 Yeah, please go ahead with second part. Thanks. So now we get what you thought you were coming here for, Grobner degeneration double-sheber polynomials. So I just want to remind everyone what a Grobner basis is. So if we have a polynomial ring, then we can define the initial ideal is going to be the ideal that is generated by all of the largest terms of every polynomial in the ideal So there is a term order, which is just a way of picking out from any set of monomials, the biggest one in a controlled way. So a Grobner basis is a very good generating set for the ideal I. So it's a generating set such that if I look at just the largest terms of that generating set, I actually get the whole initial ideal. So let's take this ideal I. And we're going to take lexicographic monomial order, where x is biggest than y, then z, then w. And everything is homogeneous. So we can just write our polynomials. We can just consider the terms like we would in the phone book and take the alphabetically first term as the biggest. So for example, our phone book has the property that w comes after z. So of course, for any monomial, the monomial is the biggest term of that polynomial. And for this polynomial here, because we're lexing with respect to x, x is largest. So x squared is the largest term of that polynomial. This is not a Grobner basis. And one way to see that it's not a Grobner basis is that yw squared is in the ideal I. There it is being a combination of the generators. But yw squared is not in the ideal generated by the biggest term of this polynomial and the biggest term of that polynomial. So it turns out that if you throw in just yw squared, now we do have a Grobner basis. So now it is the case that for every polynomial in I, the biggest term is divisible by x, y, xz, or yw squared. So why do we like Grobner basis? Well, one reason is that they're very valuable for computations. So once you have a Grobner basis, there's a way to do long division and get a definitive yes-no answer is your polynomial in the ideal. So more theoretical reasons why we might like it are that if the initial ideal can't be worse than the ideal I, so if the initial ideal is radical or cone Macaulay, then I is radical or cone Macaulay. And the last thing that we'll want to consider is that the Hilbert functions, whether standard graded or multigraded, of r mod I and r mod the initial ideal of I agree. And that's because the monomials that are not divisible by any of the generators of the initial ideal of I, they form a basis both for r mod I and for r mod the initial ideal of I. So when the Hilbert functions agree, that tells us that the k polynomials agree and the degrees agree and the multi-degrees agree. So in particular, the Schubert polynomials agree. So speaking of Schubert polynomials, let's write some down. So Bertrand and Deli and Fumine Kirilov told us ways to use these things now called pipe dreams to write down double Schubert polynomials. So what's a pipe dream? Well, we're going to write the numbers 1, 2, 3, 4 across the top. And then we're going to trace out where each of these pipes comes out on the left. So here we're getting 2, 1, 4, see, 4, 3. And we get the same thing if we did the other two permutations. And that tells us that this is a pipe dream for 2, 1, 4, 3. So pipe dreams are tilings of the grid or what we're allowed are these crossing tiles and these bumping tiles when we're above this diagonal. When we're below the diagonal, there's no information. We can just leave it blank. So the set of tilings subject to those conditions for each permutation gives us a way to read off the Schubert polynomial. And the way we do that is we look at these crossing tiles. So for example, this cross in position 1, 1 gives us a 1, 1. And this crossing tile at 3, 1 gives us a 3, 1. And this is one big term of our Schubert polynomial coming from one of the pipe dreams for 2, 1, 4, 3. Similarly, 1, 1 and 3, 3 gives us 1, 1, 2, 2 gives us 1, 1, 2, 2. That's our second big term of the double Schubert polynomial coming from the second of the pipe dreams. What Knudsen-Miller showed was that these combinatorics come from a Grobner degeneration. So if we take a term order so that the product of terms along the anti-diagonal is the biggest term of any generic minor, then the initial ideal of this Schubert determinants ideal is z11 times this product here. It has a prime decomposition in terms of these three co-dimension 2 prime ideals just generated by variables. And this 1, 1 is the same as this 1, 1 is the same as this 1, 1. And this 3, 1 is the same as this 3, 1 is the same as this 3, 1. And so it's the prime components of the anti-diagonal initial ideal of the matrix Schubert variety that are recording the information in the combinatorics in the algebra and in the Schubert polynomial. The anti-diagonal degeneration is really good. And one reason why it's really good is that all of these initial ideals are radical. So the initial ideals are radical. That means they have a Stanley-Riesner complex. The Stanley-Riesner complex is vertex decomposable. Vertex decomposable implies shellable, implies Cohen-McAulay. So the combinatorics that come from these degenerations are very helpful. More recently, there's been a theorem of concovabro, which is really my favorite theorem from the past five years, which is that we said before that if an initial ideal is Cohen-McAulay, then i is Cohen-McAulay. If the initial ideal is radical, actually, the converse is true. So if the initial ideal is radical, it cannot be worse in this way. More generally, they showed that the extremal bedding numbers agree, which tells us both that the depth of initial ideal of i is the same as the depth of i, and also that the Casanova-Mumford regularity of the initial ideal of i is the same as the Casanova-Mumford regularity of i. So if we want to study things like depth and regularity of ASMs by looking at their decomposition into Schubert's, then the anti-diagonal degeneration is maybe the right one to use, or at least a great one to use. So this is we're not out there looking for a better degeneration. This is a really good degeneration. But we will look for another one, and we'll look for it coming from bumpless pipe dreams. So what is a bumpless pipe dream? So the first thing we're going to do is we're going to think about the road of bumpless pipe dream. So the road of bumpless pipe dream is just we're going to write down the rotor diagram, but we're going to make the angles kind of bendy, so they look like pipes. And then we're allowed to do these things called droop moves. And what is a droop move? Well, we find a rectangle that has one downward-facing elbow up left, and it has a blank tile southeast, and it doesn't have any elbows anywhere in between. And then what we do is we droop by bending the wire in that rectangle. So we're going to bend down into the blank spot that we found. We can. Here's another move we can do. Here is a downward-facing elbow. Here is a blank tile. There are no other elbows in this area. So we can bend this pipe and droop like that. And now we've found a couple of new blank tiles that way. So let's see. Let's think about a droop we're not allowed to do. So we are not allowed to droop this pipe into that blank spot. And the reason is because we have this other elbow in the rectangle where we drooped. We don't have to sort of droop to the closest thing. We can droop, for example, we can take this pipe and we can droop it all the way down here if we want. And so that looks like that. So the set of tilings that we can arrive at starting from the road of BPD and doing these droop moves. This is the complete family of Bumples pipe dreams for the permutation that we started with. Anyone else talking about these would define them as a tiling of the NYN grid, certain to subject to certain conditions. There are a lot of conditions. And then it's a theorem, doodalambly and humidona, that we can get to all of them through droops. But we will just start with the road of Bumples pipe dream and we'll droop. So here is the complete set of Bumples pipe dreams for 2143. And it turned out that we can also read off, this is doodalambly and humidona, the Schubert polynomial, the double Schubert from these combinatorics. So here we have, now we're going to use the blank tiles. So 1, 1 is this blank tile, 3, 3 is this blank tile, 1, 1 is this blank tile, 2, 1 is this blank tile, and so on. So now it's the blank tiles that tell us how to write down the double Schubert polynomial. So I also have Laskew and Anna up here. And that's because Bumples pipe dreams were really rediscovering the six vertex model that Laskew used and that Anna cleaned up the details for. So these Bumples pipe dreams are new, but also they're old. So one thing that we notice is that we've gotten these two formulas for the double Schubert polynomial. So this is the one that we just wrote down from the Bumples pipe dreams. This is the one that we wrote down from the pipe dreams long ago. And if we foil and recollect terms, we can see that indeed these are the same polynomial, but they're not the same polynomial term by term for any way that directly comes from the combinatorics that we can see immediately. So there isn't a way to define, we don't find this term among any of these three. So the situation is that there are these two closely related formulas for the double Schubert polynomial. They're both coming from these tilings of grids with some squiggles in them. And one of them has this geometry attached to it, this anti-diagonal Grobner geometry. This other one is currently sort of free-floating. And at least in this small example here, there is a way to attach some geometry to it. And this is by looking at a diagonal degeneration. So now we're going to have, we're going to let Z33 be the biggest term. And it turns out that once we replace this D11 that should be there with a 0, now we do get on a Grobner basis. So these two generators are indeed a Grobner basis for the Schubert determinants ideal of 2143. And the leading term is the product of these three. So now if we take a, here's our initial ideal, and we take a prime decomposition, then what we're getting is that this 11 is this 11, this 33 is that 33, this 11 is that 11, this 21 is that 21, and so on. So now, again, we have that the combinatorics and the geometry and the Schubert polynomial are all tracking each other at the same time. And my theorem with Anna is that that is, in general, true. So if we lex from the southeast, then every time we do a step in a Grobner degeneration, and I'll say in a minute what I mean by that, then this reflects droop moves in the bumpless pipe drains. And so once we degenerate all the way down to a monomial ideal, then we get that the prime components of that initial ideal count with scheme theoretic multiplicity. We don't necessarily have radical ideals, the bumpless pipe drains of W. So here's what that looks like an example. And this is actually true for arbitrary intersections of Schubert determinants ideals. So we can get some repetition with a smaller example here. So here we have this one bumpless pipe drain for W1 for 213. And we have two bumpless pipe drains for 132. This one is contributing Z11, that's the only blank tile. This one is contributing Z22 and Z11. So we should have two copies of Z11 and one copy of Z22. And indeed, the initial ideal is Z11 squared Z22. So I said steps in a Grobner degeneration. So those steps are going to be geometric vertex decomposition. So what does that mean? This means we're going to take a polynomial ring. We're going to designate one of the variables as largest. And we're going to do a partial Grobner degeneration. So here is a generating set that we're going to declare by fiat is a Grobner basis for our ideal. And we're going to weight the y's with 1 and all of the other variables with 0. We then take the highest weight term, or set of terms in any polynomial. And these may be monomial, but they don't have to be. For example, for any polynomial that doesn't involve y at all, we're going to get the entire polynomial in our initial ideal here. So this is the ideal of initial y where y is weight 1, everything else is weight 0. So we have these ideals, c and n, that will play the role of Lincoln deletion in a vertex decomposition. So the ideal of the deletion is just all of the Grobner generators that don't involve y. The ideal of the link is all of the Grobner generators that don't involve y, together with this coefficient of y. This coefficient is anything in any polynomial that doesn't involve y. So theorem due to Knudsen-Miller-Yohm is that the ideal of initial y-forms decomposes as this ideal of the link intersect the ideal of the deletion plus n, sorry, plus y. These ideals have a co-dimension relation that mirrors the size of maximal faces in a vertex decomposition. So everything about this, in terms of size, looks like a vertex decomposition. So they use this in studying Baxillary matrix supervarieties. Here's one quick example. So this is the ideal of 2 by 2 minors in a generic 2 by 3 matrix. And we're going to take z23, which is the southeast term. We're going to take it to be largest. So the ideal of initial y-forms is this polynomial that doesn't involve z23 at all. Together with any polynomial that does involve z23, we're going to take the z23 part of it. This decomposes into this ideal coming from these coefficients of z23. And this ideal coming from the polynomial that doesn't use z23 at all, together with z23 itself. In a monomial example, a geometric vertex decomposition is just a vertex decomposition of the associated simulatial complex. So how our droop moves instances of geometric vertex decomposition? So here's our example, our nice small example, 2, 1, 4, 3 again. And here is the Grubmer basis that we saw before, just written out, for 2, 1, 4, 3. And we're going to do a geometric vertex decomposition with respect to z33. So the ideal of the deletion is going to be the only polynomial that doesn't involve z33 at all, together with z33. And the ideal of the link is coming from these coefficients of z33, together with the polynomial that doesn't involve z33 at all. So that's just the pure algebra and geometric vertex decomposition. Up in the bumpless pipe dreams, we consider two cases. So one case is that z33 is a blank tile in the bumpless pipe dream. It turns out that's just this one. And just this one gives us the ideal of the deletion here. And every time we do a droop move, we uncover a new tile. And when we uncover a new tile, that is giving us one of these variables here, because it turns out that we're already monomial. But the second case is that this tile here is not a blank tile in the bumpless pipe dream. And if this blank tile is not a blank tile, then the only option for it is to be an upward-facing alabo, because that's how the droop moves work, is if you start life as a blank tile, you can get drooped into, but then you're stuck. So these two bumpless pipe dreams here, these are determining the ideal of the link. So how do we do this in a way that gives us more bumpless pipe dreams and therefore more hyper-determinant ideals? So once we do a droop move, we can then straighten out the bumpless pipe dream by uncrossing these wires here by putting in a cross there. And what we get is a bumpless pipe dream for a different permutation of the same coxeter link, whose blank tiles have moved north-west. So the ideal of the deletion is another super-determinant ideal. It's a super-determinant ideal of a permutation that's coxeter link one smaller. This CYI, this is an ASM. And this ASM has the decomposition into, so this is the perm A. So it decomposes into an intersection of Schubert-determinant ideals of the same height. And those Schubert-determinant ideals are the ones that we get from straightening out these droop moves there. So we get an ASM that we can decompose into Schubert's whose essential cells are farther north-west. And we get the deletion, which is coming from a Schubert- of coxeter link one smaller. So we have an induction in two different senses. So we have these three recurrences. We start with a Schubert-determinant ideal. And we're going to take a lower outside corner, so an essential box that's as far south and east as we can get. And we do a geometric vertex decomposition. This geometric vertex decomposition splits, gets us partway towards the initial ideal of I. And it's splitting us into link and deletion, where link is an ASM and it splits as an intersection into somehow easier Schubert's. The deletion is another Schubert and a smaller coxeter link. On the BPDs, what we're doing is we're canceling that box, that lower outside corner. And so we're splitting the BPDs into ones that have Y as a box and ones that don't. And there's also Lascaux's transition formula, which is a purely combinatorial splitting that reflects both of these transitions, both of these splitings above. So that was the story for when we start with one Schubert. As we go through the induction, the way that we're getting multiplicity, the thing that becomes harder, is that once we want to apply induction to this case, but we have an intersection of Schubert's, and when we take the link of this ASM, we're not necessarily radical. And that's where we have to take radical and record how many copies of things we're getting. So returning to our example here, let's take a geometric vertex decomposition of this polynomial at Z22. So there are no polynomials that don't involve Z22. So this N is just the zero ideal. And this link is the coefficient of Z22. So this is Z11 square. So our geometric vertex decomposition is giving us link and deletion plus Y, which in this case is just Y. And this again is our two copies of Z11 and our one copy of Z22. So I've said that we, like ASMs, we want to know if they're co-inmecali. And I've also said that the anti-diagonal degeneration is really, really good for a lot of applications. And so this diagonal degeneration, one feature of it is that as we pass through these CYJ, we're passing through ASMs that we don't yet understand, but would like to. And so one thing that we get from looking at this diagonal degeneration is we get to start studying which ASMs are co-inmecali. So if we have R equal to one, if we just have J is one super-determinant ideal and we take a geometric vertex decomposition, we're going to get one ASM here and this ASM is necessarily co-inmecali. So I want to end with one example to give some intuition for why we're studying the initial ideals and not growth nerve bases directly. And that's because the growth nerve basis itself actually does depend on the order. So I said Lex from Southeast, but a couple of choices we can make or we can prioritize going over rows first or we can prioritize going up columns first. And those two different choices that are both reasonably described as Lex from Southeast actually give us two different growth nerve bases even for two and four, three, six, five. So these have the same associated primes and they have each associated prime, or sorry, they have each minimal prime with the same multiplicity. The multiplicity is manifest a little bit differently. This one, the multiplicity is manifesting with a square on Z two one. Here the multiplicity is manifesting with a square on Z one two. They also can have... So if we take a flat degeneration of an equidimensional variety, we get something that's still equidimensional but it can have embedded primes. And so as we go through these diagonal degenerations, we really can end up with initial ideals that have embedded primes. And the number of embedded primes is not consistent over which of these Lex from Southeast term orders we pick. So for this permutation here, if we take these two different diagonal term orders that are both Lex from Southeast, we can end up with, we do end up with different numbers of embedded primes. So this is really a theorem about initial ideals rather than Grobner bases. And it's really about the components, meaning the top dimensional components and does not capture the information that we would like to throw away from the standpoint of intersection theory. When we're doing intersection theory, components of the wrong dimension, including embedded ones, don't contribute. So this theorem is about initial ideals and it is about multiple cities. So I will stop there. Yeah, let's thank the speaker.