 All right, so welcome everyone to the Schubert seminar, the spring 2023 edition. We're happy to start with Professor Sue Cato from Kyoto University, and he's going to tell us about the geometric realization of Catalan functions. So, please go ahead. Anyway, thank you very much for the organizers, especially they're not to give me this opportunity to explain what I understand. And today I will try to explain some geometric realization of Catalan functions. Actually, I was quite, yes, I was somehow quite excited to understand some portion of this one, and write up a paper like this. But unfortunately, due to some lack of time, this original version of this paper, this is premature. So I try to update, will be updated, updated in few days. And this version, in this version, there are some, and if it's somehow inconsistent notational something like that. So, but anyway, I will try to resolve everything and try to do somehow conservative in the next, I'm really sorry about that. But let me try to explain something which I understand, and this part should be quite firm. So, start with somehow simple, simple, simply connected. So maybe I will write a little bit larger. Simply connected algebraic group over C and fix it below. And inside here, we take maximum dollars. And X equal G over B, this supposed to be alpha priority. And we put P to be the algebraic group from moving from T to C star or GM, that is supposed to set weights and inside here, we have dominant weights. And if we have a dominant weights lambda, then we have V lambda irreducible finite dimensional G module with the eigenvector of T weight lambda. So, anyway, in this, this is the basic situation and recall the basic theorem of the way. So, for each lambda, we have a line bundle, such that it's there, the global section of X of X of lambda. So, this is isomorphic to V of lambda, or there, and this happens for exactly when lambda belongs to the dominant weights, and it is there for us. And this is the basic results by Brailleville, and we are supposed to generalize this in some sense. So, our generation comes from taking cotangent bundle of this one, because cotangent bundle is a vector bundle on X, so we have a projection map, and because X is acted on G, it inherits the action of G. And also, each fiber is a vector bundle, vector space, so it admit an extra action of C star. So, this C star can be recorded by some number Q, so this is supposed to be degree one character C star. And in this situation, the theorem proved by Langey, Bromsky, Brewer shows that for lambda from dominant weight, so if you consider the graded character of H naught, of D star of X, or the pullback of X of lambda, you can generalize. So, this is supposed to be the character that records C star degree by Q. Anyway, so if we consider this one, then we get something like this in lambda, where K of G, or mu of lambda of Q, this cost of polynomial. So, cost of polynomial is a classical cost of polynomial for G equal SLN, and this is, this gives one realization of cost of polynomial in terms of the geometry of cotangent bundle. Anyway, this is something established in around 1990s. So, in order to proceed a little bit more, let me recall what was T star X. There's a vector bundle, but this is supposed to be a quotient of somehow G times the algebra of B by itself. This is supposed to be the unipotent radical of B, but by somehow taking Tilda, so Tilda denotes anti-diagonal action of B, and of course B action here by right and B action here by adjoint action, and we would like to put it together to obtain something. So, if we do this, ah, sorry, maybe if we, if we do this, then we get a smooth. So, in particular, so if we consider something, something else here, we get something different. And in this situation, a blower asked what happens if we replace D of BD with its B submodus. So, there is a worker, Shimodono Waiman, and Chen and Haiman, and that is not the situation we have considered before, but it is a closed cousin that replaces G with GLM. That is not a simple algebraic proof, but anyway, so if we do it, then it seems to use some great combina tricks in a uniform fashion. And this kind of things, but further defined by brassiac, mos, and samas, and so the claims are conjectures, consistent combinatorial part, and geometric part. And what brassiac and mos and pam proof is that combinatorial part is okay, actually for conjecture is true, and there are also some extension, no symmetric cases. So, this is the situation we had before. Anyway, as I said here, conjecture was geometric origin. So, it is better to have full account of this thing in order to see how the things are going on. And maybe this will help to understand the blower's speculation to what is the real meaning of this kind of business. So, let me try to start the actual contents. So, in order to say something, we need to say a little bit about roots, ideals, and Catalan polynomial. It is unfortunate that I totally got some confusion about the polynomial some functions, but what we call functions is in the sense of McDonald's, that is a limited polynomial. And if we specify a particular n and consider the geometricalization, it will be a polynomial. So, in a sense that the function is something a little bit too much to say, and that is a part of the result of brassiac and mos and pam. Anyway, let's try to start. So, in this case, of course, g is gln. And inside here, we take b to the upper triangular matrices. And inside here, and t is much more close. So, again, of course, x equals g over b, this is more for the probability of SLN and n to be the real algebra of b. And do you three delves this in the algebra n this something I do not define, but this is equal to identify with the B stable ideal. This, this is a representation of B, so we can consider it to be some modules. And also, in this particular case, this is in bijection is the class of sizing. So, how does it look like. So let me try to make one, only one example, and then maybe you can find what happens. So if we consider four by four matrices, and consider something some extra zero, and we consider something star here. Then, if we write it like this, then it looks like a young diagram, and if we make some upside down, that is to do, maybe. Yeah, that is to do. So, upside and down, and then what happens is that, so we have something like this. And if we consider if we take line, that goes 45 degrees. And this is something in under. So this is supposed to be also. So this is some big plus. This is supposed to be visa module. Probably, you can easily generalize to the more general situation. Anyway, somehow we record these things, or the class by side. And we would like to consider taking rise to corresponding visa module by end of time. Then, what happens with that. As I say that, so T star of X, this equal to the G times B of N. So this is supposed to be something coming from the equivalence relation. And as this is a visa module, we can consider this one instead of this one that we denote by T star of psi of X. This is a vector bundle. So this is supposed to be vector bundle. So this is equal to, and let me try to, yes, prepare a little bit notation. I'm sorry about this. But in G called GNN, the positive or dominant weight is a little bit different. And our convention is as follows. So, if you know I saw the homo, sorry, T of C star, this isomorphic to direct sum of I equal one to N of G of epsilon I. Here P plus the dominant weight is supposed to be like this. This is the dominant weight for GNN. But we will eventually restrict it to something like this. This is supposed to be the same set, but also positive. So this is usually called the dominant polynomial weight. So sorry for interrupting. Could you just remind me what kind of an object is that the capital psi? Capital psi is supposed to record big pass. Okay. Just to take pass. Yeah, just to take pass, and if we take some person collect some boxes inside here, we can think of some corresponding B module, B sum module of N. And this is N psi. Yeah, but any subside is just a big pass. Thanks. Yeah. So anyway, so, of course, we have, if you have a character, sorry, maybe I should also put like P. In this case, we get OX of lambda. Of course, some determinant character is just a character twist. But anyway, so we have a line. So we have definition. So psi is the root ideal. Actually, I'm constantly saying some lie, and this is just considered as a big pass. But the big pass and the root ideal and B sum module of N, are identical in a sense. So we use sometimes identify. So, and the dominant polynomial, right. We define the corresponding catalan. So this is the name of the name of catalan polynomial. And that is supposed to be the sum over M and mu. That is also P plus of Q of M of character B of lambda times dimension of form. Maybe I mean C star of V of mu times C of M of inside T star of psi X. Still not enough space. Sorry for pi star of psi of lambda. So this is supposed to be the action of this one, and this is supposed to be this one and define the naturally extension from G is equal SLM. And let me try to make some remark. So H of psi. So H of lambda is sitting inside of G of Q of X1 to XN. So it is polynomial. And this is not the coincidence. Because so note that usually, if we put all of X and Q to one, this corresponds to the meaning of characters, but not to this case. But so H of psi of lambda to Q is equal to X1, XN, this action number, while dimension of H0 of psi star of X of pi star of phi of OX of lambda. And this usually in general. Of course, if we take psi to be very special, it can be finite measure, but it is usually infinite measure. So what is the key to this here? So we just count polynomial representation instead of four things. And this is the introduction of Catalan functions and maybe we can take you five minutes break to explain how to make, how to understand these things. Good. Thank you. Any any questions before we take the break.