 So welcome everyone to the Schubert seminar. Today we have another three short talks by grad students this time. And without further ado, we're gonna start with Hugh Denin from the Ohio State University telling us about differential identities for Schubert polynomials via pipe dreams. Please take it away Hugh. All right, well, thank you everybody for attending today and thank you to the organizers for inviting me to speak. I'm gonna be talking about some common torques of pipe dreams and in particular how you can use pipe dreams to get some differential identities that we have for Schubert polynomials. So our story begins not with derivatives but rather with it's a tale of two operators. So I have these knobla and delta and these are gonna be linear operators acting on this complex vector space with basis index by permutations of size N. So permutations in SN. So how are they defined? Well, a knobla of a permutation is gonna be the sum over week order covers going downward. So I'm summing overall SKW covered by W. And then how many copies of SKW am I getting? Well, I'm getting K of them. So delta is gonna be a similar story except instead of going downward in week order, I'm summing over all covers in Bruchot order going upward. So I'm summing over all permutations TABW covering W. So TAB is the simple transposition swapping A and B. And then it's a little more complicated what I get. So I get one plus two times the size of this set which I'm calling CWAB, many copies of TABW where CWAB is the set of all C greater than B such that W inverse C is between W inverse A and W inverse B. So this might seem a little mysterious at the moment, but the reason we're defining things like this is because well, Krishan Gates and Ibo Gao showed that if you take knobla and delta together, they form the lowering and raising operators of an SL2 action on this vector space. So the weight spaces of this representation are gonna be the permutations of a fixed length. And the reason that they were thinking about this is because once you have an SL2 action on a post set like this, you automatically get that your post set satisfies the strongly Schwerner property, which is a regularity condition which tells you something about how big anti-chains in your post set can be. So knobla was first introduced by Stanley and conjectured to give an SL2 representation. So they explicitly constructed the corresponding operator delta, which is why it has this weird form or maybe a slightly unusual form. So I'm gonna switch gears now and head over to talking about Schubert polynomials. So to do so, I need to introduce our fundamental combinatorial object, which are gonna be pipe dreams. Or maybe I should say reintroduce since Tianyi talked about these two weeks ago in his talk. But as a reminder, so a pipe dream for a permutation w is gonna be a tiling of this staircase. So I have this staircase with these elbow tiles along the diagonal or the anti-diagonal. And then I can fill in each of these gray tiles with either a cross or a bump. So I want to have the network of pipes connect the top indices to the corresponding indices on the left. And I also want that no two of the pipes are crossing more than once. So that's gonna be this reduced adjective I have in front of pipe dream. But in this talk, all pipe dreams are gonna be reduced. So as an example, I can take the permutation. Oh. I'm sorry, there is a question by Alan. How unique is Delta given the nabla and the weight spaces? I think it's gonna be uniquely determined from the theory of like finite dimensional SL2 representations. Okay. So back to the example. So we have our permutation two, one, four, three. And we have three pipe dreams, three reduced pipe dreams for two, one, four, three, which are these first three I have. And you can check that all of the pipes connect the indices on the top to the corresponding indices on the left. This last pipe dream, it does the correct thing with the indices, but unfortunately my pipes three and four are crossing more than once. So they're crossing three times here. So I wanna ban configurations like this. Okay, so with pipe dreams, I can define now not Schubert polynomials, but this notion of padded Schubert polynomial as generating functions over pipe dreams. So my padded Schubert polynomial is gonna be in variables X's and Z's. And I get a monomial in the X's and Z's for each pipe dream where the weight is determined as follows. So I get an XI for every cross appearing in my pipe dream and the I corresponds to the row index of the cross. So every cross is gonna correspond an XI where I is its row index. And I have a similar story for the ZIs. So for every bump in my pipe dream, I'm getting a ZI where I is the row index of that bump. So you can think of these as sort of a homogenization of your usual Schubert polynomials, which is a little funny since Schubert polynomials are already homogenous, but I'm homogenizing with respect to each subscript separately. So if you count the number of XIs and the number of ZIs appearing, it should remain constant among all monomials appearing here. Right, so without the ZIs, I just have the usual definition for Schubert polynomials. So you can get that by setting all of the ZI variables equal to one. So I think it's worth going through an example. So here's our permutation from before with its three reduced pipe dreams. So to get its padded Schubert polynomial, I'm gonna take each pipe dream appearing here and then replace every cross with a corresponding X variable with the subscript corresponding to the row it's in. And I'm gonna replace every bump with a ZI where the I corresponds to the row the bump is in. And then I take a product of all of the weights and that gives me the weight of the pipe dream. So in this case, I have three monomials corresponding to each of the three reduced pipe dreams for two, one, four, three. Okay, so why am I working with padded Schubert polynomials and not Schubert polynomials? Well, it allows me to define these dual operators. So I have this knobla tilde, which I'm defining to be a sum over partial X derivatives. So I'm something over all of the partial XIs. But then I'm reweighting by a ZI. So in other words, you can think of this as decreasing the XI weight of a polynomial and then re-adding the corresponding Z weight. Right, so that's my knobla and my delta is gonna be, well, the opposite. So instead of taking partial Xs, I'm gonna take partial Zs and then I'm gonna scale back up by the corresponding XI. And it turns out that if you do this, a very curious thing happens. So Hammacher-Petchen-Expire and Weigen showed that if you take a permutation W and apply this knobla to the padded Schubert polynomial for W, you get a sum over weak order covers going downwards and you get K copies of the Schubert polynomial or the padded Schubert polynomial corresponding to SKW. So this is precisely the formula for knobla that we had in the abstract setting where we defined it just as an operator on a vector space with basis indexed by permutations. So you can think of this as a realization of that knobla from earlier. And so here we reap the benefits of working with padded Schubert polynomials because if you take these two operators together, they also satisfy the lead bracket relation that we want for an SL2 representation. So once you know that knobla has this form on Schubert polynomials, you get that delta acts the same way as our formal delta did earlier for free. So in particular, the proofs of these identities were purely algebraic, despite the formulas being purely combinatorial in nature. So I just have some non-negative integers appearing. So this begs the question, where is the combinatorial proof? So that'll be the goal of the talk today is to walk us through some examples of how this bijective proof is gonna work for each of these identities. So I guess I'll check if there are any more questions. What good? Okay, well, so to describe these bijections, I need to talk about some moves you can perform on pipe dreams. So our fundamental move is gonna be the crossover where I can swap a pipe dream like this for a pipe dream like this. And well, it's a little unclear what's going on, but if you notice, I have a cross here, which I am moving upward to this position up here. So let me highlight the pipes that are going through these crosses. And you can see that what's really going on is I'm exchanging a cross with a bump, but the pipes that are passing through these tiles are the same. So essentially, I'm just moving the crossing point of these two pipes upward. So this, the result of this is that the underlying permutation of the pipe dream remains constant, then I can move the crosses around like this. So as a consequence, I get ladder moves, which is a situation where I have a rectangle of pluses with a single plus in the bottom left. And then what I can do is I can take that plus and move it up to the top right. And you can see that this is just a special case of a crossover. And then more generally, if I have a column of pluses on the left with a row of bumps at the top and a bump in the bottom right, then I can move the bottom left cross up to the top right and shift over all of the intermediate crosses. And you can think of this as applying or a successively replying ladder move starting with the first cross here, moving it up to the top right and then proceeding in sequence. I should check if you can see my cursor. I'm not sure if that's coming up on the feed. It does, like we can see. It does. Okay, that's good. This might be very confusing without it. Okay, so let's start with the nobla bijection. So how are we gonna do this? Well, so whenever I take derivatives, I wanna think of the combinatorially as marking a particular atom giving component of my objects. So in this case, I'm taking a partial derivatives in the XIs. So I wanna think of the left-hand side as being a generating function over cross marked pipe dreams since these are the crosses that are giving me the X weights. Right, but since I'm also multiplying by a corresponding Z weight whenever I reduce the X weight, I wanna think of these cross marked pipe dreams as contributing the same weight as the normal pipe dream except the marked cross gives a modified weight of a Zi rather than an Xi. So in other words, whenever I have a cross marked pipe dream where P is a pipe dream and IJ is a cross in the pipe dream, I wanna think of the cross as giving a Z weight instead of an X weight. And then the left-hand side becomes essentially a generating function over these cross marked pipe dreams. Okay, so now if we expand out the identity a little bit, you can see that what I wanna do is I wanna start with a pipe dream Q contributing to the right-hand side of my equation. So Q is gonna come from some SKW, which is covered by W. And then I wanna produce K cross marked pipe dreams. So I'm gonna label them P1 through PK for W that contribute the same modified weight to the left-hand side. All right, so how is that gonna work? Well, I think the best way to see this is through an example. So we're gonna take the following permutation. So W is this big long permutation in S12 and we're gonna pick K equals five. So this is our starting pipe dream. So this is our pipe dream Q corresponding to SKW which lives below W in week order. And I'm gonna start by looking at columns K and K plus one. So when I do this, I'm gonna have some block of crosses up top, so rows with two crosses in them. And then eventually I hit a row that does not have two crosses in them. So I'll keep going down and then eventually I should also hit a row that has two bumps in them. So you can guarantee this is gonna happen because my pipe dream is reduced and I know that pipes K and K plus one do not cross since SKW is beneath W in week order. Okay, so my next step is to perform a generalized ladder move in this inner rectangle. So I'm shifting over these two crosses. And then the result of this is that pipes K and K plus one now bump away in this top left corner. So now I can add a cross here which has the effect of crossing five and six. And this gives me a pipe dream for W. So this is gonna be my P one. And to finish, I wanna mark this bottom left cross to go along with this pipe dream. And so you can check that the result of this is that I have shifted over the cross that was here. It's now over here. And the cross that was here is now over here. And then I had two bumps originally but now I have a bump and the marked cross. And so by checking all of this you can see that the weight of this pipe dream remains unchanged. So I had this additional cross here but because it's marked, it's contributing the weight of a bump. Okay, so this is our first cross mark pipe dream. But now we gotta find four more since K is equal to five here. So I have to find five in total. So I'm gonna describe a process that moves this cross, the marked cross over to the left one column at a time producing a new cross mark pipe dream at each stage. So there are two cases for what can happen. The first is if I have a cross directly to the marked cross, directly to the left of the marked cross, in which case I can just slide the mark over one tile to the left. And so this gives us a second cross mark pipe dream. And since I'm just sliding over the mark in a row, this is gonna change the weighting of the pipe dream. The second situation occurs when I have a bump to the left of the marked cross. So in this situation, I can find a rectangle where my bottom row is two bumps. And I can guarantee this because my pipe dream is reduced. And then I can perform a generalized ladder move to slide both of these crosses downward. So the mark starts in the top right and then ends up in the bottom left. And you can check that before and after this move, the weighting of my pipe dream is the same. Since essentially I'm moving a bump from the top right to the bottom left. Yeah, so these are the two cases for what can happen. So to finish off, we're in a case one. So I'm gonna slide the mark over to the left. This gives us our four cross mark pipe dream. This is your three minute warning. Sorry to interrupt you, but your three minute warning. Yeah, no problem. And then finally, I have a case two. So I'm gonna do a generalized ladder move, which moves the cross down into the first column. So these are our five cross marked pipe dreams. So I'm gonna quickly try to run through the delta bijection. And it's very similar in its idea. So I wanna think of the left-hand side as counting not cross marked pipe dreams, but bump marked pipe dreams, where the bump contributes a weight of X rather than a weight of Z. So what do I wanna do here? So starting with a pipe dream on the right-hand side, I wanna produce this quantity, many bump mark pipe dreams for W on the left-hand side, which I will label as P naught. And then for every C in my set, I'm gonna have a P sub C and a P prime sub C. Right, so I'm starting with a pipe dream Q for a permutation TABW that covers W. Okay, so how's this gonna work? Well, we're gonna start with our pipe dream Q and TABW. So here W is this permutation, A is five and B is eight. And our first step is to look at pipes A and B. And since I know TABW is above W, these pipes must cross at some cross in the pipe dream. So our first step is to just uncross these pipes and this will give us our P naught. So to get the marked bump, I just wanna mark the new bump that appears in the pipe dream. And then effectively I've just turned a cross into a marked bump. So here I have a pipe dream with the same weighting. So this is our first bump mark pipe dream. And then how should we interpret this set? So, right, so this is gonna be the set of all C greater than B, so that W inverse of C is between W inverse A and W inverse B. Well, this corresponds to all of these yellow pipes that begin to the right of our pipe B and that end up in between pipes A and B on the left-hand side. So I'm gonna just do the example for C equals 11. So what are we gonna do? I'm gonna describe a process that moves the bump around. You have like a one minute left before we have to put on the next speaker. So just to let you know. Yeah, so maybe I should wrap things up. Essentially, the process is gonna move the bump around until it eventually lands up on the marked pipe C. So I'm gonna eventually hit this pipe dream, which is gonna be our first pipe dream corresponding to this choice of C. And then I can perform one final crossover to get our second one. And I think I am out of time now. So I think this is a good place to wrap up. Thanks. Thank you very much.