 So maybe it's time to start the second part. And yeah, so welcome back to the second part. Alan, please take it away. Yeah, so So I'm going to take my my e pi and project it to x. And if you remember, so e pi. I said here you can reduce to the case that pi is a partial permutation. And if you're actually in one of the cables and and you're in one of these components, then you would reduce to the case that x is a permutation matrix. So pi is then the transpose of that the inverse permutation matrix times a diagonal matrix. That's the lower upper condition. And then you take. So there's a, there's, that's a, that's, that's a system of order representatives for this group action. So you hit x is a permutation matrix and you hit it with the minus cross B plus and take the closure, you're going to get a matrix you would variety. So we've long known how to compute the degree of matrix you would varieties by counting pipe dreams. So an ordinary pipe dream use only the way I want to do it today. I'm going to use these tiles in the Northwest. In the southeast, I'm going to have nothing. So the, so these five guys across the top. These are the sort of ordinary pipe dreams I want to think about. And it's, it's not mathematically any different to, to say I only have crosses and I've in the southeast I've got nothing versus in the southeast it's solid elbows. So traditionally you think about in the southeast or being solid elbows but I want to do it this way. So the definition is you have only these two kinds of tiles, and you insist that no two pipes cross twice. And so this is for 1432. So these guys 1234 are coming out as 1432 down the side. And these are the five pipe dreams where 1234 go on the top and 1432 come out the side. And so the degree of this guy turns out to be five. So that you could get from the, from the original literature on pipe dreams, but it was explained in this geometric way by me and as your Miller. I mentioned before the, the other paper I mentioned with the meeting scheme, where we degenerated the original matrix Schubert variety to a big union of coordinate spaces. So we, the way we did it was with this, this reflexing degeneration. And I mean, there's various ways to think about how we did it, but I'd write like, I'd like right now to say, we reflects the Northwest variable Z11 and then Z12 and then Z13 and then Z21 and then Z22. We just dig our way from the Northwest, we have like saying one at a time. And as we do our, our polynomials begin to be more and more monomial. And we proved that the polynomials, the ones that that Bill Fulton had shown define the major Schubert varieties scheme. We show that they're a Grubner basis with respect to this sort of term order. And when we're done, we've got a big bunch of monomials which are square free and so they're defining some reduced union of coordinate spaces. So we have one coordinate space for every pipe dream. And the coordinate spaces are very simple, you put a zero, if you've got to cross and otherwise you put nothing there's no condition that matrix entry is free. Okay, so much more recently came the bumpless pipe dreams which is another way to compute the degree or more generally the double Schubert polynomial we'll let's say the degree for right now. So it uses a bunch more tiles, oops, these guys here. There's one tile is forbidden. The, the elbows tile they call the bump. And so that one was fine for me in the pipe dreams, but it's forbidden in bumpless pipe dreams. They, there's a couple other changes. They are connecting the south to the east instead of the north to the west. So that's what you see going on here. And they still insist that no two pipes cross twice. And, and surprise, these bumpless pipe dreams there's again the right number of them, the degree of matrix Schubert for I. So there was a much tougher theorem then the analog of what Ezra and I did for the bubbles pipe just for the ordinary pipe dreams, due to two people in the audience. But if you degenerate, but you lex from the southeast, instead of rev lexing from the northwest. You don't get a union of coordinate spaces anymore, you get multiplicities so it will be a union but the each component will come with some scheming us. So we can ask what's that multiplicity, and the multiplicity is recognizing the fact that the assignment of a coordinate plane to a bubbles pipe dream isn't injective. So back in what Ezra and I were doing. There are only two tiles. So, if you knew the subspace you could figure out the tile. But, but now if you were just putting zeros where there's empty tiles, and all the other tiles are free, you can't you make uniquely figure out the bubbles pipe dream, given just where the empty tiles are. And that ambiguity leads you to having multiplicities on the components, that's their theorem, and, and, but then that means finally, I'm giving this a historical thing now. You could compute the degree by just counting the number of bubbles pipe dreams. So, that's what we have here. And you can do better than the degree you can compute the double Schubert polynomials that's an equilibrium comology calculation. Instead of just counting a pipe dream and saying this contributes one, you say contributes this big polynomial this product over all of the crosses for the ordinary case, or the empty spots for the bubbles case of the position of the cross encoded is X of its or of the tile X of the row last wise of the column, you end up with these polynomials in here and you add them over all your pipe dreams, and you get the double Schubert polynomial. All right, so now that we've warmed up with this old stuff. So by old stuff I mean this from 15 years, 20 years ago basically. And this from much more recently. Now I want to unify these and think about the EPI instead of the matrix Schubert variety. So remember what EPI looks like, you can project it to the X side, and you get a major Schubert variety on the nose, or you can project it to the Y side, and you get a major Schubert variety rotated by 180 degrees. And that's going to tie into why pipe dreams and bubbles pipe dreams seem to be off from from each other by 180 degrees. Okay. So, I'm going to now define a generic pipe dream, it's got all the times. And I don't put any restriction on pipes crossing twice. And if I cross twice you just follow them. So for example, take this guy right here. And that's this one is got the blue and green crossing twice. And for me that's just as good as this one. So these here are all of the bumpless pipe dreams for the identity. So fairly complicated already the identity is fairly complicated. And it's because, you know, with the matrix Schubert variety, the matrix Schubert variety for the identity is the entire space there's no equations very simple. So it's already very complicated, and it's going to have a complicated formula for its degree and for its every homology class. So, so there's a difference in combinatorially what the definition looks like okay we've got this combinatorial definition. There's also going to be a difference in these degree formulae so these degree formulae we just counted every pipe dream contributed or bubbles pipe dream contributed one to the degree. Three powers of two. So the powers of two have to do with how many elbow tiles there are. So there's three kinds of elbow tiles, the j tile, our tile and the bump. And those each of those is contributing a factor of two, the power of two. So, when I have the minimum number of elbow tiles. I don't want a power of two, but whenever there's any extra so these three guys have one extra these have two extra, these guys have three extra, they get me powers of two. And so the total degree of these guys is what I have here one three twos three fours two eights, it adds up to 31. So, that's going to be the, the degree of EPI. The degree of EPI will be so so first theorem I'm stating, I think a new beyond old stuff. First new theorem is here's the, here's a formula for your degree of EPI, you sum over the generic pipe dreams, each one contributes a power of two. So what's going on is we're getting into generation, but not to a union of coordinate spaces, rather to a union of these complicated varieties FD and how to define. And these FDs are defined by some linear equations, like you're used to, but also some quadratic equations. So the quadratic equations, say that these flux terms match up. So sometimes a flux will be zero, and sometimes two fluxes will be equal to each other. So remember how this, how the, what sort of group we have going on here. So when we degenerated back in this story, when we, when we did all these degenerations at every, every matrix entry either from the northwest, sorry, here, either from the northwest or from the southeast. Eventually we got something where every component would have to be invariant under the tourists that dilates every coordinate independently, so it's got to be coordinate space. So it was more interesting figuring out which coordinate spaces occur and with what multiplicity, but we knew we were going to get coordinate spaces. And all these lexogens, all these red lexogens, we were going to get coordinate spaces is only a question of which. Now, that's not going to be true here. What I'm going to be doing in a moment in this blue paragraph is I will be scaling one coordinate while anti scaling the dual coordinate. And so when I'm done, I'll have only an n squared dimensional tourists acting, but it'll be acting simplectically on this two n squared dimensional space. And so I can have so like, when I multiply x i j by y j i, that is an invariant for this n squared dimensional tourists that's acting and so I can have invariant equations of this form. All right, so these guys, they're defined by some linear and quadratic equations, and they're going to be the analogs of our coordinated subspaces. They are still at least complete intersections. So the degrees of them are just two to the number of quadratic equations that are involved. And if you want better than just then the degree, you want the green comology class, then these are the factors you're going to end up multiplying. So the b here is about scaling the x variables. The b here is about scaling the y variables. If you scale an x variable times a y variable times the conjugate y variable, then you get a plus b. All right, so I've already, I think, warned you about what the main theorem is going to be, which is, we take this guy e pi, and we do this degeneration of it. We do this lexing in one coordinate while rev lexing in the dual coordinate. And we do that, and while, and so that's doing the pipe dream degeneration on the x side while it's doing the bundles pipe dream degeneration on the y side. And we prove that when we're done, we get this union. You know, these components. So we can only prove that up to some lower dimensional junk that doesn't affect the degree of the equivalent comology class. So, I could, I guess I got, I got, but four minutes right to go to five. You're good. So yeah, I've got enough time to for a couple of things. So one is, if you take this variety e pi, and you think about projecting it to the x side, we know what you're supposed to get, you'll get the matrix Schubert variety. So, what's the corresponding co homological thing to do, where I want to somehow be able to derive the double Schubert polynomial for pi, I want to be able to derive it from this formula I have for the equity co homology class of e pi. So e pi is the sum of the class of the FDs, the FDs are the easy easy things, which are just products of those things. So, if you project to remember that I had my, this my matrix space of X's matrix space of Y's, and on the X side I had scaling and I had this coordinate. So, if you project to the X side, and get the matrix Schubert variety you can get a hold of that by looking at the be leading terms in this. So you take this formula and you say, what part of it what would I get in here, if I had the most b. And the answer is, well, I'll always use the be here always use the be here, and I want to have the fewest of these factors. So, to get the be leading term you should have the smallest number of these. And it turns out that happens exactly when you've got an ordinary pipe dream. So the ordinary pipe dreams are characterized among the generic pipe dreams, they're characterized by the ones that have minimized the number of these three tiles. And then there's a corresponding thing about projecting the other side where you say I want the a leading terms I want of course I'm going to use this I'm going to use that I want the fewest of these. And the bundles pipe dreams are characterized by those that have the have the fewest holes. So of course we know how many holes they're supposed to have it's the length of the permutation, but a bumble's pipe dream is exactly a generic pipe dream with the minimum number of blank tiles. Except they're rotated by 180 degrees. So, right, the bundles, bundles pipe dreams are in the literature they're supposed to be oriented like this. But that again has to do with the fact that when you project to the why side you don't get a matrix Schubert variety on the nose you get one rotated by 180 degrees. So, here's just me showing off that this conitorial formula lets us compute these numbers like the degree of the fourth meeting variety by summing over these, these bundles pipe dreams each one contributing a power of two. Okay, now I want to finally meant I'm going to end with something a historical that I can't. But, but bring up. What if we started the other way what if we had started with the e pies, and we didn't know anything about pipe dreams. We just said let's take these e pies and do this to generation and think about the components of what we get. All right, so we do that, and we find out the components look like this, and they have these and they've got these linear equations these quadratic equations. So, what's going on with these quadratic equations is they're telling you that for every edge in the square that will eventually be a pipe dream for every edge, some edges match up with other edges, and some edges should be blank. So just work off of that, if you recognize that the variables you should be working in are these flux variables associated to the edges, and you just use this to start massing up edges that discovers generic pipe dreams for you. And then from there ordinary and bubbles pipe dreams. So, the, the geometry is telling you, yes, you should have these pipes that are connecting. From edge e to edge e prime, or, or edge e should just be blank. All right, I'll stop there. All right, well thank you very much. I will stop the recording.