manifolds to be topological spaces equipped with a maximal atlas of charts, then you add an object which is compatible. So, the question is what is really being defined when we define something by representing a functor? Well, so first we define the functor and then we argue the functor is representable. And that's just a property, but it gives us a definition of that object. I mean, I presented things in a slightly, well, maybe more than slightly confusing way because I sort of started with, you know, the problem is there are lots of things we want to call whole CXY. There's the set, there's the set with a topology, then there's this smooth manifold which is the object representing this functor. So I sort of said, well, we already know we have a topological space and that's what we're going to use to represent the functor. So, it's not circulated despite the notation.