 But further ado, we are very happy to have Tianyu telling us about pipe dreams, bumpless pipe dreams and bruh head chains. Please go ahead. Thank you for the introduction and thanks for the opportunity, it's my honor here to tell you something about pipe dreams, bumpless pipe dreams and bruh head chains. So in today's talk I will be first telling you three different formulas to compute the, sorry, the compute the superpoinomials, we are going to compute superpoinomials using pipe dreams, bumpless pipe dreams and bruh head chains. And then we are going to use some connections to somehow jump back and forth between these three objects for superpoinomials. This time I'm not going to define these superpoinomials, I'll just tell you how to compute the superpoinomials using these three different kamikara objects. Let's just start from the pipe dreams. So this looks like a video game, so it's like you have an N, then you have a staircase grid, and then what you write game here is to put these tiles into this grid. So the rightmost tile in this grid have to be this elbow, this red elbow that looks like this, and then every single, every single other tile can be either a crossing or a bump, but keep in mind that we cannot have double crossing, meaning that two pipes cannot cross more than once. Once you have a pipe dream satisfying these conditions, you can attach a permutation to this pipe dream. To attach the permutation, you just write 1, 2, 3, 4 on top, you follow the pipes and read 2, 1, 4, 3 on the left. So these three are pipe dreams of the permutation 2, 1, 4, 3. This time, if I want to compute a superpoinomial of 2, 1, 4, 3, I first find all these pipe dreams of 2, 1, 4, 3, each pipe dream gives me a monomial. To find the monomial, you just remember a crossing in row i gives you xi. So here I have x of x1 times x1, that's x1 squared. Right here I have x1 times x2, that's the monomial, and I have x1 times x3. The last step is to add them all up, and that's your superpoinomial of 2, 1, 4, 3. We can also use bumpless pipe dreams to compute the superpoinomial, so this time the grid is a square, and every tile can have six options. This time the pipes enter from the bottom and they go to the right. And again, we don't want to see double crossings, like two pipes cannot cross more than once. This time we look at these blank tiles. Your blank tile in row i gives you an xi. So here we have x1 times x1, x1 times x2, and the next one times x3. We add them up, and we also get the superpoinomial of 2, 1, 4, 3. So these two rows are fairly diagrammatic, and I'm going to tell you the more complicated one, which is the Bruja chain formula. Let's just start from something called Bruja order. So this is a very famous partial order on permutations. So here we have the Hase diagram of this partial order. Whenever you can see an edge going from 2, 1, 3, going up to 3, 1, 2, we just write 3, 1, 3, covered by 3, 1, 2. And to describe this copper relation, we need a few notations. So this tij here is the transposition that swaps i and j. So wtij is obtained from w by switching wi and wj. We say w is covered by wtij if first i is less than j, wi is less than wj, meaning that I'm swapping a small number on the left with a large number on the right. Moreover, I want to make sure that numbers in between are not in between. This just means whenever I'm swapping two numbers, I have to make sure that every number that's physically standing between do not have values in between. For instance, 2, 1, 3 here is covered by 3, 1, 2, because if I swap 2 and 3, the one sitting between doesn't have values in between, so the 1 is happy. However, 1, 2, 3 is not covered by 3, 2, 1. That's because if I want to swap this 1 and the 3 here, the two sitting in between might not be happy about this. And once we have w covered by wtij, we can draw such a funny diagram. We can have an arrow going from w to wtij, and then we can put a label k here on top of this arrow. The k can be any number that's in between i and j minus 1. Whenever we have a sequence of permutations where each one is covered by the guy on its right, and whenever I can put the same k on each of these arrows, we say this is a k-chain. Right here, I have a 2-chain. So one way to think about the 2-chain is, you just go ahead and put a bar after the first two numbers, and then you always make sure you swap a small number on the left of the bar with a large number on the right of the bar, making sure that whenever you do a swap, numbers in between are not in between. For instance, right here, I can first swap the 1 and the 4 across the bar, then I swap the 2 and 3 across the bar, then I swap the 4 and 5 across the bar. This is a 2-chain. I made 3 swaps, so this is a 2-chain with length 3. This is actually not just an arbitrary 2-chain. This is what people call an increasing 2-chain. If you look at a smaller number being swapped, it's 1, and then 2, and then 4. So the smaller number being swapped is increasing. People call such chains increasing 2-chains. Here's one fact. If you grab two permutations, u and w, and grab a label k, and then there either exists exactly one increasing k-chain from u to w, or there exists no such increasing k-chain. So we can just use these increasing k-chains to compute Schrodinger polynomials. That's why Bedroian and Sotila, they just go ahead, and try to compute a Schrodinger polynomial of 2, 1, 4, 3. So this time we start at 2, 1, 4, 3. We take an increase in 1-chain and go to some permutation. Then we take an increase in 2-chain and go to some other permutation. And we take an increase in 3-chain afterwards. We have to make sure our final destination is the guy 4, 3, 2, 1. The permutation where the numbers are decreasing. So these three are the chains satisfying the conditions. And you can see that now we have three terms. Each term should give me one monomial. To write down this monomial, I'm going to use the way that's slightly different from what Bedroian and Sotila had, but ultimately the same. I'm going to count the number of fixed points on the right side of the bar. So for instance, this one-chain right here, on the right of the bar, it fixes the 1. It fixes the 3. It doesn't fix the 4. That's x1 squared. This guy, this chain, the second chain right here, it fixes the 1 on the right side of the bar. So this is an increase in 1-chains or have an x1. This increase in 2-chain fixes a number on the right side of the bar. That's an x2. So this chain corresponds to the monomial x1, x2. And the 1 on the bottom corresponds to x1, x3. I just add them up. And I also get a super polynomial of 2, 1, 4, 3. So now we have three different formulas to compute the super polynomials. And now let me just describe their connections. So I'm going to start with the bi-jaction between pipe dreams and bombless pipe dreams. So it's well known that these pipe dreams on the right are corresponding to these pairs of words like this. Go and find a way to somehow bi-jack these pairs of words with these bombless pipe dreams using some sort of insertion algorithm. So that's a bi-jaction between bombless pipe dreams and pipe dreams. Recently, Knudsen and Udall defined all these hybrid pipe dreams which are like a grid where this time, some rows are satisfying the pipe dream convention and some rows are satisfying the bombless pipe dream convention. So they just start from a pipe dream and gradually change the convention of each row into bombless pipe dream and eventually they just arrive at a bombless pipe dream but upside down. And what they can show is that their bi-jaction agrees with the Gaohan bi-jaction. There are also bi-jactions between pipe dreams and chains. So I'm going to give you the one from the NARS or TLA but I'm going to tell you this bi-jaction in a slightly different way. So let's just start from this pipe dream which is the pipe dream of two, one, four, three. If I go ahead and erase the first row, I get a smaller pipe dream that's a pipe dream of the permutation one, three, two. So I just write one, three, two here and then I put four, I just prepare four in front. And then let's just erase the first two rows and I get a pipe dream of the permutation one, two. Let me put four and three in front. Eventually I reach a pipe dream of the permutation one and then I put four, three, two in front. Let's just denote this four, three, two, one permutation as the W naught. So what they're getting is like a sequence of permutations and you see that to go from two, one, four, three to four, one, three, two, you just take an increase in one chain and then you take an increase in two chain and then increase in three chain. So what they can show is that this is actually a way preserving bi-jaction between these pipe dreams and these chains where the labels look like one, two all the way to M minus one. That's those chains that appeared in the Bruja chain formula of superpoinomials. Well, if you think about bumpless pipe dreams, you might ask, well, can we do the same thing for these bumpless pipe dreams? So right here, I grab the bumpless pipe dream of the permutation two, one, four, three. It's for two, one, four, three because, well, this topmost pipe ends in column two. This second topmost pipe comes from this column one and then four and then three. What happens if I ignore the last row? Well, I can still try to read the permutation. This time I will read two, one, four. If I ignore the last two rows, I can still try to read the permutation. I would read two, three. This time if I erase the last three rows, I can still try to read two, while what I'm getting here are not permutations. But we can just go ahead and append the missing numbers and decrease in order afterwards. So I get two, one, four, three, then two, three, four, one, then two, four, three, one. Well, let's just put four, three, two, one at the very end. That seems like the formula for drawing Stotilla, but not quite. So this time we have to first take an increasing three chain and then increasing two chain and then increasing one chain. You see that labels here now three, two, one, instead of one, two, three. But we can still define weights in the same way. We can still say that this guy has weight x two times x one. That's because there's two chains, six is one number on the right side of the bar. This one chain also fixes one number on the right side of the bar. So this guy has weight x one times x two, which is the same as the weight of the bumpless pipe dream I started with. And what I can show here is that this is actually a weight-preserving bijection between bumpless pipe dreams and chains that look like this. Now you might ask, well, are we getting new formulas? Because this doesn't look the same as the withdrawn Stotilla formula. Well, here's the fact. If you grab two permutations, U and W, you grab two labels, K one and K two, and you grab two numbers, A and B, you count the number of such Vs. So you first take an increase in K one chain from U to V with length A, and then take an increase in K two chain from V to W with length B. Such V, well, the quantity is the same as the V prime, such that you first take the K two chain of length B, and then take the K one chain of length A. So for those experts wondering why this is the correct, the quantity on top is actually the coefficient of super W inside super U times the homogeneous metric polynomial in the first K one variable with degree A, and then in the first K two variable with degree B. But we know that when we are multiplying two Hs, we can of course swap them. So we know this quantity, these two quantities must be the same. An open problem here would be well find a bijection between these two sets that have the same size. Well, we have this observation here and then actually I'm lying in the very beginning of the talk. So my talk is actually about two plus M minus one factorial formulas for super polynomial because you can see that for these Brach chain formulas, they come in M minus one factorial flavor. You can just go ahead and grab your favorite permutation of M minus one. Let's say I grab two, one, three, then I wanna first start with an increase in two chain, then the increase in one chain, then the increase in three chain. I just look at a lot of chains that look like this. That will give you a formula of the super polynomial. So Mounties M minus one factorial different formulas. One of them, the guy where the labels are like one, two, three, corresponds to pipe dreams. The other extreme, when the labels are look like three, two, one, correspond to bumpers pipe dreams. Now, an actual question you hear to ask is, well, can we somehow interpolate from this one, two, three chain to this three, two, one chain by doing some local swaps? If you think about this, well, that's actually asking the question, that's actually asking the same question as the open problem in the previous page. Well, this open problem can be partially solved in this very special case. But this time, I have to require one of the label here is one. The other label here can be K, but I have to make sure that the permutation W has no ascent after K. Once this is the case, I have a black box called the Nars-Gromes Diagrams. So we can just treat this as a black box filled with the Nars magic. To use this black box, we put my increase in K chain on the bottom of this box and then I put this increase in one chain on the right side of the box. The Nars-Gromes just somehow miraculously give me this V prime permutation on the top left edge. I just grab this V prime, that's a bi-jaction between these Vs and these V prime. So that solves the problem a few slides earlier in this very, very special case. However, that's actually enough for us to give a bi-jaction between bombless pipe dreams and pipe dreams. So here's one bi-jaction. Bombless pipe dreams correspond to chains that look like this. We can use the Nars to somehow grab this one chain all the way to the front. And once we grab this one chain all the way to the front, we just realize that after the one chain, all these permutations start with the number four. And if we ignore this four, actually these are permutations in S3. So actually this two chain is actually just one chain and I can still use the Nars to grab it to the front. So in general, we can just go ahead and grab the one chain all the way to the last most position and then grab the two chain and then the three chain and then the four chain. And eventually the labels will become one, two, three, four. And I conjecture that this bi-jaction agrees with the Galvan bi-jaction and I have tested this conjecture on S7. Well, now we have this picture and I have five minutes left. So let me just discuss some future directions. Well, we have these M-1 factorial chain formulas to extreme correspond to two pipe-like objects and you might ask, what about the rest? Well, if you recall those hybrid pipe dreams, you do correspond to certain chains. So you can do the same trick. You can just go ahead and erase some of these rows and re-permutation out. But this time you have to append the missing numbers on both ends. So here I'm appending missing numbers on both left and right side. So you can see that this hybrid pipe dream correspond to a chain that starts with a three chain and then one chain and then two chain. And then in general, these hybrid pipe dreams correspond to two to the N minus two chain formulas, which is actually not as much as N minus one factorial. When N is four, this quantity here is also four. So these hybrid pipe dreams will cover the first one and the second one and the last two. The two in between are actually, well, we don't know how to turn these chain formulas into pipe-like formulas. So in general, if we can have N minus one factorial different pipe-like objects corresponding to these chains, that would be really cool. Another thing people might wonder is about double-fibro polynomials. These are polynomials, which have two sets of variables, X and Y. To compute these double-fibro polynomials, you can either use pipe dreams or bumpless pipe dreams. If you choose to use pipe dreams, this time, a crossing in row I column J gives you something called XI minus YJ. So here this is X1 minus Y1, this is X2 minus Y2. So this pipe dream gives you the term X1 minus Y1 times X2 minus Y2. If you wanna do bumpless pipe dream, well, then this time of blanking row I column J gives you XI minus YJ. So this is X1 minus Y1 times X2 minus Y1. If you look at the corresponding chain, well, this time this increasing one chain fixes the number one on the right of the bar. And this increasing two chain fixes the number two on the right of the bar. That's somehow similar to this X1, Y1, and X2, Y2 thing. On the bottom, you can see that this increasing two chain fixes the one on the right of the bar. And this increasing one chain fixes the one on the right of the bar. That somehow corresponds to this X1 minus Y1, X2 minus Y1 thing. So I would conjecture that these M minus Y1 fact power formula is actually extend to double Schrodinger polynomials via the following rule. So this time you just grab your favorite permutation of M minus one, and let's say it's 213. You just grab a chain that looked like this and then if the increasing two chain fixes the number three on the right of the bar, that's giving you X2 minus Y3. If this increasing one chain fixes the number one on the right of the bar, that's giving you X1 minus Y1. I would conjecture that this is actually giving me a minus one fact power formula for double Schrodinger polynomials and this conjecture has been tested for S8. All right, that's all I wanna say today. Thank you so much for listening. Thank you very much for a very nice talk.