1/3 Noncommutative Counterparts of Celebrated Conjectures
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Tabuada, Gonçalo
Some celebrated conjectures of Beilinson, Grothendieck, Kimura, Tate, Voevodsky, Weil, and others, play a key central role in algebraic geometry. Notwithstanding the effort of several generations of mathematicians, the proof of (the majority of) these conjectures remains illusive. The aim of this course, prepared for a broad audience, is to give an overview of a recent noncommutative approach which has led to the proof of the aforementioned important conjectures in some new cases.
 Thank you so much. I would like to start by thanking Paul, Grigory, Frédéric and Arnevint for the invitation to give this course. I would like to start, I would like also to thank the ASHOES for being so accommodating throughout this academic year, in particular for setting this machinery that allows me to use this old school technology called Blackboard. And I would also like to thank the audience here for making the talk a little bit more human. So the course we'll have consists on three lectures. The first lecture will be about some celebrated conjectures and about their non-commutative. So I will always use this notation Nc for non-commutative, non-commutative counterparts of these conjectures. Then in the second lecture I'll be talking about implications of this non-commutative viewpoint on these conjectures to non-commutative geometry. And then in the final lecture, so this will be on Wednesday and then on Thursday I will talk about applications to classical job of this non-commutative viewpoint on these conjectures. So let me start with some notations. So throughout the course, little k will be always a perfect field of characteristic p, p greater or equal than zero. And in the case where p is positive, I will write wk for the ring of pt pic of its vectors. And capital K for its fraction field. So I need to invert p. So for example, if this is fp, then this is just a ring of p-adic integers and this is the ring of p-adic numbers. And then I will also write by sigma the isomorphism on capital K induced by the Frobenius on little k. So here I raise to the pth power that gives me rise to an isomorphism which I will write by sigma. And let's start with the scheme which I will assume to be smooth and proper k-scheme and of dimension d. Okay, so these conjectures are almost all of them about algebraic cycles. So let me say something about algebraic cycles. So we can consider this graded q vector space where it's graded by the dimension. So I look at algebraic cycles of two dimension i or x with rational predictions. And then if we choose to consider one algebraic cycle, we can impose several different equivalence relations in here because this is a huge graded q vector space. So there are a lot of different equivalence relations. For example, the rational equivalence. So rational equivalence says that if you have an algebraic cycle, it's rationally equivalent to zero by definition if there exists an algebraic cycle beta in the product of x with a projective line in such a way that your algebraic cycle is the difference from the evaluation of beta and on the infinite. So intuitively speaking, it says that two algebraic cycles will be rationally equivalent if you can deform one into the other using the projective line. Then there is another equivalence relation called new potency equivalence. So here you say that an algebraic cycle is new potently trivial by definition if there exists an integer in such a way that when you cross your algebraic cycle with itself a certain number of times. So let's say cross it n times. So this is an algebraic cycle in co-dimensional i on x, but on x cross with itself n times that this algebraic cycle here, it's actually rationally equivalent to zero. So if there exists an integer such that the cycle disappears, you'll say that it's new potency. So Gonzalo, can you really write really bigger because it seems that there are several, yeah really bigger, two times more maybe because there are several people complaining. And then there is a homological equivalence. So here you choose a veil homology theory. So there are a lot of veil homology theories. I'm going to consider for our purposes solely the RAM homology theory in characteristic zero and in characteristic p crystalline homology theory. And then so any veil homology theory comes equipped with comes equipped with a cycle class map. So we have a cycle class map towards the homology to ix twisted by i. And then you say that your algebraic cycle, so you can say it here, it is homologically equivalent to zero if it disappears when you're after applying the cycle class map. Okay. And then another interesting equivalence relation. It's the numerical equivalence relation. So here you look at algebraic cycles of co-dimension i with q coefficients and up to rational equivalence. And you can pair them with algebraic cycles of complementary co-dimension. And from here you can extract a rational number. So you if you have two algebraic cycles alpha and beta, what you can do is that you can intersect them. So this will give you a cycle of degree zero. And then you can simply take the degree of this. And so let's write this pairing by alpha, beta. Okay. And you say that a cycle is numerically equivalent to zero if by definition this pairing vanishes for every pattern. Okay. So we have these four difference equivalence relations on algebraic cycles. And the remark is that well if you have a cycle which is rational equivalent to zero then it is a new potently equivalent to zero. And then it is homologically equivalent to zero. And it is necessarily numerically equivalent to zero. So that implies that at the level of algebraic cycles you have all these quotients. You have these quotients on algebraic cycles when you impose these different equivalence relations. And the interesting point here is that when you impose this discussion here at the very end you get something which is finite dimensional. But here not necessarily is something finite dimensional. It can be very big. For example, there is this famous result of Mufford that if the field, the base field is large, for example, if you are overseas then if you take a surface with positive genus then necessarily your necessarily your algebraic cycles of co-dimension two in your surface the dimension is actually infinite. So this is pretty large depending on the base field and this is always finite dimension. And now there are a lot of conjectures about these different significance relations. So one of them is this famous conjecture of Grottendijk. The Grottendijk standard conjecture of type D. So it's a conjecture from the 60s. That says the following. So you feel... So Gonsalo, maybe there's a question that you could answer. So for all, I just read. So someone is confused because he asked all the relations are with rational coefficients. Yes, I'm working with rational coefficients. Yes, it's not mandatory that but in my particular talk I'm doing that and it will become clear why in a minute by the end of the talk it will become clear. Yes. Also there's a question about the fact that you start in characteristic P but you seems to work overseas. I'm working in characteristic P but my P is greater or equal than zero. Okay, can be zero. Okay. Yes. There is an algebraic equation. There is, there is but I'm not on the board. I ignore it. There is also and it will be something that will be seated here. Yes. Because algebraic equations will imply the numerically. Yes. Yes. Yes. This conjecture simply says that algebraic cycles up to homological equivalents or algebraic cycles up to the numerical equivalents are the same. So it's saying that there is no difference between these two. So that's a conjecture from the six. This is still wide open. Well, there have been a lot of people working on this. For example, is known when the case where these dimensionless are equal than two and in characteristic zero even of the national lesson or equal than four and even for a building varieties in characteristic zero and these are results of Lieberman some old results and there are many other cases. Then a conjecture of Voivotsky called the Voivotsky-Newportons conjecture. So it's a conjecture from the nineties. It says that that algebraic cycles up to newportons equivalents relation are the same as algebraic cycles up to the numerical equivalents relation. So you are imposing here an equality between these two. So in particular, these two will become also equal. In other words, a remark is that this conjecture actually implies the Grottenich conjecture and it's interesting in the sense that it does not depend on any variable multi theory and this is still of course wide open. It's known in particular cases. For example, when the dimension of X is less or equal than two, this is an independent result of Voivotsky and Claverson. Then there is also a conjecture of Dalynsen from the eighties that says that if I'm over a finite field, then in fact this conjecture says that there is no difference between algebraic cycles up to rational equivalents and algebraic cycles up to numerical equivalents. So in other words, all the difference equivalents relations are the same if you are over a finite field. And again, this is wide open is known in some cases, for example, for curves that thanks to the work of Soulet and Bruno Kahn, etc. And of course, this Dalynsen conjecture implies the Voivotsky conjecture. So these are three conjectures about these difference equivalents relations. Then another conjecture I wanted to talk about is this famous veil conjecture. This is from the forties. So it's assertive following. So we are working over a finite field. Okay. We are working over a finite field and choose an integer between zero and twice the dimension and can look at for the crystalline homology of your algebraic variety. Okay. And then this comes equipped with an action of the Frobenius. So here I have a finite dimensional capital K vector space equipped with an automorphism. And the conjecture is says the following says that if you have lambda is an eigenvalue of this operator of the Frobenius, then this satisfies two conditions. First of all, it's an algebraic number. And secondly, it's absolute value, it's complex absolute value is equal to q omega over two. And these four all possible conjugates of left. So in fact, this conjecture was of course in the forties, there was no crystalline homology. So it was not phrased like this, but it turns out to be an equivalent formulation. And it was proven, it's now a theorem, it was proven by the link that this conjecture holds. Okay. And let me say again, an interesting fact is that, well, suppose that we still didn't have the link's work, but it turns out that if your eigenvalue is an algebraic integer, if we knew that it would be an algebraic integer, then it turns out that also, so here the absolute, the Atlantic absolute value of lambda is equal to one for all conjugates of lambda and this for every L different from P, the characteristic of our base field. Okay, of course, the links result implies in particular that these numbers are in fact algebraic integers. And so we know this fact. Okay, and this motivates the study of the Asevale zeta function. So let me recall you. So the Asevale zeta function is defined as follows. So you take this product over all the close points of your, of your skin and you take one over one minus q degree of x minus s. So it's a complex valued function. So this product converges when the real part of s is greater than d, greater than the dimension. So we have s. And now if we use, if you combine the result of the link with some work of Bertolot and crystalline homology, we can rewrite this function. So let me just mention you that this number that it's here is actually the cardinality of the residue field of the point, which is a finite field. So this is what you are counting the number of points on this residue fields for all those points of your skin. And so as I was saying, if you combine the link's work with the work of Bertolot and crystalline homology, you have a homological interpretation of this function. You can write it at least, this product between zero and twice the dimension of the determinant of identity minus q minus s for Venus acting on the crystalline homology of x and then minus one power omega plus one. So this in particular tells you that this function that is defined here, so let me, so let's say we are over c. So we have our function. So we know that it's, so we know that it's defined on the south plane. So it's well defined on the south plane. It actually extends to a unique meromorphic continuation to the entire complex plane. This is what it tells you that. And moreover, the link's conjecture is going to tell you where the zeroes and the poles of this function where they are. So you can look, you see that in fact on these regions in red, this is precisely the places so where you have the poles in these regions in red. So you don't know where they are, but they will leave in these vertical lines. And in these regions in blue, these are precisely the regions where the zeroes will occur. So here you'll have the zeroes of this function. You don't know where they are. And you also observe that since this function is defined using complex conjugation, so you know that it will be periodic of period 2 pi i over the log of q. So you'll have a periodicity on this function like this. This is simply due to the fact that you are using, you have this shortest x sequence where you have 2 pi i log q z. So you have this shortest x sequence where you use, when you use complex exponentiation. So you have this periodicity. And then finally, I would like to mention one final conjecture, which is also a famous conjecture of Tate. So it's a conjecture from the 60s. It says the following. So here I will be over the finite field again. And the conjecture, we are in fact three of them, three versions. So you have one version for L prime difference from P. So here what you say is that you have the cycle class map going to the etal, the erratic homology, going to erratic homology. And these elements, in fact, they land in those elements that are fixed under the action of the absolute color group. And the conjecture, it says that, well, if you change from q-quefficient to q-l-quefficient, then this cycle class map becomes surjective. So any class invariant and then absolute color group comes from here itself. Then there is also a P version of this. So you do the same, but now you use crystalline homology. So you use crystalline homology. And you look here at the crystalline Frobenius. So these elements will land in those that are stable, that are fixed under the crystalline Frobenius. And when you change from q- to q-p, then this becomes surjective. That's the conjecture. And also there is a strong form of the conjecture which says the following. It says that, well, here, as I was mentioning, we don't know, we know that the poles live in these red vertical arrows, but we don't know where they are exactly. And this conjecture tells you that, well, there are poles, this function as poles precisely at this point, i. So there are poles in here. So here there is a pole, here there is a pole, here there is a pole, etc. And of course, because of the periodicity, then you would have infinite poles living here, etc. And moreover, it's going to tell you that, well, not only you have poles there, but the order of the pole is given by this precise number. So it's equal to the dimension of the algebraic cycles with q-peficients, but up to numerical equivalence relation. So the dimension of this q-vector space is actually the order of the pole at this precise point. And, well, and we impose this for every i between zero and the dimension. That is for all these i's. And as you see here, it's called strong-tate, because it's actually stronger than the classical-tate conjecture. So it's a theorem of tate that, in fact, the strong-tate conjecture, well, implies the classical-tate conjecture, actually for every l different from p. And moreover, if you impose the conjecture dlx, this grotten-dick conjecture, then it turns out that the converse also holds, so it becomes an equivalence. And this is also true if you have for the conjecture within the p-version of the conjecture. Okay, so these are the conjectures that I would like to establish non-commutative counterparts of. Okay, and, well, of course, let me say these conjectures are wide open in general. It's known in very particular cases. Tate-proof it in the case of curves. Nowadays, a lot of people were able to prove it in the case of k-3 surfaces. Okay, so now let me, just before finishing this commutative part, let me just mention something really quick. Let me just mention here that this function, this very interesting function also admits this function, admits a functional equation. And this is a theorem of art in and grotten-dick. So in this function, you have this relation between, you can relate the function at s with the function at s minus or d minus s. And the relation between the two is given by this. So here you have the Euler characteristic of your scheme s. And here you have a constant, which is minus the Euler characteristic, the dimension divided by two. Okay, so it's about that function over there. Okay, so this is what I wanted to say about the commutative world. Now, let's move on. So what do I mean by non-commutative geometry? So there are different people look at non-commutative geometry in a different way. Okay. And so for me, non-commutative geometry will be actually non-commutative algebraic geometry. Okay. So which is a subject that goes back to the Moscow school, let's say, goes back to Manin and its students, etc. So the idea that we put these words, so there is this standard definition of Bondow and Kapranov, that what is a differential graded category, so a DG category. So this is simply a category A, which is enriched over complexes of K vector spaces. Okay. So the home spaces are just complexes, not ordinary sets. And you have a lot of examples whenever you have an algebra differential graded algebra, then you have one of those, which is just one of those with the simple objects, with the single objects. And whenever you have a scheme, then you can look also at this category of perfect complexes on your scheme. So these are complexes of OX modules that locally are quite asomorphic to bounded complexes of vector bundles. And then this carries a canonical DG enhancement. So these are two examples. So algebra and geometry both give rise to DG categories. And this very famous example of Williamson says the following. So if you look at the projective line and you look at modules over the projective line, that category is very far from modules over an algebra. But if you go towards this drive setting, then this category is actually more equally equivalent to an algebra, and you have this algebra of matrices. So in other words, this is telling you that the projective space in this drive setting, it's actually a fine is given by an algebra, but by a non-commutative algebra. And then in this world, there is this notion of smoothness and properness. So this is good to maximum. So if you have a DG category, you call it smooth. So by definition, this is just saying that when you look to it as a bimodule over itself, then that it's compact in the category in the drive category of a bimodule. And proper simply means that well, if you fix two objects, any two objects of your category, that's a complex, you look at the cohomology, and you ask this dimension of this cohomology to be finite dimensional, and moreover, that the total cohomology to be finite dimensional. And this for any two objects, x and y. Say it again. Yeah, I'm working over a base field, it's always our little case. It can be more general, you can work over a commutative ring or even over a different base, it can be very general. And then the key remark is that if you have a scheme which is smooth and proper, then that implies that well, this category, it's actually smooth and proper in this sense. And also the converse holes. So this perfect complexes on a scheme actually reflects these properties of smoothness and properness. And so the idea here, it's now, we would like to do geometry, not with the skin, but with an arbitrary dg category, a smooth and proper one, which mimics smooth and proper schemes. So what can be done in that case? So in particular for this course, I would like to phrase, to formulate the non-commutative counterparts of all these conjectures. So if we want to do that, in particular, I would like to have some kind of non-commutative valcomology theory. So something that works not just for schemes, but in full generality. So for this, let me talk about a topological periodic cyclic homology. So this, let me put some names on the board. So this to Alencon and many others, these two, Essofold, Tuschels and many others. So let me give you an idea of this. So I suggest to look at the course of Tina last week and also of Kaledin, if you want to learn more about these things. So let's suppose to simplify that I just have an algebra, an ordinary algebra with A. What can I do? I can do the following. I can do the following construction. I can look at A, A cross with itself over the base field, A cross with itself, cross with itself over the base field. And now I can use the multiplication to define maps. I can multiply these two, that gives me a map, but I can multiply by this order that gives me a different map. So it's not necessarily commutative algebra. And here also I can multiply these two or these two or even these two. So and in fact, I get this simple initial gadget and I can totalize it. And if you totalize it, you get something that is called a Uschild homology of A. And that is something that is K-linear. So for example, the Uschild homology of our base field is actually the base field itself. And now you see that at every level what you can do, you have an action, you can promote the factors in the C-click order. So if you put that information all together, that gives you an action of the circle. So you have an action of the circle on this object. So you can do a take construction. And if you do this take construction, what you end up is that with the periodic cyclical homology of A. And this again is something that is linear, but it's more than that. It's actually periodic. So for example, if you computed over the base field, that's a ring of polynomials on one variable, of low-run polynomials on one variable, of variable being a minus 2, of degree minus 2. So you get something periodic, too periodic. So Gonzalo, there's a question, maybe I've missed it. Is there a way to define a smooth and proper relatively? So if I have a map of DG categories, I can define a smooth proper morphism. Yes, there is. I don't remember from the top of my head, but yeah, there is. And that it's also in the literature. I can dig in and then forward the definition. Yes, there is a relative setting here. And now what I want to consider is let's go to, this is an arbitrary characteristic. Let's go to characteristic P, positive characteristic. Okay. Now in positive characteristic, I can do this topological version of periodic cyclical homology. And roughly speaking, what I do is that I change the base field. And now I do tensor products over the sphere spectrum. Okay, so I get the base, which is even more initial than the original one. And I can totalize this construction and get something which is a topological version of the social homology, which is still something k linear. So if you take the thh of little k, that's a result of Boxsted. So this is a ring of polynomials on one variable of degree 2. Okay. But then it turns out that there is a relation between the two. This is actually a module over this. And so one thing that you can do is that you can take the tensor product over k. So in other words, you can think about this as one parameter deformation of official homology on this parameter here. So when you take the fiber at zero, you get the original ocean homology. That's one way to think about it. And then here you can mimic. So again, you have actions of the cyclic groups, these permutations, if you put them all together, that gives you an action of the circle. And you can do a take construction in this topological sense that's called Greenleys construction. And you get this topological version, which now no longer is k linear. So for example, when you compute it over the base field, what you get is the ring of vectors where you have added one variable of degree minus 2. So you see that from this computation and this, you see that this one is actually a characteristic zero lifting of this one. Because if you reduce mod p, the vectors you end up with your base field. And this is true in general. In fact, this is a characteristic zero lifting of the periodic cyclical model. And now here you can do something that you cannot do in algebra, which is I can invert p. And that will be important for us. Because if I can invert p, then a new feature appears here in the topological world that does not exist in the algebraic world. And this new feature is this cyclotomic Frobenius. So again, I mean a positive characteristic. So let me mention what I'm going to say. It's in this topological language, but this is originally defined by Kaledin. So Kaledin is the first person that actually make this rigorous. So let me just make a remark. Suppose that you have a commutative algebra. Okay, so if you have a commutative algebra k algebra and over a field of characteristic zero, then what you know is that these two relations hold. So if you do a plus b for any two elements, that's a p plus b p. And the product is also a p times b p. But now if you if you are no longer working with commutative gadgets, then these things don't hold. So you don't actually have a Frobenius. But the Frobenius will appear not on the algebra itself, but will appear in this realization in this. So if you have a if you have a smooth proper, smooth proper dg category, then you can look at its thh of a. And this thh of a, it's actually something called the cyclotomic spectrum, in the sense that it doesn't not only it has this circle action that I've used here, not only have this circle action that you use to define the tp, but moreover, as this map that goes from thh from a to the thh of a and the take construction with respect to the cyclic group of order p. There's a question, Monsalor. So what about the definition of tp of k and characteristic zero? Yes, how we define it. Yeah, I mean, we can do that similarly, but I'm just focusing in characteristic zero for for. But are we defining what about with the are we defining zero zero typical v vectors just to be k itself or? No, no, I'm just I just want to work here in the case when I'm working with topological. So, okay, I'm just working in characteristic zero. Okay. And so here I want I have this map, which is s one equivalent. So here we have an action of s one and here you have a residual action of s one also, or if you want to s one module, the cp that identifies with s one. And so it's this two data is called the cyclotomic spectrum. I mean, it's more general, but I'm over p. And then out of this data, what you can do is you look at thh of a and then you look at thh of a and then so you have a what you can do here is that you have a canonical map from the homotopy fixed points of the circle action to the take construction. You have a canonical map here. And this take construction is what we call the pp of a and moreover, it turns out that in this setting. So if you are working with the smooth proper digi category, it turns out that this is similar to the thh of a where you have the take construction of cp and then you take the a multiple before fixed points with respect to s one. So it's a technical result. You have this equality, but that implies that you have another map here. So you take this original map and you take the multiple fixed points with respect to s one. And now you have these two maps and it turns out that the kernel of the canonical map as a scale linear. So if I invert p that kernel disappears. In other words, this map here becomes an isomorphism. So if it becomes an isomorphism, then I can define a cyclotomic Frobenius here, which is simply you take this map and you compose with the inverse of the canonical map. And that gives you a Frobenius from tp one over p if you invert p to tp of a where you inverted p and moreover this map is an isomorphism. So the Frobenius exists here after inverting p, but it doesn't exist originally in contrast with the commutative world. It doesn't exist on the algebra itself only on the invariant. And let me make some remarks about this cyclotomic Frobenius. So first of all, it is not linear. So it's not a z2 graded map. So it's not linear here. So in contrast, you only have this relation when you change from n to n minus two. So these are equal up to this multiplication by one over p. And and moreover, these maps that you have, they are not as in commutative geometry, they are not capital K linear. They are only sigma semi linear. Okay, so in particular, if you if you are working over a finite field, one thing that you can do is to compose and get the actually K linear maps. So you simply compose the phi n r times that gives you something which is K linear. And then the relation between these different Frobenius when you change from n to the Frobenius n minus two, it's actually multiplication by one over q. Consalos, so there is a question. Don't we need something like K being perfect for Frobenius to be an equivalent? So it comes, everything comes from my assumptions on A being smooth and proper. This will imply the equivalence on the Frobenius. Okay, so now this suggests that, well, these are, you should think about periodic cyclical homology and topological Schiller homology as non commutative well conjectures. So that's what we are going to replace actually the the ram homology and the crystalline homology. But there is something that we can do something. We can do the following. You can let me just write on the side. Can you leave the blackboard? Oh, yeah, of course. Sorry. So let me just write an aside here about non commutative what you think realizations. So when we wonder, well, we can maybe extend all the commutative results to the non commutative results. And there is this result saying that well, you can look at smooth schemes. And you can go to Morale-Weyvotsky stable or multi-peak category stable A1 or multi-peak category of schemes. And then on the non commutative side, you have DG categories. And we can construct a non commutative version of this Morale-Weyvotsky category. So I'm not going to define this in this course. It's just an aside. I'm just telling you an aside. And these things are related because if you have a scheme, you can pass to the perfect complexes. And then we would like to relate the two. So here the relation is as follows. Here we have a motivic spectrum that represents a multi-peak theory. So we can look at modules over the KGL with quick coefficients. And then you see here, this is a covariant procedure. And this is a contravariant procedure. This is contravariant. So we need a duality. So we can dualize here. And then it turns out that there exists a functor here, which has a lot of good properties. It is fully faithful. It is a tensor functor. And it admits even a right adjoint. And then as a consequence of that, as a consequence of this breach between the commutative and the non commutative world in this setting, we have the following corollary. So we can do the following. Suppose that you have a realization that you'd like to be interested in. So here you have Morale, you have Wewocki category of geometric motives, and you have a realization. And what you can do, you can pre-compose with duality. So let's say that we are with quick coefficients. We can pre-compose with duality and then go to schemes. So we have this contravariant functor on schemes. And then I can modify this realization. So here I have a tight object, which is nothing but this one. And then I can somehow trivialize it by considering modules over this sum, over all possible powers of RT. So I get a modified realization. And it turns out that if I do this modified realization, when you pass from schemes to DG categories in this sense, then all these kind of modified realizations, they can be extended to the non commutative world. So you get, there exists a non commutative version of the realization. So in particular, you get this result, it gives rise to a lot of non commutative motive realizations. For example, an adic version, an odd version, a Durand-Betti version that, for example, can lead you to define what are non commutative periods, etc. And how is this corollary obtained? Well, this corollary is obtained as follows. Let me just explain this. So you have, you can do exactly the same thing in here. But instead of KGL, you can use the spectrum that represents motivic homology. Okay. You can do exactly the same thing. So I'm going to, sorry, I need some space. I'm going to erase the corollary. But so I just want to have this realization and I want to extend it to the non commutative world. And let me do it here. So you'll do exactly the same. And now the first point is that, well, this is DM, the big DM, this is a result of Rondings and Polk, post-vener. And so this composition is your motivic realization, map of Vybocki. So here you have your realization going towards your key. And the second observation is that the KGL with QP-efficient is actually this trivialization. So you get this by M to N. So you have this fact here. And so this is simply telling you that you can base change here from coefficients on the motivic homology to KGL. And then you can do the same thing on the target. So if you do the same thing on the target, you can simply look at modules on the sum of your realization of your date motive. And you get this induced map here. And so your non commutative extension, it's actually we use this functor. And then what you do is that you use the joint and then you use this induced map. And that is your way to extend any kind of realization as long as you modify it, then it can be extended to the setting. Of course, this is a bit is cheating, right? Because you are defining this using schemes. You are just using an adjoint. You are just building this functor using schemes. You look at the closest scheme associated to your non commutative motive and apply the classical invariant. So it has a lot of problems. For example, these things are not monoidal. For example, you don't have finiteness on this. So it shouldn't actually be called a realization. But it's something that it can be done. You can always extend anything in the commutative world to the non commutative world via this adjoint. It's a bit cheating. Okay, so now we are ready to come back to our original goal. And now we can actually formulate all the conjectures. And so, yeah. Yes. Yes. No, I'm getting, you have some kind of, I'm just saying that you suppose that you have a suppose that you have a realization. So it's this, so this map, this realization. And then as soon as you modify it in the sense that you trivialize the type motive, then you can extend it to any DG category by via this procedure. But of course, this is using an adjoint. So it's not a correct way to do it. What you want to do is to have an intrinsic definition that is not using actual schemes, right? Monsalou, what was the question? We did not. It was just to explain a little bit better the board. Okay. Okay, so now what's the replacement of algebraic cycles? It's the grotendic group. So now I have a smooth and proper k-linear DG category. And we look at its k0. Well, its k0 is just a k0 of its drive category, which is a triangulated category. Okay. And now we choose an element in here in the k0, and we can phrase all the similar equivalence relations. We can say, we can talk about new potents equivalence. So you say that the cycle is no potently trivial if by definition there exists an integer such that when you tensor the cycle with itself n times, then this is something that lives on the k0, a tensor with itself n times, that this is actually become zero. So you can make this definition. How about homological equivalence? Well, we now have this, our non-committative homology theories that I've explained. So you define homological equivalence as follows. So you simply say that here you have, so this comes equipped also with charm characters. So this is something that I will explain better on Wednesday. So here you have charm characters defined on the k0 to the positive part of the periodic cyclical homology and the positive part of the topological periodic cyclical homology. And you simply say that, well, it's homologically trivial if the charm character of it becomes zero. And you also have the numerical equivalence. So here, how do you define intersection of cycles? Well, what you can do is you look at the k0 and you define a pairing on k0 as follows. So if you have a module and another one and look at the corresponding classes, one thing that you can do is to look the homes on the drive category of your A, look at the homes from m to the shifts of m, and look at these dimensions, and then take an alternating sum of these dimensions for m varying on the integers. And so in this way, we extract this number and this pairing, they extract the pairing on the k0, which is neither symmetric, neither is Q symmetric. But since A is smooth and proper, it turns out that you'll have a serfometer and using this serfometer, you can prove that the left and right kernel, which are priori are different, are in fact the same. So you can actually define something numerically trivial if this pairing, let's call c of alpha beta, it's equal to zero for every beta. Okay, and then we have the similar remarks. So if you have a cycle which is a new potent with trivial, will be homologically trivial, will be numerically trivial. So that implies that on the k0, you have all these discussions, module of the new potent, module of the homological module of the numerical. And again, this is always finite dimension. Does it mean this new potency goes to zero? That means the order characteristic is zero? No, it means that beam zero here means that this pairing vanishes for every b and alpha, such that this pairing vanishes for every b. The new potency implies this the equivalency mentioned. Yes. The homological. Yes, implies the homological, yes. It's an exercise, yeah. If you have something new potentally trivial, it will be homologically trivial. Okay, and now we using this we can, we can as you expect, you can do formulate this this non commutative counterparts. So we have this non commutative grottenby extender conjecture. So what you say here, it's here, it's a conjecture, let's see, D, so this is of type D, so D and C of A. So this simply says that the k0, when you mod out biological or by numerical, they are the same. Also, you can define this non commutative version of Vojvodski, new potency equivalent, new potency conjecture. So let me write conjecture V and of C. So again, it's that the k0 up to new potent equivalence is the same as the k0 up to numerical equivalence. So once again, if these two, if these two are the same, then these two are the same. In other words, this conjecture implies the preceding one. And of course, we can also define the non commutative version of Balinson conjecture. So over a finite field, that in fact, the k0 is insensitive to all these equivalence relations. Okay, so these are the analogs. And then we can also go further and define the non commutative version of the V conjecture. So what do we do here? So again, we are over a finite field. And so if we have a smooth and proper Vg algebra, sorry, Vg category, you will look at this finite dimensional capital K vectors place, and it's equipped with this automorphism, the Frobenius zero. So it's just the cyclotomic Frobenius composed r times. And the same thing that Tp1 over P comes with this Frobenius. Here composed r times. And then the conjecture is that the conjecture is that if you take an eigenvalue, if this is an eigenvalue of the Frobenius zero respectively of the Frobenius one, then well that these numbers are Osprey numbers. And secondly, that complex absolute value is equal to one, respectively equal to square root of Q. And this for all conjugates of London. And also in this non commutative world, you have this analog of the proposition, which says that well, if there exists an integer such that when you multiply your lambda by Qn, and if you get an algebraic integer, then it turns out that the ellatic absolute value of lambda is actually equal to one for all conjugates of lambda and this for every prime different from P. So you also have this result. Of course, this conditional in this case in contrast with the commutative world. And also we have this. Consalor, how much time do you need still? I just have one and a half. It's pretty quick. Thank you. So let me just say that in the. Maybe there's a question. Is this, are these conjecture Kfury, Morita invariant? Yes, of course, yes. Yes, because anything that is Morita equivalent will actually have the same motive. Let me give you a very technological answer. They have the same non commutative motive and all these conjectures descend to non commutative motives. That's the approach that I will explain on Wednesday. Yes. And also we can talk about the non commutative asset value. That the functions. So here, of course, we cannot count points. The only thing that we can do is we need actually an embedding of capital K into C. And then if we choose this embedding, we can define this case. So the only thing that we have is the co homology. So what we can do is, is consider these functions, the determinant one minus QS and then the Frobenius at zero. But then we need to use this embedding to go towards C because here we are using complex conjugation in order that to make sense. Then you take the PA here and then you need to go to C. And then you can do the same story here. You can define it in this way. So determinant one minus Q minus S, the Frobenius one going to C on the PP1. Okay. So we have these functions, of course, over C. They are meromorphic. And what this is telling you is that, again, by by own definition, they are they are periodic with this period. And by by what our non-committative version of the Velconjecture is telling you is that the poles are actually here in this case and are actually at one half in this case. So the when we change the when we change the embedding, these functions change, but the place where the poles are actually do not change. And finally, let me just say that we also have the non-committative version of the PIN Conjecture. So I recall you that we are over a finite field. So we are over a finite field. And the conjecture is as follows. So we have this non-committative version for L different from P. So here what you do is you look at the k theory of A and you base change it to the field extensions. And then you do the both field localization with respect to topological k theory. And now you look at the pi minus one of this disabillion group. And then you do the the TL of this. So you do the the the ELADIC TATE module of this. And you ask it to be zero for every n greater or equal than one. So it's this vanishing conjecture saying that well all these TATE modules, ELADIC TATE modules of disabillion groups vanish for every possible n, for every possible extension. That's one way to phrase it. Then we have the P version of this. So here you have the churn character going towards TP zero of A. So when you inverted P and it plants on the invariance under the cyclotomic Frobenius. So when you change from Q to QP, this is subjective. And you also have a strong form of the TATE conjecture. So you say that the order at zero of this function A S. So you say that well there's actually zeros are here, but there is actually a pool here. And this pool is the order of this pool is given by the dimension of the k zero of A module on numerical equivalence. So we should like to emphasize that this thing here actually does not depend on the embedding. Because you can rethink about this if you see the definition. It's just the algebraic multiplicity of one of these operators. It does not depend on the embedding. That's what it's the order of the pool there. And then we also have this result saying that this strong form of the TATE conjecture implies the TATE conjecture itself. And moreover if you add this non-commutative version of Rotendex conjecture, this becomes equivalent. So what is the upshot of all this? The upshot of this is the the following theorem, which is the very last one and that will end there. It's just saying that all these definitions are the correct ones in the following sense. So it's the following. So if you take a smooth proper scheme, okay, and then you can do two things. You can look at the conjecture for x, where c is any one of these conjectures. So d, v, so Rotendex, Vojvodski, Bielensson, Veil, Tate, Tate, Tate, and many others, in fact. You can do this. Another thing that you can do is look at the non-commutative version of the conjecture for the associated djicothexam. And so if this conjecture holds, it turns out that this implies that the non-commutative version also holds. But what's interesting in the theorem is that this is actually inequivalent. So this is saying that these classical conjectures, all of them, that are formulated in this setting of algebraic geometry, in fact, they make sense in this much larger setting of djicothexam, smooth proper djicothexam. And if you plug, if you attack this particular kind of djicothexam, you recover the original conjecture. And this is true for these conjectures that I explained today, but there are many other conjectures for which this holds. I don't have time to explain. And now what is the idea is to explore that here now I have non-commutative techniques to try to attack the right-hand side and as a consequence, attack the left-hand side. And this is the goal of the following lectures. Okay, thank you and sorry for going over time. Okay, so thanks for the talk. So I'm sorry, so we are aware that there were serious problems with connection for certain of the attendees. So we are sorry, we will try to make it better, but unfortunately, it seems it's an internet problem. In any case, you'll have a YouTube video if you want to, if you could not follow the talk. So now we have questions. So I saved a few questions, three questions, three questions actually to, for the end of the talk. The first one is someone is asking, maybe if could you repeat what YTP is related to the Motivic stuff here? You mean Motivic stuff, you mean the commutative world, I imagine that's the question, right? Yes. So the relation is as follows. Yes, so the relation is as follows. So if you take the TP0 or 1 of your, so let me do like this. TP is too periodic and you take the TP of this TG category, you take your scheme and this one, it turns out that this gives you the positive part and the negative part, you'll get the sum of the even groups of the crystalline comology of your scheme and on the negative part, it gives you the sum of the odd comology groups on crystalline comology. So this telling you that, well, this invariant, of course, after inverting T, this tell you that, well, when you computed on this DG category, you get the crystalline comology, not the individual pieces, but up to parity. And moreover, now, you also here, so when you see that this TP comes equipped with the cyclotomic Frobenius, but this cyclotomic Frobenius, as I made this remark, is not Z2 graded, okay? It's not a Z2 graded thing. When you change to N to N minus 2, there is a scalar that appears. So let me say that I'm with P0 and P1, 0 and 1. And now I have the cyclotomic Frobenius 0 and 1 here, and they correspond to something here, and they here, they correspond to the sum of the even cyclotomic Frobenius on X, but with a certain scale multiplied by it. And here, the sum of the cyclotomic Frobenius, but multiplied by P minus T minus 1 over 2. So intuitively speaking, this is telling you that you lose the weights. When you pass from X to this category, you can recover everything up to weights. You lose the weights. But for the conjecture, that is enough. You don't lose anything in terms of the conjecture. So I want to testify here if the colors are completely unreadable. So don't use colors. Do you use colors? Yeah. Maybe the green. This is better. Slightly. Okay, just a remark. So there was another question about the relation. So I just say it again, about the relation between the NC conjecture and the usual one. And so that's on the blackboard. They are equivalent, the smooth proper case. Which one? No, the non-commutative and the usual one. So that's what you stated. And also there's a question. So do any of these conjectures have a mixed characteristic analog? Or formulation? I don't know. You want to be working over DVR or something like that? I guess so, yeah. Yeah, I mean, I haven't talked about that seriously. But I mentioned that, yeah, it's likely that you could expect something like that. Yes. Okay. So this is related, in fact. So there's another question about the relative versions of this conjecture. Are there relative versions of this conjecture? I don't know. I mean, yes, I mean, in the classical world, yes, I think there are some of them admit relative versions. Yes. I haven't explored that. Yes, that's a good question. That's a good thing for the future. So how do things work in families and things like that? Okay, so I'm not sure to understand the question correctly that knows, but I just read it. So what is a tensor m? And does it agree with the perf x to the m in the geometric case? Yes, yes, yes. When you do this, this is this. And my schemes are nice enough for which this kind of phenomena holds. Yes. So in the, I imagine that this question is about the new potency equivalence relation because on one side, it was on cycles on xn on cycles on xn and the other side was on the k0 of your a tensor m. And when a is of this form perf of x, it's yeah, there is this. Thanks. So that was that was a question of Remy van Doeben de Brun. There's also a question by Ola Sende. Is vanilla potency conjecture equivalent to the Baylinson-Soulet conjecture? And Baylinson is only over a finite field. Baylinson-Soulet. Ah, Baylinson-Soulet, maybe you can recall me what it is. Since vanishing of vanishing of motivic homogene in degrees in negative degree. Yeah, I don't remember. I have to dig in. Yeah, I think I remember that there are close connections between the two. I don't know if it's exactly the same, but yeah. Yeah, I don't remember from the top of my head. So I think there's no direct relation. Okay, if you put all conjectures. Oh, I think, I think the Baylinson-Soulet, it's actually the strong, it's actually the tight conjecture. And then you mod out by the numerical equivalence. Baylinson-Soulet? Yeah, it's not. No, it's just vanishing of motivic homogene in negative, in a simple degree. But okay, so I have to. Could you please state precisely how the LED state conjectures are formulated via TP? No, via TP, I'm formulating the TP version of this, right? So for this, I'm using the TP. It's actually, I'm using TP to formulate the P version of the T conjecture, right? And I formulated the L version of the T conjecture using KT, right? And the question is if you can do that. Yes, I mean, that can be done, but it's using these results of Thomasson that tells you that, I mean, the relation. The relation is that you have a Atya-Eirsbrug spectral sequence that goes from a latic cohomology to etalcateria. So you have something like this. So rationally, so they prove that it degenerates. So you have something like the etalcateria of X. When you complete at L and then you rationalize, that is in fact the sum of these latic cohomology groups. And so here you already have a link between the etalcateria with KT, but it's an etalcateria. But then you have this beautiful result of Thomasson that tells you that you can write this algebraically because you have completed at L. It's in fact to take the KT of U and to localize with respect to complex KT, and then if you complete at L, actually you get the same spec. And now here you have KT, so something that you can phrase for any eticateria, for any generality. And then if so you see that then the reformulation of the conjecture would be saying that if you go from the Schoen character from the K0 to here, it actually lands precisely on the things that are stable and the action of the absolute color. So I think it was already phrased like this in an old paper of Friedlander. So I have another question. So are there any strictly non-commutative applications of the conjectures or any of the conjecture? Yes, I mean this is somehow a motivation. Usually, let's say that if we divide the world in two, much more people would work on this side than on this side, I agree. And so people are motivated on trying to prove this. But on this side, these conjectures as we'll see on Wednesday, for example, will allow us to have a conditional description of the category of non-commutative numerical motives, something a bit analog to what Milne has done, described in terms of veil numbers up to a certain action of the absolute color group and things like that. So yeah, I mean we will attack, for example, this conjecture next time on Wednesday. For example, we'll prove that all these conjectures hold, if you put here an algebra, which is finite dimensional and a finite global dimension, not necessarily commutative. And all these non-commutative conjectures hold. So yeah, so we can, that's also, there are two paths. So there's Wednesday and Thursday. Wednesday, we go in this non-commutative direction and Thursday on this commutative direction. And in both cases, we are going to explore the link between these two. So there's another question about, so are there any new cases known where the conjecture holds in the non-commutative setting? What happens on Thursday, what I will explain is a way to prove this conjecture in some new cases where you don't use geometry. What you do is you prove this conjecture using some non-commutative techniques. And as a consequence, you get this conjecture for axes which were known, not known previously. If I understood correctly the question, is this more or less what he was asking? If I can, new cases, yes, on the very last talk, I will prove the classical conjectures in new cases. And the way to prove it is not using geometry, but using this viewpoint. It's not with the viewpoint. Last question, is there a paper attached to this mini course where we can read more details? Yes, absolutely. In fact, there is a survey on archive whose title is equal to the title of this course. So I think which is called the non-commutative counterparts of celebrated conjectures, if I remember. Okay, so there's no more question in the chat. It seems, are there questions in the room? No? No. Okay, so good. Thanks again. I'm going to say hello. Sorry for going over time. I got a bit confused with this time. Yeah, okay. I'm sorry for the voos who had problems with the connection. Okay, so we meet again in chapter 6.
 Thank you so much. I would like to start by thanking Paul, Grigory, Frédéric and Arnevint  for the invitation to give this course. I would like to start, I would like also to thank the  ASHOES for being so accommodating throughout this academic year, in particular for setting this  machinery that allows me to use this old school technology called Blackboard. And I would also  like to thank the audience here for making the talk a little bit more human. So the course we'll  have consists on three lectures. The first lecture will be about some celebrated conjectures  and about their non-commutative. So I will always use this notation Nc for non-commutative,  non-commutative counterparts of these conjectures. Then in the second lecture I'll be talking about  implications of this non-commutative viewpoint on these conjectures to non-commutative geometry.  And then in the final lecture, so this will be on Wednesday and then on Thursday I will talk about  applications to classical job of this non-commutative viewpoint on these conjectures. So let me start  with some notations. So throughout the course, little k will be always a perfect field  of characteristic p, p greater or equal than zero. And in the case where p is positive,  I will write wk for the ring of pt pic of its vectors. And capital K for its fraction field.  So I need to invert p. So for example, if this is fp, then this is just a ring of  p-adic integers and this is the ring of p-adic numbers. And then I will also write by sigma  the isomorphism on capital K induced by the Frobenius on little k. So here I raise to the  pth power that gives me rise to an isomorphism which I will write by sigma. And let's start with  the scheme which I will assume to be smooth and proper k-scheme and of dimension d.  Okay, so these conjectures are almost all of them about algebraic cycles. So let me say something  about algebraic cycles. So we can consider this graded q vector space where it's graded by the  dimension. So I look at algebraic cycles of two dimension i or x with rational predictions.  And then if we choose to consider one algebraic cycle, we can impose several different equivalence  relations in here because this is a huge graded q vector space. So there are a lot of different  equivalence relations. For example, the rational equivalence. So rational equivalence says that  if you have an algebraic cycle, it's rationally equivalent to zero by definition if there exists  an algebraic cycle beta in the product of x with a projective line in such a way that your  algebraic cycle is the difference from the evaluation of beta and on the infinite. So  intuitively speaking, it says that two algebraic cycles will be rationally equivalent if you can  deform one into the other using the projective line. Then there is another equivalence relation  called new potency equivalence. So here you say that an algebraic cycle is new potently trivial  by definition if there exists an integer in such a way that when you cross your algebraic cycle with  itself a certain number of times. So let's say cross it n times. So this is an algebraic cycle in  co-dimensional i on x, but on x cross with itself n times that this algebraic cycle here,  it's actually rationally equivalent to zero. So if there exists an integer such that the  cycle disappears, you'll say that it's new potency. So Gonzalo, can you really write really bigger  because it seems that there are several, yeah really bigger, two times more maybe because there are  several people complaining. And then there is a homological equivalence. So here you choose a  veil homology theory. So there are a lot of veil homology theories. I'm going to consider for our  purposes solely the RAM homology theory in characteristic zero and in characteristic p  crystalline homology theory. And then so any veil homology theory comes equipped with  comes equipped with a cycle class map. So we have a cycle class map towards the homology  to ix twisted by i. And then you say that your algebraic cycle, so you can say it here, it is  homologically equivalent to zero if it disappears when you're after applying the cycle class map.  Okay. And then another interesting equivalence relation. It's the numerical equivalence relation.  So here you look at algebraic cycles of co-dimension i with q coefficients and up to  rational equivalence. And you can pair them with algebraic cycles of complementary co-dimension.  And from here you can extract a rational number. So you if you have two algebraic cycles alpha  and beta, what you can do is that you can intersect them. So this will give you a cycle of degree zero.  And then you can simply take the degree of this. And so let's write this pairing by alpha, beta.  Okay. And you say that a cycle is numerically equivalent to zero if by definition  this pairing vanishes for every pattern. Okay. So we have these four difference  equivalence relations on algebraic cycles. And the remark is that well if you have a cycle which  is rational equivalent to zero then it is a new potently equivalent to zero. And then it is  homologically equivalent to zero. And it is necessarily numerically equivalent to zero.  So that implies that at the level of algebraic cycles you have all these quotients.  You have these quotients on algebraic cycles when you impose these different equivalence relations.  And the interesting point here is that when you impose this  discussion here at the very end you get something which is finite dimensional.  But here not necessarily is something finite dimensional. It can be very big. For example,  there is this famous result of Mufford that if the field, the base field is large, for example,  if you are overseas then if you take a surface with positive genus then necessarily your  necessarily your algebraic cycles of co-dimension two in your surface the dimension is actually infinite.  So this is pretty large depending on the base field and this is always finite dimension.  And now there are a lot of conjectures about these different significance relations.  So one of them is this famous conjecture of Grottendijk.  The Grottendijk standard conjecture of type D. So it's a conjecture from the 60s.  That says the following. So you feel...  So Gonsalo, maybe there's a question that you could answer. So for all, I just read.  So someone is confused because he asked all the relations are with rational coefficients.  Yes, I'm working with rational coefficients. Yes, it's not mandatory that but in my particular  talk I'm doing that and it will become clear why in a minute by the end of the talk it will  become clear. Yes. Also there's a question about the fact that you start in characteristic P but  you seems to work overseas. I'm working in characteristic P but my P is greater or equal  than zero. Okay, can be zero. Okay.  Yes. There is an algebraic equation. There is, there is but I'm not on the board. I ignore it.  There is also and it will be something that will be seated here. Yes.  Because algebraic equations will imply the numerically. Yes. Yes.  Yes. This conjecture simply says that algebraic cycles up to homological equivalents  or algebraic cycles up to the numerical equivalents are the same. So it's saying that there is no  difference between these two. So that's a conjecture from the six. This is still wide open.  Well, there have been a lot of people working on this. For example, is known when the case where  these dimensionless are equal than two and in characteristic zero even of the national  lesson or equal than four and even for a building varieties in characteristic zero and these are  results of Lieberman some old results and there are many other cases. Then a conjecture of  Voivotsky called the Voivotsky-Newportons conjecture. So it's a conjecture from the nineties.  It says that  that algebraic cycles up to newportons equivalents relation are the same as algebraic cycles  up to the numerical equivalents relation. So you are imposing here an equality between these two.  So in particular, these two will become also equal. In other words, a remark is that this conjecture  actually implies the Grottenich conjecture and it's interesting in the sense that it  does not depend on any variable multi theory and this is still of course wide open. It's known in  particular cases. For example, when the dimension of X is less or equal than two, this is an  independent result of Voivotsky and Claverson. Then there is also a conjecture of Dalynsen  from the eighties that says that if I'm over a finite field, then in fact  this conjecture says that there is no difference between algebraic cycles up to  rational equivalents and algebraic cycles up to numerical equivalents. So in other words,  all the difference equivalents relations are the same if you are over a finite field.  And again, this is wide open is known in some cases, for example, for curves  that thanks to the work of Soulet and Bruno Kahn, etc. And of course, this Dalynsen conjecture  implies the Voivotsky conjecture. So these are three conjectures about these  difference equivalents relations. Then another conjecture I wanted to talk about is this famous  veil conjecture. This is from the forties. So it's assertive following. So we are working  over a finite field.  Okay. We are working over a finite field and choose an integer between zero and twice the  dimension and can look at for the crystalline homology of your algebraic variety. Okay.  And then this comes equipped with an action of the Frobenius. So here I have a finite  dimensional capital K vector space equipped with an automorphism. And the conjecture  is says the following says that if you have lambda is an eigenvalue  of this operator of the Frobenius, then this satisfies two conditions. First of all,  it's an algebraic number. And secondly, it's absolute value, it's complex absolute value  is equal to q omega over two. And these four all possible conjugates of left.  So in fact, this conjecture was of course in the forties, there was no crystalline  homology. So it was not phrased like this, but it turns out to be an equivalent formulation.  And it was proven, it's now a theorem, it was proven by the link that this conjecture holds.  Okay. And let me say again, an interesting fact is that, well, suppose that we still  didn't have the link's work, but it turns out that if your eigenvalue is an algebraic integer,  if we knew that it would be an algebraic integer, then it turns out that also,  so here the absolute, the Atlantic absolute value of lambda is equal to one for all  conjugates of lambda and this for every L different from P, the characteristic of our base field.  Okay, of course, the links result implies in particular that these numbers are in fact  algebraic integers. And so we know this fact. Okay, and this motivates the study of the  Asevale zeta function. So let me recall you. So the Asevale zeta function is defined as follows.  So you take this product over all the close points of your, of your skin and you take one over one  minus q degree of x minus s. So it's a complex valued function. So this product converges when  the real part of s is greater than d, greater than the dimension. So we have s. And now if we use,  if you combine the result of the link with some work of Bertolot and crystalline  homology, we can rewrite this function. So let me just mention you that this number that it's here  is actually the cardinality of the residue field of the point, which is a finite field. So this is  what you are counting the number of points on this residue fields for all those points of your skin.  And so as I was saying, if you combine the link's work with the work of Bertolot and crystalline  homology, you have a homological interpretation of this function. You can write it at least,  this product between zero and twice the dimension of the determinant of identity minus q minus s  for Venus acting on the crystalline homology of x and then minus one power omega plus one. So this  in particular tells you that this function that is defined here, so let me, so let's say we are  over c. So we have our function. So we know that it's, so we know that it's defined on the south  plane. So it's well defined on the south plane. It actually extends to a unique meromorphic  continuation to the entire complex plane. This is what it tells you that. And moreover,  the link's conjecture is going to tell you where the zeroes and the poles of this function  where they are. So you can look, you see that in fact on these regions in red, this is precisely the  places so where you have the poles in these regions in red. So you don't know where they are, but they  will leave in these vertical lines. And in these regions in blue, these are precisely  the regions where the zeroes will occur. So here you'll have the zeroes of this function. You don't  know where they are. And you also observe that since this function is defined using complex  conjugation, so you know that it will be periodic of period 2 pi i over the log of q. So you'll have  a periodicity on this function like this. This is simply due to the fact that you are using,  you have this shortest x sequence where you have 2 pi i log q z. So you have this shortest x sequence  where you use, when you use complex exponentiation. So you have this periodicity.  And then finally, I would like to mention one final conjecture,  which is also a famous conjecture of Tate. So it's a conjecture from the 60s.  It says the following. So here I will be over the finite field again.  And the conjecture, we are in fact three of them, three versions. So you have one version  for L prime difference from P. So here what you say is that you have the cycle class map  going to the etal, the erratic homology, going to erratic homology. And these elements,  in fact, they land in those elements that are fixed under the action of the absolute color group.  And the conjecture, it says that, well, if you change from q-quefficient to q-l-quefficient,  then this cycle class map becomes surjective. So any class invariant and then absolute color group  comes from here itself. Then there is also a P version of this. So you do the same,  but now you use crystalline homology. So you use crystalline homology. And you look here at the  crystalline Frobenius. So these elements will land in those that are stable, that are fixed under  the crystalline Frobenius. And when you change from q- to q-p, then this becomes surjective.  That's the conjecture. And also there is a strong form of the conjecture which says the following.  It says that, well, here, as I was mentioning, we don't know, we know that the poles live in these  red vertical arrows, but we don't know where they are exactly. And this conjecture tells you that,  well, there are poles, this function as poles precisely at this point, i. So there are poles  in here. So here there is a pole, here there is a pole, here there is a pole, etc. And of course,  because of the periodicity, then you would have infinite poles living here, etc. And moreover,  it's going to tell you that, well, not only you have poles there, but the order of the pole is given  by this precise number. So it's equal to the dimension of the algebraic cycles  with q-peficients, but up to numerical equivalence relation. So the dimension of this q-vector space  is actually the order of the pole at this precise point. And, well, and we impose this for every  i between zero and the dimension. That is for all these i's. And as you see here, it's called  strong-tate, because it's actually stronger than the classical-tate conjecture. So it's a theorem  of tate that, in fact, the strong-tate conjecture, well, implies the classical-tate conjecture,  actually for every l different from p. And moreover, if you impose the conjecture dlx,  this grotten-dick conjecture, then it turns out that the converse also holds, so it becomes an  equivalence. And this is also true if you have for the conjecture within the p-version of the  conjecture. Okay, so these are the conjectures that I would like to establish non-commutative  counterparts of. Okay, and, well, of course, let me say these conjectures are wide open in general.  It's known in very particular cases. Tate-proof it in the case of curves. Nowadays, a lot of people  were able to prove it in the case of k-3 surfaces. Okay, so now let me,  just before finishing this commutative part, let me just mention something really quick.  Let me just mention here that this function, this very interesting function also admits  this function, admits a functional equation. And this is a theorem of art in  and grotten-dick. So in this function, you have this relation between, you can relate the function  at s with the function at s minus or d minus s. And the relation between the two is given by this.  So here you have the Euler characteristic of your scheme s. And here you have a constant,  which is minus the Euler characteristic, the dimension divided by two. Okay, so it's about  that function over there. Okay, so this is what I wanted to say about the commutative world. Now,  let's move on. So what do I mean by non-commutative geometry? So there are different people  look at non-commutative geometry in a different way.  Okay.  And so for me,  non-commutative geometry will be actually non-commutative algebraic geometry.  Okay. So which is a subject that goes back to the Moscow school, let's say, goes back to  Manin and its students, etc. So the idea that we put these words, so there is this  standard definition of Bondow and Kapranov, that what is a differential graded category,  so a DG category. So this is simply a category A, which is enriched over complexes of K vector spaces.  Okay. So the home spaces are just complexes, not ordinary sets. And you have a lot of examples  whenever you have an algebra differential graded algebra, then you have one of those,  which is just one of those with the simple objects, with the single objects. And whenever you have a  scheme, then you can look also at this category of perfect complexes on your scheme. So these are  complexes of OX modules that locally are quite asomorphic to bounded complexes of vector bundles.  And then this carries a canonical DG enhancement. So these are two examples. So  algebra and geometry both give rise to DG categories. And this very famous example of Williamson  says the following. So if you look at the projective line and you look at modules over the  projective line, that category is very far from modules over an algebra. But if you go towards  this drive setting, then this category is actually more equally equivalent to an algebra, and you  have this algebra of matrices. So in other words, this is telling you that the projective space in  this drive setting, it's actually a fine is given by an algebra, but by a non-commutative algebra.  And then in this world, there is this notion of smoothness and properness. So this is good to  maximum. So if you have a DG category, you call it smooth. So by definition, this is just saying  that when you look to it as a bimodule over itself, then that it's compact in the category in the  drive category of a bimodule. And proper simply means that well, if you fix two objects, any two  objects of your category, that's a complex, you look at the cohomology, and you ask this dimension  of this cohomology to be finite dimensional, and moreover, that the total cohomology to be finite  dimensional. And this for any two objects, x and y. Say it again. Yeah, I'm working over a  base field, it's always our little case. It can be more general, you can work over a commutative ring  or even over a different base, it can be very general. And then the key remark is that if you  have a scheme which is smooth and proper, then that implies that well, this category, it's actually  smooth and proper in this sense. And also the converse holes. So this perfect complexes on a  scheme actually reflects these properties of smoothness and properness. And so the idea here,  it's now, we would like to do geometry, not with the skin, but with an arbitrary dg category,  a smooth and proper one, which mimics smooth and proper schemes. So what can be done in that case?  So in particular for this course, I would like to phrase, to formulate the non-commutative  counterparts of all these conjectures. So if we want to do that, in particular, I would like to  have some kind of non-commutative valcomology theory. So something that works not just for schemes,  but in full generality. So for this, let me talk about a topological periodic  cyclic homology. So this, let me put some names on the board. So this to Alencon and many others,  these two, Essofold, Tuschels and many others. So let me give you an idea of this. So I suggest  to look at the course of Tina last week and also of Kaledin, if you want to learn more about these  things. So let's suppose to simplify that I just have an algebra, an ordinary algebra with A.  What can I do? I can do the following. I can do the following construction. I can look at A,  A cross with itself over the base field, A cross with itself, cross with itself over the base field.  And now I can use the multiplication to define maps. I can multiply these two, that gives me a map,  but I can multiply by this order that gives me a different map. So it's not necessarily  commutative algebra. And here also I can multiply these two or these two or even these two. So and  in fact, I get this simple initial gadget and I can totalize it. And if you totalize it, you get  something that is called a Uschild homology of A. And that is something that is K-linear.  So for example, the Uschild homology of our base field is actually the base field itself.  And now you see that at every level what you can do, you have an action, you can  promote the factors in the C-click order. So if you put that information all together,  that gives you an action of the circle. So you have an action of the circle on this object.  So you can do a take construction. And if you do this take construction, what you end up is that  with the periodic cyclical homology of A. And this again is something that is linear, but  it's more than that. It's actually periodic. So for example, if you computed over the base field,  that's a ring of polynomials on one variable, of low-run polynomials on one variable,  of variable being a minus 2, of degree minus 2. So you get something periodic, too periodic.  So Gonzalo, there's a question, maybe I've missed it.  Is there a way to define a smooth and proper relatively? So if I have a map of DG categories,  I can define a smooth proper morphism. Yes, there is. I don't remember from the top of my head,  but yeah, there is. And that it's also in the literature. I can dig in and then forward the  definition. Yes, there is a relative setting here. And now what I want to consider is let's go to,  this is an arbitrary characteristic. Let's go to characteristic P, positive characteristic.  Okay. Now in positive characteristic, I can do this topological version of periodic cyclical  homology. And roughly speaking, what I do is that I change the base field. And now I do tensor products  over the sphere spectrum. Okay, so I get the base, which is even more initial than the original one.  And I can totalize this construction and get something which is a topological version of  the social homology, which is still something k linear. So if you take the thh of little k,  that's a result of Boxsted. So this is a ring of polynomials on one variable of degree 2.  Okay. But then it turns out that there is a relation between the two. This is actually a module  over this. And so one thing that you can do is that you can take the tensor product over k.  So in other words, you can think about this as one parameter deformation  of official homology on this parameter here. So when you take the fiber at zero,  you get the original ocean homology. That's one way to think about it. And then here you can mimic.  So again, you have actions of the cyclic groups, these permutations, if you put them all together,  that gives you an action of the circle. And you can do a take construction in this topological  sense that's called Greenleys construction. And you get this topological version,  which now no longer is k linear. So for example, when you compute it over the base field,  what you get is the ring of vectors where you have added one variable of degree minus 2.  So you see that from this computation and this, you see that this one is actually a characteristic  zero lifting of this one. Because if you reduce mod p, the vectors you end up with your base field.  And this is true in general. In fact, this is a characteristic zero lifting of the periodic  cyclical model. And now here you can do something that you cannot do in algebra, which is I can  invert p. And that will be important for us. Because if I can invert p, then a new feature  appears here in the topological world that does not exist in the algebraic world. And this new  feature is this cyclotomic Frobenius. So again, I mean a positive characteristic. So let me  mention what I'm going to say. It's in this topological language, but this is originally  defined by Kaledin. So Kaledin is the first person that actually make this rigorous. So let me just  make a remark. Suppose that you have a commutative algebra. Okay, so if you have a commutative  algebra k algebra and over a field of characteristic zero, then what you know is that these two  relations hold. So if you do a plus b for any two elements, that's a p plus b p. And the product  is also a p times b p. But now if you if you are no longer working with commutative gadgets,  then these things don't hold. So you don't actually have a Frobenius. But the Frobenius will appear  not on the algebra itself, but will appear in this realization in this. So if you have a  if you have a smooth proper, smooth proper dg category, then you can look at its thh of a.  And this thh of a, it's actually something called the cyclotomic spectrum,  in the sense that it doesn't not only it has this circle action that I've used here,  not only have this circle action that you use to define the tp, but moreover, as this map  that goes from thh from a to the thh of a and the take construction with respect to the  cyclic group of order p. There's a question, Monsalor. So what about the definition of tp  of k and characteristic zero? Yes, how we define it. Yeah, I mean, we can do that similarly,  but I'm just focusing in characteristic zero for for. But are we defining what about  with the are we defining zero zero typical v vectors just to be k itself or?  No, no, I'm just I just want to work here in the case when I'm working with topological.  So, okay, I'm just working in characteristic zero. Okay.  And so here I want I have this map, which is s one equivalent.  So here we have an action of s one and here you have a residual action of s one also,  or if you want to s one module, the cp that identifies with s one. And so it's this two  data is called the cyclotomic spectrum. I mean, it's more general, but I'm over p. And then  out of this data, what you can do is you look at thh of a and then you look at thh of a and then  so you have a what you can do here is that you have a canonical map from the homotopy  fixed points of the circle action to the take construction. You have a canonical map here.  And this take construction is what we call the pp of a and moreover, it turns out that in this  setting. So if you are working with the smooth proper digi category, it turns out that this is  similar to the thh of a where you have the take construction of cp and then you take the  a multiple before fixed points with respect to s one. So it's a technical result. You have this  equality, but that implies that you have another map here. So you take this original map and you  take the multiple fixed points with respect to s one. And now you have these two maps and it turns  out that the kernel of the canonical map as a scale linear. So if I invert p that kernel disappears.  In other words, this map here becomes an isomorphism. So if it becomes an isomorphism, then I can  define a cyclotomic Frobenius here, which is simply you take this map and you compose with the  inverse of the canonical map. And that gives you a Frobenius from tp one over p if you invert p to  tp of a where you inverted p and moreover this map is an isomorphism. So the Frobenius exists here  after inverting p, but it doesn't exist originally in contrast with the commutative world. It doesn't  exist on the algebra itself only on the invariant. And let me make some remarks about this  cyclotomic Frobenius. So first of all, it is not linear. So it's not a z2 graded map.  So it's not linear here. So in contrast, you only have this relation when you change from n  to n minus two. So these are equal up to this multiplication by one over p. And  and moreover, these maps that you have, they are not  as in commutative geometry, they are not capital K linear. They are only sigma  semi linear. Okay, so in particular, if you if you are working over a finite field,  one thing that you can do is to compose and get the actually K linear maps. So you simply compose  the phi n r times that gives you something which is K linear. And then the relation between these  different Frobenius when you change from n to the Frobenius n minus two, it's actually  multiplication by one over q. Consalos, so there is a question. Don't we need  something like K being perfect for Frobenius to be an equivalent?  So it comes, everything comes from my assumptions on A being smooth and proper.  This will imply the equivalence on the Frobenius.  Okay, so now this suggests that, well, these are, you should think about periodic cyclical  homology and topological Schiller homology as non commutative well conjectures.  So that's what we are going to replace actually the the ram homology and the crystalline homology.  But there is something that we can do something. We can do the following.  You can let me just write on the side. Can you leave the blackboard?  Oh, yeah, of course. Sorry. So let me just write an aside here about non commutative  what you think realizations. So when we wonder, well, we can maybe extend all the commutative  results to the non commutative results. And there is this result saying that well, you can look at  smooth schemes. And you can go to Morale-Weyvotsky stable or multi-peak category stable A1 or  multi-peak category of schemes. And then on the non commutative side, you have  DG categories. And we can construct a non commutative version of this Morale-Weyvotsky category.  So I'm not going to define this in this course. It's just an aside. I'm just telling you an aside.  And these things are related because if you have a scheme, you can pass to the perfect complexes.  And then we would like to relate the two. So here the relation is as follows. Here we have a  motivic spectrum that represents a multi-peak theory. So we can look at modules over the KGL  with quick coefficients. And then you see here, this is a covariant procedure. And this is a  contravariant procedure. This is contravariant. So we need a duality. So we can dualize  here. And then it turns out that there exists a functor here, which has a lot of good properties.  It is fully faithful. It is a tensor functor. And it admits even a right adjoint.  And then as a consequence of that, as a consequence of this breach between the commutative and the  non commutative world in this setting, we have the following corollary. So we can do the following.  Suppose that you have a realization that you'd like to be interested in. So here you have Morale,  you have Wewocki category of geometric motives, and you have a realization.  And what you can do, you can pre-compose with duality. So let's say that we are with quick  coefficients. We can pre-compose with duality and then go to schemes. So we have this contravariant  functor on schemes. And then I can modify this realization. So here I have a tight object,  which is nothing but this one. And then I can somehow trivialize it by considering modules  over this sum, over all possible powers of RT. So I get a modified realization. And it turns out  that if I do this modified realization, when you pass from schemes to DG categories in this sense,  then all these kind of modified realizations, they can be extended to the non commutative world.  So you get, there exists a non commutative version of the realization. So in particular,  you get this result, it gives rise to a lot of non commutative  motive realizations. For example, an adic version, an odd version, a Durand-Betti  version that, for example, can lead you to define what are non commutative periods, etc.  And how is this corollary obtained? Well, this corollary is obtained as follows.  Let me just explain this. So you have, you can do exactly the same thing in here.  But instead of KGL, you can use the spectrum that represents motivic homology.  Okay. You can do exactly the same thing.  So I'm going to, sorry, I need some space. I'm going to erase the corollary. But  so I just want to have this realization and I want to extend it to the non commutative world.  And let me do it here. So you'll do exactly the same. And now the first point is that,  well, this is DM, the big DM, this is a result of Rondings and Polk,  post-vener. And so this composition is your motivic realization, map of Vybocki. So here you  have your realization going towards your key. And the second observation is that the KGL  with QP-efficient is actually this trivialization. So you get this by M to N. So you have this  fact here. And so this is simply telling you that you can base change here from  coefficients on the motivic homology to KGL. And then you can do the same thing on the target.  So if you do the same thing on the target, you can simply look at modules on the sum  of your realization of your date motive. And you get this induced map here. And so your  non commutative extension, it's actually we use this functor. And then what you do is that you use  the joint and then you use this induced map. And that is your way to extend any kind of  realization as long as you modify it, then it can be extended to the setting. Of course, this  is a bit is cheating, right? Because you are defining this using schemes. You are just using  an adjoint. You are just building this functor using schemes. You look at the closest scheme  associated to your non commutative motive and apply the classical invariant. So it has a lot of  problems. For example, these things are not monoidal. For example, you don't have finiteness on this.  So it shouldn't actually be called a realization. But it's something that it can be done. You can  always extend anything in the commutative world to the non commutative world via this adjoint.  It's a bit cheating. Okay, so now we are ready to come back to our original goal. And now we can  actually formulate all the conjectures. And so, yeah. Yes.  Yes. No, I'm getting, you have some kind of, I'm just saying that you suppose that you have a  suppose that you have a realization. So it's this, so this map, this realization.  And then as soon as you modify it in the sense that you trivialize the type motive, then you can  extend it to any DG category by via this procedure. But of course, this is using an adjoint. So it's  not a correct way to do it. What you want to do is to have an intrinsic definition that is not using  actual schemes, right? Monsalou, what was the question? We did not. It was just to explain a  little bit better the board. Okay. Okay, so now what's the replacement of algebraic cycles?  It's the grotendic group. So now I have a smooth and proper  k-linear DG category.  And we look at its k0. Well, its k0 is just a k0 of its drive category, which is a triangulated  category. Okay. And now we choose an element in here in the k0, and we can phrase all the  similar equivalence relations. We can say, we can talk about new potents equivalence.  So you say that the cycle is no potently trivial if by definition there exists an integer such  that when you tensor the cycle with itself n times, then this is something that lives on the k0,  a tensor with itself n times, that this is actually become zero. So you can make this  definition.  How about homological equivalence? Well, we now have this, our non-committative  homology theories that I've explained. So you define homological equivalence  as follows. So you simply say that here you have, so this comes equipped also with charm  characters. So this is something that I will explain better on Wednesday. So here you have  charm characters defined on the k0 to the positive part of the periodic cyclical homology and the  positive part of the topological periodic cyclical homology. And you simply say that, well,  it's homologically trivial if the charm character of it becomes zero. And you also have the numerical  equivalence. So here, how do you define intersection of cycles? Well, what you can do is you look at  the k0 and you define a pairing on k0 as follows. So if you have a module and another one and look  at the corresponding classes, one thing that you can do is to look the homes on the drive  category of your A, look at the homes from m to the shifts of m, and look at these dimensions,  and then take an alternating sum of these dimensions for m varying on the integers. And so in this way,  we extract this number and this pairing, they extract the pairing on the k0, which is neither  symmetric, neither is Q symmetric. But since A is smooth and proper, it turns out that you'll  have a serfometer and using this serfometer, you can prove that the left and right kernel,  which are priori are different, are in fact the same. So you can actually define  something numerically trivial if this pairing, let's call c of alpha beta, it's equal to zero for  every beta. Okay, and then we have the similar remarks. So if you have a cycle which is a  new potent with trivial, will be homologically trivial, will be numerically trivial. So that  implies that on the k0, you have all these discussions, module of the new potent, module of  the homological module of the numerical. And again, this is always finite dimension.  Does it mean this new potency goes to zero? That means the order characteristic is zero?  No, it means that beam zero here means that this pairing vanishes for every b and alpha,  such that this pairing vanishes for every b. The new potency implies this  the equivalency mentioned. Yes. The homological. Yes, implies the homological, yes. It's an  exercise, yeah. If you have something new potentally trivial, it will be homologically trivial.  Okay, and now we using this we can, we can as you expect, you can do formulate this  this non commutative counterparts. So we have this non commutative  grottenby extender conjecture. So what you say here, it's here, it's a conjecture, let's see,  D, so this is of type D, so D and C of A. So this simply says that the k0, when you mod out  biological or by numerical, they are the same. Also, you can define this non commutative version  of Vojvodski, new potency equivalent, new potency conjecture.  So let me write conjecture V and of C. So again, it's that the k0  up to new potent equivalence is the same as the k0  up to numerical equivalence. So once again, if these two, if these two are the same,  then these two are the same. In other words, this conjecture implies the preceding one.  And of course, we can also define the non commutative version of Balinson conjecture.  So over a finite field, that in fact,  the k0 is insensitive to all these equivalence relations.  Okay, so these are the analogs. And then we can also go further and define the non commutative  version of the V conjecture. So what do we do here? So again, we are over a finite field.  And so if we have a smooth and proper Vg algebra, sorry, Vg category, you will look at this finite  dimensional capital K vectors place, and it's equipped with this automorphism, the Frobenius  zero. So it's just the cyclotomic Frobenius composed r times. And the same thing that Tp1  over P comes with this Frobenius. Here composed r times. And then the conjecture is that  the conjecture is that if you take an eigenvalue, if this is an eigenvalue of the Frobenius zero  respectively of the Frobenius one, then well that these numbers are Osprey numbers. And secondly,  that complex absolute value is equal to one, respectively equal to square root of Q. And  this for all conjugates of London. And also in this non commutative world, you have this analog  of the proposition, which says that well, if there exists an integer such that  when you multiply your lambda by Qn, and if you get an algebraic integer,  then it turns out that the ellatic absolute value of lambda is actually equal to one for all  conjugates of lambda and this for every prime different from P. So you also have this result.  Of course, this conditional in this case in contrast with the commutative world.  And also we have this. Consalor, how much time do you need still?  I just have one and a half. It's pretty quick. Thank you.  So let me just say that in the.  Maybe there's a question. Is this, are these conjecture Kfury, Morita invariant?  Yes, of course, yes. Yes, because anything that is Morita equivalent will actually have  the same motive. Let me give you a very technological answer. They have the same non commutative  motive and all these conjectures descend to non commutative motives. That's the approach that  I will explain on Wednesday. Yes. And also we can talk about the non commutative asset value.  That the functions.  So here, of course, we cannot count points. The only thing that we can do is we need actually  an embedding of capital K into C. And then if we choose this embedding, we can define this  case. So the only thing that we have is the co homology. So what we can do is, is consider  these functions, the determinant one minus QS and then the Frobenius at zero. But then we need to  use this embedding to go towards C because here we are using complex conjugation in order  that to make sense. Then you take the PA here and then you need to go to C. And then you can do the  same story here. You can define it in this way. So determinant one minus Q minus S, the Frobenius one  going to C on the PP1.  Okay. So we have these functions, of course, over C. They are  meromorphic. And what this is telling you is that, again, by by own definition, they are  they are periodic with this period. And by by what our non-committative version of the  Velconjecture is telling you is that the poles are actually here in this case and are actually at  one half in this case. So the when we change the when we change the embedding, these functions change,  but the place where the poles are actually do not change. And finally, let me just say that we also  have the non-committative version of the PIN Conjecture. So I recall you that we are over a  finite field. So we are over a finite field. And the conjecture is as follows.  So we have this non-committative version for L different from P. So here what you do is you look  at the k theory of A and you base change it to the field extensions. And then you do the  both field localization with respect to topological k theory. And now you look at the pi minus one  of this disabillion group. And then you do the the TL of this. So you do the the the ELADIC  TATE module of this. And you ask it to be zero for every n greater or equal than one.  So it's this vanishing conjecture saying that well all these TATE modules, ELADIC TATE modules of  disabillion groups vanish for every possible n, for every possible extension. That's one way to  phrase it. Then we have the P version of this. So here you have the churn character going towards  TP zero of A. So when you inverted P and it plants on the invariance under the  cyclotomic Frobenius. So when you change from Q to QP, this is subjective. And you also have a  strong form of the TATE conjecture. So you say that the order at zero of this function  A S. So you say that well there's actually zeros are here, but there is actually a pool here.  And this pool is the order of this pool is given by the dimension of the k zero of A  module on numerical equivalence. So we should like to emphasize that this thing here actually does not  depend on the embedding. Because you can rethink about this if you see the definition. It's just  the algebraic multiplicity of one of these operators. It does not depend on the embedding.  That's what it's the order of the pool there. And then we also have this result saying that  this strong form of the TATE conjecture implies the TATE conjecture itself.  And moreover if you add this non-commutative version of Rotendex conjecture, this becomes  equivalent. So what is the upshot of all this? The upshot of this is the the following theorem,  which is the very last one and that will end there. It's just saying that all these definitions are  the correct ones in the following sense. So it's the following. So if you take a smooth proper  scheme, okay, and then you can do two things. You can look at the conjecture for x, where c is any  one of these conjectures. So d, v, so Rotendex, Vojvodski, Bielensson, Veil, Tate, Tate, Tate,  and many others, in fact. You can do this. Another thing that you can do is look at the  non-commutative version of the conjecture for the associated djicothexam. And so if this conjecture  holds, it turns out that this implies that the non-commutative version also holds. But what's  interesting in the theorem is that this is actually inequivalent. So this is saying that  these classical conjectures, all of them, that are formulated in this setting of algebraic  geometry, in fact, they make sense in this much larger setting of djicothexam, smooth  proper djicothexam. And if you plug, if you attack this particular kind of djicothexam,  you recover the original conjecture. And this is true for these conjectures that I explained  today, but there are many other conjectures for which this holds. I don't have time to explain.  And now what is the idea is to explore that here now I have non-commutative techniques to  try to attack the right-hand side and as a consequence, attack the left-hand side. And  this is the goal of the following lectures. Okay, thank you and sorry for going over time.  Okay, so thanks for the talk. So I'm sorry, so we are aware that there were serious problems with  connection for certain of the attendees. So we are sorry, we will try to make it better, but  unfortunately, it seems it's an internet problem. In any case, you'll have a YouTube video if you  want to, if you could not follow the talk. So now we have questions. So I saved a few questions,  three questions, three questions actually to, for the end of the talk. The first one is someone  is asking, maybe if could you repeat what YTP is related to the Motivic stuff here?  You mean Motivic stuff, you mean the commutative world, I imagine that's the question, right?  Yes. So the relation is as follows.  Yes, so the relation is as follows. So if you take the TP0 or 1 of your, so let me do like this.  TP is too periodic and you take the TP of this TG category, you take your scheme and this one,  it turns out that this gives you the positive part and the negative part, you'll get the sum  of the even groups of the crystalline comology of your scheme and on the negative part, it gives you  the sum of the odd comology groups on crystalline comology. So this telling you that, well, this  invariant, of course, after inverting T, this tell you that, well, when you computed on this  DG category, you get the crystalline comology, not the individual pieces, but up to parity.  And moreover, now, you also here, so when you see that this TP comes equipped with  the cyclotomic Frobenius, but this cyclotomic Frobenius, as I made this remark, is not Z2 graded,  okay? It's not a Z2 graded thing. When you change to N to N minus 2, there is a scalar  that appears. So let me say that I'm with P0 and P1, 0 and 1. And now I have the cyclotomic  Frobenius 0 and 1 here, and they correspond to something here, and they here, they correspond  to the sum of the even cyclotomic Frobenius on X, but with a certain scale multiplied by it.  And here, the sum of the cyclotomic Frobenius, but multiplied by P minus T minus 1 over 2.  So intuitively speaking, this is telling you that you lose the weights. When you pass from X  to this category, you can recover everything up to weights. You lose the weights. But for the  conjecture, that is enough. You don't lose anything in terms of the conjecture. So I want to testify  here if the colors are completely unreadable. So don't use colors. Do you use colors? Yeah.  Maybe the green. This is better. Slightly. Okay, just a remark. So there was another question about  the relation. So I just say it again, about the relation between the NC conjecture and the  usual one. And so that's on the blackboard. They are equivalent, the smooth proper case.  Which one? No, the non-commutative and the usual one. So that's what you stated.  And also there's a question. So do any of these conjectures have a mixed characteristic analog?  Or formulation?  I don't know. You want to be working over DVR or something like that? I guess so, yeah.  Yeah, I mean, I haven't talked about that seriously. But I mentioned that, yeah,  it's likely that you could expect something like that. Yes. Okay.  So this is related, in fact. So there's another question about the relative versions of this  conjecture. Are there relative versions of this conjecture? I don't know. I mean,  yes, I mean, in the classical world, yes, I think there are some of them admit relative versions.  Yes. I haven't explored that. Yes, that's a good question. That's a good thing for the future.  So how do things work in families and things like that?  Okay, so I'm not sure to understand the question correctly that knows, but I just read it. So  what is a tensor m? And does it agree with the perf x to the m in the geometric case?  Yes, yes, yes. When you do this,  this is this.  And my schemes are nice enough for which this kind of phenomena holds. Yes. So  in the, I imagine that this question is about the new potency equivalence relation because on one  side, it was on cycles on xn on cycles on xn and the other side was on the k0 of your a tensor m.  And when a is of this form perf of x, it's yeah, there is this.  Thanks. So that was that was a question of Remy van Doeben de Brun. There's also a question by  Ola Sende. Is vanilla potency conjecture equivalent to the Baylinson-Soulet conjecture?  And Baylinson is only over a finite field. Baylinson-Soulet.  Ah, Baylinson-Soulet, maybe you can recall me what it is.  Since vanishing of vanishing of motivic homogene in degrees in negative degree.  Yeah, I don't remember. I have to dig in. Yeah, I think I remember that there are close  connections between the two. I don't know if it's exactly the same, but yeah.  Yeah, I don't remember from the top of my head. So I think there's no  direct relation. Okay, if you put all conjectures. Oh, I think, I think the Baylinson-Soulet,  it's actually the strong, it's actually the tight conjecture. And then you mod out by the  numerical equivalence. Baylinson-Soulet? Yeah, it's not. No, it's just vanishing of  motivic homogene in negative, in a simple degree. But okay, so I have to.  Could you please state precisely how the LED state conjectures are formulated via TP?  No, via TP, I'm formulating the TP version of this, right?  So for this, I'm using the TP. It's actually, I'm using TP to formulate the P version of the  T conjecture, right? And I formulated the L version of the T conjecture using KT, right?  And the question is if you can do that. Yes, I mean, that can be done, but it's using these  results of Thomasson that tells you that, I mean, the relation.  The relation is that you have a Atya-Eirsbrug spectral sequence that goes from a latic  cohomology to etalcateria. So you have something like this. So rationally,  so they prove that it degenerates. So you have something like the etalcateria of X.  When you complete at L and then you rationalize, that is in fact the sum of these  latic cohomology groups. And so here you already have a link between the etalcateria with KT,  but it's an etalcateria. But then you have this beautiful result of Thomasson that tells you  that you can write this algebraically because you have completed at L. It's in fact to take  the KT of U and to localize with respect to complex KT, and then if you complete at L,  actually you get the same spec. And now here you have KT, so something that you can phrase for  any eticateria, for any generality. And then if so you see that then the reformulation of the  conjecture would be saying that if you go from the Schoen character from the K0 to here, it actually  lands precisely on the things that are stable and the action of the absolute color.  So I think it was already phrased like this in an old paper of Friedlander.  So I have another question. So are there any strictly non-commutative applications of the  conjectures or any of the conjecture? Yes, I mean this is somehow a motivation. Usually,  let's say that if we divide the world in two, much more people would work on this side than  on this side, I agree. And so people are motivated on trying to prove this. But on this side,  these conjectures as we'll see on Wednesday, for example, will allow us to have a conditional  description of the category of non-commutative numerical motives, something a bit analog to  what Milne has done, described in terms of veil numbers up to a certain action of the absolute  color group and things like that. So yeah, I mean we will attack, for example, this conjecture  next time on Wednesday. For example, we'll prove that all these conjectures hold, if you put here  an algebra, which is finite dimensional and a finite global dimension, not necessarily commutative.  And all these non-commutative conjectures hold. So yeah, so we can, that's also, there are two  paths. So there's Wednesday and Thursday. Wednesday, we go in this non-commutative direction and Thursday  on this commutative direction. And in both cases, we are going to explore the link between these two.  So there's another question about, so are there any new cases known where the conjecture holds  in the non-commutative setting? What happens on Thursday, what I will explain is a way to prove  this conjecture in some new cases where you don't use geometry. What you do is you prove this conjecture  using some non-commutative techniques. And as a consequence, you get this conjecture  for axes which were known, not known previously. If I understood correctly the question, is this  more or less what he was asking? If I can, new cases, yes, on the very last talk, I will prove  the classical conjectures in new cases. And the way to prove it is not using geometry, but using  this viewpoint. It's not with the viewpoint. Last question, is there a paper attached to this  mini course where we can read more details? Yes, absolutely. In fact, there is a survey on  archive whose title is equal to the title of this course. So I think which is called the  non-commutative counterparts of celebrated conjectures, if I remember.  Okay, so there's no more question in the chat. It seems, are there questions in the room? No?  No. Okay, so good. Thanks again. I'm going to say hello. Sorry for going over time. I got a  bit confused with this time. Yeah, okay. I'm sorry for the voos who had problems with the connection.  Okay, so we meet again in chapter 6.
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