Thank you very much. Yeah, it's a great honor for me to have an opportunity to speak here today. So, well, I don't quite know how this talk fits into the general scheme of things at the conference, but I'll be speaking about one particular paper by Masaki Kashiwara and Michel Verne. It's a paper which dates for almost 40 years back in 1978. And so it sets a so-called Kashiwara-Verne problem or Kashiwara-Verne conjecture, and I'll now explain what it is about. So, Kashiwara-Verne 78. I won't give you the original formulation of the problem, but maybe slightly changed formulation. And the idea is as follows. We're looking at the completed free Lie algebra with two generators, and we know that this Lie algebra contains an exceptional element called the Baker-Kamble-Hausdorff series. So we can write it as log e x e y. And while this is famously equal to x plus y plus one-half commutator x y plus multi-commutators, and the Kashiwara-Verne problem is about the properties of this element. So let me try to formulate it. So the question is to find an element in the automorphism group of this free Lie algebra, which satisfies some number of conditions, and I will enumerate them starting from zero. So the condition zero is as follows. We would like the image of the generators to be conjugate to the generators themselves, where g x and g y are elements of the exponential group of l. So l is graded by the number of letters x and y, so we can easily define the corresponding group. What is the ground ring of ground? Yeah, right. So let's say K is a field of characteristic zero. So that's the ground ring. Right, so then the property one says that the sum of generators is mapped to the Kambel-Hausdorff series. So the sum of these two expressions is actually equal to x plus y plus one-half bracket x y plus one. And then the second property, let me write it down and then make some comments. So it won't be immediately clear what I'm saying. So some function, some j of the inverse of f is of the form absolute value h of x plus y minus h of x minus h of y, where of course I should say something about all these letters. So let me put it here. First of all, what is j? In fact, well, especially towards the end of the talk, but already now there will be quite a lot of cheating in what I'm saying. And so j, actually that's the heart of the Kashavara-Werms theory. But perhaps I prefer not to say too much about it because it would take a lot of time. But let me say the following. So j is a one-cost cycle mapping automorphisms of L to cyclic words in letters x and y. So cyclic words, I denote them by something with absolute value. And cyclic means that, for instance, xy is equal to yx. So these are examples of elements in the space of cyclic words. And how should one think about j? One should probably think about it as m. What do you mean by cyclic words? You mean words up to cyclic words? Yeah, words up to cyclic. So we take an associative algebra. We generate a sense in y. And we consider cyclic words, meaning that this is associative algebra model of commutators. So this is a vector space. So for instance, xy is equal to yx. So this is a noncommutative analog of the logarithm of Jacobian. So imagine that you just, when you teach first-year calculus, you introduce a Jacobian of a transformation. So here we have an automorphism of a freely algebra with two free generators. This is a kind of noncommutative plane, you can think. And this j is a noncommutative analog of the logarithm of Jacobian from ordinary calculus. In particular, it satisfies, so this one, cos-cycle property. If you take j of f times g, this is f applied to j of g plus j of f. And that's exactly the property that the normal log Jacobian would satisfy. If you take a Jacobian, it would be a product. But if you take a log, it would be a sum. So that's... You don't have z for the second term? Sorry? You don't have z for the second term? No. No, no, yeah, let's take it. Unless, of course, maybe I made a mistake here, but let's take it as an exercise. Take an ordinary Jacobian of a transformation and then check that it satisfies roughly this rule. Maybe g and f should be exchanged. You should take the vector space generated by cyclic words, not the... Yes, yes? Yeah, the vector space generated by cyclic words. Sure. Okay, so that's one piece of notation. And h is something known. Right, now let's go back to the right-hand side. So h is a formal power series in one variable. And making a link to some of the yesterday's talks. So this h is one of the versions of the DuFlo function. So that's related to the DuFlo functions which were mentioned yesterday. Right. So the question is to find such an f. Of course, at first it doesn't look like a very natural problem. So let me mention that the motivation for the KV problem is as follows. The positive solution of the KV problem gives a very natural proof of the DuFlo theorem on the isomorphism of the center of the enveloping algebra with the ring of invariant polynomials for g, the Lie algebra of K, and dimension g is finite. So this is a property of the Baker-Kambell Hausdorff series, which ensures the DuFlo theorem. And as you see, even 0 and 1, these are some kind of easy properties. Now 2 looks at first a little bit esoteric, but that's basically the core of the matter, whether you can solve 2. So h is a particular function? Basically it's negotiable. I think in the original formulation of the problem h was a particular function, but if you solve it with any h, this would be good enough. So maybe one more remark. So if one wants to solve 0 and 1 together, so this is easier. So there is an easy solution of 0 and 1 together. Then solving 2 perhaps poses problems. Now this is a very concrete problem. Let me try to maybe for a second show you a bigger picture. In fact, this allows to inscribe it into a... And do they know that h should be something specific? Do they know what is h? Again, so it's a little bit negotiable. One can make it more narrow and say that h is a precise function. Just give an expression for it. But I think actually it's sufficient to solve it with any h and then it would be good enough for proving this isomorphism. So it doesn't matter if one succeeds to solve it with some particular h, so then it would be good enough. And then we know what h is one... For which h is the solution exist, for which h is it doesn't. So this is known. This it is known for which h is one solution of the kv, whatever of the overflow. If one takes h, the standard overflow function, then the problem admits solutions and this is one particular case which sweeps as well. But in principle, h is a bit negotiable. That's not unique. It's not unique. You can find solutions with other h's as Maxim showed at some point. So maybe a little bit a bigger picture. How does it fit into a scheme of things? So let me draw the following diagram. So there is a family of torsos which groups action on them. So one of them is a torso of Greenfield Associates. And it is acted by the... There are several versions of it. By the Grotting-Dectachmular Group, let me consider this group GRT1. So this is one of the Grotting-Dectachmular Groups. So then in numbers theory, there is a torso of formal MZVs. So these are solutions of the so-called double shuffle equations. And this is acted by the group. Maybe in France I should call it DMR. Now sometimes it is called DS, the double shuffle. So the group of symmetries of double shuffle. And then there is a torso of solutions of the Kashiwara-Vern problem. So there are also several versions of the Kashiwara-Vern groups. Let me call one of them. So the similar to GRT. So this is the graded version KRV. And yeah, the R, yes, was imported from here. That's right. Yeah, that's certainly a fair remark. So there is a number of links between them. So due to furusho, there is an injection from a saucer to formal MZVs and then correspondingly here there is a map between groups. And then Leila-Schnapps showed that formal MZVs give rise to solutions of KV. And here there is also an embedding of groups. And then a little bit before, Schar-Torosian and I, we showed that there is a map from a saucer to solutions of KV. And then we made it more precise with Benjamin and Rikus. So there is a direct map here and they commute. And here maybe I also write at the level of Le algebra. And here this is a group generated by a graded Le algebra GRT1, which maps to the graded Le algebra DS, which maps to the graded Le algebra KRV. And as was relatively recently proved by Francis Brown, so here there is an embedding of the freely algebra with sigma 3, sigma 5 and so on. So there is this quite amazing picture of the world where there is this part of the picture which comes from Le theory as we indicated, some freely algebras and related to the Duflo theorem. And here there is a purely number theoretic picture with the double shuffle and here there is a picture which one can say probably three-dimensional topology, right? So Drinfeldt Associates, this is the braid group with three strands and topology of three manifolds and all these pictures, so there are embeddings and the conjecture, I think one of the big conjectures is that they are actually all the same, up to perhaps minor details in definition. So they expect it to be all the same and this is confirmed by various numerical tests until maybe some not very high but reasonably high degrees. So what the groups themselves are, which kind of group? The algebraic group? What kind of group? So these are, so those would be pro-unimportant groups. So over the covers, so there is a field K that we introduced in the beginning. So I want to show you this picture but actually today I want to show you a new element in this picture so that's basically a picture of the world and I would like to add one more column to this picture. So now for a short while we'll probably forget about this story and I will be speaking about an entirely different story which was actually mentioned yesterday. That's the Goldman bracket and the one more element which wasn't mentioned yesterday called derived core bracket. So we'll now speak of some kind of two-dimensional topology and then we'll return to this picture and we try to add one more column. So that's the 2D topology part. Does it matter which field you take from that? It's expected to work over any field of characteristic zero. It's not to work over a few, maybe? Sorry? Is one called over Q? You mean like this certainly holds, yeah, this holds over Q but this we don't know what it holds, probably, maybe. Who knows? All right, so what about the 2D topology part? So let's say sigma is an oriented closed to manifold and possibly this boundary. So let's say genus G, number of boundary components is N. For most cases let me assume that there is at least one boundary component so this will be more convenient for me. So let's denote by pi, pi1 of sigma. It is a free group if N is at least one. They also are oriented, connected. And we'll be interested in the following space. So this is a span of a k of homotopy classes of free loops on sigma. So this is the same as the k span of conjugacy classes in pi. Or if you wish, it's also the same as the group algebra of pi over the space of commutators. Or this is also the same as k of k in cyclic words in pi, right? So these are all the same. Cyclic you mean because it's a free group? Yeah, because it's a free group. No, but you have to take indices of generators as well. Sorry? Here you have indices. Yeah, right. Okay, let's... It's in pi not in X. Yeah, yeah, it went a bit too far. You're right. Okay, we'll come back to it later. Yeah, you're right. So in fact, it turns out that this space, so this is a vector space of a k, it carries several exceptional structures. Is it okay if I use that one? So the first one is the bracket called the golden bracket. It was mentioned yesterday. Let me show you how it looks. Suppose this is our surface and we take two loops on this surface. Since we are interested in homotopic classes of loops, we can always arrange the intersections to be transverse. And we can always arrange such that there is only a finite number of intersection points. So let's say this is alpha and this is beta. So then the golden bracket of alpha and beta is equal to a sum over intersections of the representatives of alpha and beta. And now we do the following. Once we have the intersection point, we can construct a new free loop. We can go first along alpha and then along beta starting from this point. Right? So let me denote it alpha, p, beta, p. So that's a new loop which starts from p. But well, the initial point doesn't matter. And we multiply by sine. And the sine measures the relative orientation of the frame given by the tangent vectors to alpha and beta and the orientation of the surface. Right. So that's the first construction. And then there is a second construction called to arrive core bracket. So this construction does the following. Let me again draw an example. So this will be a sphere with three holes. I now have only one loop. And let me arrange again the self-intersections if there are any self-intersections to be transverse and to have only finite number of them. So and delta of, let's call it alpha, delta of alpha will be again sum over self-intersections. If you don't mind, I will use this fine annotation. So again, the sine and then alpha, p prime, wedge, alpha, p double prime. Alpha, p prime and alpha, p double prime are two halves or two sub-loops into which the total loop is split by the intersection point. So let me state a theorem which is slightly false and maybe I encourage you to catch me. So this is somewhat false. So first of all, due to Goldman so the bracket is well-defined which means it's well-defined on homotopy classes. I defined it on representatives, but it's well-defined on homotopy classes and it's a lee bracket. So it's quesimatic and satisfies the Jacobi identity. This is actually a very nice exercise to check it. Now the next part, due to two rives delta is well-defined called lee. It satisfies the Jacobi identity and well, we can state it in several ways. So there is a relation between the bracket and delta which says that together they form a lee by algebra and so G of sigma bracket, co-bracket is a lee by algebra. In practice, this means the following. The delta of the Goldman bracket of alpha and beta is a Goldman bracket of alpha with delta of beta plus or beta minus the bracket of beta with delta of alpha. So this is technically that's the extra relation that one needs to show. Sure, that's a good question. Here I'm cheating slightly, but not much. This means that this guy I interpreted as alpha times one plus one times alpha. So that it is well-defined. But now here there is actually, as I say, this is slightly false. If you don't see it, that's okay. We just go ahead. No, I mean. Is it trivial loops? Certainly, the trivial loops. Actually, this is false. So this is false, essentially, because if you take a small, you see you can always create a small self-intersection by creating such a piece of a trivial loop and then it will contribute here. So there are two ways to deal with it. Either you say that the class of trivial loops is central inside the Goldman bracket and then you can just factorize by it. So kill such loops, declare them equal to zero. Or another solution which is slightly better for me, one can choose a framing on the surface. And if one chooses the framing on the surface, then it's okay to have trivial loops because one can no longer create them because then one can impose some condition on framing. Sorry? Tribalization of the tangent bundle of the surface. And when you consider the bracket of alpha and delta, but if you interpret alpha as alpha times of one plus one times of alpha, this is the question of factor of two and the computation. No, no, that's just the definition. Alpha acts by derivations on veg of G. Right, it acts on this curly G and it acts by derivations on the exterior algebra. So this is, you can think of it as the action by derivations or like sometimes people write it in this way. I act, so delta of beta lives in two components, G veg G, and you act on the first component plus you act on the second component. Okay, do I get the question of terms of, okay, depending on the convention, it's a question of two. All right, very well. So why is it all of interest for us? Right, let me, perhaps I don't want to, it gets difficult what we can, yeah, probably we sacrifice the picture of the world. Okay, let me make a remark. So there is a natural filtration of the group algebra by powers of the augmentation ideal. So the kernel of epsilon mapping Kp to K by just mapping each element of pi to one. So we can now do the following. Let's complete Kpi and G of sigma using this filtration and let's consider the associated graded. So let me call it G0 of sigma. So that's the associated graded. I must say that here I'm also cheating a little bit. In truth, this is the correct filtration only say for either if you have G0 or you have only one component of the boundary but otherwise one should do something slightly more complicated. But maybe at the moment we can return to it later. But in many cases that's the correct filtration. So now if we have a filtration and then the corresponding graded space we can ask what happens with the golden bracket and derived co-bracket. And sure enough they define a Leib bi-algebra structure on this graded space. So together with the corresponding graded bracket and co-bracket, so this is a new Leib bi-algebra. Well, now one can ask the following natural questions. So we have for a given surface we have two Leib bi-algebras and the graded version. So the question is will they be isomorphic to each other or not? So let me start answering this question in a particular case. So let me start with the following situation. So I take sigma to be of G0 V3 boundary components. So this is my sphere V3 holes. So let me choose generators of pi1. Let's say capital X and capital Y. And then the claim is you see now we need to make some connection to the Kashiwara-Werner problem. So assume we have an automorphism of the freely algebra. So then it turns out that every automorphism of L induces a map from G of sigma to a graded of G of sigma. So this map looks as follows. First of all, it induces let me call it f hat. It induces the following change of variables. So since we are working on a field of characteristic zero, we can replace those generators by exponentials of small x and y, generators of a freely algebra, and then apply this automorphism f. In particular, like f can be equal to 1, right? Then you don't do anything. Just send map capital X to exponential of x. Now coming back to this picture, right? So I was saying K of cyclic words, but cyclic words in what? So assume that we have chosen some basis of pi1. In this case, the basis is x and y. So here we will be cyclic words in x, y, x minus 1, y minus 1. This standard relations that x times x minus 1 is equal to 1, y times y minus 1 is equal to 1. And if we make such a substitution, what we obtain? We obtain cyclic words, our priori infinite series, and cyclic words in small x and small y. And it turns out that this is exactly how this graded version of G of sigma looks like. So such a map induces, so the choice of an automorphism induces for us a map from this space to that space. Now the question is, will it be a... I don't understand what is the f of... Oh. L x y is what is the l-algebra or not? Yeah, yeah, sure. Is it okay? I define it. Yeah, I define it this way. That's essentially because an automorphism of free l-algebra uniquely extends to an automorphism of a free associative algebra. Okay. So now let me state some results. Does the graded l-algebra have a description in terms of connection? Wait a second. Maybe I should say that the filtered l-algebra, that's more like related, golden bracket was inspired by the T-abot construction. Right? So that's... Maybe I'll talk about it in two minutes. Okay? So... Right. So let me state a theorem and give lengthy comments about it. So let's say f is an automorphism of a free l-algebra. So then f had the map from g of sigma with a golden bracket to the graded version of g of sigma with a graded bracket. Note that I'll meet for a moment the co-bracket. It's an isomorphism. If f satisfies 0 and 1. Right? So there is 0 and 1. So these are the 0's and first Kashubar-Vernick equations. And if they are satisfied, then this is an isomorphism of l-algebras. So how do you define it? f wolf sends x to some infinite series and then how do you... So how do you define this? Oh, but everything is completed, right? Yeah, everything is completed. Yeah, yeah, whatever. Because to take the graded, I first completed it. Sorry. So let's discuss it a little bit. First of all, let me attribute this theorem. I think in some versions... So there are several versions of it. I think it was proved by Masio-Turayev and Kawazumi Kuno. And there is also an interesting version of this theorem and of the proof due to my graded student Florian Neff, which shows that this theorem is a particular instance of some rigidity in non-commutative Poisson geometry of Konsevich. So one can interpret it also in this way. Also, to some extent, we saw this... some version of this theorem yesterday in Philip's talk because... So this is... The Goldman Bracket is essentially in called symplectic or Poisson structures on moduli of flat connections. And this story essentially encodes some kind of Kerylov-Costom-Suryov brackets on residues at the... So this is a regular... what Philip was referring to as a regular situation. So there are regular singularities. And this would be, in that terminology, some kind of better picture. And this would be some kind of... Durham picture. So here, these are not answering your question. These are not connect... These are residues of connections. That's what this thing encodes. And these are holonomies of connections. Right. And, well, already that, I think, is quite interesting. So these are the zeros in the first equation. Right? Now, the second theorem. So let's now look at the theory with a core bracket. It's an isomorphism. So let's assume... let's assume that 0 and 1 are satisfied. So then this is an isomorphism, if and only if f satisfies 2. I know this is true, but I still find it very, very surprising. Why? Because you see, what is this 2 about? I didn't tell you in truth. But you see, this is about some kind of non-commutative Jacobians, right? Non-commutative analogs of the Jacobian. Now, what is this... This story is about. That's about the behavior of self-intersections of homotopic classes of curves on a 2-surface. Frankly, why is it the same? It remains a mystery to me. This is true, but I don't really understand it very well. So... Yeah, in fact, let's assume that 0 and 1 are satisfied. So then, we already know that the Lie algebras are the same. That the Lie bi-algebras are isomorphic, if and only if f satisfies 2. For a specific function h? Doesn't matter. Yeah, it doesn't matter with the function h. Right. So, as I say, this is quite a mystery to me, and this is due to Kavazumi, Kuno, Neff and myself. Just a second. Just as a formality coming from telegeometry, for example, if you have complex structure, will you give some specific effect? Yeah, I don't know. Yeah, for general surface. So, you're saying that it would give some? Yeah, no, if you consider only 0 and 3 points, it's only one complex structure. But let me just advance by one minute and we look at the other. It is not clear what, again, it is not clear what I'm confused. What is the definition of this map, the roof map, because you map the elements, how do you like to control y in the fundamental group of surface to some formal expressions, but they are not, they are in the completed universal enveloping algebra. But then you want us to go to the grade of Lie algebra. So, how do you do it? Let's first of all, confusing you was one of my purposes, certainly. So, this is achieved, right? Now, okay, let me give an example. Suppose you had an expression of this type. So, this was a cyclic word in generators of the fundamental group, right? Or maybe let's say x, y minus 1, doesn't matter. So, now we map it to exponential f of x, exponential minus f of y. And this is an infinite, a priori an infinite sum of cyclic words in letters x and y. And it turns out that this is exactly this space. This space is infinite or whatever, finite or infinite linear combinations of cyclic words in letters x and y. Right? So, this is a... Or one can apply BCH as well. But then still you need to take the series for the exponential function. No, but I don't know because for the Lie algebra, Freely algebra. Oh, but G of... So, this guy, this gadget, or that gadget is not a Freely algebra, right? So, this is a... For instance, the Goldman Lie algebra, that's not the same as L of x, y. Right? So, elements of the Goldman Lie algebra are of this sort. So, I take words, finite, or maybe if I want to complete I can also have some kind of infinite words in generators of the fundamental group. And then I take the modular cyclic rotations, right? So, the Goldman... Recall the elements of the Goldman Lie algebra, they are of that sort, right? I can take some loops in my surface. So, this is not L of x, y. And this graded gadget is not L of x, y. Yeah, but then you have to fix some combinatorial way to identify cyclic words as elements of the G sigma. Yes. Of G sigma, yeah. Yeah, so this is... One needs to prove on the way some kind of statement about the structure that these are cyclic words in letters small x and y. Yeah, sure. So, your big x is equal to the small x? Oh, that's my map. Right? Big x is mapped to exponential of small x, and then further mapped to exponential of f of small x. Right? They are not equal. These are two different... These are two different vector spacing, and I arrange a family of maps between them, which depend on capital F. And sometimes, this induces an isomorphism between lib-algebra and sometimes not. So, is it possible to just remind me what is small x? Oh, what is in... Okay, like, for instance, you can think of small x and y as generators of homology. So, that would be one consistent picture. Small x is a loop. Small x is a homology class. Corresponding to this loop. Right. Okay. So, now, let me pose some number of... So, I will still state one's theorem, and I pose a number of questions. So, maybe one question that we don't quite understand. We have some idea, but it's not completely clear. So, here, this picture, as I said, it's closely related to the... to Philip's talk of yesterday, because the Goldman bracket is intimately related to the a tier-bord structure on the modular space, and this bracket is intimately related to the structure on the residues, to the Kirillov-Kosten-Suryov bracket. Now, what about those deltas? In fact, at least I don't know what's their meaning for the modular spaces. So, this looks as a very fine problem related to de-topology, but we don't really know what this means, even though it looks very natural. Now, let me spend the last... the last five minutes talking about the surfaces of Hyogenes. So, I focused my attention on the sphere with three holes, because it is related to the... to the standard Kashiwara-Vern problem. However, there is a similar theorem, or the same kind of theorem. So, there is a similar result for arbitrary genus and arbitrary number of boundary components. In fact, not all of these problems are independent in the following sense. So, you can solve all of them if you know how to solve it for a sphere with three holes and for a torus with one boundary component. So, the only non-trivial... so, the second non-trivial example is this pro... is this statement. So, in order to solve this statement, you need to replace the KV problem with something else. So, what is the KV problem for this surface of genus one with one boundary component? Well, let me... let me formulate it. So, what needs to find... what needs to find f is going to be again, just by chance, is going to be again an automorphism of a freely algebra with two generators. Yeah, yeah, sorry. Yeah, an automorphism of a freely algebra with two generators, which satisfies, again, two equations. There will be no equation zero, but there will be equations one prime and two prime. And equation one prime says the following. f, yeah, yeah, everything is completed. So, now, instead of the Campbell-Halsdorff series, we will have the group commutator of E x and E y. So, this is some kind of symplectic... symplectic property, which was studied a lot and solved by Masiu. And then... Sorry? It's an exact formula, no correction. No, it's not. Yeah, no, yeah, that's instead of BCH, yeah. And then the second formula, the second equation will look like that. So, the Jacobian will be a cyclic word with the overflow function of the Leib bracket of x and y, and then there will be some correction terms with other functions. So, this is a fixed function, which depends on x and y. So, it turns out that such an equation... such a pair of equations, it would ensure that there is an isomorphism between the non-graded and graded versions. So, here... So, there is a following question. Before the motivation, recall the motivation for the original Kashiwara-Wern problem was the proof of the DiFluor theorem, right? So, now, we have another pair of equations, and we don't quite know what it means in Leith's theorem. Well, something, maybe not as grand as DiFluor theorem, but it would be still interesting to know is there some meaning to such a statement when the product map is replaced by the commutator map? So, I don't quite know. Maybe Professor Kashiwara and Professor Wern can help us with this interpretation. And the results for other surfaces, they can be built from these two building blocks. Well, with this question, I would like to stop. Thank you very much. Is there a question or comments? Yes. So, I just want to make for a side certain things. So, you said the definition of delta was not sufficient, it was not correct and stated because of the identity loop, and that you had two ways to fix it. You were modding out or using a framing, but then you made other statements. So, it is not clear which convention you use for that. I think, yeah. And then it is not... Okay, so this is one question, and also what was... There was another question about the filtration. You said it's not by... Okay, yeah, let me answer these two questions. So, let's say, I made a relatively strong statement in my theorems, and for that, it's more convenient to have delta defined on a bigger space, which means to add framing. Because it kind of proves... In fact, there was even more cheating. I only showed you the leaf structure and the bi-algebra structure, but there is a lot more to the story because actually it's more convenient to have double brackets of van den Berg. So, this... Okay, if one even works with leaf brackets, it's more convenient and one gets stronger statements. For instance, if and only if statement there... For that, it's better to have a framing. I think we don't know how to prove if and only if if we delete the trivial loop. So, then it gets more difficult to prove things. So, this was a question about the setup and about the filtration. If you have both handles and boundary components, then it's better to assign to A and B cycles of a handle, filtration degree one, and to boundary components, filtration degree two. So, this would make a natural filtration from which one obtains good objects. So, that's a tricky part. So, if you have both handles, then the filtration is somewhat more complicated. Can you describe the definition of denta using framing? Definition of delta using framing. Let's say, maybe in a private discussion. Okay. And about... I didn't get the status of the last... Is it conjecture or theorem or... Is this one prime to prime? So, okay, there are two statements. So, one... Let's say, there are two theorems. So, theorem one says that one prime and two prime imply the isomorphism. And the second theorem says that one prime and two prime admit solutions. And the solution, does it depends on complex structure on elliptic curve? Yeah. So, here what we do. We steal formulas from Benjamin and Rickis in his theory of elliptic associators. And so, I think probably... Okay, let's say the hope is that from any elliptic associator you can construct a solution and in this case there would be a parameter of the elliptic curve. But that we don't know at the moment how to prove. We prove it with some particularly simple formulas which probably mean that this parameter is sent somewhere to infinity or to one of the... To one of the special points. So, we don't know. Hopefully, there is a whole family of solutions which depend on a complex structure, but... Just a question related to this one. But the definition of graded, it should depend on the complex structure. No. No. Yeah. So, both topologies. So, then probably the RAM should be removed because the RAM does depend. All right. As you say. Pleasure at the identity. It's localized. Is it possible? Yeah, so the question, you said that somehow you can now extend it to arbitrary surface, but does it mean that you have some kind of compatibility between these two things? Let's say... I mean, they're both highly non-unique, right? So... All compatibility... Between sort of genus zero and genus one. I mean, you said there was some statement genus zero and some statement genus one. So, to extend it to arbitrary genus, do you need some compatibility between the two? Yeah, sure. You need some kind of... Yeah, okay. So, you need compatibility. For instance, I think you need the overflow functions to be the same. If you want to... Is it enough? Because, I mean, if you fix the overflow function... Probably. Yeah, because capital F will probably still be non-unique, right? Let's say the answer is known, but whether it's known to me right now, like, kind of, that's my guess. That's enough to manage the overflow functions. But otherwise, yes, I think... Maybe another thing to say. So, we're producing both solutions from dream-filled associators. And one can, for instance, use the same associator to produce both solutions. That would be much stronger compatibility, but I don't think it's needed for gluing. Any other question? Is there a question? Sir, is it got my bracket related to the new sheets of several constructions in a... You know, when they look at Moffittis, can they end up with curves and resolving this... It looks a little bit similar. Could be. I mean, if you start from... At the picture with the boat hoop and the Heiko algebra, the quotient of the boat hoop, someone doing whether the structure there is got my bracket. Someone, there's some sort of kind of dimension quotient which would be those sheets of Barstabou algebra. Sorry, I don't know. Yeah, why would you... One of your field theories that might make sense to have those two constructions... No, maybe you're right, but I don't know. Any other question or comments? So if not, so thanks to Anton.


So, the key images are your handle of Malleus, handle of Malleus and long process of Incus, neck of Malleus and the lenticular process, Incudo-Steptial Joint. This is the Malleu-Incudo Joint, this particular gap and the classic ice cream cone appearance.  You zoom it on one side, you go through the scroll through it, you will be able to see it every time. So, that is good enough for you to see the ocular chain is intact. Let's go a little bit into the middle here and then try and appreciate few more little bit of revision. So, we know that long process of Incus is coming here, this is the body of Incus, short process is going posteriorly into that, here is the long process of Incus and then it comes and bends a little bit and ends in a small tiny nodular area called the lentiform nodule also known as lentiform process. And then it articulates with the stapes, the stapes, so that forms the Incudo-Steptial Joint. So, the stapes has got a few, there is a head of stapes, there is a neck of stapes and there is an anterior crust and a posterior crust, so together, together this is called stapes superstructure. So, depending upon the article you read, suddenly they will say stapes superstructure. You don't know what it is, it's basically everything you see there and the foot plate is different, that is why they want to differentiate that from the stapes superstructure. And then comes the foot plate which articulates with oval window and there is an annular membrane there and that forms the steppidio vestibular joint. So, now comes to the middle ear blind areas which means the otoscopy, the surgeon will not be able to see the two important recesses which we need to comment on in our every report. The one is the sinus tympani, sinus tympani is located here on the posterior wall, so you look at this what is called the W shaped configuration of the posterior wall at the same level as the stapes. You will see this particular recess is called the sinus tympani, sometimes the ENT surgeon write it as ST colistotoma, that means they wanted to exclude a sinus tympani colistotoma. So, basically you want to look for an ocular chain, a soft tissue and a soft tissue which is hiding there. If you think colistotoma due to other reasons, any soft tissue here has to be colistotoma unless proven otherwise and then you get this top, the pyramid like projection is called the pyramidal eminence and then you got a little recess here is the facial recess because there is just posterior to it is the descending mastoid part of the facial now. So facial now, pyramidal recess. The pyramidal eminence or the facial recess, these areas are quite important because the pyramidal eminence actually has got a tunnel, inside the tunnel there is a steppidious muscle and this muscle actually gives rise to a little tendon which is very very fine hair like tendon which comes out of the tip of the pyramidal eminence and attaches to the neck of the steppies. So that is the steppidious muscle which comes there and there is a tensor tympani muscle comes in this area. If you can able to appreciate my imagination that there is a little bit is going there a small tendon fiber, I just widen the window if you can appreciate it better now maybe you can see that there is a little bit of thin tendon fiber which is going there and attaching to the neck of the malleus this is the tensor tympani and anteriorly you see the tensor tympani canal which is filled with the muscle at innovation which is little bit above the level of the corroded canal and you can see this is the tensor tympani canal and this comes out here and turns 90 degrees and goes and attaches to the to this neck of the malleus so there is a steppidious you can see this thin vis going anteriorly this is the steppidious tendon and this is the tensor tympani tendon but the blind areas are here this is the anterior epitmpanic recess that is another blind area and this is the sinus tympani.