 I will talk about non-commutative Riemann surfaces. So originally, there were two talks planned non-commutative Riemann surfaces, one and two. But Joachim Anlind has been talking about mineral surfaces. So I'm alone with this talk. And I chose the following subjects. Of course, it's a very broad subject. And I picked some of the subjects maybe in the intersection of the main themes of this workshop. First part will be about hypersurface motions and time-harmonic flows. So this is, of course, a very old paper of Jens and mine. But recently we picked it up because there is some very amusing problems in electrostatics, you will see. And which were. OK, the second part I will be talking about surfaces in R3 mainly and how to make them non-commutative. And basically, this will be centered around the third topic, the C-algebra, which was joint work with Joachim Anlind, Jens Hoppe, and Shimada. And as also Joachim has developed this further recently. Right. OK, for the first topic, the ingredients are the following. You start with a, so it's sorry for my notations. These are notations from the paper. And this is a Riemannian manifold of dimension m plus 1. And sigma will be just a compact oriented manifold dimension m with volume form rho. And we will be talking about embeddings of the surface into n embedding. And with such an embedding, you get the outward normal. And you can pull back the metric of the ambient space to the surface. So this is usually called GRS. And then you get the volume form of this. So in the formulas, I will write something symbolically square root of G. So this is the volume on S. And then, of course, what will be there very often is the quotient of this induced volume form, which depends on the embedding. And this is the fiducial volume form. And this is a positive function. So a quotient of the volume form on sigma. OK. So to just make a picture, you have this embedding space. There you have, say, sigma as a hyper surface, co-dimension 1, with the normal field. And now you want to consider a motion of the surface. So along very well known, along the normal field. So you want the velocity vectors along the normal field. And the strength of the velocity is now, this you have to choose, is a function of this quotient of the induced volume. So there is the following differential equation, dx over dt is equal to alpha of square root of G, rho, so times the normal. So this is the basic. And I forgot to say that alpha is just a positive function defined on an interval and with non-vanishing derivative. Right. So this is the equation we had considered. And now if you want to solve it, it's generally impossible. It's highly nonlinear. The normal expressed in the derivatives of x is multi-linear in general. And so but why is this interesting? There are particular cases. So if you have, say, one case I will not discuss is alpha constant. So this is just, then you can solve it. Then you take the geodesic flow, normal, and then the surface moving. So this is some iconal, leads to some iconal equation. Then the important things for some physics are this. So this is the Lorentzian membrane. And alpha s in some gauge, or euclidean membrane. And for us, very important is the case alpha s, simple case equals to s. So you can see this as a large volume limit of this. Something special seems to happen when the volume form is equal to square root of G. When s becomes 1, these cases become degenerate somehow. What's happening there? Yeah, I don't know. I mean, this thing depends on x. So it depends on time. So this is changing all the time. But you could choose rho to be? No, no, but it depends on time. Rho is fixed. All time, I see. Yeah, so that's OK. And this has been an old idea. I think we have not been the ones who invented this. So at least when I was in the physics department, my colleague Klaus Kiefer, who's in general a relativity, told me about it, and also Dominico Giulini. So sometimes when you have hypersurface motions, you can pass to a time function. So of course, this is not always possible. But you can imagine for very small time intervals that the surfaces which emerge there will give you a foliation of n in the neighborhood of the initial surface. And then so you suppose that if you have, say, if you're x, the solution. It's a different time function of this t. No, time of x on m. There are two time functions. No, no, no, time is time, but time functions. OK. Wait, wait, wait, wait, wait. So the time variable, here it's a parameter, which exists for some interval, we hope. And here the time function, I will abbreviate by tau. tau, sorry. So this goes into some open subsets of, say, the initial surface. And you suppose that this is a diffeomorphism. And then you can invert this. So the inverse has two components. And the first component is the time function. OK. And now the sport is to express this equation by, at least by an equation for the time function. And this works. So there's a theorem. And in particular cases, I think it was known before. So I'm quoting the paper with Jens, 98. And the first part is, given x satisfying this equation, and this, say, this condition, I don't know. And then what do we get for the time function? So they are quite obvious things that this is just gradient with respect to this embedding metric over the norm in the usual sense. And you can also prove that this is 1 over alpha of square root g over rho. This you get by just using that this is an inverse of this. You take the derivative and you have two equations from the derivative, which give you this. But then if you derive this equation with respect to the spatial variables, you get a second order equation for tau. And this is the following. So this is Laplacian with respect to the metric. Everything is respect to the metric in embedding space. Plus another function, beta tilde of tau minus 1. Then you get the second covariant derivative of tau evaluated. And this has to be 0. So this is a nonlinear PDE for second order PDE for the time function. OK, that's the first part. I have two functions because we have ratio of alpha divided by this square root of g divided by rho, which is function. Ah, yeah, and the beta, yeah, the beta, sorry, I forgot. Beta tilde is, of course, a function of alpha. So it's minus z d z of log of the inverse function of alpha. So it's positive, and the derivative is nonzero. Supposed, and we suppose it's invertible, of 1 over z. So this is the definition of beta tilde. OK, so will I have a chance to get it back? Anyway, OK, so and there's a converse in some sense. So when alpha equal s, this is beta tilde is 1. Yes. L'école Francaise, it's always correct. Yeah, that's right. And that's the particular case I would thank you very much for talking about. So then this equation becomes linear. It's just harmonic, that if you suppose that the gradient is nonzero, then it's the harmonic equation. So the second part is that, so I mean, if tau is a solution of this nasty equation, then, and we suppose that, and suppose, this you have to suppose that for all t in a time interval. So tau minus t is difumofic to sigma. This you have to suppose if you do this. Then finally, you take the flow of gradient t over gradient t squared and x of phi t of x0. So you fix it with the fumorphism of phi. So the phi are the very same points here. Is a solution of the first equation. OK, here to make it work, it's a bit more complicated. This is just computation. You use Moses lemma at a certain point because you integrate over one surface and you compare the two for two different volume forms. You compare the integrals. And then if they differ, you just multiply one by a positive constant. And then you know when the integrals are equal, then the volume forms are difumofic. This is Moses lemma. And this we use to get this one. OK, so this is just, anyway, in general, it doesn't help much because you have one nonlinear equation, another one linear equation. But if you now take the case alpha, particular case, alpha of s is equal to s, if T go has already said, then beta tilde of s is equal to 1. And you get just that this time function is harmonic. Now we have a linear equation. And now you can, for certain situations, explicitly solve it. And right, let's adapt just to get some amusing results. We go into N to R3 because I want to be talking about Riemann surfaces. And so this boils down to elementary electrostatics. First year, second year physics. So how do you get these tors? So one of the, for instance, if you define these tors by charges, if you have a point charge and you're looking for equipotential surfaces, charges and equipotential surfaces. So if you have a point charge, you get spheres, concentric spheres. If you have an infinite wire, you get a cylinder. And then this is one exercise, because of this exercise in the physics teaching in Freiburg. If you have a uniformly charged line element, then you can also compute explicitly the tau and you get confocal families of ellipsoids. And I give you the, so if these are vectors A and B, then I give you the formula, which I owe to Jens. I have to say why we had another formula, which was a bit more complicated. But this is quite simple. So you take a uniform charge Q. So this is just the distance, which we suppose is, say, non-zero. And then you have a logarithm of x minus A plus x minus B plus A minus B over x minus A plus x minus B minus A minus B. Yeah? OK, so explicit formula. Now, suppose you have not only one line element, and that's what we, actually this was a conjecture in our paper some more than 20 years ago, and we were not able to prove it. And suppose you have now, say, just a bunch of line elements like this with charges, OK? So what is the solution? Very simple. You have just, say, over all the edges, you have the sum like this, yeah? And now the crucial question comes about what would be the actual potential surfaces here. And of course, in general, it's hard to say. I mean, if you have this thing, if you go far away, it will be some distorted sphere. And what every physicist would immediately say that it would be trivial to see that if you go very near to it, you would have a Riemann surface of genus 2, yeah? This seems very plausible. And of course, yeah, that's the. And in fact, I tried to prove this two years ago, and then I asked a colleague in Moulouse who did potential theories or true analysts. These people don't believe in equations or in inequalities. And so we were able to prove that this particular thing is true, that the near surfaces are truly Riemann surfaces of the genus you want. So more precisely, so that's a common project with Jens. That's Nicolas Chevalier. So what you pick, you take, say, a graph made of a graph in R3 made up of line elements, and you distribute on every line element a constant charge. But you want this to be such that there are several connected components. And on each connected components, the charges have all the same sign. So it's either positive or negative. And so this is a simple case. So it even works for curves and for some varying charge distributions, but you have to be careful. They are counter examples. And then there is an open neighborhood of this graph, so such that for all points in V, the gradient, and this is the difficult thing. So the electric field, if you like, is you have a lower bound, which is the distance from the graph. And this is, well, perhaps you find a nice argument, but it is easy to bound the electrostatic potential, but it's more difficult to bound the electric field. Of course, the electrostatic potential, if you approach a charged wire, it will go up. But it's a different point of sight. Yeah, sorry. I thought of a positively charged, right. But it will increase or decrease, but you don't know whether there may be settle points like this. And as a matter of fact, you can, if you have some diabolically chosen charge distribution, so it's not uniformly charged, then you can produce a number of points where the electric field is always 0. So it's a bit more, I mean, you have to do a lot of estimates. And the near surfaces, so there is, depends on this neighborhood, of course, near surfaces are Riemann surfaces. And the genus, so if you just say, have something like this. So the genus is equal to the H1 of the graph. You don't have the same sign of charge. Here you have positive and negative. And negative. Yeah, because you want to think of a time function. I want to have surfaces at time equals minus infinity. So it will concentrate near one part and negative part. That's right. I mean, this should, in the end, I mean, the picture is, you have some surfaces for t equals minus infinity. And they develop into surfaces during the time in plus infinity. This is, and I want to have, that's why we took line charges. If you take more extended objects, then you don't get to minus infinity. The potential will be finite. So something like this, yeah? And so the genus is the first homology of the graph. And this is 1 plus the number of edges minus the number of vertices. How do you mean, I think you have to have this minus equal to gamma minus and gamma plus infinity. Because you want to say the potential is equal to the negative number to concentrate near negative part of the graph. Yeah, yeah, yeah, yeah, of course. If the number is, OK, I should have gamma plus minus infinity. If your time is, I mean, I forgot to say, if there exists tau plus, sorry, tau minus negative tau plus positive, and for all values t smaller than this, then you get the Riemann surfaces here. And for each tau plus or tau minus t is surfaces around gamma minus. And if t is smaller than t minus, et cetera, yeah? OK, right. So this was, I found this rather amusing, because I mean, this is an explicitly solvable model. You get just a sum of these functions. So if these quotients are integers, say, you can even put it into the logarithm and you get a product of rational functions with some integer coefficients. Right, and here you can study, so it's explicit, and you can study topology change, because you can give the initial, or say the primordial, the big bank topology and the big crunch topology. You can give it in advance, and then just the time function will evolve, OK? And just a remark to finish this part. So you may wonder how does the topology change is going on, because on the way it's blowing up, it's getting perhaps simpler, and then it's going down again. And there's a classical theorem. So I'm not a specialist in this, and I found very little literature on the internet. But in an old book by Kellogg, Potential Theory, it's, I think, from the 20s. The style needs to get used to it. But you can infer there just at most a countable number, countable number of singular surfaces, OK? So this is already good news. With a possible accumulation, accumulation at t equals 0. And before you ask, the number of charges is equal to 0. If you don't do this, then you would have thought that everything is compact. So that's a finite graph. So this means very far away, you would just, if there was a charge, you would have a sphere, which would come as an equi-potential surfaces running out of infinity. And if you don't like this, I mean, this sphere, you have to make sure that the number of charges is 0. So if you go far away, there's nothing. The only non-compact surface is just t equals 0. Right, OK. So you have a number, a countable number, of singular surfaces, and the only accumulation point is at possibly t equals 0. I think it would be nice to prove that this is finite. In case you have these line segments, for some, but this should be an algebraic question. OK, so this was the first part. Next, I'm just talking about some. So Mark, this theorem in Kellogg refers to equi-potential surfaces. Yeah, yeah. Yeah, yeah, yeah. I mean, it uses that equi-potential surfaces outside the charges are analytic, and you can complexify. And there are theorems from the beginning of the century, in the 20th century, where you have only a finite number of components. And so it's funnily expressed, but correct, as far as I've seen. OK, now let the second part. So just let me, of course, if your Riemann surfaces is a very, very broad thing, and of course, you think about all these pictures, and so g is 0. And then the covering spaces are just, say, c. And here you have the upper half plane. Now, in general, these two surfaces are very well understood. You know everything. You can express everything. You can make everything explicit. Whereas here, the upper half plane is also rather nice, but it's non-compact. And to go down, you need these Fuchsian groups. And just, I mean, I'm not a specialist at all, and I was just looking through the internet. So it is very hard to find explicit matrix representation of these Fuchsian groups. I mean, you get, of course, the presentation, and this is classical in generators and relations, but say, in SL2, what are matrices which belong to some Fuchsian groups? The only place I've found is a paper by Musket. I'm just quoting it in 1999, where he gives for genus 2 four matrices, which have apparently the good properties to belong to a Fuchsian group, which produces a surface of genus 2. So I mean, you can work with this, prove theorems, but an explicit thing you wouldn't get. So that's why we chose to look for surfaces in R3, because it's quite explicit. And so we fix, say, a function R3R, and we want to define the surface as just the inverse image of 0, so 0 a regular value. And then you can express everything. The whole geometry of sigma, you can express in terms of C. So you get tangent spaces. They are just the orthogonal space of the gradient. And you get an induced metric, rather scale up what you get. This has a volume form. You get even complex structure, which is, say, the normal, which is also gradient cross v. And so they are, of course, Kehler. What you don't get easily is the complex coordinates. There you have to go to isothermal coordinates, and then there you have to solve a PDE to do it. And they are not so explicit. But the complex structure as a tensor field is very simple, of course. OK, this is, of course, well known. And what I wanted to advocate is there is you can take, if you have a symplectic form, you have a Poisson bracket. And the Poisson bracket, as Joachim has already told you, does describe all the geometry. And but still, the Poisson bracket has a factor 1 over the norm of the gradient of C, so it may be a bit more difficult. So what you may do to take a Poisson bracket in R3 and of the following form, so two functions of three arguments is just like this, and these are called gradient Poisson structures. So anti-symmetry is clear, and Jacobi identity you have to compute. And they have the following nice property that C is a Kazimierz function. So C commutes with everything. OK, and this is very important. So if you just take some algebraic intermezzo, if you have C infinity of R3, and I is just C A, so it's the vanishing ideal. So C times A is equal to all the functions vanishing on the surface, that was a regular value. And then you see that C infinity of sigma is isomorphic to A mod I. And so this Poisson structure goes down to the surface. Very simple because I mean, this is a Poisson ideal, if you like. OK, this is a Poisson manifold, right. And no. And so I will come to this in the third point. So Jens' idea was to use this Poisson bracket in R3, which is quite, I mean, it's rather explicit. But before, let me give you some nice functions, examples to produce surfaces. And one class of functions is of the following kind. So it's 1 half Z squared plus Y squared plus pi of P of X squared minus C over 2. So C is a positive constant. OK, so P is a polynomial, and it's given in the form a constant alpha, X squared minus 1, X squared minus G squared. So suppose that G is 1 minus square root of C. And alpha is a constant which has to be positive and smaller equal than the maximum of the polynomial T minus 1, T minus G squared for, OK. So if you take these data, then C minus 1 over 0 is a Riemann surface of G. And you can visualize this in a following way. So if this is X and Y, and you have Z like this, so you have 1, 2, minus 1, minus 2, et cetera. So the surface is like this for G equals 2. And here's Z. So here are the holes. The holes are situated at the, if you cut the surface with the X axis, you get exactly these integer points. And the proof is by Morse theory. So you have this function, the X function, and you use the Morse function. And then you just compute the second derivative of this function and you get the result. So here you get the usual definite determinant, positive determinant and negative determinant. Right. So this is, and this will be the functions which we will be working. But just for fun, let me give you perhaps other functions which we have tried a little bit. They are simpler. But for the moment, we haven't seen whether this will lead us to nice formulas for the higher genus case. OK. So yeah, so this is quite amusing. So you can take the following functions, c of x, y, z, is equal to z square minus cosine x minus cosine y. So it's nicely separated. So one function per coordinate. And this is doubly periodic. Now it's winter, but if you go to the beach, I mean, you get these kind of matrices. So the function looks like two planes. So the 0 is always 0 is a regular value. So you have two planes. So this is the z direction. Infinite planes. And they are connected by a regular array of wormholes. Can you imagine this? Et cetera. And so this is the surface. So it's like these matrices you use on the beach. OK. And now, of course, this is not non-compact. It's too periodic. But what you do is you take GAB, you take this abelian group, A and B are integers, strictly positive integers. And they are acting on the space in the x, y direction. And if you call this surface sigma tilde, then you can mod out this group. And then you can imagine you get some kind of a closes. And this will be a Riemann surface. And you even can compute the genus of this. And this is, what was it, AB plus 1. And of course, this sigma here is no longer an R3, but an R cross T2. OK. And I say it again, so here the function is quite simple. And you can make it even, you can take the obvious function cos x plus cos y plus cos z, 0 is a regular value. And how do you get the sigma tilde, sir? Sigma tilde is just this equals to 0. This is a surface. So you prove that the gradient is non-zero on the slope. And the lower surface is the usual plane. No, no, no, no, this is connected. It was just for drawing. I mean, this is, you get at certain points, you get these are like wormholes. But it's a regular array like this. And of course, a bit rounder just to get the principle. Here, if you take this, and this is, I think, a well-known surface. I've saw it later in the book. So if you take a cube like this, then you get a sort of legal thing, which is like this. And there you connect like this, like this. I hope you can somehow imagine how the surface looks like. And then, of course, you build them together in an infinite array. And then, if you divide by the obvious group A, B, C, then you will get a compact surface in T3, OK? And if you're dividing by this, you may not get compact. No, no, you get compact here. Sure. No, no, it's fine at A times C plus B times C. It's an infinite group. It's an infinite group. So you have R3 modulo this group is R cross T2. And he'll get T3. And the genus formula is amusing. It's just like what was it? 3 A, B, C minus A, B minus B, C minus C A plus A plus B plus C. So if C is equal to 1, say, this is just the genus is 2AB plus 1. You see this if you take just one block. So A equals B equals C. Then if you identify, you get one handle like this, one handle like this, and one handle like this. These are three handles. And so the genus is equal to three. Right, so yeah, this is still open. We try to work with it. Now, non-commutativity, so this is just the surface. Yeah, I wanted to say something about star products. So this is, of course, a usual technique to work also in non-commutative field theory, to use star products. And OK, so in general, if you have a Poisson manifold, so PIG, the Poisson bracket, then a star product is written on two functions of x. It's a series of bi-differential operators. And you demand that this should be associative, very difficult. And mu 0 is just the ordinary function, product of function. And the first order commutator is equal to, now we are, say, i times the Poisson bracket. But you can also choose it equal to 1, non-zero. OK, so and yeah, of course, you have a beautiful, general existence and classification theory of these things, thanks to Maxim Koncivich. And you can express star products in terms of graphs, where you need weights, which are integrals, configuration spaces. And so there's this famous pre-print. And the thing is, so everything is done, it exists. And you also know this equivalence problem. If you have two star products, they are called equivalent, if you have a differential operator series like this. Is this a formal power series? Oh, yes. Yes, yes, yes, that's formal power series. And in general, never converging. You have Borel's classical theorem that every formal power series can be made equal to the Taylor series of a function, smooth function with compact support. And so you can hope that this converges on sub-algebra sets, but in general. Yeah, right. So explicit formulas is difficult. In general, you have only very little, so there's vial, which is, say, e to the pij, if p is constant, pi. So this, you can write it like this. And the same works on the torus, so vial's product. You get some, yeah, yeah, of course, of course. Yeah, yeah, pi, constant, constant. If not, that would be, yeah. And you get some others. So I will not write them down. For instance, you get a formula for CPN to advocate one of my early works. So with Claude M. Reich der Waldmann in 1996. So there, you get a really an explicit formula for CPN. So in particular, for the sphere, and it coincides, of course, with all the formulas before. So for S2, you have a lot of things. You have the fuzzy sphere. You have Jens's thesis. And also for Harald, this also works for the CP1n. So you get the same formula. OK, also the co-tension bundle of the Jens's sphere is also right. Now, what does it help us for this case? And so there are a lot of attempts, but they don't actually help a lot. So, of course, if you take the surface directly, so it's a two-dimensional manifold, and it's symplectic. And there you have techniques like you get a symplectic connection. You take the Kehler connection. And then it's expressible in terms of the curvature tensor. But to my best knowledge, I don't know whether you get really something explicit by this. And also you have the problem to describe the surface in some sense. So then you can do it just order by order because the third Hochschild-Comology vanishes here. You just say I have my favorite term of n until order n. It's associative until order n. So it always works. And if you look at R3 with these gradient Poisson structures, then I'd like to quote a very old result of one of my colleagues in Freiburg. So this was his PhD thesis in 1997. And he proved that order by order works, which is already amazing because we are in R3. So the third Hochschild-Comology is not zero, but there also can be a non-vanishing term. And he shows by an ingenious argument. So he constructs in the Concivich language multidifferential or polydifferential operators, which can be ordered in such a sense that the derivatives are only going down. So you don't have circles going up. And he proves by induction that finally the order by order argument works. So there are no obstructions. So the statement is for any what c infinity function tau you can construct. Yeah, yeah, yeah. Just to know what is the statement. That's the statement. It is ct, for instance? No, no, no, nothing. It's just, I mean, c infinity, yeah. You need derivatives of it, yeah. And but if you do the Concivich star product, it has the benefit that you get an equivalent star product such that c star commutes with everything, yeah. This was, I think, stated by Katanio Felder. Correct me if I'm 2099. At least I found why they. And the construction is simple. You get your formality map. So this is the equivalence transformation. I think because it's Horschel, how much I've identified out of the position, I think it's central. Yeah, yeah, yeah. What I don't know, so the Poisson center will be in the center of the star product. Perhaps also vice versa, but I don't know. This I'm, but yeah, I mean, this will be if this is your formality map and you can just. The center HH is co-optional, zero co-optional control. It's in the same, it's HH0 structure. Yeah, yeah, OK. And so, and I don't know how to do it just without these techniques here. And then naively, if you just compute the first orders, it's well. It doesn't seem to be like this. But so what is the concept H3? So what is C? You mean C? C is always this function which defines your surface. Yeah? So it's two different use of the usage of this C. Yes, guys. So what are the two different statements? Both of them are a C-based packet. A C-based bracket and C-scasinov, yeah? Yeah, a C is that this is, I'm only talking about this bracket, the gradient bracket. Two different statements. No, this is about the same, this R3 with this star product. Yeah, just a remark, this can be quantized just as it, as you don't need the conceivage techniques to do it. But if you use the conceivage techniques and additional bonuses, you get C as a quantum casimir. Yeah, that's the statement. OK, right. And this was, yeah, I wanted to say about, yeah. And now, very briefly, unfortunately, about these C algebras. Yeah, so coming back to this Poisson bracket, then C is not the same C. I think if I'm not completely mistaken, then this talk, there are two C's. One with a back, which refers to the complex numbers. And the other C, if I don't have, didn't make a mistake, should always be this function of three variables in R3. Including for this C algeba. Yeah, OK. So are these polynomials still on the blackboard? Yes, yeah. So if you take these polynomials, so you get some, you can compute the Poisson brackets of the three variables. So this is equal to z. And here you get something. You can compute this. OK, right. And so Jens Ansatz to quantize is, I mean, he says, OK, let's just replace all the x, y, z by matrices and impose these things as commutation relations. So we have, say, x, y is equal to i h bar z, et cetera. Et cetera is a bit cheating, because you may have ordering problems, and you will have ordering problems in general. OK, but the first thing is this commutator allows you, in some sense, to eliminate z from, so I will end C equals 0. So what this means more precisely, you take, say, the free algebra in the non-commutative free algebra in three variables, OK, tensor algebra, if you like, and you mod out these relations plus C equals 0. So and then, and find representations. Finite dimensional representations. So this is, of course, it is dared, in some sense, because you just take the Poisson brackets and impose them as the commutators. Coming from star products, I would say you would have to add higher orders to make it more consistent. But OK, so we tried it. And this can be said that it works very well for the genus 0 and genus 1 case. And so you get representations. And so the theorem is so form. And more precisely, yeah. So we took the following C, x square plus y square minus mu square plus 1 half C square minus 1 half C, OK. And for mu, if you set this equal to 0, this is a torus. And mu, it's a sphere. So you have the picture, say, a distorted sphere. And then you get a torus, OK. And so I'm basically, well, when I... At the end of the talk, just take two or three more minutes. Right. So let me perhaps just write down the relations. You can reformulate everything. So you have the matrix w, v is. And you get just cubic relations. So I write just one. The other one is just a Hermitian conjugate. Right. So the problem is try to find matrices which obey this. And this works. And I just give you the form of the matrices. And so the representation of w is equal to 0 and square root of e tilde 1, e tilde n minus 1, and e tilde n for, yeah. And then you get conditions you can express as like e tilde l is c mu over square root of c plus cosine 2l plus beta. So cosine theta. e to the e theta is q, q n minus 1. Yeah, basically you have explicit constructions. This is for c, square root of c. This is the torus case. And for this field, you get similar matrices. OK. Right. And now the last thing. And this was quite, if I have just one minute. So the comparison with Bérezine-Tüplet's quantization works. No, this means, in particular, for the torus and sphere case, you can parametrize these surfaces. You can actually get a map from the torus, from the usually with two angles of the sphere, round sphere, into these surfaces. And then you can do the whole Kehler quantization Bérezine-Tüplet's business on it and compare with the matrices you would get for x and y. And this you can compute. And you get for the torus almost a coincidence. And for the sphere, I think this also is due to the fact that the sphere is not really round but has this distortion. Then you get, say, an asymptotic for when n goes to infinity coincidence. Yeah. So I'm sorry, Bérezine-Tüplet's one could say some other words. But I mean, this is OK. Yeah, thank you very much for your attention. Thanks very much for the talk, Martin. I just want to have some questions or comments. So this part, I guess, with the higher-genius surfaces, is sort of seek you told us in the quest of, I mean, the open problem. That's an open problem. The open problem is to find explicitly non-commutative examples or non-commutative surfaces of higher genus, right? Yeah, but this is a nice suggestion of you. What one could try. And I forgot, of course, to mention recent work by Joachim on higher genus, where he found matrices 2 by 2 and 3 by 3 matrices for each genus. And there is another treatment, but which I didn't understand by Andreas Sükova. Before I make more comments, more questions. Other comments. Yeah, no, other comments. It's better from, I also have one or two more comments. By the way, Maxine, first, I have two comments. Yeah, about these functions of three variables. People in algebra play, yeah, we can make some polynomials in three variables, which is a very, very easy type. Yeah, like x square plus x to the power n plus y square plus z square for n type, and e, you get all three variables. Really, but this we can, people play with it and is my memory sort of, well, kind of understand how to quantize it pretty explicitly. And you should do like potential in three algevins, three variables, some cyclic word, which is x, y, z minus z, y, x, is possible for commutativity and plus a little correction for potential. And then you can see the critical points on matrices will be exactly fundamental representations. Yeah, so you're describing matrices, your algevins, like critical points for non commutative potential. Yeah, it's called a be-alge, it's called. Okay, but there's no business about this, it can go off, and yeah, this is, sorry. But what do they get for, what kind of surfaces do they get? No, they can see the some examples. They have singularities, I guess. Yeah, they're zero at a single point, but near what value, that's an interesting level sets. And also played a little bit with the same version of the three variables, and like, it's in some three squares. And when they're interesting, then we have finite dimensional representation depending on each bar, but the dimensions can be fixed. Then it means that it looks that one can make a conjecture that if you complexify your surface, and consider periods of two form divided by each bar, you get half integers. Yeah, so it could be open curves space, so it seems to be really nice quantization parameter conditions one can see from commulger of periods of two form onto complexification of one set. Exactly new singularities, it's interesting. Okay, that's good. Thanks, Martin. Questions, comments? Douglas is writing his hand, yeah? Is there something like a lower bound for the matrix dimension in terms of the genus, or can you find the object? If you are doing, I mean, I don't know the formula by heart, but of course if you do Bayesian-Turklitz quantization, then the dimension of the holomorphic section space is given by the Riemann-Roch-Helzebruch formula. So this is given for the sphere and the torus. Of course, one knows the dimensions very well. It's m plus one for the sphere and m for the torus. For higher genus, I have forgotten, but it's known you have to take this bundle of holomorphic one form. And it's tens of powers and there's a formula, but which I don't know by heart. Well, the reference you mentioned there, you find two-dimension representation for arbitrary genus, so there's obviously no lower bound. Oh. Yeah. Ah, you mean in your paper? Yeah. Okay. Okay, so I don't know. Yeah. Small questions, comments? What happens if you're considering non-financially generated in the interfaces? Like if you're taking this function in the co-science and you're not taking the quotient like this group J, B. Do you have any idea if you can get also financial matrices? We haven't actually looked at this. And my guess is it's non-compact, so you may get, say, also if you do some Kehler thing, the holomorphic section space, I would say it's infinite dimensional. But I don't have a real feeling for this, so I don't know. Ah, so concerning the first part of the talk about the hyper surface motion, you consider the case of integrable motion, and especially this equation, most of you found the Lux-Pair case, they were reproduced in integrable. You mean for this time-harmonic flow there? For the motion of the surface. I don't know. May I answer? Yes, please. Yes. There's one. Yes, Lux-Pair, no, but infinitely many conserved quantities, yes. And it's an example of a diffeomorphism invariant field theory which is integrable. I mean, both by just writing the solutions in explicit form, and you have infinitely many conserved quantities, and it's quite interesting, it relates different manifolds with similar Rucci tensor, conformally related to each other. There's a lot. One comment I wanted to make is, it's related to NAMM equations. And so this is, I mean, the scene, space, time, matrices, in your talk, we will hear, we will see again the double commutator equations. And NAMM equations also, somehow, are around, and this star equation, for the case where alpha is a linear function, where the star equation, as it is, is a set of non-linear PDEs, are the N going to infinity limit of the NAMM equations. And that's somewhat interesting. Yeah, I forgot to say. This electrostatic model is, say, the GL infinity NAMM equation and R3 equation. Yeah, so in some sense integral, you can write down a lot of solutions. I guess we need coffee and some time for prepare for the next talk. And thanks very much, Martin, for telling us lots of things. And let's thank the speaker. Thank you.