 For now, we will continue with the next speaker, which is my Mihail Tari Gracci from Rutgers. He will speak about classification of Grasmani and Schubert varieties up to isomorphism. Right. Thank you, Anders. Thank you, everyone, for the invitation to talk. Yeah. So I will speak about classification of Grasmani and Schubert varieties up to isomorphism. I should preface this by saying that the work I present here is in a joint paper with Wei Hongshu. And yeah, let's get started. So as before, my setup is a classical one. I will consider the Grasmani and variety X, which I did not buy GRM comma N of M dimensional subspaces of Cm. And my Schubert varieties are indexed by YAM diagrams, so I'll denote them lambda. And I'll require that they fit inside the rectangle with M rows and N minus N columns. Okay. So given this indexing, I asked a natural question, how does it influence isomorphism of a class of X lambda? So is the YAM diagram lambda a complete invariant of X lambda up to isomorphism? And I should say complete invariant. I mean, if I pick another diagram, let's say mu, would I get something isomorphic? And if I do get something isomorphic, I want lambda to equal mu. Okay. And this is not precise question. This is truly false for our following well-known facts. First, I can extend my Grasmani and variety. So if I take higher dimensional subspaces inside the higher dimensional vector space, then do the same YAM diagram, I get isomorphic Schubert variety. The other fact is that if I transpose my YAM diagram and it fits inside the rectangle, then I also get an isomorphic Schubert variety. And here I abuse the fact that I can extend my rectangle as much as I want, so I don't care about bounding my YAM diagram. So with this, I update my question to is the YAM diagram lambda up to transpose a complete invariant of X lambda up to isomorphism? And we answer this question in the positive. So if X and Y are Grasmanians, could be different arbitrary X lambda and Y mu are Schubert subvarieties, which happen to be isomorphic when lambda equals mu or lambda is mu transpose. All right. And most of our proof relies on the fact that we want to look at properties of X lambda, which are intrinsically geometric, so they don't change up to isomorphism. And then we try to fish out combinatorial data out of it. And this theorem is by myself and Schubert in 2022. OK. So my geometric invariance started out with the dimension X lambda, which is the size of lambda. And I'll consider, in some cases, the ranks of the child groups, which in terms of lambda, they correspond to the number of sub diagrams of a given size, and then the most amount of information I get from very visible components of a singular locus of my Schubert subvariety. So due to a theorem of Lakshmi Bai and Weyman, there's a precise description of these irreducible components and they're given by Schubert varieties by removing outer rim hooks. So looking at my lambda, I call this a rim hook. So removing it gives me a lambda 1, this is another rim hook, gives me a lambda 2, and lambda 3 comes from removing this third rim hook. OK. Yeah. So now I wish to translate these geometric invariance with Schubert subvarieties of X lambda into something combinatorial, which tells me things about lambda. So I apply induction on the dimension of my X lambda or on the size of lambda, which is the same, to obtain these smaller young diagrams from before up to a transposition because this is a statement of what I want to prove. OK. Then I also require an additional ingredient, which is fair intersection, which geometrically corresponds to the intersection of all very visible components. And again, as before, due to induction, this is up to transpose. OK. It is of interest that I know which hook was removed first and which was removed last. So I wish to order the diagrams in such a way that I can tell which hooks were removed consecutively, so which hooks were adjacent. And geometrically, this corresponds to the Schubert variety of lambda i and Schubert variety of lambda j, not being a proper intersection. So in my example, X lambda 1 and X lambda 2, don't have a proper intersection, and you can see where hooks are adjacent with touching this box. Same for lambda 2 and lambda 3, they should be touching this box, but this is false for lambda 1 and lambda 3. So at least I get some kind of ordering on my lambda i's. I could get two of these in some sense, one ordering in its reverse, but that won't make much of a difference going forward. OK. Yeah. So to summarize my combinatorial data, given a Schubert variety X lambda, I look at its dimension, there are diagrams coming from its singular locus and the special diagram lambda 0. And as an example, if I were to look at this particular young diagram, I would get 27 these diagrams and lambda 0. In some sense, I should not look at precisely this because I use induction. So transposes will be a factor. So just to transpose and reorder randomly from a point, I could also look at this data, which should be the same as this data. All right. So this is transposed and moved to the front. This is transposed and moved to the back. OK, now with all my data, I want to do a reconstruction problem to obtain lambda or its transpose. So the main ingredient is the fact that lambda 0 is very close to lambda. Lambda 0 being the intersection of irreducible components or by removing cogs is very close to lambda. In this case, I just remove these boxes. So I want to express this in terms of my previous data. The way I do this is by counting how many rectangles were removed and by a rectangle, I mean a maximal rectangle, I mean a sub diagram, which is a rectangle and happens to be maximal. So here, this is one. This is the number one. This is my third maximal rectangle. OK, the fact is that the number of components I got before is one less than the number of rectangles. So here I'd get three and here I have two components. Another fact is that when I consider lambda 0, all the rectangles of width or height 1 disappear. So counting rectangles in lambda 0 and looking at my previous count should give me a clue of what disappeared and what didn't. OK, so in this case, I started out with three rectangles, which is 2 plus 1. And here I have three lambda 0. So I had three initially and none of them are width or height 1, which actually translates to our case checking by the fact that we just add the ramp width 1 to obtain our lambda. OK, this is the simplest case. Most of the other cases go ad hoc, just like this one, to do a more complicated one. Just to prove a point, I consider a lambda with at least one rectangle of width 1. And in this case, computation say, so this is a free. So there are four rectangles inside lambda. Lambda 0 has three rectangles. Therefore I get for free that I have a rectangle of width 1. And some more case checking would say that to recover lambda, I need to union or to combine two of my irreducible components. So here's where the ordering comes in. It's always a case that I have to pick the first and the last one. All right, so in this case, it's lambda 1 and lambda 3. And since there's a transpose going on, I get two sometimes distinct cases of a transposing lambda 3 or not. But there should be only one compatible answer. In this case, we can read it off very union. So lambda 1 and lambda 3 would give out lambda transpose. While transposing lambda 3 would give something different from lambda and lambda transpose, also not compatible with our previous data. OK, the proof is more or less more case checking and so on. So I will skip that and instead focus on what kind of generalizations and similar work is being done. OK, so in 2007, Devlin, Martin, Reiner solved this problem for a class of smooth, sugar varieties in partial flag varieties. In 2020, Richmond and Slavstra solved the problem for complete flag varieties, GMRB. In 2022, myself and Shoe solved this problem from Grasmagnen varieties. And it's a work in progress. But myself, Richmond, and Shoe are solving the problem for community school flag varieties. Now, I should say that indexing by young diagrams is something very usual for Grasmagnen varieties. While it is not the case for, for example, complete flag varieties, so the combinatorial invariance there are slightly different. It's the same case for a first result for some partial flag varieties. In our current work, we are considering some postage structure, which generalizes this young diagram structure. And we assign some labels. So in particular, we're looking at some order ideals. And the postage structure kind of encapsulates both information from the diagram, but also it's transpose. So it's a much cleaner argument than we have here, much less case checking, and so on. OK, this is it from me. I think it's about five minutes left for questions. Thank you. Thanks very much for a very nice talk. And yet, we have time for a couple of questions.