Thanks, Joe. I'm going to start on this board with just a preface of generalities. Let's call this the archetypal moduli problem, by which I mean from the perspective of global analysis or differential topology or simplexed topology as distinct from algebraic geometry, where they also like to talk about moduli problems. For me, a moduli problem means this kind of thing. You have some space, which I'm calling suggestively j, but right now I'm just going to call it the space of data, which has some kind of topology on it. And for each choice, which I will suggestively call also j, in this space, we get some moduli space m of j, which also has a natural topology.  Or if we can't achieve that, maybe at least a smooth orbifold. And over the course of these two weeks, I'm sure you'll also hear things about weighted branched orbifolds with boundaries in corners, though not for me. From my perspective, well, the first thing to say about this moduli space is that locally you can identify it with some kind of zero set, which I'll write sigma j inverse of 0, except not precisely that. Often you have to divide it by some kind of symmetries. So here sigma j is some section of a Banach space bundle. And we already heard DUSA sketch one example of this. Because you can express all sorts of geometric PDEs in this way as preferably smooth sections of Banach space bundles in an infinitesimensional context. But I'm saying this is true only locally in general. So DUSA described the situation with holomorphic spheres where you can actually do this globally. But in my experience working with j-holomorphic curves, that's rather the exception to the rule. And so I will talk about some more general situations where you can do this but only locally. And these symmetries can cause quite a lot of headaches. But the first problem, of course, is you want to be able to say that this zero section here is a smooth object. And the implicit function theorem ought to tell you that if you do things correctly. But there are two general ways that you can approach that. First, the so-called abstract approach where you perturb the right-hand side of the equation. You can say something like define the moduli space m tilde, which depends on the data j, and some extra auxiliary choice that are called nu. That's the space of all u in this Banach manifold such that sigma j of u matches nu of u. So this is for some other section nu of that same Banach space bundle. And the idea is if we choose that generically, then this set of solutions will be a smooth object simply by the Sard-Smale theorem. That's kind of equivalent to the statement just expressed in local trivializations that a generic point in the target of a smooth map is a regular value of that map. So this is a generic section is also intersects this section transversely. So for generic choices of this auxiliary nu, our special section sigma j is transverse to that, which means this moduli space of perturbed objects I define is a manifold. And I even want to say more, because if we're talking about elliptic problems, this section is not just a smooth section, but also is Fredholm in the sense that its linearization at any point is a Fredholm operator. So when we have these kinds of transversality conditions, we get just not just a manifold, but a finite dimensional manifold because of finite dimensional kernels of the linearization. So that's, of course, very nice when that happens. The problem is I perturbed my equation to a different one, which might not be the one I'm actually interested in. That's problem number one. There are more problems, which I think you'll hear more about over the course of the two weeks, because this abstract approach is sort of the raison d'etre for the whole polyfield project. But it's difficult. And part of the reason it's difficult is that, first of all, this correspondence between some zero section and my actual moduli space of interest is in general only local. And it also has these symmetries. You have to deal with both of those in a sensible way when choosing these perturbations. And it means just choosing some generic section of a Banach space bundle is usually not enough. You have to choose it satisfying some conditions. And maybe those conditions are too stringent to actually apply the Sards-Mail theorem in this way. So do you want to be like invariant under the symmetry or something?  I reserve the right to not talk about that anymore, because other people will. And I'm going to talk about the alternative, which is in the title of my mini course, so-called classical or what many people call geometric perturbation methods, which is the notion that if you take a generic choice of the data in this space of data, then your actual section will be transverse to the zero section, which means, of course, the zero set of your section is a smooth manifold and finite dimensional as well, because we're in the Fredholm context. And well, the space we're actually interested in is that divided by some kind of symmetry. That, in general, will be an orbifold. So an orbifold is something that looks locally like Euclidean space divided by a finite group action. And that's what you get if you're taking a finite dimensional manifold and dividing it by the kind of symmetry groups that we're interested in. So that is minor headache number one to be aware of. We won't get manifolds in general, but we could get orbifolds if we're lucky. Much larger headache number two is that often we're not lucky, and this condition simply isn't true. And Dusse already talked about that a little bit. I'm going to talk about that some more. So that's all I will say about generality is I'm going to focus on this second, this classical approach over three talks. And in a particular context, I want to discuss symplectic field theory. So in the original propaganda paper in 2000 by Elishberg given to Holhofer, you have symplectic field theory presented as a very general way of defining invariance of contact manifolds and symplectic abortions between contact manifolds. And well, one can do it even a little bit more generally than that, which is sort of nice for certain purposes. So I'm going to talk about stable Hamiltonian structures. So first of all, if M is an odd dimensional manifold, a stable Hamiltonian structure H on M is going to be a pair capital omega lambda, which consists of the following. We have first omega is a closed two form of maximal rank. So you should imagine that as something that arises on any hyper surface in a symplectic manifold. You just take the symplectic form and restrict it to a hyper surface. It's going to be a closed two form of maximal rank. If I omit the word stable, that's what we call just a Hamiltonian structure. Now stable comes from this additional one form lambda, which satisfies following properties. First, lambda wedge, the top dimensional power of omega must be a positive volume form. So this implies in particular that you can take omega and restrict it to the kernel of lambda. Lambda is nowhere zero. So its kernel is a hyper plane distribution, which I'm going to call psi, as I always do with contact structures, because that's one important example, could be a contact structure. But in general, it's just some hyper plane distribution. And omega makes that hyper plane distribution into a symplectic vector bundle. Gives us a symplectic bundle structure. And then the other condition, since omega has maximal rank on an odd dimensional manifold, it has a kernel that's always one dimensional. That's the so-called characteristic line field, the one dimension of directions in which omega is degenerate. So what I'm going to require is that d lambda also vanishes on that same kernel. Another way of saying is the kernel of omega is contained in kernel of d lambda. OK, so this is a definition which has been around for several years. And as stated, it's not too hard to understand. But I think a lot of people don't really understand what it means. So let me talk a little bit about that. First observation is that if I now take this odd dimensional manifold and look at a small cylinder constructed out of it, so some small interval times m, this thing now inherits a natural symplectic form.  And now it's an easy exercise. That's going to be a symplectic form. As long as this cylinder is sufficiently small, it might even be symplectic for epsilon larger than that. But in general, I can only guarantee this is true for epsilon small. Moreover, what does the characteristic line field look like if you restrict? You've got a one-parameter family of hyper surfaces.  The characteristic line field of omega restricted to m sub r is independent of r. So it's the same characteristic line field for all r, all parameters in my parametrized family of hyper surfaces. That's really the original definition, which I think appeared first in the book by Hofer and Sander of a stable hyper surface. That's where this idea comes from. It's the kind of situation in which you would expect to be able to prove something like the Weinstein conjecture that asserts existence of closed characteristics of this characteristic line field. Because if you have them on some hyper surfaces in the vicinity, you have them on all of them. Excuse me. What do you mean with the first symbol of the last line? The first symbol? It's an and. That's an and, ampersand. OK. So in order to get the stable condition that you want this last condition at the kernel of omega contained in the kernel of omega, that's what implies. That's what implies this fact about the characteristic line fields, exactly. And conversely, if you are given a one parameter family of hyper surfaces that has this property, this stability of the characteristic line field, then you can always find a stable Hamiltonian structure that produces that. So that's kind of an exercise using the Moser defamation trick. How much will I lose if I just ignore this definition pretend you said contact structure? Nothing, really. That's a perfectly good point. If you don't like this definition for some reason, because it's new, just think lambda is a contact structure and big omega equals d lambda. I was going to say that soon anyway. So from this data, we have also a canonical generator of the characteristic line field, which, by analogy with contact forms, we'll call this the rape vector field. So my notation will be r sub h. And it's determined by the conditions omega annihilates the direction of the rape vector field. And then we normalize it by lambda. No, there's no reason at all why d lambda couldn't be 0. I will also say that again in a moment. So we have a stable Hamiltonian structure. It defines this hyperplane distribution psi, which might be a contact structure in certain examples, but it doesn't have to be. Could also be a foliation. We have this rape vector field. Another thing we can do is, since we have this small symplectic cylinder we built out of it, we can also just arbitrarily expand that to r times m and rewrite my symplectic structure there in the form omega phi. That's going to be d of phi of r lambda plus omega. So what's phi? Phi is going to be in the space of smooth functions taking r to that same small epsilon interval with strictly positive derivative. So this is a symplectic form also. It's symplectomorphic to the one I wrote down before. I've just stretched things. But the point is that r cross m is the kind of setting where I would like to talk about j-holomorphic curves. And I can do that now by saying, well, the space j dependent on the stable Hamiltonian structure is going to be the space of almost complex structures j on r cross m, which I will assume always to be r invariant. In other words, there's this r translation action on r cross m. I want that to preserve all my j's. I also want to say that j maps the unit vector in the r direction to the rave vector field. And I want to say that j preserves my hyperplane distribution as well as the symplectic structure. It's compatible with the symplectic structure on that. So let's say j restricted to psi is omega compatible. OK? So here's a lemma which is very easy to prove and sort of helps justify all these definitions. That is that for all choices j in this space of translation invariant, almost complex structures, and all functions phi, as I indicated above, that are mapping to that small interval with positive derivative, it's not just that omega phi is symplectic, but in fact it's compatible in the usual sense of Gromoth with this choice of j. So I can fix j and I can then allow any choice of phi I want. And that's an important detail for defining things like the energy of a j-holomorphic curve, because we then take that to be the supremum of symplectic area defined with respect to all these different choices of symplectic forms. So let me give two examples. They've already sort of been mentioned.  And then this symplectic manifold r cross m with this kind of symplectic structure I've been talking about. I'm not going to try to say these two precisely. You have to make some kind of condition on how you define phi in order to say this. But I'm just going to say one can do it. So this is what's called the symplectization of the contact manifold. I really should relax the epsilons in order to say that. I should actually at least make plus epsilon be infinity. But let's not worry about this right now. Can you quickly write down what the symplectization is? No. I'm going to say it's this. Yes? So what's the condition where you can integrate this hyperplane distribution to a symplectic manifold? Where you can integrate the hyperplane distribution. Yeah, if you integrate it and then you find a sub-manifold which has that as its tangent on board. Well, for the hyperplane distribution to be integrable, I think that's equivalent to saying that d lambda vanishes on psi. So certainly it's sufficient of d lambda vanishes altogether, which is my next example. Already in the first row, we think that the word compatible sort of looks like compact. Well, I'm sorry. I'll take this opportunity to make a blanket apology for my handwriting, as I customarily do. There isn't that much I can do about it. OK, compatible, compatible. All right, second example. If m equals the product of s1 with some other manifold w, and I take w to be a symplectic manifold. So let's say I have symplectic form little omega on w. And then set capital omega to be that plus dt wedge dh. So here, t is the s1 coordinate. And I'm choosing some function h, a Hamiltonian function, which is dependent on time in a periodic way. So this is capital omega. And I take lambda to be just dt. This is a case where it's closed. d lambda vanishes entirely. So now one can check. This also is a stable Hamiltonian structure. And it has a meaning which a lot of people will probably like. You can compute the rate vector field for this. It's simply it's the vector field in the t direction plus a vector field in the w direction, which is precisely the Hamiltonian vector field. So this is a way to turn questions about Hamiltonian vector fields on symplectic manifolds, even time dependent ones, into questions about a time independent, RAVE-like vector field on an odd dimensional manifold. And it gets better because for the class of almost complex structures that we've talked about a j-holomorphic curve in r times this is basically equivalent to a solution of the Flur equation in w. So this is a way that you can do Flur homology, the Hamiltonian Flur homology, from the perspective of a symplectic field theory. So now the setting in which I really want to think about holomorphic curves is also a bit more general than this. So let's take a symplectic manifold. So here I'm talking about co-boardisms with stable boundary. So one can define stable boundary in various ways. The main thing I want to say is there exists near the boundary of one parameter family of hyper surfaces that all have the same dynamics on them, same characteristic line fields.  Of course, I'm going to do the same thing on m minus, but the collar neighborhood will look like an opposite interval. So this is the definition of a symplectic co-boardism with stable boundary. And having done that, I can add cylindrical ends to this compact object and get a non-compact object. That's what we call the completion. So I'll denote that by w hat omega hat. It's going to depend on some choice of function phi of the sort that I use to define symplectic structures on symplectizations. And that choice of phi is not particularly meaningful. It's just nice to know that I can do it. So the definition is we take a negative cylindrical end, glue that along m minus to w, glue that along m plus to positive cylindrical end. And the symplectic structure here, we take omega phi, omega phi on each of the ends. And the original omega on w, you can choose the function phi such that, so remember, I still have this definition on the board somewhere, middle board top right. Very good. d phi times lambda plus omega, if I choose phi so that it equals just the identity map close to 0, then I can obviously glue these cylindrical ends smoothly together with these collar neighborhoods. That's what I'm doing. So with that choice of symplectic structure, and I don't really have to specify phi, I just know that I can choose one. Are you using the same phi on all ends? I don't care. Actually, I specifically want to not specify phi and allow myself the freedom to change it. So let's say j of omega and the two stable Hamiltonian structures, this is going to be the space of almost complex structures, j on the completion w hat, which I assume to be everywhere compatible with, well, let's just say compatible with omega on the original compact object and belonging to these spaces with respect to the stable Hamiltonian structures that I already defined. So the spaces of translation in very almost complex structures on the ends. So that now is going to be an almost complex structure that's compatible with omega hat phi no matter what phi I choose. Now to start talking about the actual j-halomorphic curves, any questions so far? So this you require that j's, what does it say? Compatible? Translation on the end. So what about in the cobaltism part? I mean, don't you have a factor r there anyway? When the cobaltism part, I don't have a factor of r except close to the boundary. Yeah, and I don't make any requirement there except compatible with the given symplectic form. So the kinds of holomorphic curves we're going to talk about are as follows. We take sigma j to be a closed Riemann surface. Let's actually say there's going to be two finite disjoint ordered subsets. And in particular, sigma dot always denotes sigma with the first finite ordered subset removed. That's a punctured Riemann surface. And we take that, also this subset. I don't have to worry about it too often, but it gets partitioned into the so-called positive and negative punctures. Each of those could be empty, but each puncture belongs to one or the other of those. And now I think about maps from this Riemann surface with the various points removed. Going into my completed symplectic cobaltism, I want them to be what I call asymptotically cylindrical, which means, well, let's not write down the precise definition.  Chris, you're missing a hole somewhere. I'm missing a hole. That's true. Could be worse. The opposite would sound like a medical emergency. So what I mean is, essentially, for this kind of almost complex structure I'm considering, with the translation invariant thing satisfying these conditions on the n's, I have j maps the vector in the r direction to the rape vector field, which means r times a rape orbit is always a holomorphic cylinder in the end. So those are special holomorphic curves that always exist, at least in that region, maybe not globally. And I want to consider punctured holomorphic curves whose behavior near each of those punctures approximates one of those cylinders better and better as you go toward the puncture. So to make it precise, I would say you can choose some holomorphic cylindrical coordinates near each puncture. And in those coordinates, as you go further up toward infinity on the cylinder, your curve looks more and more like one of these trivial orbit cylinders. Maybe it's time to recover this. In the picture, theta is empty or what? Theta, I haven't really said. Theta doesn't have to be empty. It's just some finite subset right now. But in the picture, it doesn't appear in the picture because it doesn't affect the picture in any way. It's just points. So theta are meant to be marked points. All right, so the actual modularized space looks like the following. It's going to be a space of tuples sigma j gamma theta u, where, well, sigma j is a closed Riemann surface, as I wrote above. Gamma and theta are these disjoint ordered finite subsets. So let me not rewrite that. Mainly, I need to say u maps the punctured surface to the completion w hat j homomorphically and asymptotically, cylindrically. And then I just need to say what my equivalence relation is. Deusa basically already said it. You could deduce completely the right answer to this from what Deusa said. I will say a tuple sigma j gamma theta u is equivalent to a tuple sigma prime phi star j phi inverse of gamma preserving the ordering phi inverse of theta u composed with phi for any diffeomorphism phi from sigma prime to sigma. Presently, the ray orbits are allowed to be arbitrarily terrible. Yeah, this definition is still OK if I allow the ray orbits to be arbitrarily terrible, but I won't have any theorems. I do have, so finally, to make some use out of the marked points, I have an evaluation map. Sending this modularized space to w hat m, where theta, the marked points are labeled zeta 1 to zeta m. So this just evaluates u at those points. And it's well-defined because of the way that I define my equivalence relation. These diffeomorphisms map the set of marked points to the set of marked points. All right, so let's finally state a couple of theorems. So theorem one says, I'm just going to abbreviate elements of this modularized space by u instead of writing out the entire tuple and saying equivalence class and everything. So if u is what I call Fredholm regular, that's a condition I'll have to explain a bit. And it also includes the condition that the asymptotic ray orbits are not arbitrarily terrible. Then a neighborhood of u in the modularized space is a smooth finite dimensional orbifold with isotropy at u, isomorphic to the automorphism group of u. So I didn't write down this definition yet. The automorphism group of u is simply the space of all bi-holomorphic maps of its domain, which fix the punctures and the marked points and satisfy u equals u composed with phi. So unless u is constant, that's always going to be a finite group. And for somewhere-injective curves, it's going to be a trivial group. But for non-somewhere-injective curves, so for multiply-covered curves, if we get this Fredholm regularity condition, whatever it is, we're going to have to deal with an orbifold rather than a manifold because of this isotropy. Is this term supposed to be essentially a tautology on the word regular? In some sense, this is a tautology on the word regular. I mean, one does have to work a little bit to see why you can make this local identification of the modularized space with something that looks like an orbifold even if you have regularity. And I will at least have time to get to that in this talk, I think. Term 2, I'm going to probably postpone the proof of this for tomorrow, but I'll state it now. Let's say let's fix an open subset u with compact closure. Then I'm going to say there exists a bare subset called j u reg. Well, it's a subset of the space of all. Oh, I needed to fix some more data in advance. So fix also an almost complex structure, j naught of the type I've described. Then there exists a bare subset. So this is a subset of the space of all j's of the type I've described with the extra condition that j matches j naught everywhere outside of this special subset u. So the special subset u we can call the perturbation domain. It's where we're going to allow ourselves to change the almost complex structure, but we leave it fixed everywhere else. So such that for all j in this j reg u, let's finish it over here, every somewhere injective curve u in the modulate space with respect to j, passing through my special subset capital u is Fredholm regular. But here bare subset is again a commuter. Bare subset literally means commuter, yes. Yeah, I never understand residual. It sounds to me like it means the opposite, so I don't use the word. Some people call it second category, which I think is also not technically right if you look up the actual definition of second category. Yeah. Thank you for that input catcher. So, yeah. Well, someday some kind of sensible general habit will prevail, hopefully. I say bare subset, and most people don't seem to mind. So the literal meaning is countable intersection of open and dense sets. Or contained, actually. If you're being strict, it contains. It contains a countable intersection of open and dense sets, very good. Maybe more useful input is though somehow you, I think you're assuming here now that the Hamiltonian structure is sort of regular. Well, that's in, oh, in theorem two. I am. That's true. Thank you. OK. So the board's out of reach, so I'm not going to write it. Let's just say this. One additional condition you need in theorem two is that the, well, let's say all the closed ray orbits for both stable Hamiltonian structures are non-degenerate. So under those conditions, theorem two holds. Otherwise, you could have trouble. And at least for if your stable Hamiltonian structures are contact forms, that's a condition which is generic. That's a standard result. Can you also do this if your ray orbits are more spot? More spot is also fine, yeah. More spot or non-degenerate are perfectly fine. OK. So let's talk a little about the functional analytics setup for a neighborhood of, let's say I have a specific element of the modulite space with complex structure J0. And the map is called U0. Just to understand the statement of theorem two. Yeah. You're not asking that it be somewhere injective inside U. It has to be somewhere inductive somewhere and pass through U. Those two things are equivalent, which is not so hard theorem. So what I really should ask for is that there is an injective point mapped into U. And it's a good point because if I were working in other settings where I have boundary with Lagrangian boundary conditions or something, then it's not necessarily true that just because the curve is somewhere injective that its injective points are dense. So then I would have to be careful. But here, injective points are dense if the curve is somewhere injective. Yeah. I think you've hidden the end of the statement. Can you write the board which is here? Ah. The statement is on the sideboard. Oh, it's this one. It's this one. This is called democratic blackboard technique. All right. So I need to actually write down some kind of Banach manifold of maps that will contain the solutions I'm interested in. This is something that's kind of standard in the case of closed maps from closed Riemann surfaces. And for punctured curves, I guess it's considered sufficiently standard that nobody really feels they need to go through the details when they write papers. But it means there is no paper where anyone does the details really, as far as I think. In Matthias Schwarz's thesis, I think there may be details about this. And that's possibly the only place. There's also this paper by the problem of traffic. No, I don't think so. No. Well, I mean, what he does in there is basically just lifted from Schwarz's thesis. So yeah. OK. So I'm going to talk about a space of sub-elev class maps from the punctured surface to W hat, class WKP. So I want to have k derivatives that are of class LP locally. Locally, that would be the definition. And it's not hard to express. But globally, what does this mean? Easiest way to say is U is of the form X along F of H, where F is going to be some smooth map, which is not just asymptotically, but literally cylindrical near the ends. I'll say that in a moment. H is going to be some section of class WKP locally along the pulled back tangent bundle along F. So in compact subsets, this is a sufficient definition. But I need to say more specifically what happens at the ends. So I understand. So the J0 that you started with is regular, right? Was that? No, I don't care. It doesn't need to be regular. So are we requiring anything for J0 to begin with? I don't have to require really anything about J0 because I'm going to perturb it in a certain region. And I'm only looking at holomorphic curves that pass through that region. So near each puncture, in holomorphic cylindrical coordinates, I can choose coordinates ST, which are in a half cylinder. I'm going to just call it a positive half cylinder if it's a positive puncture, negative half cylinder if it's a negative puncture. And then the requirements are that the map F just looks like Ts plus a constant gamma of Tt plus a constant in the cylindrical and r plus minus times m plus minus. So here, c1 and c2 are just some real constants. And what do you mean by gamma plus a constant? I'm about to say that. Oh, sorry. Well, gamma will be valued in s1. You can add constants in s1. I thought gamma was an m. It was a m, actually. Very good point. Is that better? Thank you for that. So gamma is going to be a t-periodic orbit of the ray vector field. And then what I say about h is simply that in these cylindrical coordinates, where also I'm assuming I've chosen some kind of sensible asymptotically translation invariant trivialization of this tangent bundle, I want to have h on that end b of class w, k, p on the cylinder, the half cylinder. And so I haven't said anything about what k and p are. I'm not really going to make any requirement about that, except k times p should be greater than 2, because then the Sobolev embedding theorem tells me w, k, p injects continuously into c0. So these maps are at least always going to be continuous. If they weren't continuous, I'd have real troubles making this a good definition. So you don't require any kind of exponential of k for h or anything just? I'm getting to that. I'm getting to that. So I'm confused because if we're looking at a neighborhood of u0, shouldn't we be writing things like neighborhood in the form x sub u0 of something? Yeah, but I'm going to, well, u0 is going to be a map of this form, and I'm going to consider other maps of this form that are close to u0. I'm still, I mean, I haven't localized yet. This is still global. Wouldn't want to have some sort of general way-tech error. Can you explain what's being ruled out by this list of conditions? Well, I'm ruling out maps that don't have decay at infinity to some very sort of, well, f is a very controlled map. It has this precise cylindrical form. I'm only considering maps that have asymptotic decay to a map of that form. So I don't want them to move around non-trivial amounts arbitrarily far away. Augustine kind of preempted the next little variation I have to put on this. And I won't be able to explain why just yet, but let's just say it. For a parameter delta, I can let, I'm going to call this b. This is the actual Bonnack manifold that I'll work with. So this is going to be, again, u equals x belong f of h as above. But the condition on h will be slightly different. I'm going to require that the norm of h, not the norm of h itself, but the norm on the half cylinder of e to the delta s times h is finite. So that enforces some exponential decay in addition to the decay that the WKP norm already requires. So I'm not going to tell you yet why this is necessary. I will tell you that eventually. I will tell you why it's possible, which is called this fact, though I guess it's sort of a hard theorem. First version of it proves by Hofer-Visatsky and Sander. And there's been other versions in various contexts since then that if the asymptotic orbits are non-degenerate, then for all delta sufficiently small, all asymptotically cylindrical, or let's just say, all curves in the modular space I've defined belong to this Bonnack manifold. That is, the map involved in defining that curve belongs to that manifold. It has the asymptotic exponential decay as long as delta is sufficiently small. Hey, Chris, I'm in the back row. I can't quite see. Are there absolute value signs around that s in the exponential? No, but there should be. Thank you. Let's make it a plus or minus. I want a minus sign if I'm at a negative puncture. The point is that it enforces asymptotic decay, or exponential decay, rather than exponential growth or something. What can I sensibly say in the zero amount of time I have left? You have, like, three minutes. My question stands. That would be remarkable. So in that last HLVZ theorem, the delta doesn't depend upon the components of m of j, your own? Yeah, it does. Yeah, that to say this properly, I would have to say that given a specific set of asymptotic orbits, there is a delta sufficiently small so that all curves that are asymptotic to those orbits satisfy the decay condition. So let me try and at least get to the definition of Fredholm regularity in some form. So what I'm going to do now, I've defined a class of maps I want to work with. I also have to worry about the complex structure and the domain, because that's not fixed. It's not some sort of God-given data to work with, but I allow it to vary in different points in this modular space of different complex structures. Now, I'm looking at just a neighborhood of some U naught with complex structure J naught, which is it still visible somewhere on the board? No. There it is. So let's just worry about the complex structures that are close to J naught. And one way to do that, I don't have to consider everything in this infinite dimensional space of complex structures. But if I want to cover all of them that are not equivalent by defumorphisms, I can think about Teichmuller space. That's at least a nice smooth manifold. So what I'm going to do is say, let T be a smooth finite dimensional family of complex structures on sigma, which I can, if I want to, I can even assume that they are fixed near gamma and theta. Another thing, I can just take a model surface sigma and fix punctures in one particular place, or several particular places, fix also the marked points in particular places. I don't need to allow them to move around. Because if I want to get all the different conformal structures up to equivalence with these punctures and marked points, all of them can be represented by some complex structure that have the punctures and the marked points in that particular place. That's just a question of choosing a defumorphism. So I'm fixing gamma and theta in place on sigma, but I'm letting the complex structures vary. And this is going to be a smooth finite dimensional family that parameterizes a neighborhood of the equivalence class of J0 in Teichmuller space. What's that? I don't understand you just said. Suppose that these like W, the ends are empty and sigma is the sphere. Yes? Then you're not varying the complex structure at all, because you can, but you're saying that you can get any configuration points that you want. Yes, I can. But I have to vary the complex structure. Because a complex structure is just a section of the tangent bundle. I mean, automorphisms of the tangent bundle are all defumorphic, but you can suddenly vary them. And also, they're not all defumorphic if there are more than three marked points. No, I'm sorry. Even on the sphere, I have to do this sometimes. OK, so I'm parametrizing a neighborhood in Teichmuller space, which is the space t sigma dot theta of all complex structures on sigma divided by the identity component of the defumorphism group preserving or fixing punctures and marked points. So it's a classical fact that this is a finite dimensional manifold. So I can choose some sort of finite dimensional space of complex structures parametrizing it. And there are various sensible ways to do this. We'll talk about tomorrow if I have time. There's a quick lemma. Let's also assume, just for simplicity at the moment, that the Euler characteristic of the punctured surface minus the marked points is negative. That's, Dussel also talked about this a little bit. That's the stable situation. So that's the situation in which the automorphism group of the domain with its punctures and marked points is finite. It's not an absolutely necessary assumption for what I'm talking about, but it does make a few things So I can in particular say there exists a choice of this object T. I'm going to, in the future, call this a Teichmiller slice since it parametrizes a neighborhood in Teichmiller space that is invariant under the action of the automorphism group of the domain. So having made this choice, let me finally write down the point of theorem one. We now can define a smooth section of some Bonnack space bundle. The domain is going to be the product of the Teichmiller slice with my Bonnack manifold of maps. Section is called d bar j. So the fiber over j comma u is going to be the space of wk minus 1p sections of the bundle of complex anti-linear maps from T sigma j to pullback tangent bundle. I hope that's the right number of parentheses. If not, somebody will surely let me know. And we're going to take j comma u to the obvious thing, Tu plus big j composed with Tu composed with little j.  And so finally, so how about changing the t for d? My own PhD students need to be especially careful in moments like that. I go back and forth. So the lemma is that the automorphism group of u acts on the 0 set of this section by the obvious phi star j u composed with phi such that a neighborhood of u in the moduli space is in bijective correspondence with a neighborhood of the equivalence class. So I should have said neighborhood of u naught in the moduli space with the equivalence class, this automorphism group of u naught, j naught u naught in the 0 set modulo of that action by the automorphism group. So now, what does this really say? By the implicit function theorem, mj is going to be an orbifold near u naught with isotropic group equal to the automorphism group if that 0 section is a manifold, which is going to be true if the linearization of the section is surjective. So that's some linear operator from the direct sum of a finite dimensional space tangent to a space of complex structures with, well, let's just write it as tangent space to the Bonnack manifold, which is some space of sections of the pullback tangent bundle of class w, k, p. Yes? Sorry? That should be a data on your e in the definition of e. Oh, you're right. Thank you very much. This is the moment where I say I thought you people cared about mathematics. OK, basically, so I wanted to at least end with having written down this operator, right? So Fredholm regularity means, specifically, I need to finish the sentence and say it is surjective. So Fredholm regularity means just that. This operator is surjective. I'll talk more tomorrow about what one needs to do to make sure that that operator is surjective. The result is then we identify local neighborhoods in the moduli space with this thing, which is a finite dimensional manifold divided by a finite group. I'm not going to say anything about the proof of the lemma except that if you stare at it long enough, it becomes obvious. It's not that hard. So I'll stop there for now. So there is office hours. So I again suggest longer questions, get postponed until then. So do you have a short question that you're willing to ask for the whole one? So good to practice. I'm just going to show that something is regular, right? What do you need to put into the slides and bury your complex structure in it? Right. OK, that's a good question. If you want to show that something is regular, well, the answer is you. I've never been in a situation where I had to actually choose a tifemailer slice concretely. But it has to be there in the theoretical setup, because if it's not there, it would obviously be sufficient if the linearization of this operator just mapping this space of sections to this space of sections is surjective. That would suffice. That would imply that this one is surjective. The problem is that that's not true as often as you want it to be. You have this additional finite dimensional stuff in the domain which sometimes actually makes a difference between an operator with a finite dimensional co-coronal and one that's surjective. So in fact, so if I have time to talk about automatic transversality tomorrow, hopefully, it's a really crucial detail there. The criterion that used to prove automatic transversality does not work if you leave out this whole discussion of tifemailer space, because you need to have this freedom to vary the complex structure in order to get as much transversality as you need. Do you have an actual example of the curve which is regular in this sense and not regular if I take down the tifemailer slice? I could probably come up with that. Ask me in office hours. And you have a discussion tomorrow. And I have a discussion tomorrow. In addition to your lecture. You can certainly find examples if you take four items up as a pie of tifemailer. Definitely need to vary types of expressions.  If I took all of them, this operator would not be Fred Holm. That would be bad. So here you're thinking of fixing big J, which is the complex structure on the target. So far I'm fixing big J. And your space of J here, I was using a curly J to mean variations of J on the target space. But you're fixing that and making your J as a finite dimension. Well, I have two curly Js. Well, you have two questions. Only one on this board. You see, your curly J there is a type of a slice. But you could, in principle, vary the other J on the target space if you want it to. Yes. You mean I'm talking about finite dimensional families or you're just generic? I mean, in year 0 and 1, you were saying that you were taking generic elements where you were allowed to vary in the target with an open subset, U of your target, right? That curly J is not actually in this limit. Here you're somehow assuming you fixed that. Right. No, I mean, the question of why you get transversality for simple curves for generic J I will address tomorrow. So far, all I've talked about is what criterion actually gives you smoothness of the moduli space and how does one go about proving that? So one thing to say is, why do you need to vary J and type on the spaces? For example, if you look at a space of tori in CQ2 and look at it and say, you know, there are clouds of degree C embedded in tori, you look at the induced complex structure on the tori. Does that change it? So if you fix the complex structure on the tori, it's not necessarily going to be regular because when you move your points around and you move the curve around, J on the domain naturally changes. So you have to have that variation in Without further ado, I propose we break for lunch. And at 2 PM, I'll make some short remarks about organizational things like finances and stuff. So I'll see you at 2. Thank you.


Craft presents the Great Gilder Sleeve. Each week at this time, Craft presents from Hollywood, California, Harold Perry as the Great Gilder Sleeve, written by Leonard L. Levinson. We'll hear from the Great Gilder Sleeve in just a moment. But right now, here's a question for you. Did you know that you can overeat and still be undernourished, that it's not so much the quantity of food you eat as the kind of food you eat that's important? Well, it's true. And that's why you should learn all you can about the right foods to serve your family. Wholesome protective foods that provide the energy and real nourishment your family needs. So it's important to you that delicious parquet margarine made by Craft is one of the right foods. And that it's so economical you can serve your family all they need. Yes, parquet margarine is a protective food that's packed full of wholesome nourishment. Parquet is one of the best sources of food energy you can serve. And you all know how important vitamin A is. It's truly a protective vitamin. Well, parquet margarine is rich in vitamin A. There are 9,000 units in every pound. And don't forget, parquet is the margarine with the delicious flavor, whether you use it at the table for baking or for pan frying. So ask your dealer for delicious economical parquet margarine made by Craft. Just ask for parquet, P-A-R-K-A-Y. And now let's visit our friend the great gilder sleeve. Mr. Gilder Sleeve, report of a state of Marjory and Leroy Forrester miners submitted by Throckport and P. Gilder Sleeve Guardian. Well, it looks very neat, Ted. Should impress Judge Hooker. Is it complete? All but the name of that firm that leased the 12th Street property. Oh yes, let's see. What was that firm? Oh yes, the swanky-hanky shoppy. Thanks, I'll just fill that in. It was a 99-year lease, wasn't it? Yes, 99 years with monthly options. Oh, Marjory. Hi, Marjory. Hey, that's a new dress, isn't it? Yes. What do you think of it? Stop asking him questions, Marjory, or he'll charge the estate for giving a legal opinion. I'll go in the other room, young man, and tend to your paper. Did you want me? Yes. All ready to go to court, eh? What about your brother? Where is Leroy? Well, I've sent him to change his shirt for the third time. Uncle, I wish we'd make him get rid of that printing press.  Now tell him to hurry. I don't want to be late. This is important and I'm getting jittery about it. Oh, now relax and take it easy, Uncle Throckmorton. Relax? Ted says the report is in fine shape. Why, there's nothing to be excited about. Now, where's this? No, by George. Come to think of it, I've done wonders. If I do say so myself. In fact, I will say so myself. I've done wonders. I think so. Why, since you arrived in Summerfield a month ago, you've straightened out all of our investments, rented that vacant property, and even put the kitchen on a budget. Why, Judge Hooker should be very pleased. I hope so. I made up my mind to demonstrate to that old... What is it Leroy calls him? Uh, old Leroy. What, Uncle Lawrence? What was it I told you not to call Judge Hooker? Picklepuss. That's it. I made up my mind to show that old Picklepuss that a competent businessman could administer this estate properly. Why? Because you can't put anything over on me. Excuse me, Mr. Gilsley. Yes, Bertie? Where did you buy them bananas? Well, from a man in a truck. They were bargains, too. The store's won 30 cents a dozen, and he only charged me 25. Well, he done gypped you. There was only nine bananas. Yes. Yes. As I was saying, Marjorie, you can't put anything over on my type of businessman. We have a certain alertness. Oh, great Danes. What's wrong? Look at the time. We'll be late for court. Oh, but court stays open until five o'clock, Uncle Moore. Yes, but we can't just drop in whenever we want, my dear. It isn't a barbershop, you know. We have to be there when the judge is ready to take us.  Yes. You see, I don't want to arrive late and have trouble with old cucumber face. I've got to get back here and pack up my bag so I can take the night train. The night train? Yes. Oh, I forgot to tell you. I'm going back to Whistlow Vista this evening. Oh, Uncle Moore. I thought you were going to stay here and live with us. Well, I am. That's why I'm going back to Whistlow Vista. The cell or lease my house there. You are? Oh, that's wonderful. Oh, Leroy. Yeah, what's this? Uncle Moore is going to sell his house in Whistlow Vista and come back here and live with us. Oh, boy, gee, that's wonderful, Uncle Moore. Take it easy.  No, no, no. Thanks. Just the same. You haven't got that many shirts to spare. Oh, Ted, is it time to go? Yes, we should hurry down to the courthouse. Everybody ready? Leroy, Marjorie, Ted, Birdie? Birdie is not going to court with us, Uncle Moore. I know that. I just want a glass of water. That ham I had for breakfast. Water, Birdie? Yes, Mr. Gilles, please. Bring in it. Good. Anybody else want any water? No, thank you. Here you are, sir. You better hurry, Mr. Gilles, to say it. One second while I drink this. Thanks, Birdie. Now I'm ready to go. Oh, excuse me. I must have drunk too fast. Wait a minute, Ted. I think Uncle Moore has a hiccup. A hiccup? No, I'll be all right. Oh! Maybe you better do something about it. No, no, no. I'll be all right.  Well, what about court? You know, Judge... Sir? Sir, look, there are hill ways. Yes, I can have it put over until tomorrow. But I got to get back to Whistler, Mr. Ted. I couldn't do that. Or could I? Why not, Uncle? No, no, no. I'll be all right in a few minutes. Oh, isn't this silly? Don't try to talk, Uncle Moore. Just sit quietly for a few minutes and rest. Rest? All right. I'll rest. Yes, that seems a little better. Spoke too soon. Better button it up again. Yeah!  It never fails. It doesn't? Well, what is it? Drink a glass of water. Oh, but Marjoram, my dear, don't you remember? That's how I got him. Drinking water. No, but you didn't drink slowly. You've got to take nine swallows of water and not breathe in between. Not breathe? What am I? Fist? Now, Uncle Moore, just cured. Proud of me. Sure, you know a hare from the dog that bit you. Uh, Ted, this is hiccups, not hydropobia. Well, I'll get a glass of water. You better get a pitcher in case one glass four won't do it. What are you trying to do? Drown them? One more swallow and I'll fly back to Capistrano. Oh, help me out of this rocking chair. I'm getting seasick. Sure, Uncle, if you can keep on hiccuping for another two hours, you'll get your picture and strings and things. Yeah, I'll keep quiet. Mr. Gilda Sleep, I just remembered something that'll take care of those hiccups. You do, Ted? What is it? It's simply a matter of breath control. Say, Peter Piper picked a peck of pickled peppers without taking a breath. Oh, I...  Well, all right. Peter Piper picked a... No, no, it's slower like this. Like this. Peter Piper picked a peck of pickled peppers that way. Peter Piper picked a... picked a... picked a... Oh, more water, Birdie. There we have, Mr. Gilda Sleep. Thanks. Well, who's got the next suggestion? Step right up, don't be bashful. Gilda Sleep, the guinea pig. That's me. I wouldn't be bringing it up, Mr. Gilda Sleep, except I know it'll work positively.  Well, all right, I'll try anything once. If you got a cold knife, Birdie. Yes, sir. I brought one right here with me, Mr. Gilda Sleep. Unbutton your collar, Uncle. There. Ooh, it sends the shivers up and down my spine. Where'd you get that knife, Birdie? I had it in the refrigerator. You never can tell when a nice, cold knife comes in handy. Hey, Uncle Mark! Little Mark, I was outside talking to my pal, Piggy Banks, and when he had to hiccups, I... Say, what's that knife doing in your back? Did they operate on you? Well, no, LeRoy. Birdie suggested cold silver against the back of my neck. Oh, that won't work. It will, too, you LeRoy. I've been watching your Uncle since he tried it, and it won't. Well, I, George, come to think of it, I have it. This is wonderful, Birdie. Thanks very much. It'll remind me to give you a dollar. Ted, let's get started for the courthouse. But Uncle Mark, a cold knife against the back of the neck cures nosebleeds, not hiccups. Why, is that right? Oh, I thought so, too. You mean to say it's not good for hiccups? Oh, jumping jelly beans. Ooh, they've come back again. Forget about that dollar, Birdie. Say, Uncle Mark, I know a sure cure I can't miss. No, LeRoy. It's my turn this time. I just remembered a remedy. But that isn't fair. I spoke up first. Say, whose hiccups are these? Yours or mine? Okay, go ahead. They're your hiccups. What's your remedy, Uncle Mark? Well, I'll take a cold shower. The shock should stop me. It sounds logical. Well, it won't hurt at any rate. We've got to do something so I won't keep speaking out of turn in court. I'll get the car out of the garage. I'll bet that shower doesn't work, Mark. Now, my idea is to scare Uncle Mark. What for? Well, that's an absolute positive cure for hiccups. How do you know? They cured Piggy Banks when he had them something terrible. How'd he get them? Drinking a whole bottle of pop at one go. Honest, his family tried everything. Then his kid brother put a string of firecrackers in his pocket and lit the fuse. That did the trick all right, all right.  No, Piggy wasn't wearing a coat. See, if I could only think of something super to pull on for more, I'd better scare the hiccup right out of him. Now, you wait a minute, LeRoy. Don't you do anything drastic? For me? When did I ever do anything? Say, how's about it if I put ketchup on my head and stagger into his room and fall down on the floor? D-Roy's Forester now, don't you dare. Well, all right, let's see. What else would frighten those Donald hiccups? In the shower, Birdie must be using the hot water in the kitchen. Oh, this water is ice cold. Oh! Where's my clothes? No time for clothes. In my bathrobe, where'd I put it? Never mind, here's a big towel. All right, come on, LeRoy. Which way shall we go? No, no, no, Uncle Mort, go back. Why? Because you still got the hiccup. What's that got to do with the house being on fire? It isn't on fire. I was just trying to scare you. Scare me? What's a big idea? She only meant it for the best. I was just trying to frighten the hiccups away. If I ever ran out of the house like this, I'd frighten the neighbors away. I'm awful sorry, Uncle Mort. Say, you better get back in the shower. There's a big drip on the carpet. Who, me? You clear out of here now. As soon as I get dry, we're going down to the court. Hiccups or th- No hiccups. He must be waiting in the car. Good. I hope the judge doesn't mind. Okay, buddy, stick him up.  LeRoy, come out of that closet and put back that water pistol. Oh, that didn't work either. You can't frighten me, LeRoy. I'll go out and get in the car, young man. I told you, LeRoy. Just you wait. I'll figure out a scheme that'll make Uncle Mort forget all about those deeps ahead. Get in the car, everybody. Lovely day, isn't it? Too bad I can't appreciate it. Maybe being in the fresh air like this, my hiccups will... Oh, no, they won't. Oh, and the car is doing it, too. Boy, every time you hiccup, P.P., your foot goes down on the gas. You think so? Yes. You want to stop and let me drive? No, we haven't got time. I can't help it. I'm afraid they're getting worse. The idea of driving down the street like a jackrabbit and jumping the signal. Where's your driver's license? Well, it's like this, Officer. Oh, it's like that, is it? A hiccup. Yes, he got it drinking a glass of water. Water, huh? Well, that's original, anyway. He's been hiccuping for hours. Show the officer how you've been hiccuping, Uncle. See? Never mind. I've heard him before. Officer, we're in somewhat of a hurry. We're rushing down to the... I get it, to a doctor. Well, come on. What? Oh, of course. That's it. This is Dr. Officer. That's not Dilly-Dally. See? You got it bad. Follow me. I'll clear away for you. Thank you. Same thing happened to my sister's kid two months ago. You know how we cured her? Made her eat a quart of ice cream fast. Oh, yes. Well, I'll put that down on my list. Hurry, Uncle Morse. Shall we go with you?  Going up? Yes. Is there a doctor in the building who specializes in hiccups?  Well, you might try Dr. Simard on seven. Get him, please. You've got them bad, mister. Yes. I know something that will cure them in no time. You do? What is it? Eat a quart of ice cream fast. Yes. Seven blocks. There's Dr. Simard's waiting room. Four doors down. Thank you very much. Oh, this fuss over these silly hiccups. Well, at least I'll get rid of them for sure now. Dr. E. E. Simard. Throat, chest, and stomach. That should cover hiccups, I guess. Did you see the doctor? Oh, I just dropped in to catch up on my last year's reading. Well, you should do something about those hiccups. Now, a quart of ice cream eating fast. I know, it's a short cure. But I want some competent medical advice. Is the doctor busy? Not at the moment. Now, if you'll step in here and disrobe. I don't want to disrobe. I want to see the doctor. But if the doctor is going to examine you, you... He keeps his clothes on, doesn't he? Yes. Then I'll keep mine on, too. Where is he? Step in here. Doctor, this gentleman wishes to consult. Ted, Ted, Miss Wood, how many times have I told you that patients must disrobe? They've been all through that, doctor. There's no need for me to do that. You can see what's wrong with me.  Thank you. You can close your mouth now. Well, it didn't take me long to diagnose this case. No? No. You're suffering from an intermittent, uncontrollable, diaphragmatic spasm causing a sudden inhalation which is interrupted by a spasmodic closure of the glottis. I am? Yes. Well, what does that mean, doctor? You hiccup. I know I hiccup. I can hear myself. How do I get rid of? Ah, no, don't get excited. I have a painless and invaluable cure. Oh, you have? What is it? Eat a quart of ice cream fast. That'll be $5, please. Oh! Judges' chambers are down at the end of the hall, TP. Oh, jumping jeeps. Look at the clock. This is a thump of a time to show up. No. I've eaten so much ice cream, I sound like a good humor man. Easy, Uncle Moore. You take it easy, young man. Don't say anything to the judge. Here we are. Ted, did you send the financial report down this morning? Keep quiet. Now, don't worry. That report went down early. It should make a wonderful impression. Well, come on, let's go. Yeah, might as well face Old Picklepuss. Careful, Leroy. Anything you say will be used against me. Come on. Hello, Judge Hooker. At last. I was ready to go home. What are you hiccuping for, Gildersleeve? For about four hours now. Uncle Mark's been suffering all day long, Judge. Yeah, maybe if you could frighten him. Quiet, Leroy. Quiet. Yes, Judge. I'd have been here sooner, except for that. Well, I'm glad you sent down your report, Mr. Gildersleeve. Gave me time to study it. I'm pleased with what I found. See, that's swell, Judge. I thought you could do a good job for these children. You did? Well, thank you, Judge Hooker. In that case, we can leave. Come on, Leroy. Come on, Marjorie. Come on, Ted. What's your rush, Gildersleeve? Take it easy.  I never knew nerves came with such large bundles. Very good. I wonder if it would be all right with five zaps and for my duties for a while, Judge. I have some business to wind up in Whistlow Vista. Sure, sure. Go right ahead. Take all the time you want. Only be back next week. Yes, I see. Well, thanks, Judge. Excuse it, please. Poor Uncle Mort. You can't travel in that condition. I bet Mr. Fowler at the drug store would have something to relieve you. Why don't us two go see him, Mort? And if he can't help, why not a couple of other guys that can? All right. If you'll excuse us, we'll run along. Certainly. Goodbye, then. Goodbye, goodbye.  Oh, sure. Hey, Leroy. Well, Ted, let's get started. Oh, I can't stop. Hey, those hiccups must be annoying. They are, indeed they are. By the way, I know a sure cure for hiccups. What, you two? Oh, this one never fails. All you need is a brown paper bag. A brown paper bag? Well, that takes the prize. Shall I go out and get one, Judge? No, no. Here's a bag. Wait a minute. Wait till I dump the apples out. Apples? I teach a class at law school, and the boys always bring me apples. Now, let's try this remedy. Oh, no, don't bother, Judge. I think they're stopping now. Now, let's make sure. All you have to do is to breathe in and out of this bag. Understand? I understand what to do, but I don't understand why. Well, you will. Just put your face in the bag. Fine. You look better already. Now, go ahead and breathe. Oh. The principle is this. Normally, you exhale carbon dioxide and inhale oxygen. I see. But this way, you inhale the carbon dioxide. You've already exhaled. I see. Is that clear? No. Well, if you stop inhaling oxygen, you'll stop hiccuping. It's really very simple. So are you. You ought to be all right, but now, how are you feeling? Worse. Oh, my, and I almost had him licked. Strange. It's never failed before. Let me see. There's a hole in the bag. Oh, take me home, Ted. I'm going right to bed. Oh, stop that. There's a pillowcase. The ball is here while I'm here. Oh, true. Now, when we go out, if anyone asks, who we are, we're the laundry man. I got you, Rick. Shall we take this silver cup, too? Let me see. Yeah. I wore it to Throckmorton Peak, Gildersleeve, first place potato, race, annual picnic, Gildersleeve, Gettlewicks. Hey, I guess he was the whole way. What do you say? No, no, no. It's more trouble in its work. Oh, yeah. Too easily traced, huh? Hey, you sure there's no dough laying around? No, no. I looked every place, even behind the wallpaper. You think we should, Dad? Think we should take any more clothes lefty? No. I'm wearing three of this guy's suits already, one on top of the other. I'd hate to have to run from some copper this way. That's too bad. We didn't snag any dough. Well, let's get going. OK, you take the DJ. Say, look. Get away from that window.  Quick, out the back way. Wait a minute, wait a minute. Is he alone? Yeah. Let's get out of here. No, no, no. Wait.  That's high behind his curtain. But where? Get snappy. But where? Boy, get him up. Oh, I see. You must be a friendly Roy. He puts you up to this, eh? What'll that boy think of next? I says for you to get them hands up and keep quiet, too. I'm sorry, mister, but it didn't work. I still got him. You see? I read. Oh, you brought a friend, eh? What's the matter with this guy? You know, I've got the hiccups. You see? Look, you look. This is a gun in my hand. Yep. And I've got a good notion to let you have it. No, thanks. I wouldn't know what to do with it if you did.  No, I don't want yours either. You're asking for it, mister? I am not. I don't want any guns. I'm afraid of guns. Who sometimes are loaded. Shall I plug them, Red? Well, I don't think that'd cure me either, mister. Stand out of the way, Red. I'll show this model it. Well, very realistic. Blank, say. You missed. Move. I'll try again. Don't do that. You'll have a brick up in town here. Oh, gangbusters. What are you gonna do with a guy like that? I know, I know. Lefty, you stick your get in his ribs and I'll frisk him. Okay. Now hold still, will you? This time I can't miss. Now cut it out. Hold still. Sorry, I'm tickly. You stop it. Here, give me that gun. No, no, no. Well, you get away from there. Hurry up, Red. I can't, I can't. All right, all right. Let's take his pants off. What? Yeah. That way we get his roll and he can't follow us either. Okay. Oh, no, you don't. By George, that's carrying things too far. Fine friends Leroy has taken my...  Grab him, Red. Oh, look out for those flowers. I warned you. Now you see what... Oh! That's right, Lefty. Tom and a paw. Oh, you wanna fight, eh? Well, all right. Wait a minute, boys. Wait a minute. Stop. Stop.  Yes. You can quit now. My hiccups are all gone. More tupperware. Yeah? Grab his leg. Grab him. Get away from me, Red. Oh, Leroy should have never done this. That's right, Lefty. Sit on him. What is this? Wearing my new blue shirt suit. This is the last roll.   Get him. No, no, no. Now look what you've done. You've clunked your little partner. I left him. Hey, Lefty. Speak to me. Yeah. Speak to him, Lefty. Hi, George. He's out cold. Give me that gun before you do any more damage. Oh, no, you don't. Ouch, my foot. Oh. A minute, I saw those tight shoes. I knew you had corns. Hey, mister. Mister, please. What? Please don't point that gun at me.  That's a very good idea. A couple of blanks might teach you not to go on blanks. No, no, no. I don't want. I don't want.  Well, at last you're here, young man.  What friends? Who are these guys? Come, come, Leroy. Stop pretending. It's all right. My hiccups have disappeared. Oh, sleeping on the floor. That's one of Leroy's friends, and he's not sleeping. Hey, don't try to sneak out, Red. Gee, I... Come back here, Red, and tell Leroy what you did to me. Well, Uncle, I never saw these men before in my life. Ah, Silver, we're doing an appeal case. Uncle Mort, these guys are burglars. They are? What? They weren't fooling? And to think that I... I've got the hiccups all over again. The great gilder sleeve will be with us again in a few minutes. But first, I want to say that being a mere man puts me at quite a disadvantage in talking to you housewives, especially you housewives who are really good cooks, because so many of you are probably already using delicious parquet margarine made by Kraft, and already know from your own experience that it's a grand-tasting, economical margarine that can be used in many ways. Yes, you know, for instance, that parquet margarine is really delicious for table use and good for your family, too. You know that parquet makes cookies, cakes, and pie crust taste better because it's a genuine flavor shortening that adds its own delicate taste to all baked foods. You know that parquet margarine seasons hot vegetables to a queen's taste, makes pan-fried foods taste better, too, and it doesn't spatter or stick to the pan. And don't overlook the fact that parquet margarine is a highly nutritious energy food and a year-round source of vitamin A. Yes, you housewives who use parquet know how good it is, but some of you listeners probably haven't tried parquet margarine yet. Well, if you haven't, try it. Yes, ask your dealer for a pound or two of parquet margarine tomorrow. Just say parquet, P-A-R-K-A-Y. It's made by Kraft.  Oh, thank you, Marjorie. Say, Aunt, how soon will you get back from Mousseful Vesta? Not until Wednesday. Oh, you'll be gone that long? Yes. I've got to put my house up for sale, and I also want to be on hand to greet my two little chums, Fibre McGee and Molly, when they return from their vacation Tuesday night. Say, maybe Fibre McGee and buy your house. No, no, no. From my past experience with Fibre McGee, he went on a trip to Mousseful Vesta in Toronto. From my past experience with Fibre McGee, he wouldn't buy the place. He'd just borrow it. Good night. Good night. The original music heard on this program was composed and conducted by William Randolph. This is Jim Vannon speaking for The Kraft Cheese Company and inviting you to be with us again next week at this same time for the further adventures of The Great Gilders League. This is the...


The Quave twins win Mzanzee over. Ever since twins Viggy and Virginia Quave made their debut on the past season of reality show, Idle's essay they became firm fan favorites. Each week they won more fans over with their singing abilities and versatility. Die Hard fans were convinced that they would emerge victorious at the end of the competition but things took an unexpected turn when Virginia was eliminated from the show in November. Hours after Virginia was axed, her sister Viggy announced that she had decided to withdraw from the competition as well. Shortly after Viggy's announcement of her departure from the show, social media was filled with speculation that there was more to her withdrawal than met the eye. The rumor mill went into overdrive, with claims that the twins could have been bullied. However, the hot of PR and publicity of Mnet's local entertainment channels, Philly Koop Hecke dismissed the speculation. Viggy withdrew from the competition on her own accord and her decision is fully supported by her family and the channel. A few weeks later it was also confirmed that the twins had declined an invite to be at the finale of the competition. According to So-Anne, the twins' father and spokesperson Lawrence Quabe said the decision not to participate in the finale was taken by the family and fans. He didn't want to reveal the reasons behind the decision but confirmed that they were not going. This is a decision that was taken by the family and their fans that they should not go to the finale. Despite all the drama surrounding their exit from idols, the twins have a bright future ahead of them. In November, it was confirmed to Chvisa L.I.V.E that the twins had bagged a record deal with DJ Tira's afrotainment. Then on November 29, the twins answered fans' prayers and dropped their debut track, Hamba. Now the twins want to focus living their dream and making music.