 Alright, David, please go ahead with the second part. Okay, great. So let's see, there was there was probably one more thing I wanted to mention. So during the break, if people missed it. Oh, yeah. So Alan asked, so I don't have a good formula for you, I'm afraid, but Alan asked what what's the relationship between between these S lambdas versus the kind of double shares, which I'll write with a little S lambda and let me write them like this. So these are meant to be symmetric, kind of they involve these symmetric functions via the C's and they have some double parameters via the Y's. And there is a change of basis kind of formula, you can work out between these two. And probably kind of lying behind us, there is a formula due to lambli and shimizono, which, which expands a double Schubert, a back staple Schubert in terms of double shares. So SW as a sum of stuff, times these double shares. And and it has a similar flavor to this one. So I want to kind of be clear about that, in that they give you some way of expressing certain pipe dreams of enumerating a set of pipe dream, a bumpless pipe dreams and taking their weights and something like that looks here shows up here. But so our formula is is maybe requires less qualifications about which pipe dreams they are once I show them to you. Okay, so yeah, that was something I wanted to go back to. And the other comment I wanted to make was that by instead of fixing we, we did fix C equals C zero zero. And I'll keep doing that for the rest of the talk. But you could fix any other KK or in fact, maybe KL for any other integer K. And you get a similar formula to this. In fact, it's almost exactly the same. You can just have to change your conventions about where your pipe dreams start. So and then simultaneously, you can do this for any permutation of the integers where by that I mean it mixes up only finite the many numbers. Okay, but that's beyond what I'm going to talk about the next 15, 20 minutes. Okay, so let me talk now for the rest of the time about bumpless pipe dreams. And fairly quickly, we're going to move away from these enriched or batch stable super polynomial and just into the specialization where, oh, I didn't mean to erase, but that's what's going to happen. We're going to get rid of the Cs. Okay, so bumpless pipe dreams. So these things, these are gadgets which are similar in some sense to the pipe dreams that have been thought about since at least the 90s. But in some ways, they're richer, although I guess in some sense there's an equivalence. I expect most people here have seen them before. So let me just kind of go through the features of it. You've got a permutation, W, and you want to connect, you've got some pipes, and you want to connect them from the east edge to the south edge, and you're allowed to have different kinds of connectors. So I guess I'm using red. So you can have a connector like that, you can have a connector like that, none of them showed up like that there. You can have a crossing of a pipe, you can have a blank pipe, and you can have straight pipes, and you can have a blank pipe, I guess all straight like that. But the key thing you can't have is a bump. Like two pipes cannot come together like that. So that's why that's the reason for the name. And so here to see which permutation they correspond to, you just start at one and then figure out which entry it ends up in. So this one ends up at three, that one ends up at four, that one ends up at two, seven, six, one, five. And what else about these? Well, each one of these gets a weight, and the bumpless pipe streams are very easy to assign a weight to. I guess the ordinary pipe streams were too. But here you just product over the blank boxes. So I tried to line up this product so that it roughly matches the order here, but here's x1 plus y1. So these are the x1s, x2s, x3s. So let me be a little bit that way, x1, x2, x3, etc. This would be y1, y2, etc. And so here I see x1 plus y1, x1 plus y2, x2 plus y1, and so forth. There are all the contributions. Okay. And there's a, so this is a formula. So this is first stated in Landli Shimizono's paper, but there's a wonderful proof by Anna Weigand using transition, which makes the whole thing very simple. But so the theorem is that the Schubert polynomial is the sum over all bumpless pipe streams for a given permutation of the weight of the pipe, the weight of the diagram. Okay. So again, I expect most of you have seen something like this before. If you're seeing this for the first time, then you're going to see a bunch more pictures like this in a moment. Okay. There's some moves you can do on the bumpless pipe streams. And the first of these was introduced by Landli Shimizono. It's called the droop move. And I guess I skipped over this, but this one here is called the rotha diagram, bumpless pipe stream. And that's just because it is the rotha diagram. The blank boxes here are right there in the rotha diagram for that, for this permutation. Okay. But then you can start playing with things. And so here's a droop move. I'm going to go from here and I'm going to droop this pipe down into there. And so that's what it's going to look like. So I'm going to droop it down. And when I do that, this empty box gets moved up and it ends up looking just like this pipe stream right here. So that's the first move. There's another kind of move which I'll call a drip. And this is like the small droop. Oh, I guess I should go back and say, what are the rules? When are you allowed to do a droop? You're allowed to do a droop when you've got a elbow tile like this in this corner. You've got a blank tile in the southeast corner. And you've got no other elbows at all in the rectangle. So again, elbow tile in the northwest, blank in the southeast, and no elbows anywhere else in the rectangle. So for example, like right here, according to the strict rules of drooping, I would not be able to droop that guy down to there because, look, there's an elbow right there. So people have written put other moves on pipe streams that do allow that kind of droop but which is a good thing to do. So that has to be allowed in what we're calling drips. But basically a drip is one where you only move within a two by two rectangle. You're only allowed to move in a two by two rectangle. And it basically is just one where you take any kind of elbow here and a blank box there and you just swap them like we did here. So this two by two rectangle, this kind of picture got swapped. So for the drip moves, you could have had an elbow in the corner, in the other corners too. That would be okay. Okay. And lastly, what we're calling a drop, which is another kind of small droop. But this one necessarily moves across. So here what I've got is I'm going to drop this elbow down into that empty one. And the rule is I should have an elbow, I should have a blank box and I should have no other blank boxes in the rectangle that they form. And I do that. And I just do the drip. And that blank box ends up jumping up to where the elbow was. The key thing about the drop move is that it always ends up moving one of the crosses. So that crossed right there got moved over to this cross right here. Okay. So those are the three moves that we'll use. And yeah. So I want to say I want to introduce a new kind of notion of an adjective for BPDs. So a bumpless pipe cream is called flat if whenever there's a northwest corner. And by northwest corner, I mean kind of of blank boxes, of blank boxes, of blank tiles. So any northwest corner should have a cross directly up to the left of it. And what we think of with the reason that this is a, yeah, sort of roughly this is because I don't want to allow any drip moves, any small drips to let these boxes drift to the northwest. So we're going to be moving boxes around and these crosses will block those from drifting any further northwest. So I kind of just alluded to it, but the drift class of a bumpless pipe dream is all of the other bumpless pipe dreams that are attained from it by these drips. And there are inverses, which you might call undrips. So you could, you could drip down, taking one step to the southeast, or you could undrip up, taking one step up to the northwest. And the condition of being flat is that you can never undrip it. It's as undripped as possible. And so, yeah, each drift class has its unique flat bumpless pipe dream. So those ones are kind of canonical representatives for drift classes. And from that observation, you get, Daoji has a question on the chat, and I'll address it in a second as soon as I write out the formula. Just observing that you can break up the set of all bumpless pipe dreams into drift classes together with the formula of Lamely Shimuzono for Schuber polynomials, that tells you that, well, you can chunk together bumpless pipe dreams into their drift classes and then just sum those up and then, therefore, you have a formula for Schuber polynomial as a sum of drift class polynomials. And then, of course, that just pushes the question down the road to, well, what can you say about these drift class polynomials and the answers? Well, we have Tableau formulas for them. So sometimes we can say quite a lot. Okay, Daoji asks, in the drop move, can they involve non-adjacent columns? Yes, they can involve non-adjacent columns. I don't, yeah. So the example that you didn't have one, but that can certainly have. So you can generate all the, so, yeah, one question about this is, in order for this kind of formula to be useful, you'd need to have a way of, well, figuring out how to get all of the drift classes or, equivalently, how to get all of the flat bumpless pipe dreams. And you can generate all the flat bumpless pipe dreams by starting with the Rotha pipe dream and doing a sequence of drop moves and undrips. So you drop and what you end up with might not be flat, so then you flatten it. So in this case, I had to flatten it by moving that one up one step and this one up one step, and then I get something flat. Okay, and so it's a little lemma that says that this kind of composed procedure start with the Rotha diagram, drop and flatten in all possible ways, you get all possible flat diagrams and therefore you generate all drift classes. Okay, so that's part of our story here. So here's an example of the one that we were looking at before. So three, four, two, seven, six, one, five. In this case, there are 10 flat diagrams, and they're, in this case, there happened to be 30 total bumpless pipe dreams. So this is this groups things together a little bit. Now the ones we were looking at before, let's see, there's the Rotha one, and I think the two that we looked at before. I'm not seeing any moves at the top of my head that are against non-adjacent columns, but truly there are some. So that's fine. You can get somewhat more efficient. So if we do the permutation one, three, two, nine, eight, seven, six, five, four, that has over 163,000 bumpless pipe dreams, and 21 of them are flat. And using Julia, which I'm going to blame Anders for convincing me to try to start doing, it takes essentially no time to compute the flat ones, and a little bit of time to compute all of them. So I'm about done, and what I'll do in just one minute is do a little demonstration of how these things are computed, maybe a little advertisement for anyone that wants to help me develop this a little bit more. But let me wrap up by, I alluded to a Tableau formula, and let me kind of illustrate that briefly. So here is a different permutation. It's a permutation in S8. And we would, and this is a flat diagram, right? There's, in this case, there's only one corner that's worth worrying about, and this one, and there's a cross up to the northwest. And we'd like to figure out, we'd like to somehow get the polynomial, which enumerates everything in its drift class. And the rule is, well, you just write down Tableaus. But you have to be a little careful with the Tableaus, because if you kind of look at it, this one, this box here, it can drift, but then it can't drift past this box here. So whatever the number is in there, it's got to be at most that number. The numbers, I guess, by the way, here, I'm going to, I should write it. So I'm going to record the row. You have a little bit of a choice here. You could record columns of the drift. So if I've got a box and it's drifting to row four, then I'll write down a four. So here's the picture of the drift configuration. This box drifted down to row four. There it is. This box drifted down to row five. There it is. So this box drifted down to row four, two. But you see, this box can never drift to a lower row in this box. So its number, its label should always be bounded above by that one. So that's the kind of Tableaus formula that we have. And yeah, right. Here's kind of the maximal southeast drift Tableaus. Is there any relation between the one and the four? This one and this four. That's a good question. In theory, there could be. In this case, there's not. That one is blocked by that cross. So yeah, there's a, there's a bunch of rules that you can come up with that tell you how to come up with these labels. But for example, that cross blocks that one. Thanks. If you forward stabilize a drift class, what symmetric function do you get? I'm going to start with the Schubert as a sum of drift classes. I'll take that equation and stabilize everything. Schubert turns into a Stanley. Okay. I see. So you, I see, stabilize that way. Yeah, that's a, that's a good question. And I don't think I know the answer to it. So in the cases when there's a single component, it's very easy. It's just going to be some version of a flagged Schubert function. Or I guess then if you forward stabilize it, it'll just be one of the double super symmetric shares. But in general, when you got multiple components, I do not know the answer to the talking maybe. I'm going to change my sharing. If there are any other questions about the slides, now is a good time to ask, but I'm going to put up one other thing before we quit, before I quit talking. Rebecca, do you have a question or? No, okay. I have a question. Sorry. So, so, well, maybe you don't need a slide. So is that bijection between Tableau and bumpless pipe dreams? Like when you fill in this Tableau, is that just a representative for the bumpless pipe dream that is in its class? Yes, that's all it is. Yes. All right. Or equivalently the drift configuration that tells you how far but the key thing is that the crosses in the bumpless pipe dream cannot move under these moves. They're not allowed to move. Okay. So we're about out of time, but I wanted to just advertise this. Well, I guess I need to do one more thing. Sorry. This story that Anders got me involved in, which is working in Oscar and Julia. And so for example, if I want to, I can generate all of the bumpless pipe dreams rather quickly. So here's, let's take one, let's take W2176543. And let's take the rocket diagram of that. And so here's a little ASCII image, or depiction of the bumpless pipe dream. And let's take all of the bumpless pipe dreams for this one. So that this one's real reasonably small. Let's see how many there are. There's probably not very many. Okay, 594 of them for that. Okay. I'm not going to, we're running out of time, so I'm not going to do a longer one. But the, for fairly large examples getting up into 80,000, it takes about 40 seconds to generate all of the bumpless pipe dreams. And so the one other thing I wanted to show off was a little gadget for generating and kind of jumping around with these things. So you can play with these droop moves. Here's the Rotha diagram. And if I want to see all of the droops of that, I just click on it and there they are. Now I have to say I bounded it at 20. So there might have in fact been, they're probably, let's see, in this case, there might have been just, there might have been 21. So I had to come up, I had to cut them off somewhere. But if you want to do drops, there's many fewer drops. And so there you go, there's just two of them. And you can just navigate through all of the, all of the moves on bumpless pipe dreams by clicking on a diagram. So if I click on this diagram, it will give me all the droops of that one. And there they are. And if I click on this one, I get all the droops. And eventually there won't be any droops. So let's see what happens there. There were no droops for that one. Okay, so that that was just a little bit of fun, but it's hopefully useful to for kind of exploring these moves. Thanks everyone for listening. Thanks very much, David, for a very nice talk. Let's thank the speaker.