 Can you all see? Yes, it looks very good. All right, let's go right ahead then. Yeah, our last speaker today is Gabriel Houdel from Cornell, who will speak about interpolating between ordinary and bumpless pipe dreams with hybrid pipe dreams. All right, please go ahead. So this is about ordinary pipe dreams, bumpless pipe dreams, and hybrid pipe dreams, all of which are methods of computing super polynomials. And they have a lot of interesting combinatorics around them. But the first half of this talk roughly is going to be reviewing ordinary pipe dreams and bumpless pipe dreams. And then we'll talk about hybrid pipe dreams, which give another way of computing super polynomials. So in each case, we're going to talk about examples first. So here are some examples of pipe dreams. And some things to notice about them are, one, they're each a squared grid of tiles. These tiles have these pipes running through them, making these lovely curves. These pipes are moving from the top of the diagram to the left of the diagram. They're connecting between adjacent tiles. And we've labeled the left of the diagram. So we've labeled the edges along the left side, 1 through n, and we do so starting at the top. The numbers along the top of the diagrams we think of as the inverse of a permutation. So each of these diagrams, each of these pipe dreams, is associated with a permutation. And we write the inverse of that permutation on the top in one line notation, which means 1, 3, 5, 2, 7, 4, 6 in that barbecue sample. That's pi inverse of 1, pi inverse of 2, pi inverse of 3. So if, for that example, pi would be 1, 4, 2, 6, 3, 7, 5 in the same pitch. Yeah, also another important thing is that pipes aren't allowed to cross twice. So here's some non-examples of pipe dreams in our notation. And there's some things that are different from some other presentations. So technically, you can have pipe dreams where pipes cross twice, but those are thought of as non-reduced pipe dreams. We're only talking about reduced pipe dreams. In the second to last example on this slide, we have pipes which are coming from the edges. And in the original way of introducing pipe dreams and talking about it, you always had that. But in our framework, we're not thinking of those curvy pipes as being important. Yeah, so to define them more concretely, a pipe dream is a filling of a square with the tiles given so that pipes don't cross twice. Every pipe is connecting some label i on the top to a label on the left. And our left labels go 1 to n, and our top labels are an inverse of the permutation. So this is a little bit different from the standard way of defining these because of this first property here, this curvy tile, the r tile, and the horizontal tile, never appear in any pipe dream. But we're going to define pipe dreams to be allowed to use them, even though they won't, because it's going to help us later. So we'll denote the set of pipe dreams for pi as pd or pi. And we'll define this set weighty of d for a pipe dream d, which tells us the locations are weighty tiles. And the weighty tiles are going to be the crosses and the horizontals. You can use this to compute double shoe republicans. So the double shoe republicans for pi is the sum over all the pipe dreams for pi. For each pipe dream, we're going to take a product over the weighty tiles. And the terms we product are x sub the row of the weighty tile minus y sub the column. So we have variables x1 through xn and y1 through yn. If you plug in zeros for all these y's, you recover the normal shoe republicans that you might be used to. So now moving on to bumpless pipe dreams. These are a bunch of bumpless pipe dreams. Also, in the examples of ordinary pipe dreams, this is all possible three-by-three pipe dreams, three-by-three pipe dreams. And over here in our bumpless examples, we have all of the possible three-by-three bumpless pipe dreams. So bumpless pipe dreams are a little similar to pipe dreams. They have these pipes which connect the top to the right-hand side now. The right-hand side is labeled starting at the bottom, so it goes one through n, but the one is at the bottom. Pipes don't cross twice in these reduced bumpless pipe dreams. One thing that if you've seen these before it might be a little weird is these are upside down from the standard. That's also because it helps us with our generalization. And yeah, so that's what bumpless pipe dreams are and we can define it more precisely if we want to. Bumpless pipe dreams have all of these tiles, but they don't have the bumpless pipe dream, sorry, the bump tile, which looks sort of like a hyperbola. Pipes don't cross twice. They connect some label I on the top to an I on the right and we're associating each one with a permutation and the inverse of that permutation we write along the top. So we can just as before talk about the set of all bumpless pipe dreams for pi and define the weighty set of a bumpless pipe dream, which is going to be the positions of our weighty tiles. And in this case, our weighty tile is just blanks. So for both pipe dreams and bumpless pipe dreams when we have weighty tiles, we're shading them in. So that's how you can tell from the pictures. So also you have to be careful where these eyes here, the rows are given by the row labels, but the columns are not given by the column labels they're given by the physical positions. So this lets us compute double-shooper polynomials from bumpless pipe dreams. So wonderful, we have these two formulas for computing double-shooper polynomials and they're very combinatorial, very nice, lovely to study. So single-shooper polynomials are a little bit simpler and for the purposes of this talk, we're gonna focus on those. We can define these X monomials and Y monomials or mon X of D of pipe dream or bumpless pipe dream and mon Y of D of pipe dream or bumpless pipe dream and mon X is your product over the weighty tiles and just do X of the row and mon Y is product over the weighty tiles just negative Y sub the column. And it's somewhat clear from the formula for double-shooper polynomials that this thumb. What you get if you sum up these X monomials over all of the pipe dreams for pi is the single-shooper polynomial, which is what you get when you plug in zero for all of the Y variables. And you can actually get the same thing if you sum over the pipe dreams for pi inverse if you use now some of the Y monomials and put in negative X1 to your negative XN for Y one is required. And you can see that by thinking about transposes. So if you take a pipe dream and you transpose it, that's switching the rows of the columns. So the X's and Y's get switched. So there's this clear relationship between the X monomials of the transpose and the Y monomials of the original or the Y monomials of the transpose and the X monomials of the original. And when you transpose a pipe dream, you now have a pipe dream for the inverse of that permutation. Same thing with a bumpless pipe dream. Also every formula on this slide holds for both pipe dreams and bumpless pipe dreams. So because pipe dreams and bumpless pipe dreams both compute single-shooper polynomials and each one is contributing a single monomial, there must be a bijection between pipe dreams and bumpless pipe dreams for a given permutation which preserves that X monomial. It has to exist on a set theoretic level. That said, finding it and describing it is a very non-trivial task. And it was done recently by Gallant one. It's a cool paper, I recommend you check it out. They also further show that the bijection that they construct commutes with a combinatorial monk's rule that was described in another paper by one. And by weight preserving here, we mean that it's commuting with this monax function. So this is really great, but it's a little complicated to compute this bijection and prove that it works sometimes. So the goal of this work is to understand that bijection is the composition of a series of simpler ones. So we're introducing hybrid pipe dreams to get a clearer look or just more of a look at the connection between pipe dreams and bumpless pipe dreams. So these are a bunch of hybrid pipe dreams. Notice that the last two examples, the second to last one is a pipe dream and the very last one is a bumpless pipe dream. So this is the generalization. In each of these, we still have the inverse of a permutation is written along the top and one through N is written along the left and right edges. Some of these rows are essentially rows of an ordinary pipe dream and some are like rows of a bumpless pipe dream. So the idea of hybrid pipe dreams is some rows are ordinary, some rows are bumpless. Ordinary rows get to use ordinary tiles. Bumpless rows get to use bumpless tiles. An ordinary row has a number on the left and a bumpless row has a number on the right. The numbers, so to label these numbers, first we're gonna label the numbers on the left of the ordinary rows and we'll go one through K starting at the top. And then we label the numbers on the right hand side of the bumpless rows and we go K plus one through N starting at the bottom. So this matches before how with ordinary pipe dreams we labeled from the top and bumpless ones we labeled from the bottom. Also as before, pipe stone crosswise. So as a reminder here are the ordinary tiles and here are the bumpless tiles and here's written out the definition of a hybrid pipe dream which is due to Alan Knudsen. Yeah, so each of these has some particular type. There's some rows which are ordinary rows and some are bumpless rows. So we introduce this notion of the type of a hybrid pipe dream which tells us which rows are which. So it's a string of O's and E's where if the first character in it is an O that means the first row in the physical sense is an ordinary row. Yeah, and we associate permutations with these in the same way as before and we'll use HBD of pi comma tau to denote the set of hybrid pipe dreams for pi with type tau. Here's some more examples going through hybrid pipe dreams for the permutation one three two and a couple of different choices of tau. I'm not going to focus too much on this right now but here's some more examples. And we can use these to compute cheaper polynomials in pretty much the same way as before. So in bumpless rows the blank tile is weighty and in ordinary rows the horizontal and the cross are weighty and we'll let this weighty of D be defined as the positions of these weighty tiles. We have to be a little careful. The I part which is telling us the row tells us the label of the row that a weighty tile is in where in the J part which is the column tells us the physical column not the label of the column. It's a little bit different but that's again mirroring what we have before. So it turns out with any type of hybrid pipe dream so if you fix the type you can compute double sugar it's using the same formula we've been using. And this is cool because it means we now have two to the N formulas to compute double sugar polynomials for pi. And we can prove this essentially geometrically. So the great thing here is that they have hybrid pipe dreams have some lovely combinatorics and we can prove the equivalence of all of these formulas for double sugar polynomials in a combinatorial manner. So I'm not gonna get into that but there is a bijection between these decorated hybrid pipe dreams which preserves weights in a way that proves you get the same formula from doing this with any type of hybrid pipe dream. But I'm going to focus more on two special cases and really just the last one on this slide. So the first special case is for any two types of hybrid pipe dream, tau one and tau two there are strings of O and B telling you which rows are ordinary, which are the formulas and any permutation. There's a bijection between the hybrid pipe dreams for that permutation of the first type and the hybrid pipe dreams with the second type for that same permutation that preserves X monomutes. So this is a combinatorial proof that you can, your formulas for single sugar polynomials in the two hybrid types are the same and a special case of this result would be saying that pipe dreams and bombless pipe dreams can be the same thing. The second result said something very similar but now we're preserving the Y monomutes instead of the X monomutes. And I'm gonna focus on this one because it's a little bit easier to use and we can show that it implies the same overall result about pipe dreams and bombless pipe dreams. So suppose that this theorem is true, it is. I'll talk a little bit more later but let's suppose it's true to work from there. We have some piece of Y now that goes between hybrid pipe dreams or really, sorry, it goes from pipe dreams to bombless pipe dreams. It's a bijection and it commutes with this mon Y function which tells you the Y monomutes. If you have that, you go from that starting points then the X monomial of the transpose of this bijection applied to the transpose of a pipe dream. That X monomial is the same as your original X monomial. So we now can use this B Y to construct a bijection from ordinary pipe dreams to bombless pipe dreams which is going to commute with one X and thus prove the equivalence of formulas for single-sugar polynomials. So how do we do it? Here is an outline basically. The first part is we're going to define it for switching the type of the last row. So your last row of your hybrid pipe dream can be ordinary or bombless and we're going to define this bijection for two hybrid types where the only difference is in that last row whether it's ordinary or bombless and we're going to do that in a way that only changes the last row of a hybrid pipe dream. Then the next part, the second part of this is we're going to talk about how to swap to adjacent rows. So you have somewhere in your hybrid pipe dream an ordinary row over a bombless row and we're going to find a way to switch the order of those so that we have a bombless row over an ordinary row. And having these two pieces allows us to go between any two hybrid types and at the bottom we sort of see an example of getting from all ordinary to all bombless where the final O turns into a B and then we switch that O to B and we keep going. So to switch the last row, it's very simple. There's really only one way to do it. There's not really any freedom in this last row whether you're in the ordinary setting or the bombless setting. In the ordinary setting you have to swoop to the left and in the bombless setting you must swoop to the right and when you switch from swooping to the left to swooping to the right the horizontal tiles you had turn into blank tiles and the blank tiles would turn into horizontal tiles and those are the weighty tiles and they're staying in the same row in the same column. So even the equivalent weights are staying the same. Switching to rows a little more complicated. So here is two examples not saying how we're doing them yet. So on the left side of the screen is our first example on the right hand side we have a second example. On the top we're just looking at those two rows and on the bottom of the screen we show whole hybrid pipe streams that contain those two rows. So we start with having an ordinary row on top of a bombless row and the ordinary row has label one on its left and the bombless row has a label four on the right and then once we swap the rows we have this bombless row on top with a four on the right and the ordinary row on the bottom with one on the right and it's similar on the right hand side of the screen. And in each of these examples you can note that there's a single weighty tile and it's staying in the same column. So since it's staying in the same column you're preserving these Y weights. So this is how we do it. It's a bunch of rules. Basically you take your row and for each two by one column we have a rule which says what should I do if I see this two by one column? And here are all the rules and that's kind of it. The one thing to note here is that it does matter how these pipes are connected. So you don't need to know about things outside this two by one column except that you need to know whether two pipes appearing in your two by one column are the same pipe. Aside from knowing that small piece of information about connectivity this row-swapping bijection is very local. And if you stare at these long enough which you don't have time for that but if you did it's not that hard to see that they're gonna preserve connectivity that it's going to be a bijection and that it proves this second theorem here that we have a wide weight preserving bijection. That's all I have. Thank you for listening and thank you for having me here. Yeah. Thank you for a very nice talk. I wonder how many rules do you have? I think it's 23. Okay. And it's one rule for every possible two by one column that you could have in pipe dream or sorry, you could have with ordinary and then bumpless or bumpless than ordinary. Okay. Yeah. Maybe we could have another quick question for Gabriel.