 Before we start the talk, let me just remind everybody that we are hoping to do some student talks in some format at some point in the rest of the semester. And we hope we're still hoping for nominations of students who have something good to speak about. I guess especially graduates graduating students would be good to be presented here, but but really anybody who has been doing anything interesting and shoot but calculus and surrounding areas would love to hear about them and hopefully put them on the screen. So I haven't gotten any nominations so don't hold yourself back. Yeah, that's about it. Then today. Yeah, we will have a talk on on on a board with a camera and right now the zoom is pinned for everybody to show this one. There was some reason it was pinned before and then it stopped working. If it stopped working again you don't have this screen as the main one you can right click on a fabulous little screen and then you can pin it for yourself. Just in case anything goes wrong again. And that yeah, it's a pleasure today to introduce Pasha Juliowsky from University of Minnesota, who will tell us about bumpless pipe dream or SK and superstructure constants. Please take it away. Okay, hi everybody. Thank you very much for inviting me. I'm a bit sick today so you might see me like being slow or coughing or I don't know, hopefully I don't have to interrupt this but hopefully everything will be okay and this is joint work with the G1. Okay, and I will start by telling you a certain story for the case of grass manion. I will try to tell you how we try to do the same story for complete flag right. Okay, so for the grass manion there is this very famous map called RSK, which you can feed into it a word. W, and it will give you insertion tableau and recording tableau of w and q of double. Okay, my markers could be better story, but I'll try to so can everybody see this. Yeah. Okay, so this is an example would be something like this if you start with word which is actually also permutation is this case to see one. It will map into insertion tableau. 132 and recording tableau. I will not. I will not be reminding you how to do this map. Many people have told that I should assume that audience has some background. So this is one of the very famous kind of tools. Why we are interested in today is that it allows to do one so once one has this map it's nice properties allow one to derive at least one version of so called little wood research on right so RSK. Some school gives a sugar structure constants in the in the grass manion case. Shubert structure and I will remind you what it means constants. Okay, let me I'll be reminding it to people who probably know this much better than I do but just just for the completeness of the talk I'm very briefly remind you. How this works. So in the collage ring of grass manion we can multiply the so called sugar classes. Which form a basis for this ring and expensive and from from some consider geometric considerations we know that the expansion is a positive expansion. So this is a kind of same. I mean, same basis sugar classes. Right and the little research on problem or structure structure constant problem us what this is and what those numbers are and in the case of this money and this is a sole problems there are many ways to describe them. Here I'm going to show you one way to describe them. So first of all, let's slightly. Let's talk about this problem. Everybody can hear me by the way I love being interrupted if anybody has any questions. So, so, but of course probably most of you already knows that we can instead of multiplying sugar classes we can reduce this problem to some problem inside symmetric functions ring. We can ask to multiply two sure functions. They expand positively in terms of basis of sure functions with the same coefficients. And then here's how we can use RSK to solve this problem. Let's pick T to be a standard young tableau of shape of shape lambda of particular shape and let's denote bracket T the sum over all words. W such that Q of W is T. And we will just sums the words themselves here like that. Okay, everybody disappears. Everybody's still there. Okay. For some reason I stopped seeing people. Okay, so. Yes. Okay, so I don't know if it's a curious question how would you call this thing. I would call it the crystal. If you think about it. If you pick any such double you and you start acting on it by crystal operators, you will get more such double use crystal operators preserve recording tableau. So this, this way to do say just let's take all words with particular recording tableau is like saying let's take the crystal without without saying what the crystal operators are just some cool. Like, like, marks Amazon has crystal for dummies this is crystals for people who don't like arrows. Right. We don't, we don't describe edges we just immediately describe all the crystal by saying it's every single she starts with this recording tableau. Okay, and now let's try to lift kind of this line over here to the level of crystals. So here I'll have to erase something. Well, I will keep this part. So let's do the following. Let's pick T lambda of shape lambda. And let's pick R mu of shape mu two standard young tablets shape. And then we're going to just multiply two such sums of words like this to lambda times. R mu. So, and we will multiply just by concatenation, by concatenation. And we will decompose it as a sum over some other crystals like this. Some other, I guess, standard young tablets as of associated crystals. And it's not, I mean, the first question is, it's of course not obvious that it does decompose this way. It turns out it does. So any such product you can again, right in the basis of such things. And then if you just look what are the shapes of those standard young tableaus you and you forget the tableaus themselves as a shape. You get your formula for sure functions. So here I'll give you an example. And let's do an example and we will see how it works. This is probably very familiar to everybody. Also, you know, there are so many variations of little research and rule that I think there is maybe some benefit of showing you explicitly an example. So let's do an example. Whereas this guy will be this standard young tableau. And then this guy will be actually not this, just a single column standard young tableau. You could ask, well, what is, what is bracket T something like this one plus two plus three. Let's do everything in just three variables. So there are three words in alphabet once or three which instead was this recording tableau which a one two and three. So that's a kind of a trivial thing. And then we can also do what is bracket are in this case all the words which insert was this recording tableau in alphabet 123 and they would be 21 plus C one plus 32. So first if you know RSK you can easily tell that right so if you first insert two and then bump one into two it will bump two down, etc. Okay, at this point I want to ask, does everybody understand what I'm doing, like is there anything I can repeat or explain again. I picked two standard young tableaus. So some for each of them I did not bracket that some of all words in some fixed alphabet say in this case once or three of all words which insert was this recording tableau. Okay, makes sense. Okay, so now let's just make a table like we will make a little table so hopefully it will be so here my 213132. Here's my 123. And I just can continue them right so I do 121131132, who 21231232, and then 321331332. In this case it's easy to convince yourself that it just splits right so all of those. I guess eight of them will give us so as for this shape shape to one. And then this one guy three to one is the only one which gives us. The red color if it's red it's very difficult to. Okay, well let me try. Let me just write it with normal color right so. So, you get two standard young tableaus here. One of them is 123. And the other one is like this 123. And those eight over here are all the words in this alphabet which have this recording. And this guy is the only word in this alphabet which has this recording. And from which we of course concludes that you know as one box times as two boxes equals this plus this. And to state a rigorous little literature some rule we would say something like this we would say that the coefficient. See lambda menu you need to determine that you need to pick one standard young table of shape lambda one standard young table of shape new, and one semi standard table of shape new. You can count the number of pairs of words, such as the first one has recording lambda, this tableau shape lambdas the second one has recording tableau this particular tableau shape new new. And once you can continue them they insert the insertion tableau is your choice of semi standard tableau shape new. So once you make see such choices you get the rigorous little research and rule out of this picture. Okay, so now I will try. I will try to tell the same story for flag variety and ultimately I will fail, because if I didn't I would have solved you know the big problem but, but it will be interesting to see in which cases I succeed and maybe why why do I fail in general and maybe it can still be fixed some hope. Okay, so first of all what will be the words in this world. The words will be in words in by letters. Ai, such that a is lesser equals and I, and words are just words is this alphabet so you know a one by one a two by two. And soon I will explain where this comes from what why this is the correct choice of words. Right, I mean I can show you, I guess an example immediately. Something like, let's see what examples I have here. Later we will see example, like 11231224 something like this is a word in this world. Okay. So now we have a map this generalized RSP you to doji and myself. Which you feed into it such a word. And it gives you again a pair p and q. And I need to tell you what is p and what are p and q so he will be a bomb plus pipe dream. And you will be defined by lamb Lee and she was on her. And you will be a decorated chain in your brother, the creative chain. And I will carefully tell you what those mean, the creative chain. And then the serum is again that this is abjection. Once you properly define what what is also objects what is also objects and what is also objects this is abjection. So let me tell you about those two things. So first of all, what are bombless pipe dreams. So bombless pipe dreams or bpds. As I said they come from the work of lamb Lee and she was on probably many of you have already seen them that just some kind of pilings with six allowed pieces. To draw those pieces pilings of kind of a board or a part of positive quadrant. And such that, such as a form pipes, those pieces once you connect them. And the pipe number I, I travels from infinity I to pie of I infinity where pie is certain permutation. So for a given permutation double users a finite list of bombless pipe dreams with with this permutation. And okay so let me just give you example I like, I mean, it takes a lot to define things carefully but giving example is often easier. The serum of, you know, lamb Lee and she was on a says that if you sum over all bombless pipe dreams, a particular w sample weight of this bombless pipe dream you get Schubert polynomial. Okay, so let me give you an example so if we have Schubert polynomial. Let's say of permutation C one five to four. I computed it or maybe that would you help me I don't remember we have this example. So it's just this x v square x one square x two x v plus x one x two square plus x one cube x three plus x one cube x two. But for example, this term so each of the terms should correspond to some bump with pipe dreams so here let me draw for you a couple of them so this one would correspond to something which looks like, like this. Okay, so if your numbers was 12345 you will see that those are 31524. So those two are in the first row and those two are in the way. Yeah, those two are in the third row. Second row is occupied by this line. So you have x one squared and x three squared and that's this term. Or the last one, for example. Let's see, let me show you the last one. So this term over here, what one plus five Jim gives his term or a little trickier to draw. It's better to go on the grid but it takes longer to draw it on the grid. And in this case the empty spots are here, here, here. And that's why you have three of them in the first row, one of them is in the second row. So you get x one cube x two. And you can tell that this one you can get from this one by some kind of drooping the procedure of drooping where you could, for example, take this part and kind of droop it to be to go this way so this will not be an empty spot anymore. But this will be an empty spot. Okay, so this is object number one just as a P side the insertion side of the map. Now let me briefly tell you about the side the decorated chains in order to create it. We are change. So this will be just chains from identity permutation. Going up also a way to the desired permutation. Each pi I plus one is the single transposition times pi I, the length of each pi I plus one has to be strictly one beggars and lengths of pi I. And the last thing, those steps, each step has a decoration which is some number k one k two. Okay see all the way to, I guess, KL. And the condition on on those numbers is that if, if, if this step is obtained by applying T alpha beta and as in the alpha has to be less than equals and k and less than beta. Okay, and again, hopefully once I show an example it will become kind of clear. So, let's do an example. Let's take the following chain. One, two, three, four. Right so identity permutation of size for less than was parameter two, then one, three, two, four. Less than with parameter three. Then one, three, four, two. Less than was parameter. Then one, three, five, two, four. That's so we can, I guess we can put five at the end of the first realize just assume it's stable right so it's so like one to see four is also the same as 12345. This is the first step we are we touch the five. And the last one, this one is less than was parameter chosen. One, five, three, two, four. But that's the permutation we had in the previous example one, one, five. Wait, is it. No, it's not. But whatever it is some permutation. So what happens here is look over here in the first step we swap those two. And we swap them so that there is only one more inversion right so in each step we swap them so there's only one more inversion. And this, this number has to be in between this is a second gap. This tells us mark the second gap in this word. And the gap must be in between those two things that we swap. So it's the only place where we actually have a choice here is this step over here, because here we are swapping. What are we swapping we are swapping four and five over here. And we could have marked is a one of those two gaps so we mark gap number four but we could have also got marked gap number three. So for this step the decoration could have been three or four. So it's pretty restricted right so for once you choose which, which two things alpha and better you're swapping down to actually, well, possibly that many choices for what this declaration is. But you would get a different next line if you had taken the other choice I assume. So the permutation would still be the same right I could just change it to three. So the only which is different is over here I would get one of those case to be different. So the permutation part will say the same but the decoration would change. But it would be a you know overall it would be a different the current chain. Yeah, I'm just asking if you have a chain. Does that determine the numbers you put on in the circles. No, no. So, and that's exactly the example of this. Here I have a chain of permutation is brought water. Each next one is one transposition applied to previous one. And in many cases it does this there means this number it's here I could could only choose two, but there will be some places where you can choose different numbers. Like to go from here to here. I swapped those four and five. The question says that I just need to choose a gap between those four and five but there are two choices here I could choose this gap or this gap. I see, which would correspond to putting either three or four in here. Okay, thanks. Yeah. So this is actually so we kind of rediscovered this object but it was in some sense already present in the word by Christian Leonard. Okay, and then one more thing I promise to you. So again just was like was the usual escape I will actually not show you how to do our escape it's a bit technical it's not super horrible but you basically to insert the next by letter you need to start drooping those pipes in a bumpless pipe dreams in certain way. So there is a precise formulation in our paper. But, but let's just sort of agrees that there is such a map with the properties I will I will show you an example like that. But first of all, why, why do we have I promise to explain why those by words correspond to Paris P and Q where P is bumpless pipe dream. And the Q is a chain, decorated chain. And you can think of this by Jackson as combinatorial proof of of monk's room. So, more specifically let's say you multiply a bunch of Schubert polynomials corresponding to single adjacent transposition right so as KL times, etc. And this is just as K one, where Sigma as K is just X one plus plus X K. This product, you know, you can expand it in terms of some sugar polynomials. I mean with multiplicities doesn't have to go. And basically this is by Jackson. So it gives you by Jackson between monomials on the left and monomials inside those Schubert polynomials on the right so in the fall so when we have a some kind of by word. Let's use the same one I used before so 11231224. What does it means that it's 23 it means. This was X to coming from X one plus X two plus extreme. Right, that's that's what it means it means that over here we had as K three I guess, sigma sigma S three, and we picked my nominal accident. So each of those is like a choice of one X in each of those and then once we expand it in this way. Anyway, so this can be made precise and there's a similar thing for just for the usual escape for the usual escape you just think how you take one box to the power and it's some of as lambdas. And the usual escape can be viewed as by Jackson between monomials here and monomials here. Okay, okay so that's just some kind of for general hand waving, but but if you didn't follow this part that's that's okay so the important thing that there is such a map. It can be defied precisely and in fact there's two versions of this map and this will be important for us. Just like in the usual case there is left insertion and right insertion. So, in our world there's also left insertion of such by words and right insertion of such by words. Okay, I think I talked for half an hour and I'm supposed to take a break so let's take a break. Okay, thanks very much. Questions before we go on to a break.