 We'll come back and yeah, please continue. Okay, hi everybody. I'm trying to draw for you as an example of insertion. So I'm almost done. So it's just to show you an example. If you were inserting this by word from the right, right? So if you're using right insertion, first we would insert one one and it would bring us from here to here. And you can see what happens. We kind of find correct position and droop it. And then we kind of glue them there and now we have this guy, right? So, and all other steps are similar. And if sometimes it takes more than one drooping, there are certain rules. But at that we get this insertion and this is the final boundless prime dream, right? So this is our P over here. And the chain, the recording chain is this piece is something like this. So this is one, two, three, four, five. Two, one, it just records which permutation each of those is. So two, one, three, four, five. Four, three, five, four, two, five. And then the last one, back then I'll do it here. Four, three, one, five, two, four. So you can check that this is permutation three, one, five, two, four. And so in this is the decorations, they're just the decorations we had on our letters, right? One, three, two. But the insertion process depends on those decorations, right? So depending on what this number is, the insertion will go differently. Okay, so now let's just try to do the same thing which worked for us in the Grasmanian case. Or let us define for specific decorated chain T, let us define here some over all by words W whose recording chain is this particular T. And we just, if we sum the weight of such by words Ws which is still the same, I mean the usual weight you would estimate is then the serum by, you know, the serum is by doubt GME is that you still get the Schubert polynomials, T still equal sigma of W. As long as this chain ends, you know, the top element is W, Schubert polynomial. So that's okay. So in fact, this kind of suggests that we should treat the collection of such by words as some kind of complete flag variety and analog of crystal. Again, we don't know how to define it locally so that, you know, you just define local moves and all together they is a connected component will give you, you know, Schubert polynomial. But globally you can just write all correct rights. You can cut out the set by just saying that the recording tableau should be this particular decorated chain and this cuts out. And in fact, we have some, you know, work in progress on what happens if you try to build crystal operators on such a set. So I can say something about it if we have time. Your notation seem to be suggesting that those chains, decorated chains is the right analogy for Tableau maybe. Yeah, so those decorated chains they are exactly right analogy for Tableau. In fact, let's do an example. Let's try to multiply two Schubert polynomials the same way as we did before. So let's take two decorated chains just like, you know, in the beginning I showed you I took two standard Young Tableaus, one and two, one and one, two, right? So now let me pick two decorated chains and I'll again pick simple ones. So one, two, three, four, then two, one, three, four, two, so two, one, four, two, three, right? So one is for permutation, one, four, two, three and the other one, three, four, one, two, four, three, two, one, four, three. Okay, now I will try to create a Tableau similar to the one I had at the beginning or maybe I don't know if I have enough space here. Well, I'll try. So, okay, can you see this grid? Okay, so here are three words by words which apparently is a complete collection of by words which have this first chain, two, two, two, two, two, one, two, two, two and one, two, one, two. And the second one is three, three, one, one, two, three, one, one and one, three, one, one. So maybe you could move the screen a little bit back because the bottom row is missing. No, this is the bottom row, I didn't. Yeah, we can see it. Can you see this thing? No. Oh, I can see it on my screen. Well, really, oh, oh. Do you have some kind of thing at the bottom? I just need to do, okay, okay, maybe other, if other people can view it, that's fine. Yeah, I can tilt it a little bit here, okay, here. Better? No, no, I can see it also, yep. Okay, and what we do just like before, we are trying to multiply them by simple concatenation. So here I would write, you know, three, three, one, one, two, two, two, two, et cetera. But instead I will do something more informative. Like once we get such a byword, we write in the cell, we can insert it, right? We can, and I will just tell you permutation which this byword gives once you insert it, right? And those permutations look like this, two, four, three, one, three, two, four, one, four, one, three, two, four, one, three, two, two, five, one, three, four, two, five, one, three, four. Three, four, one, two, four, two, one, three, five, one, two, three, four. Right, so everybody understands what they did here. For example, how do they get two, four, three, one? I take three, three, one, one, two, two, two, this word of length four, I insert it and then I forget, you know, it's actual bumpless pipe, jammer, chain, I just remember permutation and I'm telling you this permutation. So, and this turns out to be, it's pretty close to how product of those two Schubert's, you know, sigma one, four, two, three times sigma two, one, four, three should decompose into Schubert's. So, but there are some mistakes sort of there. So this table is not there. So this should have been two monomials in the Schubert of two, four, three, one, but this one is wrong. This one is actually correct. Those two together give us two monomials in Schubert of four, one, three, two. This one is correct. It should be in this product. This one is correct. It should be in this product. And this one, again, it should have been three things with permutation two, five, one, three, four, but this one is wrong. So sometimes there are two places in this table where we didn't get what we should have. If we got the correct permutation in those two, then the trick would have worked, but it doesn't. Okay, so does it make sense? What I'm saying? So like, for example, Schubert of four, one, three, two has two monomials. One of them is x one cubed x three and the other one is x one cubed x two. And this Schubert does occur in product of those two Schubert's. So this chunk is correct. It's perfect. It tells you what it should be there and it's there. So almost everything is correct except there are two mistakes. And you could stop there and you could just say, well, this doesn't work, game over. But I'm going to tell you that in some cases it works. Okay, yeah, question. You're saying that your product has five Schubert polynomials. That's right. This product will have five Schubert polynomials with coefficient one. You can see which ones here. They should be two, four, three, one, two, five, one, three, four, three, four, one, two, four, one, three, two and five, one, two, three, four. And some of them we get perfectly and some of them, you know, we get almost all monomials perfectly except for one. Okay, so now let me try to convince you that there is a reason why this didn't work in general. And I want to say that the reason is as follows. So what's different between this world and the world of the usual Plactic, you know, Plactic Manoid, whatever, crystals, Tasmanian case, right? So what's different? And I claims that the difference is that by words in general are not associative. So can we get the camera a little bit up now? Yes. Great, thanks. Yes, perfect. Okay, so I want to say the following, by words are not associative. And here's what I mean by that. If we have some kind of, let's think about Tasmanian world first, right? So we have some kind of word or permutation or I don't know, something like this. We can just insert it, right? We can use right insertion to insert it into something. Or we could have done left insertion and just read it from this direction and inserted it. Or we could do something even more sort of complicated. We could just say, I will start here. I will first have empty, you know? And then maybe I will insert this three from the right and then I will insert this three from the left and then I will insert this one from the left and then I will insert this four from the right. You know, in each time, you know, once I insert this three, I will have three, then I insert this three, I will have another three, I will insert this one. It will bump something out, you know, et cetera. I think that's the right way too. Like this was column insertion. So and then they insert this four. So I don't know. So a property of the usual RSK insertion, in fact property of two RSK insertions, the right one and the left one is that doesn't matter what choices they make. I can decide to insert towards any, you know, goal, like I can insert, put my goal over here, you know, my empty and just insert everything from the right or I can put it here or I can put it in the middle and choose arbitrary ordering, which I, you know, alternate left and right insertion. The resulting tableau will still be the same. It turns out that this is not true for bywords. So if we pick this, for example, this byword we got here by concatenation C3 or 111222. Yeah, I guess this guy where it didn't work. You know, one way to think about the order in which we do insertion is to draw something like a binary tree. So if I was to draw a tree like this, it would mean that I, you know, I start with this guy, insert him first then I insert this guy into him then I insert this guy into him then I insert this guy into him. Or I could, you know, I could start on the other end. I could right insert, right? So this tree would give us the insertion of all the wording from the left. So I guess it would be left insertion. I would first insert this guy then I would insert this guy in the result, et cetera. Or I could pick something which is sort of in the middle. So something in the middle would look something like, I don't know. So I would first insert this guy then this guy into him then this guy into the result and then this guy from the right into the result. Something like that. So there are all those choices you can make on how you insert things. And it turns out that in general, they're different. Meaning that if you take this monoid of things which insert into a particular bumpless pipe dream, it's not an associative monoid. It matters in which order you do those operations. So, however, there is one case. There's one case, or rather one kind of, I don't know how to say, subset of cases where this monoid is associative. So here's a serum again by the gene me serum on if indices of your words, right? So those numbers, three, one, two, two. If those numbers are weakly decreasing, in other words, if A1 is greater or equal, sorry, I1, greater or equal than I2, greater or equal than et cetera, greater or equal than Im, then this word A1, I1, et cetera to An, In is associative. In the sense that it doesn't matter in what order we do those insertions, we will always get the same result. So, and this suggests that we can multiply Schubert polynomials using this technique as long as we can find the decorated chains which you can just concatenate without breaking this condition, the weakly decreasing of indices. Right, so if you want to compute sigma w times sigma pi to Schubert polynomials, the question for us is, can we find a decorated chain for w and decorated chain for pi such that, this decorated chain has indices I1 through In and this chain has indices J1 through Jm and they together, they just form a single non-decreasing sequence once you can concatenate them because if this is true, then we have no problem. And then in this case, we can in fact prove that this kind of table drawing and then whatever actually does give you the combinatorial rule for structure constants of Schubert polynomials. So the question is, when does this happen? When can you find a decorated chain for w and decorated chain for pi such that the smallest thing in the decorated chain for w is bigger than or equal than the largest thing in the decorated chain for pi? And this is, so the so-called separated dissent case. Separated dissent case. So which was generally due to Knudsen and Zinjustin, the definition. But it's in some sense, it generalizes the Grasmanian case. It's also generalized as a case, Lenard considered which is, so okay. So what's the definition? So the definition is that all dissents in w are bigger than some or equal than some k and all dissents in whatever pi are smaller than or equal than k, right? So if all dissents on w are just one dissent which is at position k and all dissents in pi are also just one dissent at position k, then that's just a Grasmanian case, right? That's a case of Grasmanian. So when each of them is a Grasmanian permutation with the same k, but here we allow a lot more. We allow w to have more dissents. They just have to be to the right of k and we allow pi to have more dissents. They just have to be to the left of k. So, and in this case, you can just find two chains for w and pi such that they, if you concatenate them, you don't violate this non-decreasing condition. And from this, you somehow can prove that this gives you a little resource on the rule. So everything works out just fine because this manoid in this case is associative because it's actually well-defined thing to write a byword without specifying in what order you insert into it. Okay, this is like pretty much all I wanted. So the little resource on the rule would look sound like this. Again, just like at the very beginning for Grasmanian case, we would say, let us count the number of pairs of bywords such as the first byword has this first chosen recording chain and the second byword has this chosen recording chain. And once you concatenate them and insert it, you will get this chosen bumpless pipe, GM4 for, you know, for the Schubert you want to compute coefficient of. And if you count the number of such pairs, this is a structure constant. Okay, this is the end of my talk. I will stop. Thanks very much. Let's thank Pasha for your next talk.