 Okay, so let's start to welcome everyone to the Schubert seminar. Today, we're happy to have Nikola Peran from Mekalpolitechnik telling us about the positivity of Grasmani and Quantum K theory. Please go ahead, Nikola. Thanks, Leonardo. Thanks to the organizers for the invitation. I'm happy to be in this Schubert seminar. So I will report on joint work with two organizers of the seminar. So maybe this is the reason for my invitation. I don't know. So and also with Pierre-Manuel Chappu, who is in Nancy. So I'm going to focus on Grasmanians, but this applies to some other cases. I will explain this. So, okay, let's see. Yes, so I will start with the basic Schubert calculus for at least for illustrating the fact that this is a Schubert seminar. And then I will focus on, so I will, the Schubert calculus will be will be a bit large. So I will include quantum cohomology in Schubert calculus. And then I will focus on quantum K theory, then state the results. And maybe I don't know exactly give ideas of proof. I don't know where I can, where I can go. Okay, so I start with really basics. So I was told to start with basics. So let's see. So the main character is a is a rational projective homogeneous space. So a quotient of a reductive group by a parabolic subgroup. Okay, so the you have, you have the usual, the usual characters in this, in this topic where you fix a parabolic subgroup. Okay, then you fix a Borel subgroup inside your parabolic subgroup. And then you fix a maximal torus inside this Borel subgroup. And you can also look at the opposite Borel subgroup, which is unique Borel subgroup. So that the intersection of your Borel with opposite is maximal torus. And with these subgroups, you can, you can produce lots of combinatorial gadgets. I am going to recall and that most of you already know. Okay, so the basic example is to, to look at GLN. And then you have the Borel subgroup, which is given, for example, by upper triangular matrices, maximal torus by diagonal matrices. And the opposite subgroup will be given by lower triangular matrices. And once you have such a Borel subgroup or opposite Borel subgroup, you can look at the Borel decomposition, which is simply the decomposition in terms of p orbits or the decomposition in terms of p minus orbits. These two are conjugates. So some of them are, they are the same up to conjugation. And you get just a discharge union of finitely many p orbits. So you have only finitely many and they are indexed by some subsets, some nice subset of the variable. Okay, so I guess, I guess this is, this is well known to everyone. Okay, once you have all these gadgets, well, you, you can do a bit of geometry and go to, to cohomology. So you, you take the closure of the p orbits. These are the Schubert varieties. If you take the closure of the opposite orbits, then you get the opposite Schubert varieties. So they have nice properties. For example, the dimension of the Schubert varieties, the length of the corresponding variable element. And this is also the co-dimension of the opposite Schubert variety. Okay, so this is very classical. And the main point in, in looking at Schubert varieties and opposite Schubert varieties is not only on, on co-dimension and nice combinatorics, the fact that these are like canonical choices of sub varieties in general position. But this is something, then something very convenient because you have like these varieties which are almost fixed, but still they like represent something which are in general position. And therefore any intersection theory will be very nice using these sub varieties. Okay, so I will, I will, I will present several avatars of, of intersection. So cohomology, I will not talk about equivalent cohomology, but they are also very convenient for equivalent cohomology. But okay, so let's start with the basics. Co-homology, so once you have these varieties which are per-person varieties in, in the homogeneous space, you can look at their Schubert classes, which are simply the cohomology classes and the Schubert classes form bases and actually form dual bases for the cohomology. So, so actually these two bases, that's what I'm explaining right after, is that these two bases are actually the same. It's just that you have a parametrization of, of the indexes. So you can just do some bijection on indexes to get the classes of Schubert varieties and then the classes of opposite Schubert varieties. Okay, and, and they have, they are in duality for, for intersection. So this is called Poincaré duality. And as, as is explained below, if you look at the opposite Schubert variety, actually the opposite Schubert variety is nothing else than the translate of a Schubert variety for, of the dual Schubert variety. And, and these W zeros somehow explain the fact that they are in general position. So you have to translate by the general element. Okay, so this, this is classical Schubert calculus and using this you can compute lots of things on co-moral G and intersection products. So you can define, especially, at least compute the cup product in terms of these Schubert classes. So yeah, I expressed the cup product in terms of opposite Schubert classes, but as we have seen, this is the same as with Schubert classes. It's just better for, for the degree reasons. Okay, so if you multiply any two Schubert classes, you get an expression in terms of Schubert classes because these are, these form a basis and you can expand them with, with coefficients. And these coefficients are actually solution of enumerative, geometric enumerative problems. So they count the number of intersection points of the translate of three Schubert varieties in general position. So you, you pick your first opposite Schubert variety, you pick a general translate of the second opposite Schubert variety. And then for the last one, because you have quantum ideality, you have to take a general translate of the dual class of Schubert variety. Okay, so, so in particular, what you get is that these numbers are always non-negative integers. So, so a very classical problem and difficult problem and, and some people in the audience claim that this is like a non-really possible to have a nice expression to this problem is to give a combinatorial enumerative formula for all these now, these intersection numbers. So in general, yeah, I'm not only restricting to the case of Grasmagnans, for which we have nice solutions for these numbers. Okay, so this was Schubert calculus, classical Schubert calculus. And now I want to, to go into some avatars of Schubert calculus. So I go in two directions. There will be two directions. There's a quantum direction, which I'm going to talk right now. And then there is a k theory direction. And I'm going to talk about a bit later. Okay, so, so to do quantum Comodji, what you do is instead of doing intersection on the variety, you do intersection on families of curves. So the moduli space of curves. Or if you want to, to see this as an enumerative problem, you, you are not counting points on the variety anymore, but you are counting rational curves joining points in the variety. Okay, so you fix an effective curve class D. You can think about D as a degree. As soon as we go to Grasmagnans, this will just be a degree since a bigger group of the Grasmagnans you see. Okay. And you look at the moduli space of stable maps. So this conservation moduli space of stable maps. So as a set, it's just the set of all morphisms from a curve of students zero, which is only, which has only nodes as singularities. So it's just a tree of P ones with some mark points P one to P N, which are just smooth mark points and the morphism from C to X. And the only thing you want to, to fix is the class, the commodity class of the push forward of your, of your curve, which should be D. Okay. So this represents the curve or you can, you can, you can think of the, the concept of moduli spaces as a family of, of spaces are very big space with lots of irreducible components. Okay. So in the case of, okay. And you have, and you have, you need to, to ask that F satisfies the stability condition. So the stability condition is, is simply that there is no, what there are finally many automorphism for your map. Or if you want for rational, for, for stable maps of general zero, this is just that when some irreducible component of C is contracted by the map F, then there must be at least three mark points or three, or three singular points of the curve. Okay. So, so this moduli space is very nice for several reasons. So the first reason is that there are lots and lots of maps from this moduli space to other spaces. So the, at least there would be, there would be three that I will, I will really need, which are more than three because I have end points, but the, the evaluation maps. So each time you have, each time you choose a fixed point, then you can map the, the stable map to its value at that fixed point. So you get lots of map to X. Okay. So this will be used to, to pull back classes from X to the moduli space. And you also have lots of sub varieties in this moduli space, which are very nice, which are given by curves, which have reduced source. I will talk about them a bit later on. And for X, so this is defined for any projective, smooth projective variety or any, even probably any variety, but, but for, for X rational homogeneous space, this space is, is very nice, this space of stable maps. So at least for us, it is, it is irreducible. It is even rational and it has nice singularities. So it has rational singularities. Actually it has, it has quotient singularities. So this, this is almost a smooth, smooth variety and rational, so very, very, very, very positive. Well, lots of geometry in it. Okay. So we want to use this moduli space to define a new product, which is a deformation of the homology product. So first you, well, because it's a deformation, you need to, to extend a bit the space on which you're going to, to do the product. So the space is just to take, you take the homology of, of the variety and you tensor it with some, some, okay, let's say some Laurent series or the power series of the integers. Actually we don't really need the power series here. We, we will only need the polynomials, but for the moment, let's, let's just define it this way. So yeah, I assume pica rank one, but it's not really important. You can, you can view q as a, as several variables and then, and then definition works for, for any homogeneous rational projective space. Okay. So this is just a space. It's just a tensor product. And you define the, you define a product on this, on this vector space or on this z module as follows. So it's going to be the deformation of the classical homology product to you define, you define the product on, on, on a basis. So you choose two true bad basis, which are here opposite true bad basis. And you pull them back to the modular space using the evaluation maps. Okay. You multiply these two bases on the modular space and then you push forward. Okay. When you do that, obviously this depends on the modular space you chose. So what you do is you choose all the possible modular spaces. So you, you are summing over all the effective classes of curves. So all these D and to remember the fact that you are summing on all these days, you add this quantum parameter, which depends on D. So all these quantum parameters will depend on the effective class. So if you have, if you have pica rank one, you just have one variable. If you have higher pica rank, you have more variables. And then you decompose the power in terms of, of these different generators of pica pool. Okay. So this is the product. And so one of the main result of the series, the fact that this product is, is very nice. So especially it is associated because this is absolutely not obvious from this definition. And then the fact that it's commutative and graded is, is not difficult. You need to, to give the right grading to, to the quantum parameter, but then the commutativity just follows from the commutativity of the, of the product on the modular space. Okay. So this is, this is quantum chronology. And it's easy to see that this quantum chronology, the deformation of chronology, because when you take the modular space of curves of degree zero, this is just the space X itself. And these evaluation maps are just identity maps. And so what you get is just the cup product that you push forward. So it's just a cup product without, without curves. So you are, you are getting back the cup product for q equals zero. And it is a deformation of the, of the, the chronology. Okay. So this is a mark on the fact that you don't really need to have power series here. So if you have a final variety, the, the dimension of the modular space is going to be, is going to be linear on the degree. So, so if you, if you take a degree large enough, then the modular space is so big that the, the interest is a cup product is of too large dimension. And then the push forward will be of a variety of dimensions strictly bigger than the variety X. So the push forward in chronology is going to vanish. And therefore the, the term will be zero for the large enough. For non-final varieties, you, you need to, to do, to be more careful on, on convergence. Okay. So again, when you, you have this quantum chronology, you can look at the, at the stricter constant of the product. And this stricter constant of the product are very nice enumerative geometric properties. So if you, if you look at degree zero, then you get back comologies. So you get back the numbers we were talking before, and which are counting points in intersection points of three shubard varieties. And if you go in higher degrees, and what's happening is that the same kind of techniques using Kline-Mann's versatility imply that these numbers are actually just counting the number of rational curve having geometric properties. Okay. So what, what do you have? You look at the, the set. They're just, you can, you can do this just as a set. The set of all maps from, you can even assume that the source of the map is, is irreducible. So the set of all maps from P1 to your variety, such that the first mark point, which will be zero in P1, is mapped in a given opposite shubard variety. The second mark point, which you choose to be one, is mapped to a general translate of the second shubard variety. And then the last mark point, you map it to the, again, the Poincaré dual of, of the third shubard variety. Okay. So you have this set of, just a set of maps. And if this set is infinite, then actually, because you are doing push forward in comology, you just get that this coefficient is zero. And if it's not infinite, then the number of maps is exactly the coefficient you're looking for. So, so these coefficients are really enumerative. They're really counting curves. And in particular, these coefficients have to be non-negative integers. Okay. So again, you get, you get, you get non-negativity or positivity in quantum comology. Okay. So I, I prepared a small example. This is a three-dimensional quadratic. Okay. So this is the homogeneous space for the symplectic group or for the orthogonal group, depending the one you prefer. Okay. And the shubard varieties in, in the symplectic, in, in, in the three-dimensional quadratic are just given by, okay. The old variety acts itself. Then in, in minimal dimension, you have just a point. Then in co-dimension one, you have the orthogonal of that point. So the orthogonal of that point, this is for example, the set of all lines passing through that point. Or if you want, it is also the orthogonal of that point. Or if you want to view this geometrically, it is a tangent space passing through the, at that point of the three-dimensional quadratic. So it's a two-dimensional cone over a, a conic. Okay. So this is a, again, a shubard variety. And the last one is just a line, which is for example, passing through the point x. So these, there are just four shubard varieties in this situation. Very simple. And, and if you want to, so I, I, I, I'm, I'm going to do two basic computation in quantum comology and in comology. So the first one is just to take two hyperplane sections in your, in your quadrics. So two, two-dimensional quadrics. And if you do this, you have, you are in dimension three. You intersect with two hyperplanes. So you get a, a, a conic in dimension one. Okay. So this conic in dimension one in comology, this is just twice a line. So that's what you get here. And I'm claiming that there is nothing in quantum comology more than in comology. So there are no q terms. And this is because if you take a general point in the quadric and you look, for example, if you start to look for, for, if you, if you're counting curves, you start to count lines. And if you pick a general point in the quadric and you count how many lines are passing through this point and meet the two hyperplanes, then actually any line passing through the point are going to meet the two hyperplanes, whatever the hyperplanes are, just because it's called dimension one. Okay. So the, the, the number of such lines is, this is the number of lines passing through a point, but the number of lines passing through a point, as I was saying, this is exactly a one dimensional variety because it's given by the, the ruling of the, the cone, the quadratic cone, which is a tangent space to the point in the quadric. So there are infinitely many such lines. So there are always infinitely many lines or curve of degree D passing through these two hyperplanes and something. And therefore all the quantum parameters, all the, all the quantum converging product will be zero. So there is nothing really interesting in this situation. But if you go in co-dimension one more, then you look at a line and an hyperplane in that quadric. Then, then, okay, the intersection is very easy. It's just intersection of an hyperplane and a line. This is one point. So this is a homology class, a homology product. But if you look for lines passing through a general point, no. The lines passing through a general point, they are just, they just cover the, the, also going to have that point. The also going to have that point is going to meet the line you chose in exactly one point. So this fix a line, and this line is going to meet the hyperplane in one more point. So it's exactly one line, which meets a general point, the line and the hyperplane. So that's the coefficient one, which appears here. And since you have to take the point-care ideal, the dual to the point is a unit, and you get this quantum parameter. So this is, this is the way you, you can do it. This is just doing geometry and counting. So of course, when you go to higher dimension and more complicated varieties, this is more, more difficult. But in principle, you can just count rational curves. Okay. So, okay, I go on with a quantum k theory. I don't know when I can do a break. So I could do a break now. Or I go on with definition and result of quantum k theory, and then I do the break before results and idea of proof. We can do a break now. It's almost half essentially. So let's have five minutes break now. Let's have a break now.