 Thank you so much. I would like to start by thanking Paul, Grigory, Frédéric and Irving for the invitation to give this course. I would like to start, I would like also to thank the HOS 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. I will always use this notation and see for non commutative, non commutative counterparts of these conjectures. And then in the second lecture, I'll be talking about applications of these 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 these 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 greater or equal than zero. And in the case where P is positive. I will write WK for the ring of PTP called V factors and capital K for its fraction field. So we need to invert P. So for example, if this is FP, then this is just a ring of patic integers and this is the ring of patic numbers. And then I will also write by Sigma. So this is a morphism on capital K induced by the Frobenius on little K. So here I raised to the pith 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 case scheme. Of dimension D. Okay, so these conjectures are almost all of them about algebraic cycles. So let me say something about right cycles. So we can consider this graded cube vector space where it's graded by the dimension. So I look at algebraic cycles of dimension I X with rational predictions. So we choose to consider one algebraic cycle. We can impose several different equivalence relations in here because this is a huge graded cube vector space. So there are a lot of different equivalence relations, for example, the rational equivalence. The equivalence says that if you have an algebraic cycle it's rationally equivalent to zero by definition, if there exists an algebraic cycle better in the product of X with a project of line in such a way that your algebraic cycle is the difference from the relation of beta 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. There are a number of times so let's say cross it and time so this is an algebraic cycle in in of co-dimensional I on X but on X cross with itself and times that this algebraic cycle here it's actually rationally equivalent to zero. If there is an integer such that the cycle disappears you'll say that it's new. So, Gonzalo, can you really write really bigger because it seems that bigger several yeah really bigger two times more maybe because there are several people complaining. And then there is homological equivalence. So here you, you choose a veil homology theory. So there are a lot of them homology theories I'm going to consider for our purposes. So the rank homology theory in characteristic zero and in characteristic P crystalline homology theory. And so any very homology theory comes equipped with comes equipped with a cycle class map. So we have a cycle class map towards the co-mology to I X twisted by I and then you say that your algebraic cycle. 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 the algebraic cycles of co-dimension I with coefficients and up to rational equivalence and you can pair them with algebraic cycles of complimentary co-dimension. And from here you can extract a rational number so 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. Okay, and you say that a cycle is numerically equivalent to zero if by definition, this pairing vanishes for everybody. 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 rationally equivalent to zero then it is new potently equivalent to zero. And then it is homologically equivalent to zero. And it is necessarily numerically equivalent to zero. And 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 difference, these different equivalence relations. And the interesting point here is that when you when you impose this, this quotient 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 algebraic cycles of co-dimension two in your surface, the dimension is actually infinite dimension. Okay, 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 this difference equivalence 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 Gonsalow, maybe there's a question that you could answer. So for all, I just read so someone is confused by 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. So there's a question about the fact that you start in characteristic P but you seems to work over C. I'm working in characteristic P but my P is greater or equal than zero. Okay, can be zero. 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, yes. So this conjecture simply says that algebraic cycles up to homological equivalence or algebraic cycles up to the numerical equivalence are the same. So it's saying that there's 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 or equal than two. And in characteristics here even of dimensionless or equal than four, and even for a billion varieties in characteristics here and these are results of Lieberman, some old results and there are many other cases. So it's a conjecture of a Votsky called a Votsky new potence conjecture. So it's a conjecture from the 90s. It says that get algebraic cycles up to new potence equivalence relation are the same as algebraic cycles. It's a conjecture to the numerical equivalence 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 difference between the two. And it's interesting in the sense that it does not depend on any vehicle multi theory. And this is still, of course, wide open. It's known in particular cases, for example, when the, when the dimension of x is less or equal than two, this is an independent result of Votsky and Cleverson. Then there is also a conjecture of Baylinson 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 equivalence and algebraic cycles up to numerical equivalence. So in other words, all the difference equivalence relations are the same. This is a finite field. And again, this is wide open is known in some cases, for example, for curves, that thanks to the work of Suley and Bruno Kahn, etc. And of course, this Baylinson conjecture implies the Votsky conjecture. Okay, so these are three conjectures about these difference equivalence relations. Then another conjecture I wanted to talk about is this famous veil conjecture. This is from the 1940s. So it's a certain 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 co homology of your algebraic variety. 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 this for all possible conjugates of left. And in fact this conjecture was of course in the in the 40s there was no crystalline co homology so it was not phrased like this, but it turns out to be an equivalent for relation. And it was proven. It's now a theorem it was proven by the lean that this conjecture. Okay, and let me say. Again, an interesting fact is that, well, suppose that we still didn't have the links 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 London is equal to one for all conjugates of London 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. So this motivates the study of the asset value that the function. So let me recall your. So the asset value that the 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. So it's a complex valued function. So this product converges when the real part of us is greater than D greater than the dimension. So we have. Yes, and now if we use. So if you combine the result of the link with some work of Bertolo and crystalline homology we can rewrite this function so let me just mention you that this number that it's here it's actually the cardinality of the residue field of the point, which is a finite field. So this is what you're 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 links work with the work of Bertolo crystalline homology you have a comological 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. That'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 overseas. So we have our function. So we know that it's. It's defined on the south plane. So it's well defined on the south plane. It actually extends to unique meromorphic continuation to the entire complex plane. This for what tells you that. The links conjecture is going to tell you where the where the zeros and the poles of this 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 this vertical lines. And in these regions in blue. These are precisely the regions where the zeros will occur. So here you have the zeros 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 two pi I over the log of q. So you'll have a periodicity on this function like this. Okay, this is simply due to the fact that you are using. Short is a sequence where you have two pi I log q that. So you have this short is a sequence where you use when you use complex exponentiation. So you have this. Okay. And then finally, I would like to mention one final conjecture, which is also the famous conjecture of Tate conjecture from the 60s. It says the following so here I will be over a finite field again, and the conjecture. Three of them. Three versions. So you have one version for L prime different from P. So here what you say is that you have the cycle class map. Going to the et al, the electric homology. Going to a lot of ecology. This elements in fact the land in those elements that are fixed under the action of the absolute color group. And the conjecture, it says that well if you bet if you change from q coefficients to q L coefficients then this cycle class map becomes surjective so any class invariant and then absolute color group comes from here itself right. Then there is also a P version of this. So you, you do the same. But now you use crystalline co homology. So we use crystalline co 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. So it says the following. It says that well here as I was mentioning we don't know we know that the, the polls live in this red vertical arrows but we don't know where they are exactly. And this conjecture tells you that well, there are polls. This function as polls, precisely at this point I so there are polls in here. There is a poll here. There is a poll here. There is a poll, etc. And of course, because of the periodicity, then you would have infinite polls, leaving here, etc. And moreover, it's going to tell you that well no, not only you have polls there but the order of the poll is given by this precise number. This is equal to the dimension of the algebraic cycles with q coefficients but up to numerical equivalence relation. So the dimension of this q vector space is actually the order of the poll at this precise point. Well, and we impose this for every eye between zero and the dimension that this holds for all these eyes. And as you see here, it's called strong tape because it's actually stronger than the classical tape conjecture. So here of tape that, in fact, the strong tape conjecture well implies the classical tape conjecture actually for every L different from P. And moreover, if you impose the conjecture on the effects this growth and the conjecture, then it turns out that the converse also holds so it becomes an equivalent. And that's also true if you ask for the conjecture within the P version of the conjecture. Okay, so this is, these are the, the conjectures that I would like to establish non commutative counterparts. Okay, and well, of course, let me say these conjectures wide open in general, it's known in very particular cases, take proof it in the case of curves nowadays a lot of people were able to prove it in the case of K3 surfaces. So let me, just before finishing this commutative part, let me just mention something really quick, let me just mention here that this function is very interesting function also admits this function admits a functional equation. So this is a was a theorem of our team and grotting it. 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 other characteristic of your skin as. This is the function at S minus which is minus all the characteristics of the mansion 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. There is a different commutative geometry in a different way. So for me, non commutative geometry will be will be actually non commutative algebraic geometry. So, which is a subject that goes back to the Moscow school, let's say, goes back to my name. So the idea that we put this word so there is this standard definition of Bondow and Capron of that what is a differential graded category, so a DG category. So this is simply a category. A, which is enriched over complexes. Of cave at the 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 we have one of those, which is just one of those with a simple object with a single object. 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 as a morphic to bounded complexes of vector bundles. And then this carries a canonical DG enhancement. So these are two examples of 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. If you look at modules over the projective line that categories very far from modules over an algebra. But if you go towards this drive setting, then this category is actually more equally equivalent, more equally equivalent to an ulcer and you have this ulcer 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 ulcer but by a non commutative ulcer. And then in this, in this world, there is this notion of smoothness and properness. So this is good to Maxime. 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 by module over itself, then that it's compact in the category in the drive category of a by modules and proper means that's well if you fix two objects, any two objects of your category that's a complex, you look at the co homology and you ask this dimension of this co homology to be finite dimensional and more over that the total co homology to be finite dimensional. And this for any two objects x and y say it again. I'm working over a base field it'll case always our little case. It can be more generally can work over a commutative ring or even over a set of different ways I can be very general. And then the key remark is that if you have a scheme which is smooth and proper. That implies that well, this category. It's actually smooth and proper in the sense. And also the converse holes. So this perfect complexes on a scheme actually reflects these properties of smoothness and properness. So here it's now we would like to do geometry, not with the skin, but with an arbitrary digi category, a smooth and proper one which mimics smooth and proper schemes right. So what can be done in that case. So in particular for his course I would like to phrase to formulate the non commutative counterparts of all these conjectures. I want to do that, in particular I would like to have some kind of non commutative value co homology theory. So something that works not just for schemes but in full generality. So for this let me talk about topological periodic cyclic homology. So this, let me put some names on the board so these two to Alan con and many others these two as a whole. 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 Colletin 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 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. In fact, I get the same potential gadgets, and I can totalize it. And if you totalize it you get something that is called a shield homology of a, and that is something that is scaling here. So for example, the issue of 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 secret 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 a periodic cyclical knowledge 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. And one variable of variable being a minus two of the green minus two. Okay, so you get something periodic to 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 digital 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 begin and then forward the definition. Yes, there is a relative setting. 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. Now in positive characteristic I can do this topological version of periodic secretology 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 books that so this is a ring of polynomials on one variable of degree to. Okay, but then it turns out that there is a relation between the two. This is actually a model 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 who shield homology on this parameter here. So when you take the fiber at zero to get the original shield 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 Greenwich. Construction. And you get this topological version. Which now no longer it's K linear. So for example, when you computed over the base field. What you get it's the ring of vectors where you have added one variable of degree minus two. 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 not P, the vectors you end up with your base field. This is true in general in fact this is a characteristic zero lifting of the periodic cyclical mode. 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. This new feature is this cyclotomic Frobenius is this 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 was originally defined by Caledon so Caledon is the first person that actually make this rigorous. And 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 characteristics zero then what you know is that these two relations whole so if you do a plus B for any two elements. That's a P plus BP. And the product is also a P times BP. So 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. 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 the circle action that you use to define the TP, but moreover as this map that goes from THH. To the THH of a and the take construction with respect to the secret group of order P. There's a question. So what about the definition of TP of K and characteristic zero. Yes, how. Yeah, I mean, we can do that. Similarly, but I'm just focusing in in characteristic zero for for. 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. Okay, I'm just working in characteristics zero okay. Yeah, that's it. And so here I want I have this map, which is as one equivalent. So here we have an action of S1 and here we have a residual action of S1 also, or if you want to have one module, the CP that identifies with S1. And so it's these two data is called the cyclotomic spectrum. 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 so you have a what you can do here is that you have a canonical map from the amount to be 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 TP 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 amount to be far fixed points with respect to S1. So it's a technical results you have this equality, but that implies that you have another map here. So you take this original map, and you take the amount to be fixed points with respect to S1. 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 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 community 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 are working over a finite field, one thing that you can do is to compose and get 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 a multiplication by one over Q. So there is a question. Don't we need something like a being perfect for Frobenius to be an equivalent? So it comes, everything comes from my assumptions on 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 acyclic homology and topological shill homology as non commutative veil conjectures. So that's what we are going to replace actually 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 an aside. Can you leave the blackboard? Oh, yes, of course. Sorry. So let me just write an aside here about non commutative, what do you think realizations. So when may 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 category. You can look at modules over the KGL with quick coefficients. And then you see here, this is a covariant procedure and this is a counter-variant procedure. This is counter-variant. 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 we'd like to be interested in. So here you have Morale, you have Wajwocki 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, motivic realizations. For example, for example, a Nordic version, an Odge version, a Durambati 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 do the exactly the same. And now the first point is that, well, this is the end, the big DM, this is a result of Rondings and Paul Ostweiner. And so this composition is your motivic realization, or a map of a Vodsky. So here you have your realization going towards your key. And the second observation is that the KGL with QP efficient, it's actually this trivialization. So you get this by M to M. 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 just simply come look at modules on the sum of your realization of your date. And you get this induced map here. And so your non commutative extension, it's actually we use this function. And then what you do is that you use the joint. And then you use this in this 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 function 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, you do these things are not monoidal. For example, they you don't have finiteness on this. So it shouldn't actually be called a realization that is something that it can be done. Okay, you can always extend anything in the commutative world to the non commutative world via this adjoint. It's a bit cheating. Yeah. 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 you have some kind of. I'm just saying that you 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 tape motive, then you can extend it to any tg category by via this procedure. But of course this is using an adjoint so it's not a correct way to do it you what you want to do is to have an intrinsic definition that is not using actual schemes right. That's a good 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 grotten the group. So now I have a smooth and proper. Dg category. And we look at the case zero well, it's case zero just the case zero of its drive category which is a triangulated category. And now we fix we choose an element in here in the case zero, and we can phrase all the similar equivalence relations we can say, we can talk about new potents equivalent. So you say that the cycle is new potently trivial. If by definition there exists an integer such that when you answer the cycle with itself and times. And this is something that leaves on the case zero a answer it with itself and times that this is actually become zero. So you can make this definition. So what about homological equivalence. Well, we now have this. Our, you know, committed in the homology theories that I've explained. So you, you define homological equivalence as follows. So you simply say that here you have. Also we turn characters. So this is something that I will explain better on Wednesday. So here you have char characters defined on the from the case zero to the positive part of the periodic cyclical analogy and the positive part of the topological periodic cyclical analogy and you simply say that well, it's homologically trivial. If the character of it, 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 K zero and you define a pairing on K zero as follows. So if you have a module. 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. Okay, and look at these dimensions and then take an alternating some of these dimensions. And so in this way we extract this number and this pairing, they extract the pairing on the case zero which isn't neither symmetric neither is Q symmetric. But the since a smooth and proper, it turns out that you'll have a surf after and using the surf on that 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 be called C alphabet, it's equal to zero for everybody. Okay, and then we have the similar remarks. So, if you have a cycle, which is new potent, the trivial will be homologically trivial will be numerically trivial. So, so that's implies that on the case zero, you have all these, these quotients module of the new potent module of the logical module of the numerical. And again, this is always finite dimension. Zero. No, it means that being zero here means that this parent vanishes for every be an alpha such that this parent vanishes for every be. The importance implies this. Yes, the homological. Yes, it's an exercise. Yeah, if you have something new potent lead reveal it will be homologically trivial. Okay, and now we using this we can we can as you expect you can do formulate this is non commutative counterparts. It's non commutative grotting the standard conjecture. So what you say here it's here it's a conjecture let's see D. So this is of type D. And C of a. So this simply says that the case zero, when you mod out by homological or by numerical. They are the same. Also, you can define this non commutative version of a vote ski. The importance equivalent importance conjecture. So we write conjecture V and C. So again, it's that the case zero up to new potent equivalent is the same as the case zero up to numerical equivalence. So 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 Baylinson conjecture. So over a finite field. In fact, the case zero. It's 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 very conjecture. So what do we do here so again we are over a finite field. So we have a smooth and proper DG algebra, sorry, did you category, you 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 are times. And the same thing that you want over P comes with this Frobenius. 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. Well, that these numbers are all track numbers. And secondly, that complex absolute value is equal to one, respectively, equal to square root of q, and this for for all conjugates of London. And also in this non commutative world we have this analog of the proposition, which says that well if there exists an integer, such that such that when you multiply your lambda by q and, and if you get an algebraic integer. Then it turns out that the electric absolute value of London is actually equal to one for all conjugates of London is forever prime different from P. Okay, so you also have this result of course is conditional in this case in contrast with the commutative world. And also we have this. Hello, how much time do you need still? I just have one, one and a half is pretty quick. Let me just say that in the, in the Maybe there's a question is this is this conjecture k fury morita invariant. Yes, of course, yes. Yes, because anything that is more equivalent. We'll actually have the same motive. Let me give you a very technological answer of the same non commutative motive and all these conjectures descent 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. 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. 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 see 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 see. And then you can do the same story here. You can define it in this way. So determinant one minus minus the Frobenius one. Going to see on the TV one. Okay, so we have these functions, of course, overseas. We are metamorphic. And what this is telling you is that, again, by, by all definition they are, they are periodic with this period. And by, by what our non commutative version of the well conjecture is telling you is that the polls are actually here. They are in the same place 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 polls are actually do not change. And finally, let me just say that we also have the non commutative version of the paint conjecture. So 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 commutative version for all 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. You have both field localization with respect to topological K theory. And now you look at the time minus one of this. This is a billion group and then you do the, the TL of this so you do the, the, the eladic hate module of this. And you ask it to be zero for every n greater or equal than one. And this vanishing conjecture saying that well, all these tape modules, a lot of the tape modules of these civilian groups vanish for every possible and forever possible extension. That's one way to phrase it. Then we have the P version of this. So here you have the chair character. So when you inverted P and it lands on the invariance under the cyclotomic Frobenius. So when you change from Q to Q P, this is subjective. And you also have a strong form of the tape conjecture. So you say that the order at a zero of this function. 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 case zero of a modular numerical equivalence. So we should like to emphasize that this thing here actually does not depend, does not depend on the embedding. Okay, because you can rethink about this, if you see the definition is just the algebraic multiplicity of one of this operator and 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 the this strong form of the tape conjecture implies the tape conjecture itself. And moreover, if you add this non-commutative version of Rotendix conjecture, this become equivalent. So what is the upshot of all this, the upshot of all this is 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 cells. So it's the following. So if you take a skin, a smooth, proper skin, a skin that's 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 Rotendick, 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 digiculture. 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 an equivalent. 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 digic categories, smooth, proper digic categories. And if you plug, if you attack this particular kind of digic category, 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. So 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 actually 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 TP 0 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 on the positive part on the negative part, you'll get the sum of the even groups of the crystalline co-mology of your skin. And on the negative part, it gives you the sum of the odd co-mology groups on crystalline co-mology. So this telling you that, well, this invariant, of course, after inverting TG. This tells you that, well, when you compute it on this DG category, you get the crystalline co-mology, not the individual pieces, but up to parity. And more over now, you also here. So when you, 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 multiple scalar 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. 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. Slightly. Okay, okay, 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. Formulations. I don't know what 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 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 is there 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 some of them admit the relative versions. Yes. I haven't explored that. Yes. That's a good question. That's a good thing for the future. Yes. So I'll do things work in families and things like that. Okay, so I'm not sure to understand the question correctly that mouse, but I just read it. So what is a 10 so M and does it agrees agree with Perth X to the M in the geometric case. Yes, yes, yes. Yes, yes. When you, you, yeah, you, you, when you do this. This is this. I'm, I'm my schemes are nice enough for which this kind of phenomena holds. Yes. So in the I mentioned that this question is about the new potents equivalence relation because on one side that it was on cycles on X N on cycles on X N and the other side was over K zero. Of your a sensor and and when a is of this form Perth of X, it's, yeah. Thanks. So that was, that was a question of Remy van doben. There's also a question by Ola Sunday. Is the new potents conjecture equivalent to the balance and Suley conjecture. And the balance and is only over a finite field. And the balance and Suley, maybe you can recall me what it is. It's a vanishing of vanishing of motive emoji in in degrees in a negative degree. Yeah, I don't remember. I have to begin. 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, I don't remember from the top of my head. No, I think there's no. Direct relation. Okay, if you if you put all conjectures, I think I think that they, the Suley. It's actually the strong is actually the take conjecture. And then you move out by the numerical equivalence. Yeah, it's not. No, it's just vanishing of a motive emoji in in the simplicity in negative simplicity degree, but. Okay, so I have to. Could you please state precisely how the electric take conjectures are formulae formulated via TP. No, via TP. I'm formulating the, the, the, 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 take conjecture, right. And I formulated the, the L version of the take conjecture using KT, in fact, right. So the question is if you can do that. Yes, I mean, that can be done, but it's using this results of Thomas and that tells you that mean the relation. The relation is, is that you have a tier, here's Brooke spectral sequence that goes from a latic. So you have something like this. So rationally, so they're proof that it degenerates. So you have something like the at all K theory of X, when you complete at L, and then you, you rationalize that is in fact, the sum of this like co homology groups. And so here you already have a link between the etal co homology with K theory, but it's an etal K theory, but then you have this beautiful result of Thomas and that tells you that you can write this algebraically, because you have, you have completed at L. It's in fact to take the K theory of you and to localize with respect to complex K theory. And then if you complete at L, actually, you'll get the same spec. And now here you have K theory. So something that you can phrase for anything particularly for any generality. So if so you see that then the reformulation of the conjecture would be saying that if you go from the shunt character from the K zero to here, it actually lands precise on the things that are stable and action of the absolute color. That's one way to phrase it. So I think it was originally already phrased like this in an old paper of Friedlander. So I have another question. So are there any strictly non commutative application of the conjectures or any of the conjecture. Yes, I mean, this is somehow motivation. Usually, let's say that what if we divide the world and to much more people with work on this side and 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, we 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, describe it in terms of veil numbers up to a certain action of the absolute calva 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 committed. 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 community. And in both cases, we are going to explore the link between. So there's another question about, so are there any new cases known where the conjectures, the conjecture holds in the non commutative setting. What, 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. You use using some non commutative techniques. And as a consequence, you get this conjecture for access, which are were known, not known previously. If I, 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 view point is not with the view point. 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. I think which is called non commutative counterparts of celebrated conjectures, if I remember. Okay, so there's no more questions in the chat. It seems other questions in the room, no. No. Okay. So good thing. Thanks again. Sorry for going over time. I got a bit confused with this time. I'm sorry for those who had problems with connection. Okay, so we meet again in at six.