Diff-equivariant index theory
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Rodsphon, Rudy
In the early eighties, Connes developed his Noncommutative Geometry program, mostly to extend index theory to situations where usual tools of differential topology are not available. A typical situation is foliations whose holonomy does not necessarily preserve any transverse measure, or equivalently the orbit space of the action of the full group of diffeomorphisms of a manifold. In the end of the nineties, Connes and Moscovici worked out an equivariant index problem in these contexts, and left a conjecture about the calculation of this index in terms of characteristic classes. The aim of this talk will be to survey the history of this problem, and explain partly our recent solution to Connes-Moscovici's conjecture, focusing on the part concerning `quantization'. No prior knowledge of Noncommutative Geometry will be assumed, and part of this is joint work with Denis Perrot.
 I'm going to ask the organizers for your participation here. So, it was a wonderful conference. Oh, you speak up. Okay, so I'll try to speak up. So, many ideas in this conference. So, I'm going to bring some contribution here to bring some other ideas that haven't been seen in the conference either. So, let's say, half of the talk here would be a sub-survey, some historical part on this comment here. So, it's motivated a bit of what we're doing there. And the second half will be my own contribution on this comment. So, which is going to be with the new problem. So, motivation. So, let's start with some things. So, start from the index here. So, we start with some vibration. And over some base peak. So, let's say, everything is close many faults. So, compact fibers and compact base and the one around. And let's say you have this, you don't have the vertical tangent bundle. So, in the kernel of the bar. Okay. So, from here, I think you already know, but for the sake of bringing some motivation to what you do. Let's recall what this family is next year. So, let's say you have the x and b, which is some families of three decompriers on fibers. So, I don't know fibers, I would say. So, from here, you will know that if you take the symbol of t, this defines some element in the k-therey of the dual bundle to the vertical tangent bundle. And from here, the index takes values. So, it's not a number, so it's not an integer. Once you have the point, here it's a k-therey element. So, it's something which is in the k-therey of the base. Okay. So, roughly speaking. So, here, the index of d, it's more or less speaking, just the element in the kernel of d-co-kernel of the x. So, more or less speaking, so it's not quite true, because there are jumps. So, these things is not quite the vector bundle, but you can modify this thing a little bit by some compact propagation so that you can make everything like vector bundles and make sense of this thing. Okay. So, from here, so here's just the index coefficient. So, you can move over co-molotage to get some conflict formulas. Just some trunk hacker. Once you do that here, so here you land, sorry, so here you land in the co-molotage of them. Let's see if we can get it in the c. So, I'm just telling myself, I'm telling myself what is about to be. So, I'm hiding it in the trunk hacker. And here you land, you take the trunk hacker, simply, in the co-molotage. So, I should say, here, co-molotage of the base. And you have a formula here, which says the formula thing. So, you should take the trunk hacker, so here, trunk hacker, of the index of t. So, what we get here would be a differential formula on b. So, a co-molotage customer on b. So, you just integrate over the fibers. And then here you have the trunk hacker of the symbol. So, it's an index there. And you partially integrate over the fibers. So, here from here you get something which is already some differential formula on b. And then you pair it with the top class. So, top class of this number here. So, you can just complexify the practical tangent problem. Okay, so, this is things that you already know. So, now I just want to make a sort of little observation. So, before that, let me do some little discussions on co-molotage. So, co-molotage, so, rather speaking, so, k3, it's a generalized co-molotage theory. So, for that you have an associated JRS whole logic theory. So, what you do. So, what you can do. Here? No, on the fiber. So, the fiber on the fibers here. So, here you can put some x here. So, it depends on the, so, x here is an event in the base. So, it should form the base. So, it depends on x here. Okay, so, let me come back to co-molotage. So, it's simply a dual theory to co-molotage in a sense. So, in a sense you have a, so in a point sense here. So, you have a pair which goes this way. So, if you have some thing here. So, let's say, the k-there of b. So, here is the k-there of b. So, here is co-molotage of b. So, here is. Then, you have a pair between two instance in integers. And simply, so, it's a very simple pairing actually. So, let's say that you have an n-n there. Represented by some vector problem over b. Then you can pair it with some elliptical parameter. So, represent these guys here. They are simply elliptical parameters over b. So, let's say, so, take a co-molotation that's said, let's say, what should I take here? Okay, another q. Okay, and here, the method goes from here to there. Let's take the twist in the next of q. So, you twist the, you twist q by the vector model e, and you take its index. So, that's the index of the integer. So, here it's a two-spice. Okay, so, what we can do now from here is the following thing. So, a very simple operation is that you can kind of continue to follow the square here. And so, let's say you have the index of b, and it's here, here. So, you can pair this thing with, Kevon will do simply, and get a number. So, let's do that here. So, now what we can do here is to, so you can pair, you can pair index of b, which is in k here at b, with, and then you get the right amount of base. So, we have to say something q in the kernel model of b. So, from here we get a number, index of b, paired with q. And it's actually very simple to enter practice, so not very simple. But what you can show actually is that the index that you have here, so the integer you have here, represents exactly the, the index of a native pickup right there over the two-spice for the vibration. So, basically what you do is very simple, dramatically. So, you have something which goes along the fibers, and it's going to be on the fibers. You combine with something which is over the base. So, you kind of combine the two, and you get something which is at the total space. So, this observation here is kind of the starting point of KK here. So, that's all in here. So, just a little note here, I don't want to say. So, KK here actually, so, so you have a right-close neighbor actually, of two KK here, which is maybe more familiar to people there. So, super connections basically, which is, and this bit has used this technology for many, many years. So, there are two, it's like two phases of the same coin, if you'll. So, one which is kind of topological, which is a little bit of a corpology. So, that's really the starting point of this technology of Kasparov. And actually, so, one note I wanted to say, so I said, so it's the fact that I think Kukun had in mind KK theory when he invented his super connection for medicine. So, it's kind of clear if you go over his notebooks. So, his notebooks, you can see at some point, he's already tried to understand KK theory. And a few years later, there is this super connection thing showing up at some point. So, I really think there are some, these two are very, very tied together. I just wanted to mention that. So, now, let's see a very easy consequence of this principle here. What happens here? So, some observation now. Which goes back to the digital rigs. So, digital rigs, there is a theorem that says the following things. So, now you just put spin structures where are cooperating, if you'll. So, it says the following things. So, if you have the agent of N, which is nonzero. So, I'm not making a mistake here because there are some specialists of FSC matrix, who discovered curvature matrix. But, simply here, so it's a simple case. So, there, one consequence of this here is the N carries no positive scalar matrix. That's something well known right here. But, now let's see a very simple principle of what happens here. So, now let's take this over columnology. So, if you take this over columnology, what happens? So, it means you'll be simply, so what you see here. So, you have one index which is associated to this guy here. So, it's all clustered on the fibers. Or, 18-news-along fibers. And, one guy here, which is associated to 18-news-along base. So, what you can see here, actually, it's kind of just a sort of multiplicity property of plastic clustered symmetry. But, actually, you can say even more. So, what happens here is that, now, if you combine, this observation here, with this principle here, that if you combine one elliptic operator on the fibers and one elliptic on the base, you get the total index over the total space. So, you also have something which is kind of not really trivial. So, you get the fibers themselves. So, it's not really trivial, but it's a simple observation you can do here. So, you'll see, if you want. So, now, what you want to do is move on to another trajectory from here. And see what you can do actually. And how we can kind of extend this principle to something a bit more general. Not just a bit more general, actually. So, two, Ncg. I'm not sorry that I don't understand that. I mean, if you have a trivial fiber model, the metaphor M as a fiber, sorry, let's say F is a fiber and we use the base. And, if B as a positive scalar curvature metric, that guarantees that M as a positive scalar curvature metric. So, the fiber is not required to have. As it goes here, so if you take the index from here, so what happens here, it means here, you take the agent is here, along very impossible corresponding classes here. So, if you pair the two, and the product of the two is zero. Yeah, I'm sorry, I was thinking so. Yeah, it's not simply agent is times the other agents, which is zero. So, it's very simple. That's why I was saying that's very something integral. So, you have to, there's something with corresponding classes you have to work out here. But it's not so difficult. Okay, so from here, Ncg. So, now our goal would be to carry out this principle. So, carry out this. And then we would have to say singular situations. So, my singular situations here, I'll talk about just one today, which is foundations. So, we're going to have foundations. So, what we're going to see is that, what we're going to do here is completely true, is the base of it here, by the space of a foundation, which is completely singular actually compared to the space we have here. So, that's one thing. Actually, very simple. So, you can even carry out this to singular foundations. So, singular foundation, I just want to say a few words, but there is, let's say two things you can consider here. Two main words I would say. So, one which is from a scandalis, Android address, scandalis. I think it's about 2005, 15, sorry. And it's a leaf-wise next term for very general singular foundations, basically. So, we have one thing recent of Casparov. Casparov. I shouldn't scratch his name because he's my mentor, I think. If he sees the video at some point. Okay, so Casparov. Casparov, I think it's okay here. So, very recent work too. So, I think it's a new pretty minimum of work that he carried out in like 15 years or something like this. And he finally published something. So, let's see, 2005, 15 I think. And here it's a, so where the first works here are. So, where the first works here, it's kind of leaf-wise. So, it's Casparov, it's transverse. So, it's not for anything, it's for Casparov, although. So, it's for relations that are just defined by the attributes. But, in this case, it's kind of already very interesting because you can take our legal actions from there. So, actions of even non-compact verbs on non-compact manifolds. So, yes, I think we'll leave the video on here. And one consequence of this work here is something kind of very interesting. So, it says at the end that, so you know already that the index of any electric operator can just reduce that to the study of direct operators, or spin-sea direct operators, no, more. So, we'll see that later. But, so, one consequence of this work here and a little more work that I'm trying to do currently is that the model of transverse elliptic operator on four relations defined by group actions, it simply is an operator we've seen all along the week. They form direct operators or government type operators that come. So, what's at stake in this? In this theorem at the end, it gives a very simple model of what you should study for transverse elliptic theory. So, that's something very, very strong here. So, but now we're concerned, we just focus on relations because it's already very interesting. So, in particular here we'll see a problem that does not reduce to the case of direct operators, using direct operators. And we see a problem which we can try to cope with it. Okay, so, relations. So, so, more or less, you can carry out everything I've said before. So, more or less, you can replace vibration by relation. And you can replace also the base B with m of f, so here m of m f is a relation. So, this space here is highly singular. But we'll see how we can kind of desingularize it using non-diffusing geometry. So, that's a very general principle that has guided the field for a long time. So, let's just see a bit of transverse geometry. Okay, so, naively let's say, so you want a base for your relation. So, naively you could just take a transversal. Okay. So, take a transverse, so w here, it's a complete transversal. So, complete, you have to meet every leaf at this once. It can be disconnected by the way. So, second point here. Now, if you just naively take the transversal, you see that it's not good enough to reflect exactly the transverse geometry of the relation. Because compared to, so, when you have a vibration, things are kind of easy because you have the one fiber that goes to the base only once. But in this case here, when you have a relation, it can come back to, if you go around the leaf, it can go back to transversal again. So, there is a group which can close this thing, which is the homing group. And that's a separate of the thermophysicism of the art, right, the transverse. So, now we begin to see exactly why we call this title homing. So, why we call this talk here in different thermophysicism, the current, next theory. And now the idea here is to define a thing. So, if you can define spaces, so you would like to look at the orbit simply of this compassion. So, now, but it's kind of singular. So, just this thing right here, so you just consider, so, instead of the base, you replace this with, not the base, but I mean, instead of, so the base, you would like to see that as a portion, but it's kind of singular. So, instead of this here, you look at this actually, gamma is what we call homing groupway. So, here it's registered transverse. But actually what we get at the end, so the groupway that you have here is more equivalent to the homing groupway. So, it's simply either you take the whole formation and you get the whole homing, or you just try this to one transverse, so they knew, let's say, first return maps. So, it's equivalent. And so, in non-competitive geometry, what you do is to, instead of studying spaces like this, you look at the, so, you study functions. But this thing here is kind of singular. So, you replace this by the convolution and problem. So, that's the convolution and problem. Okay. And from here, you can do the same thing as what we've done before. So, one scanner is here, 10 different things. So, 82. So, it says that you have an index morphism here, which goes from the k theory of, so the k theory of, so that's the v-wise tangent bundle here. And there, so you have something, so now you see what to do. So, instead of taking the base, as in the case of vibrations, you just take this guy here. But you have to consider transverse instead. So, k theory up and k theory down. Okay. So, you have some next morphism like this way. So, you should have k-mology on the left. Should it be k-mology on the left? No, it's k theory. No, on the left. Left here? You can put this. Yes. You can put k theory, k-mology instead also. So, you can basically, they call that k-tart, you know. It's not an issue here. So, just for our purpose of simplifying the talk also. So, if you know, you can take the geometric chronology if you want it to. Perfect fine. Okay. So, from here, now you can just do the same thing as before. So, but it will be just a bit more complicated to, I would say, carry out some logical formula. So, from here, like I said, but there are some, you have kind of some circles of ideas. So, at some point, so just remember before we pair some elliptical pattern on fibers and elliptical pattern on base. So, here you'll be able to carry out the same kind of principle. So, you'll see that a bit there. But now there is a theorem of code if you want the formula. So, that's a bit more complicated. So, first, to carry out the principle I've just mentioned before to you. One, we talk about Fermi's in this theorem. So, there is something a bit more complicated here. The first thing that was done was code, actually, in 82. And he worked entirely in the chronology instead. So, here is a foreign thing. So, the theorem is kind of one of your first things. So, let's see here. So, let's say you take any x in the k-thoracic style. So, from here you can take a transfer pattern. So, here, that's something from the Bohm-Ecquan. So, you have a transfer pattern which takes this here in the homotopy quotient. So, here we have k-thoracic. So, we have a quarenthalmology. Here, a quarenthalmology of, so it's simply this thing. So, that thing here is that you're also bounded over a classifying space. And you take the homotopy quotient. So, it stands here, your thing, it kind of faces the whole space, the whole space of the equation. So, you can make a parallel with what we have in the case of the Freibese next theorem. So, we have a Neutronk-Apar here. And given, so I just hope I'll have enough room. So, you take any choice here of class which is there. So, you want to pair it with here, the quarenthalmology of this thing. So, it's all common, it's the right thing to call it here with that, as you can see. So, you have x here. And then for any, so here you have a choice of the quarenthalmology just there. What is it? Gamma, you're right. It's the group. It's a homomy group. Okay, I mean, m is w, the quarenthalmology. Ah, sorry, it's w. Surprise w. So, I'll change to something called m to be there. You're right, thank you. You have a choice, let's take any w, so for any w which is in the ring, directed by, say, trac glasses, and get full successes. So, for any choice of such a guy here, so you can find various, and then if you have a man here, say, ta, which goes from the k3 of the convolution at the point here, to c. So, that's where it's there. So, we have some men this way. So, for some reason. So, here how we're trying to get to c. And here we have this map here, which goes to c. Such that, so here it's, let's call it omega because it depends on omega. Such that you have the four-line thing, which is true. So, call this map here, w. So, if you just form this way, and form this way, you get equal x. Let's just write this down. So, we have ta omega of mu of x equal to trac of x, apparent with the castle omega. Now you see that's exactly the same kind of emerald as what we had before. So, you pair basically one nif wise, the index of one nif wise operator with some transverse Pompereon class or Tram class or any other kind of class you want. So, actually it's a bit more than just Pompereon classes or Tram classes. It's also secondary classes that are in different fluxology. So, that was something which is crucial to us for the pair. So, this is the four-line thing. So, how do you contract this map here basically? So, it's kind of not too difficult. So, you have this little Tram cacher that goes from the kp right there to C of c of infinity of this actual component here. So, it's here. So, click homology.com. So, here now you have a map which is important for us. So, it's called a five. So, it's a, it's required by code. And you can also pair this thing. There. So, see, homology of the action book, right? And now the trace here is simply the four-line thing. So, you make a choice of the guy there. And you hit that against someone which is here that comes from the chain character out there. And you just go this way. Yeah, is that clear? Okay. So, this will be from your important for us at some point. So, now on consequence of this there is a four-line thing. So, some queries. So, it's the four-line thing. So, it's the same thing. You put some spin structures where I've already. So, one is that if you just take same thing. So, if the h is of n is none other. So, n carries, so it's f here. So, the leaves actually. The leaves carry no p-semi-fers. So, we have our same proof now without using non-commitative geometry. But actually it was still so of a chain. But it was still fundamentally. It was like waking. So, what's fundamentally is perfect to read the same principle as here. So, you have to find some anything of our time that you can couple. And you can pair with the defy the electric operator. So, that's really the same principle actually. And because it's you don't have basically this tool here. Things are technically more complicated. But it's essentially the same principle. So, that's one corner here. So, the meaning of this as before is that there is no kind of global family of metric leaf-fers. Yes, yes. Yes, exactly. It's only the global thing. Exactly. So, you have to you have something non-compact. But there are still containing some compact things. You can control everything. Second thing which is also non-fragile. But kind of reformulates things in the sloppy way. So, it stays the following thing. So, when you have a full issue you have something called could be only classes. Which are classes which they don't really try classes. They are not the primary classes. They are secondary classes and they live there instead. So, in the sloppy way this term can say also following thing. So, the non-fragile thing of global only classes. Let's say GV classes. It measures how far one can equip. Equip. Equip. Let's say the solution of F here. With a transfer scenario. So, it's kind of very strong what it says here. So, it says for example that if you take a remaninflation. You don't see any good in your classes actually. That is something that is very strong already. And at that time it was even more strong at that time. Because at that time people were kind of worrying about finding some good interpretation of the volumetric variation of the GV class. And here this term here of cool kind of blue everything. Certainly. So, I think it was, I think to me it's still one of the most convincing interpretations of this class. Okay. Now we move on to the main topic of the talk. And I hope I arrived at this moment when I'm using quantization. Because it's the subject of the conference. So, here you see there's no quantization so far. But I hope to you. So, we should have time. So, 10 minutes, etc. Maybe a minute. Yes. Yes. Ah, so you're chairman. You were chairman before. Okay. So, now we go in here. So, the purpose of Koni Moskvich's work. So, now you want to carry out everything. So, let's say. To Camology. So, if you want to carry out everything to Camology. So, that means here you want to instead work as in Comodic there. So, you have to find some A-people right there in the conscious direction already. So, that's the first going in. So, you want to construct. So, that will give some class here. In, so you get something in the Camology. Of the convolution network. So, I should say here. W. Okay. And so, actually immediately here. So, because you have something which is in Camology. Your index before was something in k-theory. So, you can pair the two. And the pair of the two here will give directly some additive map as in here. Okay. So, now as you have an active map here. So, it's kind of a sort of inverse problem. So, you have an additive map which is given by pairing with the somebody here by the transverse here. So, then can you find a class which is associated to this guy. And, so now that's something. So, two. So, we have code snap here. This goes from phi from the current Camology of W. And to the, it's a quick Camology. So, here it's actually a void. So, W. So, this is the pattern of the convolution or the flow reconstruction or the question. That's a good question here. So, it's not known what's the range of the classes you can get from Camology. So, that's a good question in terms of where to put some. I hope to answer something. Probably the end. So, here we have a, we want to have a map here. Okay. And now you want to find some case. So, if you can construct such a name on here. So, then find the necessary W. So, same notation as in here. The number is, there are some difficulties. And you will reach soon on the subject of quantization. But before that, you just mentioned the point. I think so. Okay. So, what will you do here? You get something like this. So, natural thing to do would be to take some gamma equivalent adiabatic operator on W. So, that's a problem because in general, it's not a reaction here. So, reaction is something just by the formalisms. So, you have no reason to preserve an instructor or whatever. So, gamma doesn't preserve anything. And there are new gamma adiabatic. Gamma, sorry, gamma invariant. Adiabatic operator. Okay. So, there's some solution of code here, which is the following thing. So, here you have a replacement of W by, a vibration number of W, which is the matrix on W. So, you want to say matrix here, it's a reminded matrix. And now, by the Thomas-Arofism here, you will pull up the problem, so pull back the problem to, instead of W to N. That's the first step. And once you do that here, what you can find here, so that's the work of Konrad Moskowicz, you have. So, we can construct some cube, which is here, the Keho-Molzik element on this bundle of matrix here. So, there are some bundles, but I'm also going to put that here. So, Keho-Molzik is a hub, we have that first. So, it's not completely, it's not completely, so there is something here, which is a bit subtle. So, it's not completely a diptych, it's actually a hypodiptych. So, hypodiptych, hypodiptych symmetric operator. So, we have something like this. So, what's good with this operator here is that it's not completely invariant on the reaction of the properties, it's almost invariant, so up to this principal symbol, if you take a right motion of principal symbol. So, actually you have to take also the principal symbol and the sub-principal symbols at the same time. So, that's one thing. And now from here, we have some Schoenkraken, a kind of cube, which is dancing the C-Komologi of M log gamma. Okay, so now answer the two, that's a complete answer, but so one part of my answer is that the C-Komologi worked out, so two. So, you have some pair, so C-Komologi, 98, very hard work by the way, which says it's a formula thing, so the Truncator of Q here, so let's say you suppose also that the lift detection on the model of matrix is free, so we'll discuss about that if you want later, it's not actually transparent. So here, so Truncator of Q is free the range of five, so that back here, so you have a Truncator of Q that's there, and actually what they show is that the Truncator is the range of this map here. So it means you should be able to find a plus, which is associated to this thing. So now I'm questioning, so what's the frame rate? Now frame rate, so you have some partial results, nice to meet you, so the same paper. So, partial results, which is here 90 low dimensions. What happens here, so partial results, and actually they have to come up with some bad conclusions at the end. So what they wanted to see actually, so from the constraint, so you see already you could catch either Trunclasses or different classes. So they wanted to see, they do want to see secondary classes appear, but actually what happens is that they don't see them. So, these examples here, so that x is some conjecture, which is a following thing. So, it's some full email, in the profiling classes, different from all the other classes of this homogene question here. So, right there, is that the conjecture is true. And since I may not have time to talk about quantization, so unless you allow me to, but German. I'll dedicate to the organizer. One minute? We can't discuss it in there, so yeah, so thank you very much. Now, the audience can't ask the question. So can you give us a four minute version? Three, four. The formula for this one here? No, the version of the quantization. Okay, so I can talk about that. Okay, so quantization is a following thing. So, first you need to understand why this formula here can't be registered in the case of the arc of letters. So that's one very simple thing, so that starts from theorem of that here, chess part. It says the following thing. So, if you take an A, you can write a P. So, then the next of P is simply given by some, the pairing between the symbol of P. That's okay, three elements on the quantization model. So, k theory of the quantization model here. With the do-bo direct to this star, this star here. So, which shows exactly why you can reduce the symbol of the dns of any of the co-writers to some stuff like this. So, now they think, what's the problem? So, here you need to make the choice of the most complex structure or spin c or whatever you want. So, there's a choice here that you have to make. And here's the problem because now you have a group action that doesn't preserve anything. So, in particular, they do the one preserve and do your choice here. So, your choice is only valid if you want to work out the current and next terms in case you preserve this structure. So, that's not quite the canonical here. But now here's the idea. So, this time is simple. So, this thing here, it's on the front of the star. But now, this time is simple. So, what you're back to do here would be to trade almost complex with a symplectic. So, you find some of your co-writers which would take into account the symplectic structure. So, actually, it's a bit more than that in my case. It's the real improvisation actually. So, the natural improvisation of the cot engine here. And so, one good choice of spin structure that could go on with this is that something to spin c as here. So, you would like to trade spin c with spin n. So, that's kind of if you just think of two minutes, it's very obvious why you should do this. Because what happens here? So, you want something which is equivalent for any formalisms. So, in terms of, you look at bundles, let's say, in terms of structure groups, that means your structure groups should be something that contains gel, the gel linear group. So, this is not possible because if you take a linear group or something, it's compact. But if you take here, so this is here, it's the double cover of, so this is the more double cover of the unitary group. This is here, double cover of SLNN, which contains this general linear group, up to preservationist. So, now here's the operator, the overall composition will come very soon, so I'll try. So, the operator, let's say, if you have any postponement. So, here's the default you should have to do, so you have spin n and point step. So, what's a good quadratic form? So, quadratic form here. So, just simply pair this out, again, with the kin. So, here it's one forms, so here basically you have some bundles like this. And you can pair the two by just taking a turn of one forms and make a fit. So, simple as that here. Now you have 54 generators, so, so, key formulations. So, you have let's say psi and psi bar here, equal to one and psi equal to, so, square here, is equal to zero. Okay, so let's say let me do R2 here. Okay, so, no, R, okay. So, then here was the operator, so, maybe so you can see one thing here. So, you have spinners or this thing, so, you have one of those spinners, which is simply the X-ray forms. So, you have spinners, and then you can represent them in your spinners this way. So, this is here the X-ray product with the X, I here, so, psi here, psi bar, is the internal product of the X-ray. Okay, so, now here's your operator, so, you're iterating that to the, so, here you have a bundle over M, but now iterating to the content in bundle, so, that you have enough dimensions to build a dark operator. Now you have an operator here, which is D equal to... Yeah, this formula, the dopo by... I'm cheating a bit. It's on the tangent, it's on the manifold. The manifold is the cotangent bundle. Yeah, so, here it's a tangent bundle. So, now you're iterating. So, now you have, so, now you're iterating to the cotangent model, and I have an operator like this, C i, D over X i plus C i bar, D over D p i. So, here X p is the coordinates on the psi i. And now you see here what happens is that you have a hyperbolic operator. That's kind of the thing here. So, you can carry out this construction to arbitrary manifolds instead, so close to the accounting. And what happens here is that you should have some additional term when you lift this thing to the cotangent bundle, because this thing is quite kind of the cotangent bundle, but you can lift it by a choice of connection. And this thing here, it kind of, it still remains equivalent in the reaction of a group, because what you do, so when you have any deformism on M, you can lift that to some symptomatism on this time. And even more, you can do also preserves of your optimization. Okay, so now here's the point. So, now the problem is that you want to work out index theory or calculations of characteristic classes here from this operator, which is hyperbolic. And there is nothing you can do in terms of analysis. So, here in terms of analysis, what you do. So, there's a bit long here. And you finish here now. Sorry for that. 30 seconds. So, analysis here would be replaced by deformation compensation. So, what happens is the following thing. So, if you formally do the same calculations as what you can do with the Hickard method. So, here, if you square this thing, you have some hyperbolic processing. If you do formally the same calculations as what you have for Hickard method, you can still see the top class showing up. Actually, you can still extract the top class from here. And so, if you trade this thing here, at this formally, you can extract your top class. And now there is something that saves you to go back to analysis at some point, which is the fact that when you have content in bundles, you can deform them to semi-classical solution check items instead. And from here, actually, you can carry out this whole analysis to...
 I'm going to ask the organizers for your participation here.  So, it was a wonderful conference.  Oh, you speak up.  Okay, so I'll try to speak up.  So, many ideas in this conference.  So, I'm going to bring some contribution here to bring some other ideas  that haven't been seen in the conference either.  So, let's say, half of the talk here would be a sub-survey,  some historical part on this comment here.  So, it's motivated a bit of what we're doing there.  And the second half will be my own contribution on this comment.  So, which is going to be with the new problem.  So, motivation.  So, let's start with some things.  So, start from the index here.  So, we start with some vibration.  And over some base peak.  So, let's say, everything is close many faults.  So, compact fibers and compact base and the one around.  And let's say you have this, you don't have the vertical tangent bundle.  So, in the kernel of the bar.  Okay.  So, from here, I think you already know,  but for the sake of bringing some motivation to what you do.  Let's recall what this family is next year.  So, let's say you have the x and b,  which is some families of three decompriers on fibers.  So, I don't know fibers, I would say.  So, from here, you will know that if you take the symbol of t,  this defines some element in the k-therey of the dual bundle to the vertical tangent bundle.  And from here, the index takes values.  So, it's not a number, so it's not an integer.  Once you have the point, here it's a k-therey element.  So, it's something which is in the k-therey of the base.  Okay. So, roughly speaking.  So, here, the index of d, it's more or less speaking,  just the element in the kernel of d-co-kernel of the x.  So, more or less speaking, so it's not quite true, because there are jumps.  So, these things is not quite the vector bundle,  but you can modify this thing a little bit by some compact propagation  so that you can make everything like vector bundles and make sense of this thing.  Okay. So, from here, so here's just the index coefficient.  So, you can move over co-molotage to get some conflict formulas.  Just some trunk hacker.  Once you do that here, so here you land, sorry,  so here you land in the co-molotage of them.  Let's see if we can get it in the c.  So, I'm just telling myself, I'm telling myself what is about to be.  So, I'm hiding it in the trunk hacker.  And here you land, you take the trunk hacker, simply, in the co-molotage.  So, I should say, here, co-molotage of the base.  And you have a formula here, which says the formula thing.  So, you should take the trunk hacker, so here, trunk hacker, of the index of t.  So, what we get here would be a differential formula on b.  So, a co-molotage customer on b.  So, you just integrate over the fibers.  And then here you have the trunk hacker of the symbol.  So, it's an index there.  And you partially integrate over the fibers.  So, here from here you get something which is already some differential formula on b.  And then you pair it with the top class.  So, top class of this number here.  So, you can just complexify the practical tangent problem.  Okay, so, this is things that you already know.  So, now I just want to make a sort of little observation.  So, before that, let me do some little discussions on co-molotage.  So, co-molotage, so, rather speaking, so, k3, it's a generalized co-molotage theory.  So, for that you have an associated JRS whole logic theory.  So, what you do.  So, what you can do.  Here? No, on the fiber.  So, the fiber on the fibers here.  So, here you can put some x here.  So, it depends on the, so, x here is an event in the base.  So, it should form the base.  So, it depends on x here.  Okay, so, let me come back to co-molotage.  So, it's simply a dual theory to co-molotage in a sense.  So, in a sense you have a, so in a point sense here.  So, you have a pair which goes this way.  So, if you have some thing here.  So, let's say, the k-there of b.  So, here is the k-there of b.  So, here is co-molotage of b.  So, here is.  Then, you have a pair between two instance in integers.  And simply, so, it's a very simple pairing actually.  So, let's say that you have an n-n there.  Represented by some vector problem over b.  Then you can pair it with some elliptical parameter.  So, represent these guys here.  They are simply elliptical parameters over b.  So, let's say, so, take a co-molotation that's said,  let's say, what should I take here?  Okay, another q.  Okay, and here, the method goes from here to there.  Let's take the twist in the next of q.  So, you twist the, you twist q by the vector model e,  and you take its index.  So, that's the index of the integer.  So, here it's a two-spice.  Okay, so, what we can do now from here is the following thing.  So, a very simple operation is that you can kind of continue  to follow the square here.  And so, let's say you have the index of b,  and it's here, here.  So, you can pair this thing with,  Kevon will do simply, and get a number.  So, let's do that here.  So, now what we can do here is to, so you can pair,  you can pair index of b, which is in k here at b, with,  and then you get the right amount of base.  So, we have to say something q in the kernel model of b.  So, from here we get a number, index of b,  paired with q.  And it's actually very simple to enter practice,  so not very simple.  But what you can show actually is that the index that you have here,  so the integer you have here, represents exactly the,  the index of a native pickup right there  over the two-spice for the vibration.  So, basically what you do is very simple, dramatically.  So, you have something which goes along the fibers,  and it's going to be on the fibers.  You combine with something which is over the base.  So, you kind of combine the two,  and you get something which is at the total space.  So, this observation here is kind of the starting point of KK here.  So, that's all in here.  So, just a little note here, I don't want to say.  So, KK here actually, so,  so you have a right-close neighbor actually,  of two KK here, which is maybe more familiar to people there.  So, super connections basically, which is,  and this bit has used this technology for many, many years.  So, there are two, it's like two phases of the same coin, if you'll.  So, one which is kind of topological,  which is a little bit of a corpology.  So, that's really the starting point of this technology of Kasparov.  And actually, so, one note I wanted to say, so I said,  so it's the fact that I think Kukun had in mind KK theory  when he invented his super connection for medicine.  So, it's kind of clear if you go over his notebooks.  So, his notebooks, you can see at some point,  he's already tried to understand KK theory.  And a few years later, there is this super connection thing showing up at some point.  So, I really think there are some, these two are very, very tied together.  I just wanted to mention that.  So, now, let's see a very easy consequence of this principle here.  What happens here?  So, some observation now.  Which goes back to the digital rigs.  So, digital rigs, there is a theorem that says the following things.  So, now you just put spin structures where are cooperating, if you'll.  So, it says the following things.  So, if you have the agent of N, which is nonzero.  So, I'm not making a mistake here because there are some specialists of FSC matrix,  who discovered curvature matrix.  But, simply here, so it's a simple case.  So, there, one consequence of this here is the N carries no positive scalar matrix.  That's something well known right here.  But, now let's see a very simple principle of what happens here.  So, now let's take this over columnology.  So, if you take this over columnology, what happens?  So, it means you'll be simply, so what you see here.  So, you have one index which is associated to this guy here.  So, it's all clustered on the fibers.  Or, 18-news-along fibers.  And, one guy here, which is associated to 18-news-along base.  So, what you can see here, actually, it's kind of just a sort of multiplicity property  of plastic clustered symmetry.  But, actually, you can say even more.  So, what happens here is that, now, if you combine,  this observation here, with this principle here,  that if you combine one elliptic operator on the fibers and one elliptic on the base,  you get the total index over the total space.  So, you also have something which is kind of not really trivial.  So, you get the fibers themselves.  So, it's not really trivial, but it's a simple observation you can do here.  So, you'll see, if you want.  So, now, what you want to do is move on to another trajectory from here.  And see what you can do actually. And how we can kind of extend this principle  to something a bit more general.  Not just a bit more general, actually.  So, two, Ncg.  I'm not sorry that I don't understand that.  I mean, if you have a trivial fiber model,  the metaphor M as a fiber, sorry, let's say F is a fiber and we use the base.  And, if B as a positive scalar curvature metric,  that guarantees that M as a positive scalar curvature metric.  So, the fiber is not required to have.  As it goes here, so if you take the index from here,  so what happens here, it means here,  you take the agent is here,  along very impossible corresponding classes here.  So, if you pair the two, and the product of the two is zero.  Yeah, I'm sorry, I was thinking so.  Yeah, it's not simply agent is times the other agents, which is zero.  So, it's very simple.  That's why I was saying that's very something integral.  So, you have to, there's something with corresponding classes you have to work out here.  But it's not so difficult.  Okay, so from here, Ncg.  So, now our goal would be to carry out this principle.  So, carry out this.  And then we would have to say singular situations.  So, my singular situations here, I'll talk about just one today,  which is foundations.  So, we're going to have foundations.  So, what we're going to see is that,  what we're going to do here is completely true,  is the base of it here, by the space of a foundation,  which is completely singular actually compared to the space we have here.  So, that's one thing.  Actually, very simple.  So, you can even carry out this to singular foundations.  So, singular foundation, I just want to say a few words,  but there is, let's say two things you can consider here.  Two main words I would say.  So, one which is from a scandalis,  Android address, scandalis.  I think it's about 2005, 15, sorry.  And it's a leaf-wise next term for very general singular foundations, basically.  So, we have one thing recent of Casparov.  Casparov.  I shouldn't scratch his name because he's my mentor, I think.  If he sees the video at some point.  Okay, so Casparov.  Casparov, I think it's okay here.  So, very recent work too.  So, I think it's a new pretty minimum of work that he carried out  in like 15 years or something like this.  And he finally published something.  So, let's see, 2005, 15 I think.  And here it's a, so where the first works here are.  So, where the first works here, it's kind of leaf-wise.  So, it's Casparov, it's transverse.  So, it's not for anything, it's for Casparov, although.  So, it's for relations that are just defined by the attributes.  But, in this case, it's kind of already very interesting  because you can take our legal actions from there.  So, actions of even non-compact verbs on non-compact manifolds.  So, yes, I think we'll leave the video on here.  And one consequence of this work here is something kind of very interesting.  So, it says at the end that, so you know already that the index of any electric operator  can just reduce that to the study of direct operators,  or spin-sea direct operators, no, more.  So, we'll see that later.  But, so, one consequence of this work here and a little more work that I'm trying to do currently  is that the model of transverse elliptic operator on four relations defined by group actions,  it simply is an operator we've seen all along the week.  They form direct operators or government type operators that come.  So, what's at stake in this?  In this theorem at the end, it gives a very simple model of what you should study  for transverse elliptic theory.  So, that's something very, very strong here.  So, but now we're concerned, we just focus on relations  because it's already very interesting.  So, in particular here we'll see a problem that does not reduce to the case of direct operators,  using direct operators.  And we see a problem which we can try to cope with it.  Okay, so, relations.  So, so, more or less, you can carry out everything I've said before.  So, more or less, you can replace vibration by relation.  And you can replace also the base B with m of f, so here m of m f is a relation.  So, this space here is highly singular.  But we'll see how we can kind of desingularize it using non-diffusing geometry.  So, that's a very general principle that has guided the field for a long time.  So, let's just see a bit of transverse geometry.  Okay, so, naively let's say, so you want a base for your relation.  So, naively you could just take a transversal.  Okay.  So, take a transverse, so w here, it's a complete transversal.  So, complete, you have to meet every leaf at this once.  It can be disconnected by the way.  So, second point here.  Now, if you just naively take the transversal, you see that it's not good enough to reflect  exactly the transverse geometry of the relation.  Because compared to, so, when you have a vibration, things are kind of easy because you have the  one fiber that goes to the base only once.  But in this case here, when you have a relation, it can come back to, if you go around the leaf,  it can go back to transversal again.  So, there is a group which can close this thing, which is the homing group.  And that's a separate of the thermophysicism of the art, right, the transverse.  So, now we begin to see exactly why we call this title homing.  So, why we call this talk here in different thermophysicism, the current, next theory.  And now the idea here is to define a thing.  So, if you can define spaces, so you would like to look at the orbit simply of this compassion.  So, now, but it's kind of singular.  So, just this thing right here, so you just consider, so, instead of the base,  you replace this with, not the base, but I mean, instead of, so the base,  you would like to see that as a portion, but it's kind of singular.  So, instead of this here, you look at this actually,  gamma is what we call homing groupway.  So, here it's registered transverse.  But actually what we get at the end, so the groupway that you have here is more equivalent to the homing groupway.  So, it's simply either you take the whole formation and you get the whole homing,  or you just try this to one transverse, so they knew, let's say, first return maps.  So, it's equivalent.  And so, in non-competitive geometry, what you do is to, instead of studying spaces like this,  you look at the, so, you study functions.  But this thing here is kind of singular.  So, you replace this by the convolution and problem.  So, that's the convolution and problem.  Okay.  And from here, you can do the same thing as what we've done before.  So, one scanner is here, 10 different things.  So, 82.  So, it says that you have an index morphism here, which goes from the k theory of,  so the k theory of, so that's the v-wise tangent bundle here.  And there, so you have something, so now you see what to do.  So, instead of taking the base, as in the case of vibrations, you just take this guy here.  But you have to consider transverse instead.  So, k theory up and k theory down.  Okay.  So, you have some next morphism like this way.  So, you should have k-mology on the left.  Should it be k-mology on the left?  No, it's k theory.  No, on the left.  Left here?  You can put this.  Yes.  You can put k theory, k-mology instead also.  So, you can basically, they call that k-tart, you know.  It's not an issue here.  So, just for our purpose of simplifying the talk also.  So, if you know, you can take the geometric chronology if you want it to.  Perfect fine.  Okay.  So, from here, now you can just do the same thing as before.  So, but it will be just a bit more complicated to, I would say, carry out some logical formula.  So, from here, like I said, but there are some, you have kind of some circles of ideas.  So, at some point, so just remember before we pair some elliptical pattern on fibers  and elliptical pattern on base.  So, here you'll be able to carry out the same kind of principle.  So, you'll see that a bit there.  But now there is a theorem of code if you want the formula.  So, that's a bit more complicated.  So, first, to carry out the principle I've just mentioned before to you.  One, we talk about Fermi's in this theorem.  So, there is something a bit more complicated here.  The first thing that was done was code, actually, in 82.  And he worked entirely in the chronology instead.  So, here is a foreign thing.  So, the theorem is kind of one of your first things.  So, let's see here.  So, let's say you take any x in the k-thoracic style.  So, from here you can take a transfer pattern.  So, here, that's something from the Bohm-Ecquan.  So, you have a transfer pattern which takes this here in the homotopy quotient.  So, here we have k-thoracic.  So, we have a quarenthalmology.  Here, a quarenthalmology of, so it's simply this thing.  So, that thing here is that you're also bounded over a classifying space.  And you take the homotopy quotient.  So, it stands here, your thing, it kind of faces the whole space,  the whole space of the equation.  So, you can make a parallel with what we have in the case of the Freibese next theorem.  So, we have a Neutronk-Apar here.  And given, so I just hope I'll have enough room.  So, you take any choice here of class which is there.  So, you want to pair it with here, the quarenthalmology of this thing.  So, it's all common, it's the right thing to call it here with that, as you can see.  So, you have x here.  And then for any, so here you have a choice of the quarenthalmology just there.  What is it? Gamma, you're right.  It's the group.  It's a homomy group.  Okay, I mean, m is w, the quarenthalmology.  Ah, sorry, it's w.  Surprise w.  So, I'll change to something called m to be there.  You're right, thank you.  You have a choice, let's take any w, so for any w which is in the ring, directed by, say, trac glasses,  and get full successes.  So, for any choice of such a guy here, so you can find various, and then if you have a man here,  say, ta, which goes from the k3 of the convolution at the point here, to c.  So, that's where it's there.  So, we have some men this way.  So, for some reason.  So, here how we're trying to get to c.  And here we have this map here, which goes to c.  Such that, so here it's, let's call it omega because it depends on omega.  Such that you have the four-line thing, which is true.  So, call this map here, w.  So, if you just form this way, and form this way, you get equal x.  Let's just write this down.  So, we have ta omega of mu of x equal to trac of x, apparent with the castle omega.  Now you see that's exactly the same kind of emerald as what we had before.  So, you pair basically one nif wise, the index of one nif wise operator with some transverse  Pompereon class or Tram class or any other kind of class you want.  So, actually it's a bit more than just Pompereon classes or Tram classes.  It's also secondary classes that are in different fluxology.  So, that was something which is crucial to us for the pair.  So, this is the four-line thing. So, how do you contract this map here basically?  So, it's kind of not too difficult.  So, you have this little Tram cacher that goes from the kp right there to C of c of infinity of this actual component here.  So, it's here.  So, click homology.com.  So, here now you have a map which is important for us.  So, it's called a five. So, it's a, it's required by code.  And you can also pair this thing.  There.  So, see, homology of the action book, right?  And now the trace here is simply the four-line thing.  So, you make a choice of the guy there.  And you hit that against someone which is here that comes from the chain character out there.  And you just go this way.  Yeah, is that clear?  Okay. So, this will be from your important for us at some point.  So, now on consequence of this there is a four-line thing.  So, some queries.  So, it's the four-line thing. So, it's the same thing.  You put some spin structures where I've already.  So, one is that if you just take same thing.  So, if the h is of n is none other.  So, n carries, so it's f here. So, the leaves actually.  The leaves carry no p-semi-fers.  So, we have our same proof now without using non-commitative geometry.  But actually it was still so of a chain.  But it was still fundamentally.  It was like waking.  So, what's fundamentally is perfect to read the same principle as here.  So, you have to find some anything of our time that you can couple.  And you can pair with the defy the electric operator.  So, that's really the same principle actually.  And because it's you don't have basically this tool here.  Things are technically more complicated.  But it's essentially the same principle.  So, that's one corner here.  So, the meaning of this as before is that there is no kind of global family of metric leaf-fers.  Yes, yes.  Yes, exactly.  It's only the global thing.  Exactly. So, you have to you have something non-compact.  But there are still containing some compact things.  You can control everything.  Second thing which is also non-fragile.  But kind of reformulates things in the sloppy way.  So, it stays the following thing.  So, when you have a full issue you have something called could be only classes.  Which are classes which they don't really try classes.  They are not the primary classes.  They are secondary classes and they live there instead.  So, in the sloppy way this term can say also following thing.  So, the non-fragile thing of global only classes.  Let's say GV classes.  It measures how far one can equip.  Equip.  Equip. Let's say the solution of F here.  With a transfer scenario.  So, it's kind of very strong what it says here.  So, it says for example that if you take a remaninflation.  You don't see any good in your classes actually.  That is something that is very strong already.  And at that time it was even more strong at that time.  Because at that time people were kind of worrying about finding some good interpretation of  the volumetric variation of the GV class.  And here this term here of cool kind of blue everything.  Certainly.  So, I think it was, I think to me it's still one of the most convincing interpretations of this class.  Okay.  Now we move on to the main topic of the talk.  And I hope I arrived at this moment when I'm using quantization.  Because it's the subject of the conference.  So, here you see there's no quantization so far.  But I hope to you.  So, we should have time.  So, 10 minutes, etc.  Maybe a minute.  Yes.  Yes.  Ah, so you're chairman.  You were chairman before.  Okay.  So, now we go in here.  So, the purpose of Koni Moskvich's work.  So, now you want to carry out everything.  So, let's say.  To Camology.  So, if you want to carry out everything to Camology.  So, that means here you want to instead work as in Comodic there.  So, you have to find some A-people right there in the conscious direction already.  So, that's the first going in.  So, you want to construct.  So, that will give some class here.  In, so you get something in the Camology.  Of the convolution network.  So, I should say here.  W.  Okay.  And so, actually immediately here.  So, because you have something which is in Camology.  Your index before was something in k-theory.  So, you can pair the two.  And the pair of the two here will give directly some additive map as in here.  Okay.  So, now as you have an active map here.  So, it's kind of a sort of inverse problem.  So, you have an additive map which is given by pairing with the somebody here by the transverse here.  So, then can you find a class which is associated to this guy.  And, so now that's something.  So, two.  So, we have code snap here.  This goes from phi from the current Camology of W.  And to the, it's a quick Camology.  So, here it's actually a void.  So, W.  So, this is the pattern of the convolution or the flow reconstruction or the question.  That's a good question here.  So, it's not known what's the range of the classes you can get from Camology.  So, that's a good question in terms of where to put some.  I hope to answer something.  Probably the end.  So, here we have a, we want to have a map here.  Okay.  And now you want to find some case.  So, if you can construct such a name on here.  So, then find the necessary W.  So, same notation as in here.  The number is, there are some difficulties.  And you will reach soon on the subject of quantization.  But before that, you just mentioned the point.  I think so.  Okay.  So, what will you do here?  You get something like this.  So, natural thing to do would be to take some gamma equivalent adiabatic operator on W.  So, that's a problem because in general, it's not a reaction here.  So, reaction is something just by the formalisms.  So, you have no reason to preserve an instructor or whatever.  So, gamma doesn't preserve anything.  And there are new gamma adiabatic.  Gamma, sorry, gamma invariant.  Adiabatic operator.  Okay.  So, there's some solution of code here, which is the following thing.  So, here you have a replacement of W by, a vibration number of W, which is the matrix on W.  So, you want to say matrix here, it's a reminded matrix.  And now, by the Thomas-Arofism here, you will pull up the problem, so pull back the problem to, instead of W to N.  That's the first step.  And once you do that here, what you can find here, so that's the work of Konrad Moskowicz, you have.  So, we can construct some cube, which is here, the Keho-Molzik element on this bundle of matrix here.  So, there are some bundles, but I'm also going to put that here.  So, Keho-Molzik is a hub, we have that first.  So, it's not completely, it's not completely, so there is something here, which is a bit subtle.  So, it's not completely a diptych, it's actually a hypodiptych.  So, hypodiptych, hypodiptych symmetric operator.  So, we have something like this. So, what's good with this operator here is that it's not completely invariant on the reaction of the properties,  it's almost invariant, so up to this principal symbol, if you take a right motion of principal symbol.  So, actually you have to take also the principal symbol and the sub-principal symbols at the same time.  So, that's one thing. And now from here, we have some Schoenkraken, a kind of cube,  which is dancing the C-Komologi of M log gamma.  Okay, so now answer the two, that's a complete answer, but so one part of my answer is that the C-Komologi worked out, so two.  So, you have some pair, so C-Komologi, 98, very hard work by the way, which says it's a formula thing,  so the Truncator of Q here, so let's say you suppose also that the lift detection on the model of matrix is free,  so we'll discuss about that if you want later, it's not actually transparent.  So here, so Truncator of Q is free the range of five, so that back here, so you have a Truncator of Q that's there,  and actually what they show is that the Truncator is the range of this map here.  So it means you should be able to find a plus, which is associated to this thing.  So now I'm questioning, so what's the frame rate?  Now frame rate, so you have some partial results, nice to meet you, so the same paper.  So, partial results, which is here 90 low dimensions.  What happens here, so partial results, and actually they have to come up with some bad conclusions at the end.  So what they wanted to see actually, so from the constraint, so you see already you could catch either Trunclasses or different classes.  So they wanted to see, they do want to see secondary classes appear, but actually what happens is that they don't see them.  So, these examples here, so that x is some conjecture, which is a following thing.  So, it's some full email, in the profiling classes, different from all the other classes of this homogene question here.  So, right there, is that the conjecture is true.  And since I may not have time to talk about quantization, so unless you allow me to, but German.  I'll dedicate to the organizer.  One minute?  We can't discuss it in there, so yeah, so thank you very much.  Now, the audience can't ask the question.  So can you give us a four minute version?  Three, four.  The formula for this one here?  No, the version of the quantization.  Okay, so I can talk about that.  Okay, so quantization is a following thing.  So, first you need to understand why this formula here can't be registered in the case of the arc of letters.  So that's one very simple thing, so that starts from theorem of that here, chess part.  It says the following thing.  So, if you take an A, you can write a P.  So, then the next of P is simply given by some, the pairing between the symbol of P.  That's okay, three elements on the quantization model.  So, k theory of the quantization model here.  With the do-bo direct to this star, this star here.  So, which shows exactly why you can reduce the symbol of the dns of any of the co-writers to some stuff like this.  So, now they think, what's the problem?  So, here you need to make the choice of the most complex structure or spin c or whatever you want.  So, there's a choice here that you have to make.  And here's the problem because now you have a group action that doesn't preserve anything.  So, in particular, they do the one preserve and do your choice here.  So, your choice is only valid if you want to work out the current and next terms in case you preserve this structure.  So, that's not quite the canonical here.  But now here's the idea.  So, this time is simple.  So, this thing here, it's on the front of the star.  But now, this time is simple.  So, what you're back to do here would be to trade almost complex with a symplectic.  So, you find some of your co-writers which would take into account the symplectic structure.  So, actually, it's a bit more than that in my case.  It's the real improvisation actually.  So, the natural improvisation of the cot engine here.  And so, one good choice of spin structure that could go on with this is that something to spin c as here.  So, you would like to trade spin c with spin n.  So, that's kind of if you just think of two minutes, it's very obvious why you should do this.  Because what happens here?  So, you want something which is equivalent for any formalisms.  So, in terms of, you look at bundles, let's say, in terms of structure groups,  that means your structure groups should be something that contains gel, the gel linear group.  So, this is not possible because if you take a linear group or something, it's compact.  But if you take here, so this is here, it's the double cover of, so this is the more double cover of the unitary group.  This is here, double cover of SLNN, which contains this general linear group, up to preservationist.  So, now here's the operator, the overall composition will come very soon, so I'll try.  So, the operator, let's say, if you have any postponement.  So, here's the default you should have to do, so you have spin n and point step.  So, what's a good quadratic form?  So, quadratic form here.  So, just simply pair this out, again, with the kin.  So, here it's one forms, so here basically you have some bundles like this.  And you can pair the two by just taking a turn of one forms and make a fit.  So, simple as that here.  Now you have 54 generators, so, so, key formulations.  So, you have let's say psi and psi bar here, equal to one and psi equal to, so, square here, is equal to zero.  Okay, so let's say let me do R2 here.  Okay, so, no, R, okay.  So, then here was the operator, so, maybe so you can see one thing here.  So, you have spinners or this thing, so, you have one of those spinners, which is simply the X-ray forms.  So, you have spinners, and then you can represent them in your spinners this way.  So, this is here the X-ray product with the X, I here, so, psi here, psi bar, is the internal product of the X-ray.  Okay, so, now here's your operator, so, you're iterating that to the, so, here you have a bundle over M, but now iterating to the content in bundle,  so, that you have enough dimensions to build a dark operator.  Now you have an operator here, which is D equal to...  Yeah, this formula, the dopo by...  I'm cheating a bit.  It's on the tangent, it's on the manifold.  The manifold is the cotangent bundle.  Yeah, so, here it's a tangent bundle.  So, now you're iterating.  So, now you have, so, now you're iterating to the cotangent model, and I have an operator like this, C i, D over X i plus C i bar, D over D p i.  So, here X p is the coordinates on the psi i.  And now you see here what happens is that you have a hyperbolic operator.  That's kind of the thing here.  So, you can carry out this construction to arbitrary manifolds instead, so close to the accounting.  And what happens here is that you should have some additional term when you lift this thing to the cotangent bundle, because this thing is quite kind of the cotangent bundle,  but you can lift it by a choice of connection.  And this thing here, it kind of, it still remains equivalent in the reaction of a group, because what you do, so when you have any deformism on M,  you can lift that to some symptomatism on this time.  And even more, you can do also preserves of your optimization.  Okay, so now here's the point.  So, now the problem is that you want to work out index theory or calculations of characteristic classes here from this operator, which is hyperbolic.  And there is nothing you can do in terms of analysis.  So, here in terms of analysis, what you do.  So, there's a bit long here.  And you finish here now.  Sorry for that.  30 seconds.  So, analysis here would be replaced by deformation compensation.  So, what happens is the following thing.  So, if you formally do the same calculations as what you can do with the Hickard method.  So, here, if you square this thing, you have some hyperbolic processing.  If you do formally the same calculations as what you have for Hickard method, you can still see the top class showing up.  Actually, you can still extract the top class from here.  And so, if you trade this thing here, at this formally, you can extract your top class.  And now there is something that saves you to go back to analysis at some point,  which is the fact that when you have content in bundles,  you can deform them to semi-classical solution check items instead.  And from here, actually, you can carry out this whole analysis to...
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