 Thank you so much. Thank you for all of those that returned to this third and final lecture online and also here at Yashua's. So this is lecture number three. And the title, it's Applications to Geometry. So let me recall you what we have done so far. So we have been working always with the perfect base field of characteristic P, P greater or equal to 0. And in the first lecture, we saw this theorem saying that if you have a smooth proper case kin, then you can look at the conjectures, the celebrated conjectures for x. This non-commutative counterparts for the associated DG category of perfect complexes, we saw that there is no difference between these two. And also that we can more generally consider not solely the DG category, but also the corresponding non-commutative child motive, even with rational coefficients. And this, for any C, the conjecture, the one of Grottendijk, Weyvotsky, Veilinsen, Veil, and all these variants of the Tate conjecture. OK, so this is what we have seen so far in today. So what's the idea today? So today what we will explore is the following. So we want to find, we can look at these non-commutative motives, even with rational coefficients. So given any two x and y smooth proper algebraic varieties, you can look at the corresponding non-commutative child motives, even with rational coefficients. And also I can look at the conjectures for x and the conjectures for y. And I would like to find today a relation between these two. I want to find a relation in such a way that that relation will imply an equivalence between these conjectures. So that's the main idea for today. So what can we do? First of all, the first case, the very first case where we can find this relation is when we have a derived invariant. So suppose that the categories of perfect complexes on x and on y, they are even equivalent. Equivalent via Fourier-Muckay equivalence. Then in that case, well, that implies that the corresponding motives of x and y are equivalent, are isomorphic. And of course, if you just use the above string of equivalence, that's going to imply that the conjectures of x and of y do not change. So the conjectures x and y for x and y do not change if x and y are derived equivalent. Let me make some remarks about this. So first of all, it is known that if the canonical or anti-canonical bundles of x and y are ample, then it's a very interesting result of bondale n or log that in that case, the varieties themselves are isomorphic. So it's a very strong restriction that we have here under this assumption. But of course, there are cases of non-isomorphic algebraic varieties, for example, and a billion variety. And it's dual, a billion variety, for which the corresponding derived categories of x and y are equivalent. And this was this pioneering work of Mukai. So our formalism that we, our results that explain automatically imply drive the invariance of these classical conjectures. But so this is a very strong relation. And there are many other cases where we actually have a relation between these non-commutative motives. And one machinery to establish this relation is called homological projective duality. So we usually write HPD. And this is a theory that was developed by Kuznetsov. So let me explain how does this work. So the idea is the following. So you have an algebraic variety equipped with a map towards a projective space. You can think as being an inclusion. But it doesn't need to be an inclusion. Then you can look at the associated line bundle, where you pull back along this F, the line bundle on this projective space. And then what you want to do is to study hyperplane sections of x. So you have hyperplane sections of x. And if you want to study them, you can study them all together simultaneously. So you can first look at the incidence quadric inside of the projective space across the dual. So you can look at this incidence variety. And then you can look at the universal hyperplane section. And this is simply the fiber product of x with this incidence quadric over the projective space. And this sits inside x cross the dual projective space. So this carries information about all possible hyperplane sections of x. If you fix here a point, you are thinking about an hyperplane. And so when you look at the corresponding fiber over the projection over PV, that corresponds to the hyperplane section. And so our goal would be to understand this category of H, the category of all hyperplane sections. And the result is that suppose that we understand the category of x itself. So suppose that the category of x has a very specific form. It is what is called left sheds decomposition. Suppose that it admits a left sheds decomposition. So that means simply that you have a sequence of triangulated categories inside in such a way that when you twist them using this line bundle. So here you have a 0. And then you twist one with this line bundle, et cetera, until ai twisted i times. Then you get the semi orthogonal decomposition, the classical semi orthogonal decomposition of your variety. Suppose that you are in such a setting, then you can say something about this drive category. So inside of it, what do you have? Well, so let me get some extra room here. So what you have here is that all these blocks will also exist here. So you have this. And then, of course, you need to do the box product with the perfects on the dual projected space. And then it turns out that there is something orthogonal that is still missing. So here you have an extra category that appears. And this category is called the HP dual category of X. So what I've done so far is something very formal. You start with a scheme. You keep it a map towards a projected space. And suppose that this scheme has a semi orthogonal decomposition of this very specific shape, like a left shape's decomposition. For example, that's what happens with the projected space. For example, we'll have o, o1, twist it by 1, o2, twist it by 2, et cetera. Then you can say something about this category of the universal hyperplane section. It has all these pieces in its interior, box time this part. And then something mysterious, something orthogonal that appears here. OK. And now what's interesting about this theory is when this very abstract category is HP dual category, it's in fact geometric. Sometimes it turns out that this category, it's geometric. What do I mean by that? So I mean that sometimes this category here, it's actually equivalent to perfect complexes on a different algebraic variety. And sometimes it's not actually just an algebraic variety, but it's twisted by a certain shape of algebra. So not very far from being an algebraic variety. And in both cases, we call this y. We call it the HP dual variety, dual variety of x. And in particular, it comes equipped with a map towards the dual projected space. And suppose that we are in such a setting, then what you can do is that if you have a subspace of the dual of v, well, then you can consider its orthogonal, which is just the kernel of this induced map. And then here you can do two things. First of all, you can do the linear section with respect to this subspace. So you just take your original x and do the fiber product over the projective space with p, v, or orthogonal. Or you can do the linear section of y. So where you do the fiber product over the dual projective space with p of v. And now the theorem, which I call the HPD invariant, stays the following. So what was upstairs was the derived invariance of the conjectures. This is the HPD invariance of the conjectures, stays the following. Suppose that these linear sections are, in fact, smooth. Suppose that they have the expected dimensions in the sense that the co-dimension of this linear section is equal to the dimension of l. And also that the co-dimension of this linear section is equal to the dimension of l perpendicular. And also let's assume that our conjecture, the non-commutative version, is true on this largest piece, on this a0. Yes, because you see this thing is defined in this sense. So you can think about x inside the projective space, for example. And then you are taking a look at the hyperplane section. So and you assume that this conjecture for the largest piece holds. So some remarks about these assumptions. So first of all, the first two assumptions, they hold for a generic choice of your hell. So they are quite mild. And the last assumption, it holds, in fact, in all the cases, in all the examples so far, because in all the examples, these categories here that appear in the left chance of the composition are quite simple. In particular, they often admit full exceptional collections. So assuming this, it turns out that the conjecture for this linear section and the conjecture for this linear section, they are equal. So this is the content of the theorem, simply saying that, OK, under these assumptions, the conjectures do not vary. So what is the idea here? So just a brief idea of the proof. What's behind of this is that when you look at the non-commutative motive of this linear section and the non-commutative of this linear section, they encode. So there is a common piece inside of them coming from this mysterious C. So you have a certain category here, which is common in both of them. And then you have extra things, extra factors. But these extra factors here, they all, the conjecture, this conjecture holds for all of these pieces. Because these pieces are the pieces that are coming from this one. They are, in fact, direct summons of this one. And since the conjecture holds here, it will hold for direct summons. These pieces are not necessarily the same. But well, this tells you that, well, the conjectures hold here, hold here. So in fact, the non-commutative conjecture for this linear section is just the non-commutative conjecture for this non-commutative gadget. The non-commutative conjecture for this linear section is just the non-commutative conjecture for this. So you get an equivalence between the non-commutative conjectures of these two. And then you use the fact that the non-commutative or classical conjectures are equivalent. So somehow, you want to relate two geometric things, XL and YL, both of them live in the world of geometry. And what you do is a zigzag. You go towards this mysterious non-commutative thing. You go there, and then come back. You pass from X to L, from L to Y, via a zigzag, via something truly non-commutative. OK, so that's the idea behind. So now there are a lot of HPD dualities in the literature. So there are a lot of HPD dualities in the literature. Let me just give you an idea of two of them. So we have examples of HPD dualities. This, for example, is a Clifford duality, the Grassmanian-Pafian duality, the Spinner duality. The Determinant Hall duality, and many others. So I propose to look at two examples to look at this one, where in that case the HPD dual variety will be of the first form. So we twist it by a certain shift of algebras. And then the second case where the HPD dual will be actually geometric, given by just an ordinary algebraic variety. So let's look at the Veronese Clifford duality, so HPD duality. So this is built from work of Kapranov and Kusnetsov. So what is the idea here? You start with a vector space of dimension V. And you look at V, the sin2 of V of W, sorry. So this is your V of the theory. And then your algebraic variety is very simple. It's just a projective space on W. But then the map here, it's a non-trivial map, going to P of V, in other words, to P of sin2. And the map, it sends W, in other words, it's the Veronese embedding. So this is my algebraic variety. This is my projective space, and this is my map. So in this case, if you look at the universal acroplane section, and you project to the dual projective space of V, in other words, you project here. It turns out that what is this is, in fact, a quadratic vibration. So here, let's simplify and work nothing characteristic to. So what you have here, you have a quadratic form, and above it, the fiber is precisely the associated quadratic. So in fact, this thing, in this particular case, it's a quadratic vibration. And if it's a quadratic vibration, I can build out of it a shift of algebras, which I denote like this. So let me explain what it is. So let's make a parenthesis here. So suppose that you just have a quadratic form on a vector space towards your base field. Suppose that you have a quadratic form. Then what can you do? If you have a quadratic form, you can look at the tensor algebra of V, and then you could mod out by V tensor V. If you do this, you just have the exterior algebra. But then what you can do is that you can quantize this by deforming it by this quadratic form and get this new algebra, which is called the Cliffer algebra of your quadratic form. But you see that the way these relations are set up, it's not graded, but it is Z2 graded. So you have an even and odd part. So you can look just at the even part. So you can do this for any quadratic form. So here you have one of those on the base, and then you can vary that, and actually you can get a shift on this space. So you actually have a shift of algebras by doing this procedure over the base. You should be non-degenerative. No, it can be degenerative. Oh, okay. So that is precisely... So let me just say that the notation, let's denote by Z, just the singular locus. So inside here, the singular locus of the singular locus of your quadratic fabric. Okay, so above those points, you actually have singular quadrics. And now it turns out the key point is that the HP dual variety in the story, HP dual variety of X is the following. It's just this dual projective space, but equipped with this shift of algebras. So we are in that setting where the HP dual variety is not just an algebraic variety, but it is perturbed by a shift of algebras. And then what can we do? If we have a subspace on my dual, then I can do, as before, I can take linear sections here or linear sections on the HP dual. And if I do sections here, that corresponds to do the intersection of the quadrics, the quadrics in my projective space, which are parameterized by L. Okay, so intersect as many quadrics as the dimension of your space L. And on the other side, what you do is just that you do a restriction. And you restrict here this shift to the P of L. Okay, so we know that to solve the conjecture for intersection of quadrics is to solve the non-commutative conjecture for this gadget. So let's understand this gadget a little bit better. So it turns out that this category with this shift of algebras, it's equivalent to a category perfect complexes on the on a certain double covering of the projective space and this with an azomai algebra. So what's happening here is that you have the projective space and you have this double cover, twofold cover of this, which is defined. So what do you do? You take the spec over P L and then you take the center of your even peripheral. Okay, so this is what you get and you get the twofold cover, which is ramified on this singular locus. And that is what's happening when D is even. Okay, D upstairs is even. So D it's actually equal to the dimension of the quadric that you are considering, the quadrics that you are considering. In the odd case, in the odd case this category is equivalent. Now what you have, it's a root stack as we saw yesterday. It now equipped with a certain shift of algebras which is also an azomai shift of algebras and this root stack that you have above your P L, it's in fact this two root stack where what you do is you get the singular locus intersect with P L over P L. So it's the root stack that we saw yesterday. And so as a corollary, so as a corollary what do we get? We get the following that the conjecture for intersecting quadrics, that's equivalent while we know that it's the conjecture for I of L. And so this turns out to be just the classical conjecture for this two-fold cover of the projective space, this in the even case. And here it becomes the conjecture becomes the non-commutative conjecture for this perfect complexes on this root stack equipped with this shift of algebras. So what does that imply? That implies that, for example, when D is even, in other words, when you are intersecting even dimensional quadrics, so this conjecture about intersection of quadrics holds when, for example, when you are intersecting three or less quadrics for the conjecture of Grotten-Dicken-Weiswitzky, when you are intersecting two quadrics for the other conjectures of Bellinson and Tate, and of course for any dimension for the Vail conjecture. So of course you already knew that Vail conjecture is true, but here you give you like an alternative way to see it. And also you get, for example, when D is odd, so when you are intersecting odd dimensional quadrics, that the conjecture for this intersection holds when, for example, when the dimension, when you are intersecting two quadrics and moreover, for example, when you are working over a finite field or over an algebraically closed field, because in that case is the Broward group of curves disappears. So in that case what you get is that this, this sheaf of Azomai algebras goes away, you just get a root stack, and then root stacks we saw yesterday how to manage them. Okay, so this is just an example. It gives you an alternative way to prove these conjectures for intersection of quadrics in an alternative way to the one that was done by Miles Reed in his thesis which was geometric. In the even D case, you don't need to consider to worry about the Azomai algebra. Because we saw yesterday how to attack the Azomai algebra, when you have an Azomai algebra, the motives with Q-q coefficients are the same, with or without an Azomai algebra. But when you have an Azomai algebra over a root stack, that is not the case. But over an ordinary scheme, there's no difference, but over a stack there is a difference. So now I propose that to look to the other example of the determinantal varieties. So that's the example where the HPDUO variety will be actually a variety. So determinantal HPD. So this was established by Bernard Dara, Bolo, Menzi, and Tanzi. So how does it work here? So you have two vector spaces, W1 and W2, of dimensions m and n. Let's assume for simplicity that m is less or equal to n. And then what is our V? Our V is this tensor product of the two. And then I choose an integer r between 0 and n. And out of this, I can look at the variety rmn inside the protective space, consisting it's the determinantal variety, m by n matrices, which rank is less or equal to r. So let me maybe give you a down to earth description. What do you do? You look at this matrix of indeterminances. So you have a matrix like this. I'm going to xn1 xmn. And then what you impose here, the relations that you impose is that the r plus 1, r plus 1 minors are actually equal to 0. So you have all these variables and you have these relations and that's going to give you a protective variety. Okay? For example, so are those varieties that are defined by equations that are coming from minors? So, for example, if you fix the rank to be 1, then these are nothing but the saver varieties. And for example, 1, 2, 2, this is just you look at those quadruples such that this relation holds, this determinant relation holds. And that's a surface in P3. Okay? And in the same way that you can do this determinant, define this determinantal variety, you can define it's dual. So you have something inside the dual projective space. Okay? So it's the determinantal variety of m by m matrices. But now we co-rank greater or equal than r. And it turns out that these things are in fact singular. So we cannot work with them, but they admit the very canonical resolutions of singularities called springer resolution of singularities. So we have this one admits a resolution of singularities. And the key fact is that x and y are HP dual to each other. So that implies that if I, whenever I have a subspace in the dual of v, I can take one linear section or another linear section. Okay? And now what's so interesting here, you see the following. So we get this corollary consisting of two parts. So first part, so let's suppose that, so we see that in this definition, we have four parameters. We have m and m, the size of the matrix, that are the rank. And we also have the possibility of choosing my L. And so we have the dimension of L. So we have four parameters. And now let's assume that we have the following relation between these parameters so that this is, so this number that it's here, it's actually the dimension of this linear section. And the dimension, and similarly, this number here are m plus n minus r minus 1 minus dimension of L. This is the dimension of the other linear section. And so let's suppose that they have dimensions less or equal than one. Then in that case, what does that imply? That whenever the parameters satisfy this inequality, well, you get for free that the conjecture for the other linear section holds. And so let's explore this. So let's explore, for example, the first implication. So the first implication, let's explore it, for example, in the case of Segel varieties. That means that r is equal to 1. Well, when r is, so in the case of Segel varieties, they are already smooth, so we don't need the resolution. So we are really taking linear sections. And so in that case, that number becomes m minus n minus 2 plus the dimension of L less or equal than 1. Then I know that the dimension of XL, in that case, is equal to m plus n minus 2 minus the dimension of L. So what's interesting here is the following. So suppose that I enlarge, so I'm in this case, and suppose that I enlarge the size of my matrix in the same way. So I replace n by n plus i and n by n plus i. And I keep the dimension of L fixed. So if I do this, this number does not change. But in contrast here, this augments us by 2i. So you see that here you have infinite number of infinite family of algebraic varieties, all of them satisfying the conjectures. And they are of arbitrary high dimension, simply because the dimension of this is fixed and the dimension of this is changing. And the conjecture for this one is true. So is it geometrical that you are taking ample line bundles, tensile for twisting the ample line bundles with high enough power? Does it mean that? Because this 2i comes in this dimension of XL, because maybe I'm not very precise, but is it geometrical in this you have enough embedding because you're working with several varieties and you're embedding this variety to some big variety space? It's not that sure with that. We can discuss after this. It's not... And the other example, for example, of square matrix. So in that case, that means that the dimensions m and n are the same. In that case, this you have this. But in this case, the dimension of the dual variety is 2rm minus r square minus 1 minus dimension of l. So again, you see that if you change... So here m not even appears, so when you increase the dimension of your... This number does not change, but this increases by 2ri. So again, you see another family of algebraic varieties, all of them satisfying the conjectures and they are of arbitrary identity. And interestingly, these cases, it's not known to the experts how to prove the conjectures for these cases using just algebraic geometry, just classical algebraic geometry. Another interesting remark is that, for example, in what regards the veil conjectures, well, they are known, of course, thanks to the limb, but you see here that in the case of curves, so this is the dimension of yl. So I'm assuming that it's a curve. So for curves, the veil conjecture was originally proved by veil using elementary intersection theory of divisors on surface. So here you are bootstrapping this result of veil to higher dimensions without using the limb, for example. So it gives you an alternative proof of the veil conjectures that doesn't use the limb's techniques. And of course, these results are much more refined. This number one is here because we know that all the conjectures that I'm talking about, they are true for curves. But if you want to be more specific, for example, if you want a veil votsky conjecture, then you could put a tool here, et cetera. So to the biggest board, for curves you have to show some veil positivity in the intersection number. You have to show that this is positive. Yeah, that's the... How does this for the curves means this for the veil positivity here? No, it is exactly the point. The way that I prove it is like the drawing that I mentioned to you is that the conjecture becomes from a curve in this higher-dimensional variety. The conjecture is equivalent to this non-commutative gadget. And then you come back. So you do this exact, this step where you use the non-commutative world. So it's a not geometric proof and that's the interest that you can go to higher dimensions. Okay, so in the remaining of my time, there are many other examples, many other HPD dualities, many other cases where you have proven the conjecture in this perspective for new varieties. But just to finish this lecture and the course, I wanted to talk a little bit about the Riemann hypothesis. So let me recall you that if you have a smooth, proper scheme over a finite field of dimension D, okay, then we defined in the first lecture it's also the LZ function as the product over all closed points of 1 over 1 minus Q degree of X minus S. So this is actually the cardinality of the residue field at that point. And this converges when the real part of S is greater than D. And then we saw that by combining the links work on the veil conjecture with the work of Bertolou in crystalline homology, you have this homological interpretation of the Asa-Vael Zeta function using the Frobenius and acting on the crystalline homology. So we wrote it like this, omega plus one. So in what follows, I will, these pieces here, so this piece here, I will be noted by the Asa-Vael Zeta function of X of weight omega. Okay, this will be my notation. Gonzalo, it's hard to see what you have written on the bottom. This red chart. So it is the Asa-Vael Zeta function of weight omega. Okay, just this piece. So the ordinary one, it's an alternating product of Asa-Vael Zeta functions with different weights. So this is over a finite field and that's what we saw last time. So now we go to a number field. So to simplify, let's just go to Q. So we can go to Q. So take now, it's a smooth and proper Q-skim, okay, of dimension D. Then we know that there exists a smooth model, a smooth, proper model over the, so away from a finite number of bad primes. And so let's assume that we have good reduction. In other words, let's assume that there exists a smooth, proper model. So X i over each one of these bad primes. So localize that P i. And this for i between one and n. And if this is the case, if we have good reduction, we can do the following. So if you choose an integer, a weight between zero and twice the dimension, what you can do is do the L function of this weight of X by definition. You just do the product over all primes different from the bad ones. And here you do the Aseville Z function of this weight on the fiber. So this stands for the fiber of this model at P. So that will give you an algebraic variety of graphonite field. And then you multiply by the remaining bad primes. So, and here you do, by this I mean the fiber of this model at the prime P i. Okay. And then it turns out it's a result, a theorem of Sarah that in fact, this function actually converges on this health plane where the real part of S is greater than omega 2 plus 1. And it's more over different from zero in this region. So you have the following picture. So over C, you have omega 2 plus 1, omega 2 and here in the middle we have this one. So you know that in this region it converges and it's not zero. Okay. And now what regards this L function so there are associated with this conjectures. So the first one is the conjecture M of X that says the following. So whenever you have a weight between zero and twice the dimension first of all this function admits a unique meromorphic continuation to C the entire complex plane and moreover this extension has no poles has no poles in the critical stripe. So consisting on the complex numbers whose real part sits between so here you have this critical stripe here in this interior the conjecture is that there are no poles so not only you can extend it to the meromorphic function but there are no poles here in this critical stripe and then if this is the case you have the conjecture R of X so maybe it's called a generalized Riemann hypothesis that says that if this function vanishes at a certain complex number and this complex number belonging to this critical stripe then necessarily this number lives in this vertical line so it is necessarily in other words all the zeros of this function they are here so zeros they are necessarily on this line and why is it called the generalized Riemann hypothesis it's because it generalizes the classical Riemann hypothesis so if you take let's see the following example let's take for X just a point so per Q the simplest example that you can think of which is of dimension 0 then of course that you have that there exists a smooth proper model smooth proper model even on the entire over Z which is speck of Z of course and in this case when you take there's only one way 0 so you look at the corresponding L function here and this is just a product over all primes of this Asavel Zeta function of the fiber at P so this fiber well you know that it's precisely the speck of Fp so it turns out that this is just a product over primes of your 1 over 1 minus P minus S well and that you know that you can rewrite in terms of Dirichlet series as fault so it's the classical Riemann Zeta function which we'll see so in this case what's happening is that you have 1, 1 half and 0 and so you know that this function it's well defined and norm 0 on this region first of all and secondly what we know about this function is well the conjecture M of X in this case it holds there's no problem here so this function actually extends uniquely to the entire complex plane and there are no 0s on this critical stripe and then the other conjecture this well this maybe it holds maybe it doesn't but what it's called it's nothing but the classical Riemann hypothesis okay so the simplest example is the Riemann hypothesis and of course this can be more general you can think you can take here a number field and then you would also have a model which is just given by the integers okay and so in that case here what you have it's a product over all primes and then you need to do the product over those prime ideals above P and it becomes 1 minus the norm of that prime ideal minus S and so you can rewrite this in terms of the sum over all possible ideals in this ring of integers of 1 minus the norm of I S so what you get in that case is the classical that can set the function which is also known to for this conjecture to hold and this conjecture is something called the extended Riemann hypothesis in that case Bansala we have received a question that's not a smooth model at the ramified primes yes you can do that also what I'm talking about in terms of finality continuation the bad primes don't count for the third function exactly right it's just a finite amount of information to put them there to correct where the zeros will land etc if you don't have a good reduction well you still can do it but in that case you need to use comology inertia etc no it's not I don't need it and that's a key point here I'm not using the primes at infinity and making it complete I don't need it because when you are putting the primes at infinity what you are doing with the gamma function is that this when you do the extension there are zeros that appear here and then you put the gamma function in such a way that the poles of the gamma function will kill these zeros so that the only possible zeros will land here but the way I'm phrasing it is that I'm just considering zeros in the critical stripe so I don't actually need to complete it so you don't need this big zero is completed at infinity that's why I don't need it the way I phrase it so my point is now is that this still works in the non-commutative world so now we can talk about the non-commutative women hypothesis so if you have now a smooth proper dg category over the finite field so we saw in lecture one so we saw in the first lecture that you can define this as a zeta function associated to them ok and now if my a is over q so a smooth proper dg q-linear category so this still works still exists a smooth proper model a smooth proper model which I'll write like this over so away from the bad crimes and I will assume as above that we have good reduction I will assume there exists enough smooth proper model over my localization of p i and this for every i between 1 and n and in that case we can just use what we have done in the first lecture to define a l function so in fact as usual it's not just one but two so I have this l function defined so I do the product over all crimes different from these bad ones and I do the as above zeta function for this fiber here and then I multiply by the the remaining bad primes I do this as above zeta function p i and where this is the fiber in the sense that it's doing the drive transfer product over fp width and similarly here I'm doing this model and I'm changing it to fp over zeta and you can do of course the one version of it in a similar way for every prime it's this again and I do the product also I put in place the bad primes and this it's this and now what's the point here is that we have a non-communicative counterpart of series result so the theorem is that let's assume that let's assume that the non-communicative well conjectures hold for the fibers hold for these fibers okay this for every p different from the bad primes for i between 1 and n suppose let's assume that this holds then in that case this l function it converges on this off plane whose real part is greater than 1 and it's moreover non-zero on this region so I have c and I have one half and one and here I have this region where my function is infinite product converges it's well defined of course I don't need this assumption because this is just a finite amount of information and similarly here in the one case this function converges on this half plane where it's greater than 3 over 2 and it's moreover different from zero so here again over c so I have one half one and three over two so it's in this region here that my function is well defined okay and of course in this setting we can also we can also conjecture that we have meromorphic continuation so we can conjecture that first of both of these functions both of these l functions they admit a unique meromorphic continuation to c and secondly also that these extensions respectively of l0, l1 so it has no poles in this critical stripe going from the real part between zero and one respectively when the real part is between one half and three over two so on this critical stripe on this critical stripe I conjecture that there are no zeros there there are no poles there and then we can also phrase the conjecture of Riemann that says the full length so suppose suppose that this function vanishes respectively this function vanishes with a complex number in this critical stripe in this critical stripe respectively here and this implies that this number it is necessarily leaving on this line necessarily leaving on this vertical line so the zeros must be here the zeros must be here okay and so what is the point of all this is the following so I think I need to finish so it's my very last blackboard so let me just mention that the result is that if you take X a smooth proper scheme over Q this conjecture holds then you can look at the Riemann hypothesis for it or you can look at the non-commutative version of the Riemann hypothesis for this so this conjecture to hold will imply that that this conjecture for perf of X will hold also and so I can talk about this and it turns out that these conjectures are equivalent as in the first lecture okay so possible applications of this of course we cannot use it as I did today to prove new cases of the Riemann hypothesis because there are not even cases of the Riemann hypothesis the Riemann hypothesis over a point it is the classical Riemann hypothesis but you can use it to establish connections between two algebraic varieties saying that in fact the Riemann hypothesis for them which are very different algebraic varieties are equivalent so for example suppose that one of them would be zero-dimensional and the other would be higher-dimensional this would say that the Riemann hypothesis for the higher-dimensional it's equivalent to the Riemann hypothesis for the zero-dimensional but for the zero-dimensional it's just that that can set the functions so you can somehow reduce the complexity to get consecutive functions which are very well studied okay I think I'll stop here thank you and sorry for going over time okay many thanks for a wonderful mini course any questions or comments the last point you were making when you stopped after that here so here so you see in the first lecture we had this result for the conjectures right exactly this result I'm saying that the Riemann hypothesis I could so put it in this package where it can be extended to the non-committative world in such a way that when you plug this kind of DG categories coming from schemes you recover the original conjecture so now I'm saying that if we played the game of today of having two very different algebraic varieties with geometrical are very different but in terms of the motives are related then I can relate to the conjectures so this I could relate the Riemann hypothesis for two different algebraic varieties suppose that one of those algebraic varieties is zero-dimensional for example then this would say that the Riemann hypothesis for the other one which could be higher-dimensional is equivalent to Riemann hypothesis for dead-kin set of functions of course this doesn't allow you to prove the Riemann hypothesis because the simplest case of a point is the classical Riemann hypothesis but allows you to establish connection between Riemann hypothesis of very different algebraic varieties another question from Remy is the reason that you are only considering L sub-node and L1 that you have tape twists what's happening is that if you the L-node of the perf of the scheme it's actually going to identify a product of all the even weights of the classical L function on X but here you need to shift it by this so you grab all the L functions for all even weights and you shift them to plus them precisely in that position there that's the relation that you have and you have a similar relation for one and now it will be a product of odd and now it's a shift by 1 minus 1 over 2 so this is the link between the non-commutative world and the commutative world as usual what you do is that you trivialize the tape motive so somehow you get lost of the individual pieces you only keep track of the parity thank you another question from anonymous attendee last talk you mentioned that in the non-commutative world there is no tape twist you want to say that such an object does not exist in the non-commutative world or just that you did not find a good way to define it yeah I I would say that it doesn't exist it makes no the tape twist comes from what's happening with the projective line and the projective line by additivity is just two points somehow it is not there any other questions or comments if not done, grand merci for a wonderful lecture mini course many thanks indeed thank you for having me, thank you