 All right, so welcome back to the second half of the Schubert seminar with Nicola Perin telling us about positivity of Grasmani and quantum K theory. Let's go ahead. Okay, thanks. So, so now I start with quantum K series, so I will start with with first case theory. And then and then go to quantum. So K series is another avatar of homology. So now what what you what you do is you look at current shifts on your writing that you look at isomorphism classes of current she's on your variety then you look at the Z module they generate. And you mod out by the relations given by short exact sequences. So if you have a short exact sequence and the middle term is the sum of the two extreme terms. And actually, if you are on a smooth variety, then you can resolve any current shift by locally free shifts. So actually you can replace the fact that you were working with current shifts with working with locally free shifts, which is sometimes more convenient, sometimes not depends what you want to do. But but this is still it's still very convenient to have these two possibilities when you have a smooth writing. And know that you have this this group, you can define several things on this group so you can define you can define some which is just given by by the direct sum. Okay, and you can define the product. So yeah, I define the product for only for classes of vector bundles. So then in that case, the product is just given by the tensor product. And if you want to define the product on classes of query on shift, you have to you have to use the right function. So you have to use tor tor functions, but it's it's just just doing the same with resolutions. Okay, what you also have to replace like the degree of intersection in in Comology is the degree in care theory, which is just given by by this integration formula. It's just alternating some on the dimension of the Comology group of your of your current shift. Okay, and once you have this, you can define the you can define the duality because then you can you can do the product and compute the degree. So I will talk about duality right after. And this this case here is is for tutorials, you can define pullbacks. For example, on local if you shift, you just take you just have usual pullbacks, and you can define push forwards. So for push forwards, having a local if you shift is is not helpful, but having current shift that it's useful. It's if you have a proper map, then you can just define the push forward by alternating. Okay, so I forgot some signs. There should be some some signs. So you want to to have alternating sums on on push forward. So also this is like really very similar to what you have in Comology is likely is likely more is likely more precise. And especially you can you can again use Schubert varieties or Schubert variety or opposite Schubert variety with same convention indices for Schubert varieties, exponent for opposite ones and then define their their classes. So they are current shifts or you result them and you get you get Schubert classes in K theory. And these Schubert classes, they form basis of the case here. The slight difference with with Comology here is that the Schubert bases are not open are not dual for the intersection. So so you have to you have to do some perturbation because because K theory is really not only viewing what's happening at just the zero level of intersection. It also sees everything what is in in higher dimension. So if you have two varieties, which do intersect, for example, if you have two Schubert varieties in and so Schubert variety and an opposite Schubert variety, their intersection will be either empty or will be a nice variety, which is called the Richardson variety, which is rational. And therefore, if you compute its degree, you will you will get that the product, the product and then the degree is always equal to one. So so you will not have that Schubert varieties and opposite Schubert varieties are in duality for for this intersection. You need to remove the boundary terms which are coming from the smaller Schubert varieties in in your Schubert variety. And therefore the duality is given with the ideas of the boundary of the Schubert variety. I will I will I won't need this much, but this is really an illustration of the fact that K theory really sees more geometry. And this will be reflected also in the in the quantum definition. Okay, so this is this is K series. So you have dual basis. And so I prepared the also a small example if you if you look like a three dimensional quadric. You again want to compute the product of two hyper hyper plane classes. We just have these three dimensional quadric to hyper plane classes when you do the intersection, the intersection will, if they are in general position will really be the class in K series. So this intersection is a quadric, sorry, a conic of dimension one, so a smooth conic. But then you want to express the smooth conic in terms of the Schubert classes and the smooth conic is not a Schubert variety. So what you do is you deform this conic to the union of two lines. So there is a there is a one parameter family of conics with the union of two lines. And once you look at this union of two lines, you have a nice exact sequence which for which you have the sum of the structure of shift of the two lines which maps to the intersection point. And the union of the two lines in their meeting is exactly the kernel of this map. So what you get is that the class of a conic is just twice the class of a line minus the intersection point because these are really two lines, but minus the point because you have to count the intersection point only once. Okay, so this is an example of what's happening in K series, classical K series. And we want to do the same as before, but using K series instead of commology. Okay, so the naive, the naive try is really you do the same. So you extend the definition of K series to quantum K series by tensor product with a power series in, in Q. Okay, you're, you're viable. And then you define the product as before. So you take two classes, you put them back, you multiply on the modular space and you push forward. Okay, so when you do this, the problem that you don't get something which is associative. Okay, so you don't really see why because I didn't really explain how you get associativity in quantum co-mology. But associativity in quantum co-mology comes from, from these very nice sub varieties in the modular space which are given by curve with two irreducible components in the source. And then using that you can, you can prove some by induction. You can prove that there are some very nice combinatorial properties and formulas which induce the fact that the product is associative. When you do that in K theory, the problem is that K theory is seeing more than co-mology and you don't only see what's happening in the right dimension. You should also see the geometry in, in higher co-dimension. And the fact that, for example, the classes of sub varieties, so the classes of stable maps which have non-reduced source, they do meet. And you have to take into account the fact that if you have sources with D1 and D2 and then sources with D1 prime and D2 prime, when you intersect these two, then you are going to get some intersection that you have to take into account. So this is, this is not associative because of, of that. And you have to, you have to correct the, the product and to do it, you have to take, to take care of all these maps which degenerate, so which have a source which is degenerate. So I keep the notation for, for the group. The group is, is fine. Only the product have to be corrected. And to correct the product, you, you have to look at all possible sequences of effective classes of curves. Okay, so I look at the sequence which has R plus one terms with a first term which is D0, which will be some of the main term. And then a tail which will be of degree D1 to the R. And so I have to assume that the, the sum of all this degree is D to, to be in my moduli space. And I look inside the moduli space, the locus of, of chains of irreducible curves, which are all rational. So I forgot to say rational, but they all have to be rational. So you look at chains of irreducible curves, C0 up to C2, they just meet in one point of the only chance. And the degree of the component C0 is D0 or the degree of the map on that component is D0 up to the component of the last one is the air. So that's, that's what you look and you want that the first two mark points are on the first component and the last one on the last component. So you have a chain, you look at all these possible chains and you take the closure in the moduli space. So of course in that, in that sub variety, you don't only have chains with R plus one components, you may have like much more components, but the general element is of this form. Okay, and you have to take into account all these sub varieties in the moduli space to compute the product. So what you do is you define the product again using the quantum parameter and the term in degree D is a very big thumb. Over all possible decomposition of the degree D in terms of chains of curves of degree D0 up to the air so that the sum is D. And then when you put a one, when you have a irreducible curve, and then you put a minus one, when you have two irreducible source with two components, and then you put a minus two to the minus one to the square when you have three components and so on. So we have alternating signs. And what you do is you just pull back the two classes, you multiply on that, you multiply these two classes and you restrict to this sub variety of curves with chains of degree D0 up to the air. Once you did that, you push forward and you get a class in the k-series of the variety and you sum with signs all these classes. So that's the definition of the product and the main result of the series due to give and tell, at least for homogenous spaces, is that this product is commutative, which is obvious, and associative, which is really the part of the story. Okay, so you get this new product and really what you have to do is just taking into account the fact that these boundary devices in the modular space are meeting. Okay, and again, again, if you look at the case where there is no quantum parameter, so the degree is zero, then what you are doing is just your modular space is X itself and you are doing nothing else than pulling back and pushing forward by identity. So you are just doing classical k-series. So these are basic stuff on k-series or quantum k-series. So k-series or quantum k-series are just filtered by dimension. And if you look at the associated graded algebra, then you get back either comology or quantum comology. So the graded algebra associated to quantum k-series is quantum comology. So this is a finer refinement of quantum comology. And what this means is just that the leading term in quantum k-series is the quantum comology term. Okay, so here are some examples. I'm not going to explain the computation because you need more geometry to do the computation, especially in this very special example. So I chose the example of the quadratic because here you see that, okay, if you look at comology and quantum comology, there is nothing happening. So these two are explained. So the intersection of two hyperplane sections in the quadratic of dimension three are just giving a conic. And this is also what you get in quantum comology, nothing else than classical comology. So there is no quantum deformation here. While in k-series, we have seen that this is twice the line minus the intersection point. And if you go to quantum k-series, then actually you see that you have more terms than what's happened in quantum comology. So this is something which will not happen in Grassmannians, but which may happen in general, is that you have no q-term in the comology, but you do have q-terms in k-series. So I will at least say later on that this is not happening for the quantum k-theory of Grassmannians, meaning that the q-terms in the quantum comology and in quantum k-theory are always the same. But it's not true in general. And you see the quadratic is a quite simple example, but still it's not true. Okay, and then when you have quantum comology or quantum k-series, so for quantum comology we have seen several properties. And you may ask, are these properties of quantum comology still true for quantum k-theory? So the first question you can ask is whether the sum in the definition, which is a huge sum with lots of terms, whether this is still finite. So it's absolutely not obvious that this sum is finite a priori because there are infinitely many terms which are non-zero. But actually the sum is finite, so this was proved quite recently, and the sum is finite and it's because there are lots of cancelling in these alternating sums. So there are lots of plus one minus one which cancel out. So the sum is finite, you can do the definition of a polynomial and not only power series. And so the question I want to talk about is positivity. So if you write quantum k-theory product in the basis, in the fabric basis, so this is a fabric basis also because if you look in comology and you look at effective classes, then for homogenous spaces all the effective classes form very nice monoids and they are generated by Schubert classes. So any effective class is a non-negative linear combination of Schubert classes. Schubert classes are really the basis you want to look at when doing geometry. And if you do this product in Schubert classes, you expand, so you get lots of coefficients. And the conjecture is the following one, so this formula might look complicated, but it's easy to explain. So this statement is in quantum comology, so you can obviously this statement tells you something already in classical k-theory if you take d equals to zero. And this statement, this conjecture is a theorem in classical k-theory, was proved by Breon. So Breon proved that this positivity assertion holds. So what is this positivity assertion? So the conjecture, it simply means that if you start to do the product, since it is filtered by dimension, the first term is always a term in comology. So in quantum comology or if you are in classical k-theory, it's a classical comology term. So this term, we already know that it is non-negative because it's an innumerative count. So this term is always non-negative and then the signs are supposed to change each time you go in higher co-dimension. So that's what this conjecture says. So in the example of the three-dimensional quadric, the product of two hyperplane classes, you start with this classical comology term, which has a two, so it's the class of the conic. And then even in classical k-theory, you see that you have one more term with a sign, so the sign changes. And then if you go in quantum k-theory, then q has the same degree as a point. So these two have the same degree, so they have the same sign. And then you can go in higher degree and you get a plus sign. And this goes on until the end. So that's the conjecture. And now I want to... Nicolás? Yes? In the statement of the conjecture, maybe you said it, but is that some of the links or the product of the links? Yeah, it's the same, actually. It's the same parity? It's the same, but it's shorter. But it's the same. But this is again under as willing to write things as short as possible. So at the end, it's the same as the sum. I don't think it's the same. It's the same if you put them inside the parenthesis. If you put... If you put u, v and w, if you take the product of the group elements. Oh, okay. That sounds better. So that's my mistake. And under this version is correct. Sorry. Okay, thanks. Sorry, can I ask back in the definition of quantum k? So you weren't taking the whole modular space, just the closure of this chain subset? Yeah. And what's something that's not in that closure? I'm confused because this chain case includes there being one curve, right? Well, the one curve is when air is equal to zero. Indeed, you have this one curve here. So you do it for all the compositions. So the first term in the sum is the case where you have a new decibel curve. So that's the leading term. It's going to be the quantum comb or g term somehow. Not really, but it's the leading term. And then you have to correct by terms where you forget about curve which are irreducible. Like if you have a chain of a curve which is of the grid D minus one and a line. This will not contain irreducible curve in its closure. I see, I see. Okay, I got it. And then you have to count these ones also. And then again, you have to count those which decompose more and more. I see. You see, you are doing the intersection. And then you restrict this intersection to that special sub-variety in the modular space. And you have to compute, you have to compute the Euler characteristic there. Or you have to push forward this sub-variety of chains of curves. So the first one is meeting the two sub-varities you are starting with. And the last one is going somewhere. And you are somehow looking at where it's going. It's a bit more than where it's going. But more or less it's where it's going. So where is the third point in such a chain where the first component is meeting two given opposite trope varieties? Can I think of it as if I used the structure sheaf on the modular space, I would get the wrong answer. And this is a way of correcting it to the class of some different fundamental K-class on the modular space? Yeah, yeah. So I think of it as when I try to prove associativity, then what I want to do is to decompose curves to do some induction. And when I do that, when I do that in comology, what I'm doing is just I see the two... So if I have two possible decomposition, then these two are divisor in the modular space or sub-varities in the modular space. And when I do the computation, the formal computation in comology, what I do is I do as if these two varieties were not intersecting. And the problem is in K-theory these two are intersecting. So when I do this, when I do as if they were not intersecting, I am replacing the... As for lines, the conic given by the union of two lines by two disjoint lines. And this is fine in comology, but this is not fine anymore in K-theory, so you have to correct that. And so that's my way of doing this. Okay, I think I got it. Thanks. Okay, so that's where I was. And I want to explain the results, but the results, this is easy. So the results is that this conjecture holds true for grass manians and actually a bit more. There are a few more cases where the conjecture is known. So for all minuscule spaces, so for example, the three-dimensional quadric is not a minuscule space, but it also holds for all-dimensional quadrics, which are not minuscule spaces. This is because we can do all computation by hand some more for all-dimensional quadrics. This is simple enough as I explained in the case of three-dimensional quadric. And so these results on minuscule and all-dimensional quadric, they're all in the same paper with Anders, Pierre-Manuel and Leonardo. There's a question. No, no, it was just me having a comment. Thanks. It's all good. And then there are two more cases which are not minuscule or co-minuscule anymore. So the first one was put by Rosset and Weyong-Chu for the incidence point hyperplane in the projective space. So again, you have this positive conjecture and lots of nice combinatorial formulas. And very recently, I put with Vladimir Benedetti and Weyong-Chu again that this also holds for isotropic grass mania of lines in the symplectic space. So these two are adjoint or co-adjoint varieties that are still very, very simple homogeneous spaces. Okay. So this is what we know for the moment. And now I'm going to try to explain a bit how you can prove this. So I have a few more minutes to do that. So let's try. Okay. I want to go deeper into the definition. So the first thing I want to do is to reinterpret the definition more geometrically so that really you start to understand a bit better of what's going on. So the first thing you do is you forget about the fact that you have to sum on degrees. So you focus on the degree D case. Okay. So you look at the term in degree D. So that's this part that I'm going to write a product in degree D. Okay. So this is a big sum of all the composition of D in terms of a sum of smaller degrees. So here I'm in a grass mania. So these are just integers. So I have this big sum, which is an alternating sum of classes in k theory, which are really like products for the moment. So I want to to interpret this class as a variety. So what you do is you look at this modular space. You look at the inverse image by the evaluation map of the first true but variety the inverse image by the second evaluation map of the second true but variety. MDXUXV. This is really the set of all stable maps with three markpoints where the first markpoint is in one true but variety and the second markpoint is in the opposite. So you have you are joining these two opposite super varieties by curves of degree D. Okay. And then you can do the same, but instead of just having curves of degree D, you, you, you look at chains of curve of degree D zero to the air. So you fix the composition and you do the same. So you have the two super varieties and you want to join them by chains. Okay. So it's not, it's not exactly joining them by chains. You have to be careful. So what, what you ask is that the first term of the chain is joining the two super varieties and then the tail is going somewhere else. Okay. So for example, if you start with the first degree zero, then you fix a point in the intersection and you look at points in the tail, in, in a curve of, in a tail of degree D one to the air, which are starting from the intersection of your two super varieties. So this is, this is one, one situation. So at all these varieties and a general statement. Well, again, this is again a variation on, on Climans-Versality. It implies that the, the key theory class you want to compute, which is this product is actually a geometric class. It's just given by the, the structure or sheaf of this sub variety in the variety of stable maps or of all stable maps with sources. Given by this decomposition, where the first component is passing through the first two bed variety and the opposite one. This is very geometric. And so this replaces this big sum by another big sum. But now you have only one geometric object. This is this modular space of chains. And you have to compute the push forward by the third evaluation map. The push forward maps these curves to the third point, which is the locus of all points wiped by all these kind of curves with chain of curves with first one meeting the first two super varieties. Okay, so this is the first reduction formula, which is, which is rather easy. And then you, then you want to, you want to, to compute. You want, you want either to compute the push forward by the evaluation map. If you remember, so these classes are just pushed forward of some variety by an evaluation map. So you want either to compute this class or to say something about about this class and to say that it has some positivity properties. So I will use two statements. The first one is a statement by Kola, which says that if you have a proper map, which is subjective from Y to Z, such that the general fiber is commonly clear trivial. So all the communities are zero except for degrees zero. And if the two varieties have rational singularities, then the push forward of the structure ship of the top one is the structure ship of the image. Okay, so this, if you have this, this situation to the evaluation map we had before, then it's just mean that the push forward we are looking for is simply the structure ship of the image, which is the locus of all these points by these, by these curves. And the second statement, which is, which is a statement, which the brilliant used to prove the positivity in classical case theory, is that if you have a birational map into a homogeneous space. So this year, this axis, the homogeneous space we are working with. And if you have such a birational map where, where the source variety is nice enough. So for example, it's going my colleague with rational singularities. I think my colleague is enough. Then actually, if you expand is the class of the push forward, it's going to be, well, you, at least you're going to, you're going to have some positivity properties. So you're going to know the science. Okay, so this is a, this is a positivity property. But this is, this is what you would do. Then if you want to do the intersection in case theory, you just look at intersection of two super varieties, you get the Richardson variety, you do the push forward of the Richardson variety, which is just inclusion. And you get the expansion and expansion as alternating science. So that's the proof of brilliant when we want to use it in our situation. So this gives you two ways of computing the push forward. Either you don't compute the push forward, but you get positivity. If you have this birational map. And if you don't understand, for example, if you don't know that the image is as rational singularities. Or you know the image as rational singularities and you're able to really compute the class of the push forward. Okay, so I don't know how much time do I have. Yeah, you can go a few more minutes, maybe five minutes. Let's see, let's see. I try. So, so, so few more reductions. So we want to we want to compute this system. So this is this is thus an alternating sum of push forwards. Okay, so, so this alternating sum as I was saying, so you want to look at the evaluation map. So you want to understand this map. So this map is is is just you look at the stable maps and you do the third evaluation. So this is just a set of points which are on a tail of curves of this form. Okay, and the expectation. This is not true for the expectation. Okay, the first the first part is true. That's what I explained. This is this is like a climate. And then what if you if you want to apply color serum than the expectation. If this map as for example, general fiber, which are comodically trivial, then the expectation is that the push for the class of the push forward will be the push forward the class of the image. So it will be the class of these lockers of points in the variety, which are covered by these curves, this chain of curves. Okay, so, so actually what you can you can prove is is the following. So this is a first simplification which which is not always who but which works in any situation is that you can you can simplify a lot. The very big sum which is at the top in terms of a sum is only two terms. So what I wanted that using color serum is if you know that as soon as two points are connected by a rational curve of the greedy then it is connected by a chain of lines. So this means that any two points which are connected by some curve of the greedy they can be connected by any kind of chain. That's what this means. Then actually in the in the very big sum is very big alternating some only two terms remain the first two terms. So the curves with only one irreducible component and the curve with two irreducible components one of the great D minus one and the tail of the green one. And the reason why is that actually the the target. So so you can always replace the modulus space push forward by by the image, the image with always be the same because the image will not depend on the form of the tail. And then if you look at the possible tails, let's say we have a tail of degree two, then you will have a plus one for tails which are given by a conic. And then you will have a minus one for a chain of two lines. These two terms will be the same they will cancel. If you have a tail of degrees three, then you have an irreducible curve of degrees three, you have two curves of degree one and two and two and one. So there are two minus ones. And then you have one more term with the chain of three lines, which is a plus one. And if you sum all these terms together, you again get zero. So so doing that you you you see easily that this is this is formal computation. Actually, only these two terms are left if the geometry of your variety is nice enough. So any two points connected by a curve of degree is connected by a chain of the lines. This simplifies a lot the description of the formula and you only have two terms. So we want to understand these two terms. Okay, so I finish with the key statements. So we have these two terms and we want to understand whether they are positive or not. So you look at the evaluation maps. So here they shouldn't be an underline. So there is one one evaluation map for irreducible curves and one evaluation maps for a curve with two terms, a curve of degree D minus one and one. Okay, and what what we can prove is the following. This is a general general statement. This is the main state main technical statement. So the first thing is that if the evaluation map, the evaluation map with irreducible source is barational, then the same is true for the evaluation map with curves with two with two irreducible components. And the image of curves with two irreducible components, because it has to be it has to be smaller because the two maps are birational and and curves with irreducible source of co-dimension one in the in the modular space. Then what you get is that the image is also of co-dimension one. So the the fibers being being of the same dimension, you get that these two are different but with respecting co-dimension. Okay, so these these are always birational so you can use brilliant and you can compute the science. And you see the science of gamma D will be given by brilliant and the science of gamma D minus one one will be also given by brilliant. They will be opposite because they are in co-dimension one but because of the minus sign in the formula where they will compensate and add together to get positivity. So this will will give positivity in the case where the map is birational. And if the map is not birational, then actually both maps are not birational and both maps are have ecologically trivial fibers. So to compute the images you just look at the images. But because you have maps with fiber of infinite dimension, in the case of irreducible sources, you have an irreducible an infinite family of irreducible sources. And then doing a bed and break, you get that in the limit. In that family you have curves of degree D minus one one. So the images are the same and therefore the two terms are concealing out and there is nothing. So either you are birational and you get positivity or you have nothing so you get the you get the result. And actually what you get even in this situation is that because quantum homology has no term, if the map is not birational you get that the quantum terms appearing are exactly the same in quantum k theory and in quantum homology. Okay, I think it's time to stop. Thank you. All right. Thank you very much for a beautiful talk.