Noncommutative Manifolds, Chern Characteristic & Homology

0:00 The module of the North between K and K-O, then, you have to find out that the order to go to the North between what is the ear of the North. What you do for the North is to keep the fundamental class in this K-O-N. Now, if you do something like K-O-Rooter-Key, and if you do the cycles of K-O-Rooter-Key, you have to get the class. And in order to be fixed, you want what is the cycle, the curve of the cycle. You need sort of the class, then what you find out is that it's exactly a matrix activity. And this gives you the notion of matrix . So what I was saying is the following. When I was talking about the Fresnel operator, I didn't want to be specific. It turned out that you have to do the right choice. And the right choice for the freedom operator is that it should be self-adjoint unbounded. So it should be self-adjoint unbounded in the operator D. So D is equal to this adjoint. And the freedom property is simply the following. is that if you take the resolvent of D, so the resolvent of D is compact. Okay, you have to know that compact, so, I mean, to certify you, you can say that D inverse is compact. Of course, in general, D is not invertible, but you put a deeper x and then D inverse. Okay, so compact really means small, in the sense that the operators which are compact but really are exactly the properties of anything basically You know, if you multiply a compact panel by a mountain panel, it will still be compact Ok, so now what happens in the following is that so what is an auto-community geometry which contains both the information that it's a manifold and that you have a ds so it is when it's called a spectrum of people so the algebra is a function of space So the algebra is the algebra function of the space. And what you have then is the algebra is represented completely in the space. And in this inner space you have an unbounded self-algebra operator, D. And the only property of D essentially is that it has bounded commutations with the algebra A. So dA is bounded for any A in the algebra.

2:30 A plus is compact. Now what is the simplest case? So of course you have to know that in the Riemann case you can go here. Now what is the passage if you want from the Riemann case to this situation? Well, the algebra is of course the algebra of states. with both functions and non-tri. The upper space is the upper space of two spinors. For this, I need to have k-orientation. But I mean, by passing to T for large amounts, you can get away with this, without a problem. And finding the operator of T is a direct plane. So T is a direct plane. Okay, so at first, it might look like a very strange way The first one is that it is very easy to recover the distances between points just from this operator of theorem formulation. I mean, the distance between two operating points x and y is just given the derivative distance. I mean, this is the derivative distance, just given by the supremo of f of x minus f of y. I mean, how a function can be pre-finite of an x and y, with the condition that the complication of d with f is smaller than 1. Now, what is the intuitive way of thinking about it? The intuitive way of thinking about it is that we are looking to offer ds. So what is ds? The formula for ds is the following. So let me contemplate this for a while and explain what I need I need to write this formula as ds equals this So let me explain this So first I don't have to explain this You see the operator of d has dimension which is inverse of the next as can be seen by this formula so when I look at the inverse because the result of the small part is to be a very small operator that's why it's an infinitesimal when I write this formula I refer to the notation by physicists of what they call the propagator of Fermi and I will come back to that notion so physicists have a notation for the inverse of the dirham operator

5:00 and they denote it by this and that's why I call it the infinitesimal line so now let me pause make you look very strange. Let me pause and it's not a joke, but you know, it's an explanation at a very intuitive level. Explanation at a very intuitive level is the following. You see, when people were asked at the end of the 18th century to give a proper definition of the unit of length, the meter. Okay, so I mean they were asked, so perhaps the slide was asked what should be the definition of the unit. And of course the first reaction was to give some astronomical definition of the meter. But they wanted to have something to do. So in order to have something to do, they said that the meter should be defined by measuring the diameter of the Earth. They went out very far, measured this, and eventually they got this meter, which was deposited in variance, which is a proportion. So, this is the definition of ds that would go here, where you would take general distances, you would take a path, and you would make measurements like this. But now, of course two centuries past, and physicists have a much better definition of the unit of length. Now you can look in the metrology as well, what is their definition of the meter length? And it's not the meter proportion of the diameter of the earth at all. What is it? It is exactly the wavelength which is corresponding to the transition between two states of the caesure. Now, if you think a little bit, you'll find out that it corresponds exactly to this definition, with the Dirac magnet. Because the Dirac magnetor corresponds to the one-particle at the communion of Hermes. And, I mean, as I mentioned, if you look at transitions, you get frequencies, not wavelengths. And when you invert them, you get appropriate wavelengths. So what I am proposing is the transition from this notion of geometric space to that notion of geometric space. I mean, there are several advantages. The first advantage is that it doesn't require the algebra to be computed by movements.

7:30 For instance, for the non-computative torus, which I mentioned before, there is a fairly straightforward operator of T and so on. The main thing, one problem with Riemann's definition is that we have only the square root. Dirac has solved the problem of the square root because Dirac found is Dirac's operator as the square root of Laplacian. Laplacian would correspond to D squared, but Dirac's operator corresponds to the square root. Choosing the square root is choosing the k-o-molity from the task. It's choosing the orientation of k-o-molity. And this is also essential. So what I am saying is that the two problems, what is manifold and what is linear, are a common answer. And this answer is given by the system. Now, among the many features, this answer is one of them that it fits exactly with what problem I was saying yesterday. It also embodies infinite dimension. I mean, if you consider the performance of 3D, and if you consider for B the operator, which is called the supercharged plane, and for A the algebra, which corresponds to open strings with a point marked in the middle, then you will find out that the axioms that I am going to mention are actually 50. So what is really doing, even in string theory, when one is talking about the extended ideal manifold, they do fit in this framework, they are in three dimensions. And what happens in infinite dimension is extremely interesting. You see, the usual notion of infinite dimension are very flat. But if you want to fit with this notion, you have to construct the universe space and the operability. And that is a very delicate analytical problem, which, for instance, in the case of conformal pencil, is still open. There are many examples where it can be done, but in general, it's quite a hard problem. It's like the problem of constructive fixer, if you really want to get it at this level Ok, so I didn't mention the nuance with K-O So what happens is that K-Omology, this is the description, if you want, of K-Omology cycles in a complex case when you call the ko, the nuance with ko is that what you need is a real structure on diverse space so what you need is a real structure and this real structure J, so what it would be to be an anti-unitary operator

10:00 so anti-unitary operator in diverse space H and what turns out is again that there is a coalescence of different concepts there is first a concept of what is called in physics the charge complication which comes from the fact that the spin of the rotation of spin n is actually is prompted to its contract gradient representation so there is a charge complication and there is another one which is the Tomita operator Tomita's operator j in the Serbian factors So in the theory of algebra at the level of pressure theory, maybe only when the algebra is represented in the first place, it turns out that there is a very general understanding of what happens in the relation between A and its commitment. And Tomita, I mean in a fairly general context, has known that there was a way to find the operator J, anti-linear image such that when you take J A J inverse okay we call this A opposite for instance then you will have the computation that this element will commute with the element space so it will satisfy that A opposite will be equal to 0 before any A B may be now in terms of the correct formulation of the conditions on the operator D and on this evolution of j, are exactly this condition. So, first, this condition, what I did tell you, it tells you that you not only have an orientation of a, but you also have an orientation of the opposite direction. You see? In fact, not only the algebra a is acting in inverse space, but in fact, a tends to a opposite direction, just because of the orientation of j. It's acting in h. Okay? and then what you have the main condition which is saying that you are dealing with a square root of the square that the square root is sort of linear is the order one condition and the order one condition is that D commutator A opposite commutator B is equal to 0 for any AB okay, I mean it's very easy to check is an operator of order 1. It is less easy to check them, for instance, in the example

12:30 that Graham was mentioning. In the example Graham was mentioning, the operator J is the operator which exchanges the two sides of the string, and it turns out to be anti-linear in a natural way to satisfy exactly the coin. Okay, so, and then once you have this, you have a replacement, because it's not the algebraic, but the algebraic, you have a replacement for Poincaré duality, using the Viber and Caspar-Ovster. So there is a new formula exactly in Poincaré duality, meaning the fact that you have Poincaré duality in K-1. Now, this I have not checked in the case of the Poincaré duality, but there is a natural element in K-0 which should be duality. All right, so this is the first point I want to make if you want to talk about it. The point is that passing, if you want to form, only is you want to be trans of the cycle to a specific representative of the cycle. Okay, it's an extremely important point because here it gives us a metric, it gives us a correct motion metric and we're going to do this linear range. Now, of course, we have, in general, we have the Chiang-Hak, which goes from case theory of A to the cyclic homology, so HP star with the lower star of A. So let me fix my notations here. So I want to consider this by means of the little g capital Debye complex. So I mean, I remember that as a chain complex, I mean, what you have are the tensor powers of A, A times A times A times L times, I mean for the chains. And, well in fact I want to consider reduced chains, so I consider A tilde, A tilde will be the quotient by the units. and then we have the two natural operators which allow to form a bi-complex the first is the Orchid form of the operator, so b is just the Orchid form

15:00 so of course b is such that b of a0 tensor up to an I mean this is a sum with alternate signs, I don't want to write it down, of A0 tensor and then you contract AJ, AJ plus 1 and in the last one you put the appropriate sign and you put ANA0 tensor up to AN-1 which means that you have a worksheet complex with coefficients in a certain bimodule, which is A itself and you have the other operator, capital B and capital B is as follows, so it's obtained by taking one tensor, so it's a sum and then you put alternate signs of course and you put sign in permutations so ai times ai plus one, you will always keep the scientific harmony of the elements okay, I put ai minus one and you put the appropriate sign which is minus one ai times perhaps n minus one, okay ok, these two operators on ticombute, of course SS5B equals minus B, SS5B, B squared equals zero, capital B squared equals zero and they define, so they define if you want a corresponding periodic analogy, of course there is filtration which defines a corresponding group HC7 of A So, of course, this H is again over R. I mean, there is a natural filtration of HB star, right? I just want to make one comment about your actual talk. So this is just a parenthesis. It's about the non-commulative torus. So, you see, when you take the non-commulative torus, this non-commulative torus is like an elliptic curve. And it turns out that there is an analogue of the Jacobi theorem, that you have the isomorphism between the elytic and the jacobian but in a very strange way, instead of involving the odd cobalt, it involves the given cobalt and it puts in this modulus theta so what happens is the following what happens is that you consider HP star with the lower star of the algebra okay, of an algebra now, because the algebra is an evolution, you can view it as a real vector space, not complex vector space you will find an R2. Now this HP star contains, naturally, the K-theory. Just because of the chunk factor. So the K-theory matches to it. So because the K-theory matches to it, you have a natural lattice, this square, and this R square.

17:30 So you can consider the quotient. The quotient of HP star by this lattice, by the K-theory. Now, because of the filtration of the cyclic homology, this torus now inherits a foliation. It inherits a foliation because the cyclic homology is filtered. And the theorem is that the algebra of the foliation is isomorphic to the original algebra. And moreover, the case theory classes, which are elements in this lattice, when you consider it as a transversal, which is associated to a case theory class, you recover exactly the case theory class. So it's perfectly identical to what happens for the Jacobian, but it's even. So over the same video that this filtration of the scientific oncology by the degree of the classes is extremely important. It is very similar to the filtration in the case of elliptic curves and the incompatibility of the urge filtration with the lattice, with the natural haldinger lattice. Okay, so anyway, so we have this y-complex and the formula for the Chang character is is that the nth component, if you want CHN of E, modulus of normalization, is just equal to E-1.5 So what is E? So E is, like the important, E squared equals E. And at first I will write the formula as if E was in the algebra. But in fact, E belongs not to the algebra, but to Q by Q matrices over the algebra, to have Q of A. And so what you have to do, you have to take the trace of this expression, then you have to take, if you want, Eij, or EI0I1, let me say, minus 1 alpha delta I0I1, then I have to take EI1I2, etc., up to EI and I0, and I have to sum, of course, over the indices. So it turns out that this is the cycle in the dv-by-complex and this is the formula for the Chiang character. The formula for the Chiang character in K-omology is much more subtle. So when you look at the Chiang character CH now going from K-omology of A to cyclic homology to HP star of A,

20:00 Then the formula is much more interesting because it will give you, in particular, the formula for the Pontiacian classes of the non-competitive space. Because what is contained in the variability for the non-competitive space are the information about the Pontiacian classes. So the formula is much more subtle than the local formula, which is the following, which is that you get a cyclic cocycle which is dual to zero both and this cyclic cocycle is designed by an explicit formula phi n of a0 an which is a sum with some rational number of coefficients cn k and then there is the following expression integral of a0 dA1 up to dA n okay so d is the operator that defines the metric or if you want to define the structure and there are some parentheses here k1 to kn okay and then there is like the ds or d inverse if you want ds to the power minus n minus 2 I don't want to return and claim this formula but this quantity is really the important one And this is the residue, which is a generalization of the Wojcicki residue, in the general case. So, I mean, the important part about this formula is that, well, it is locally computable. So what it gives you, it gives you a purely operator theoretic formulation of the local pontriac in class, in full general. When you take a manifold, it will give you the H in this, very simply, but I mean, as soon as you go to more delicate examples, the computations become triangles, but they give you local answers. And so I don't want to explain too much about this formula, except for one thing, which is that the residue which is here sort of stops, it vanishes above the dimension of the manifold. So the dimension is a very subtle concept. The dimension in this non-computative situation is the following. First of all, there is an upper bound, if you want. There is an upper bound of the dimension.

22:30 And the upper bound of the dimension is just governed by the growth of eigenvalues of the operator ds. So if you want to, what you do is you take the eigenvalue of the value of ds and it is a vehicle of n to some inverse power minus alpha and this inverse power covers a dimension If you take a manifold of dimension n, if you look at the inverse of the Dirac parameter then the eigenvalue will decrease, but it will decrease all the slower the bigger the dimension is So in fact, this number alpha turns out to be one over the dimension. But it only gives you an upper bound in the dimension. And the proper notion of dimension, when you analyze things, you find out that the dimension is in fact a spectrum. So in fact, in this theory, you find that dimension is a spectrum, and it's a subset of the complex numbers. So this subset of the complex numbers does admit, in general, an upper bound, which is related to Rastor's dimension. but in general it will admit other points in the case of ordinary manifolds these points will be integral points which are below the ordinary dimension but for instance if you take a cantorset the dimension will start by the Osdorff dimension but it will have complex points which are located on the imaginary line above this Osdorff dimension and these complex points will be related to the scaling ratio of the cantorset So the notion of dimension becomes much more interesting. Now what happens is the following is that when the number of terms here is above the dimension, then this formula vanishes. The corresponding term phi n vanishes. So what you get in this formula, you get in fact a cosycle, so if you want, phi n, family phi n, is a cosycle in the bi-complex. I mentioned just in one word that in the case of infinite dimensions there is still a cosycle and there is still an infinite cosycle in the complex and it doesn't need support

25:00 but here it has finite support Alright, so now what happens is that there is a case where there is a tremendous simplification so if you want the index formula tells you that the index of the operator T with coefficients in the band of E what is this? I mean this is just the index of the sudden random operator which is obtained by compressing the operator T by the projection E so the index of this operator is given by the pairing between the chain character of E in a cyclic homology so zero star of E of the operator D, if you want, in cyclical biology. Of course, as you know, this formula is the index formula, the index theorem in the ordinary case, but I mean it has much more power because it's applicable to all operator theoretic situations. Okay, for instance, . Now, this formula is simplified drastically in the situation. So if it happens that the lower-term classes, that the lower-term components, that the CHJ of E are zero, are zero as chains, for J smaller than the dimension, it might be there is a J over 2 it depends how you label the J so two J's the above formula simplifies then the above formula simplifies and you only have the last step so what you get if you want the index of D index of D to the integral, to this residue, of the chain characters, the n-th component So here it will simply give e to the minus one half times d commutator e to the power n and then you will have this d to the power minus n So let's take n even, n equals 2n

27:30 okay so I mean the first thing that you see then is that somehow in this situation where you have a k-theory class all of those chain classes, chain character classes, vanish below the dimension, then the formula simplifies the dimension. Now, the problem which I sort of asked myself is the following, is that somehow one should be able to formulate the first problem of classification, which is if you want to look for algebra which contain in their matrices an independent e we don't get to cubic matrices over a and which satisfies that the chain characters i mean chg of e is project zero for any j like zero one two and so okay and so okay i just and by formulating this program i thought i could find any competitive solutions okay i mean it's easy to see that there are solutions I will explain that. But why did I think that there would be only commutative solutions? Because somehow this is saying that the space is really looking like a sphere and the projection is like the bot generator. You know the bot generator is the bot generator of the k-0. It does, of course, because of the convergence of the sphere, it has no lower chain components. So I was trying to find spheres. So now, one can easily work with simple, very simple cases The first case one can look at, which I looked at Let me start with even simpler cases You can start with a one dimensional case And then what do you find? You find an operator u and then to find that the corresponding correlation for the operator V is U x D commutator V is equal to 1. Now, I ask you to do a formal exercise. I ask you to look for representation of this relation in linear space and prove that you will get the ordinary geometry of the theorem. Maybe what I claim is that this equation is the defining equation for the ordinary geometry of the theorem. The algebra is generated by u, the metric is defined by d, there is only one irreversible representation, defined in this relation, and you find the average of the other set.

30:00 But you go there in dimension 2, so in dimension 2, then I was looking for projection e belonging to 2 by 2, okay, and satisfying the relation that, so this e minus 1 half, so if you want, okay, is equal to 0. So when you look at this relation, you find that it is a two-by-two matrix, 1, 2, 1, and if you take the algebra generated by the matrix elements, just writing that this square is equal to e, you find that the algebra has to be smooth functions of the two spheres. In other words, what I am telling you is that I am telling you that 2 by 2 matrices across the good sphere is exactly the 3-folder of the margin-like variable of the sphere and 2 by 2 matrices. And I am going to do this relation. So is this something very strange? you see that you get the error of 2 by 2 matrix functions by an incredibly simple presentation because it's generated by E with E squared equals E and 2 by 2 matrices so then I went to 4 by 4 I went to 4 by 4 and so M4 of A a projection E belongs to M4 of A and satisfying now two conditions so the first condition was the same as before component CH0 of issue 0 and then the second condition was that the first component of the chain contact is 0 and when you write this condition you find that E minus 1.5 times ok, let me find it with Eij you find this condition here minus that line 0.1 answer e i 1 i 2, answer e i 2 i 3, e i 2 i 0 so what you want is that this sum is equal to 0 in the algebra a times a times a because that's exactly the definition of 0 so you look for solutions in this equation and what do you find? well the first thing that you find is that there are commutative solutions And if you write this E as a 4 by 4 matrix as T, so T is the diagram matrix, Q is a quaternil, so alpha, beta, minus beta, star, alpha, star, then the conjugate quaternil, Q bar, here, and here you put 1 minus T, 0, 0, 1 minus T.

32:30 Okay, I mean, it's a computation which takes some time to show that this relation is identically satisfied. And then you look for the similar equation to the one I have written here, and the equation now is e minus 1 half times in dimension 4, and e to the 4 is equal to the grain, this is the arithmetic equation to the one here and here. In the same way, in the same way, in the same way, if you think the algebra corresponding to this solution, then the algebra is thinking T of S4. And the solutions to this quartic equation in T, in the matrix, are exactly all matrix on the fourth sphere, which have the round volume form. So what you find is that, you know, now you can get some handles on all these matrix which will, of course, interfere when you want to be radical. You get an angle on this matrix. They are exactly given by a solution of this part of the equation. So the equation I have written is an equation for the matrix on the fourth sphere. The reason why they have the fixed volume form is that the volume of this, you can combine it with the previous index theorem, and you find that the integral of this multiplied by gamma, and this gives you exactly the integral of ds to the 4. So now what happened is that I got a paper in February last year, one year before, in which I mentioned this commutative solution and the fact that all matrix could be obtained by solving this particle equation. But then in the fall I went to a meeting in Italy and I met Giorgio Landi, and they were working with some quantum groups and some instantons and so on. And then he mentioned that one could do something to try to find a solution which is to replace this by what you get by the SU2Q, So we tried to work it out. We found that it was not a solution. It was not fulfilling CH1 or Q2.

35:00 In the computation, we got terms of the form Q2 minus 1 and Q2 minus 1. After a while, we realized that it was not Q2. you do it more cleverly, it's not just q-squared, it's q-q-bar q-q-bar minus q-squared so eventually what we found there is a way, which is done by common rules, because it's different there is a way to find non-commodative solutions by taking a complex number of problems And eventually, so this gave, of course, non-community four spheres, the corresponding matrix, but eventually it led to the following theorem, which is a favorite of our theorem, which is that any Riemannian manifold, such that the isometric group of Riemannian as the inferior equal to 2 admits admits a one-parameter a one-parameter isospectral deformations isospectral deformations to non-committative geographies and so this theorem tells you that you know, I define what was non-committative manifold and so on, and you could say well, because there will be very few of these No, that's not true. Actually, there is a general cell, which follows from the proof that we have, that we could find on cognitive solutions, that says that, you know, instead of being the exception, I mean, there are many, many, many manifolds, and in fact, not only that, but they have countercours of similarities, of isolates. And if you start with a symmetric space of compact types or sufficient rank, then you can deform it to an alternative geometry, and you still have a compact quantum group of resonance. So they really deserve the name of geometry, so they really deserve to be studied. Now, I didn't have time, of course, to talk about what I really wanted to talk about, which will appear rather soon, I think. And it's joint work with Richard Barclay.

37:30 And so what we did, if you want, we actually classified the non-computative 3 spheres. Namely, the non-computative 3 spheres, now you have the vanishing of the chain component CH1 of a unitary element of k1, not of k0, and so what we have found, we have found the complete general solutions to these long-computative spheres, and they turn out to be a three-parameter deformation of the ordinary commutative sphere, S3, and a part of the theory coincides with the Schianian algebra, which was discovered long ago by Schianian in the theory of the converters. But now it allows to do geometry of this linear algebra to have the non-competitive geometries and so on and so forth. So in 19, I just want to say, of course I won't have time to explain about this, but you see, the beauty of the theorem is the following is that then, when you find the non-competitive matrix, you find out that because of Morita equivalence of the algebra n is itself, and algebra is always Morita equivalence itself. I mean, in the in the non-pogritative case it's not simple but in the non-pogritative case it generates new matrix from old and replaces a matrix D by a new matrix which is D plus A plus J A G inverse where J is a real structure and A is any element of the form sigma A and D B R now it turns out that if you take even extremely simple examples of matrices over an ordinary space the new matrix will generate the gauge theory. And it turns out that not only... So, the gauge theories, if you want, the gauge bosons, appear exactly in the same relation to the gravitas, if you want, to the gravity part, as the inner automorphisms are with respect to general automorphisms. Okay? And you can even go further and you can even find what is the action and the action really is that it can be strongly about its principally spectrum, its pure spectrum and it only depends on the spectrum of the operator. So I think that there is really a lot to be learned in the sense that because the simplest problems which are the problem of classification of spherical neurons in dimension 3 generate really extremely interesting algebraic computations and so on. Okay, so I think that's it.

40:00 Any questions, comments? When you were carefully skipping over the S3, I was thinking, have you managed to approach the Poincare conjecture this way? No, no, I mean, you know, there we are in total new land, you see, in the sense that... I was thinking about, you know, what happened with the manifolds and I oversimplified the problem by only looking at the spherical log. Now it turns out that, okay, in dimension 3, they can be classified. But even to classify their matrix, it's not at all a theory of force. Why? Because the matrix now, you know, for instance, when I was petrolling, I used petrolling a Riemann manifold to get a non-competitive one. I was only perturbing the matrix, which were invariant under a certain isometric. So you could say that you have less matrix in the perturbed case than you had before. In fact, that's not true, because when you perturbed the moncomunative, you have these inner fluctuations. So there is a trading from gravity to gauge theory. By making the moncomunative, you trade a part of the gravity to the gauge theory. Now, of course, the reason why string theorists and so on are interested in this idea is that they did appear, the new commutative fellows appear, totally unexpectedly, from these M-theory considerations. So, in a certain way, it's not something... What I am saying is that commutative geometry is sort of not isolated at all. And when you look at things from a certain standpoint, you find out that space-time, because of the presence of gate-series, it tells us that it doesn't want to be considered as commutative. It tells some element of commutative liberty. Instead of putting these cycles to equal zero, if you wrote them as downward, would you get more degrees of liberty? Okay, this is an extra question. This is a true way to formulate the question. It's a little bit artificial to say that the chair class has vanished. You want some conceptual meaning. Now the conceptual meaning is the following. The conceptual meaning is that two exact sequences, due to Max,

42:30 that the algebraic case theory, mapped by the relative algebraic case theory, goes to the topological case theory. And as you know very well, this is a long exact sequence. But this long exact sequence is in fact sitting over another long exact sequence, which is a long exact sequence where here you take the psychic homology HCN, here you take the transport of HHN, here you take the denis trace map, here you take the channel, and here you take a kind of a relative channel and it goes into Hc and here there is the operator S of periodicity transposed and this should go Now, the correct question, you see, here I was asking that the lower Chiang-Kat is managed. But in fact, the correct question about the K-Series class is that it should come from a algebraic K-Series class. Because if it comes from an algebra in K0 class, then its cyclic homology chain character comes from an oxygen homology chain character. And that's a very strong problem. It's the exact value of the law process. So in fact, I believe that the correct formulation of the problem I was addressing, conceptual formulation, is in fact to write the element as common from algebra in K0. and this gives a lot more freedom this gives a lot more freedom, because it's much harder because then, ok, you have exactly this freedom of representing cycles in K0 but Marx has proven itself that the form of that class in K0 the bot class, which you can obtain by doing a little disk somewhere in K0, that class does come from a form of K0 For the non-sperical case, then you really have to involve the trans-project of the trans-project of the trans-project of the trans-project of the trans-project of the trans-project.

45:00 You told us that the toros is a morphic Jacobian. Is there some derivative theorem that says that you can have two derivatives of a morphic Jacobian? Yeah, well, okay. This is related to the Michel Armand theory of different morphisms of the circle, for instance. You see, because you can construct non-cumulative torus by different morphisms in a circle. And so it depends on the nature of the rational number of theta. If theta is a good diaphragm number, then it should be true. If it has good diaphragm coordination, like the program ratio. But if it's a lubid member, then it should be true. It's also related to the fine nature of psychic commodity. You know, because the second colloquy of the one who can get the toes, not the period, but the imperialized one, also depends on the diaphantine nature of the earth. Depends very strongly on the diaphantine nature of the earth. Thank you. Thank you. Okay, now we're going to see the example of the spirit in front of you, so I'm excited because we have some solutions. So, that means not to be able to... Yeah, sure. Yes, there is something that... Right. Yeah. Oh, yeah? Ah, no, no, no. I understand, I understand. Yeah, sorry.

47:30 Of course, you would have to have what is this? What type of algorithm? Can I always take this? Sure. Let's take that. Do you have a website? No, but if you look at FTH and click on my name then you will get but you have to put FTH not only mass Google code I don't know it's much easier to go to FTH FTH and mass there are two things usually we put things in mathematics but I also put them in FTH sorry I'm not I'm sorry I'm just not quite catching what we doing is this this is the FTH oh right I'm sorry I understand yes of course I do this I'm sorry I just wasn't hearing clearly so sorry sir so just click on your name under that right okay fine got it okay fine so I thought you said that your reference to the um ENS I was missing I thought you were able to cite out the ENS that's what I was doing sorry I'm not sure yeah but you might get a lot of junk you That's true, but I think you might find something, but you might find something after Severa, because they don't order them. I try to find myself, and I found directly in my way. so that on its own should be enough to find all the references including this new this one is not yet no no okay I thought oh that's extremely thank you I'm sorry yes well I can run and copy it if you'd like no but I have especially an entire thank you I'm not sure if you answered the question.

50:00 I'll see you next time.