 Okay, it's five o'clock on my clock, so why don't we start the second half? Okay, great. Wonderful. So I want to now go to Monk Rules, which is the main point of this paper with Tom. So let's first review for the Schubert case, how I want to think of them. So in the Schubert case, I have my polynomial ring and it has a basis, a special basis, which we want to focus on, which is these Schubert polynomials. And now this polynomial ring has favorite generators also. So these x1, x2 generate the polynomial ring. So just from that perspective, you have a desire to compute, you feel like you'll understand your ring if you compute the generator times the favorite basis element and re-expand it in terms of your favorite basis. So whatever you feel like your favorite generators are and your favorite basis, you want to compute this. And that's the point of view I want to take for the McDonnell case, is that now I'm working with Laplace polynomials, x and inverse. And I have my favorite basis, which is this basis of electronic McDonnell polynomials. And of course, if I think about this ring, then x1, x2, up to xn, and x1 inverse, x2 inverse, up to xn inverse, they might be my favorite generators. So the question then is, can you compute xj times emu and xj inverse times emu and get re-expand in terms of your favorite basis. And so that's the way we took it. Now, of course, there are several choice, different people may have different choices of favorite generators. So another favorite choice in this classical setting is to take the Schubert polynomial corresponding to a simple reflection. You could use that set of generators and compute. And the same is true in McDonnell polynomial theory is that there are other possible choices for favorite generators, like you can take the e epsilon j's times emu and the e minus epsilon j's times the emu. And in the paper, we did several cases of choices of generators, but for today for the talk, I'm just going to do one. And the idea is the same for all the natural choices that we chose, but somebody else may have other natural choices. So how do we do this? So I want to go back to those operators. So xj be the operator on polynomials given by multiplication by xj. So I want to really think of this operator that when I act on a polynomial, what it returns for me is just little xj times that original polynomial. And then the theorem is that we prove is that this operator acting on polynomials, I can rewrite that operator in terms of things. So I should have copied, let me go back here and copy the action of my favorite operators. So we've got that on the screen. So these are the things that act well on McDonnell polynomials. The tau i checks, they act nicely. The tau pi checks, they act nicely. And the yi's, they act nicely. I can track combinatorially the action of these operators on McDonnell polynomials. So my job is somehow clear from that perspective is I've got to take this operator, which doesn't act so nicely, and I have to rewrite it in terms of these tau checks. And so that's what we did is we rewrote that and the answer is you sum over subsets of one through N. And here the condition is that that subset must contain J. So if I'm trying to multiply by xj, then the subset I choose must contain J. And then there's a factor that's full of tau checks, which I call tau cj. I mean, I walk through slowly through these things once I get them written down. And then there's a factor which I call fcj, which is a function of y's, which depends only on mu and not on J, which is a function of y's. But the philosophical point is that these parts depend on y and over here I know how y acts on my McDonnell polynomials. And these parts depend on tau and points one and three over here they tell me how the tau is act on McDonnell polynomials. So once you prove this, you get corollary, which is xj acting on emu is equal to the sum over subsets of one through N, such that subset contains J. And then there's a factor fmu cj, which is a constant, it's going to depend on qt. There's a factor weight mu of c, mu of c. And then there's a factor, and then there's e of what I call rote mu of c. And so this is, this is my basis element here, basis element. And these, these guys will be functions of q and t. And so that'll give an expansion of xj and emu. All right, so let's copy this because we're going to have to refer to these. Let me explain the combinatorics, just getting busy. Okay. So, so let me explain how I think about this we'll just do an example. So I'm going to do N is 11 and J is seven. And so one term in this sum depends on a subset of one through 11. So I'm going to choose for my subset two, three, five, seven, nine and 10. And very important is going to be the complement of that subset, which in this case is one, four. Let's see, is that what I wanted? I'll better do the same example I had in my, oh, so no. Two, five, seven, nine, 11, one, three, four, or six, eight, and 11. Okay, did I get the compliment? Correct. Okay. All right, so, so how do I think about this? So I think about a circle like this. And at the top. And this is like a roulette wheel that so there's something that marks where, where the, where you, where you turn over where you go past. And then I've got me one. So I'm drawing now a picture of me, because I'm thinking about the action of XJ on email. I'm going to draw a picture. And I've got me three, me, me one, me three, me four, me six, me eight, and me 11, and the other ones are my subset. So this is me two, me five, me seven, me nine, and me 10. So that's the, and over here is my, this is, this is my J. So this is me J in this example. So that's a special point because that's telling me which XJ I'm trying to multiply by. All right. So, now I want to tell you what this wrote me was, because this is the somehow the first thing when you start doing computations is okay how do you classify these, these terms that actually appear in the, in the expansion. And so wrote me is this guy. So this is wrote me see. And what I do is I draw the same picture again. That's my, my first marker should have done that dotted. And then I've got me one. This is before me three, me four, me six, me eight, and me 11. So the, the compliment set this compliment set just stays the same in wrote, wrote. But then these guys that are these other spokes of this wheel. So the way I think about it is I think about these red muse as the runners. And there's a relay race, right. And so what happens is that the new to runs to the next spot before hands off the baton, the, and then the new five runs to the next spot. And then the, the new to the next player, the runners to the next runner. So what happens is that the new to ends up down here, and the new five ends up over here, and the new seven ends up over here, and the new nine just ran one spot only. And the new 10 did so the new 10 did something interesting, because the new 10 ran past this. But kept running all the way to the new to so so so what happens is that when the new 10 runs past the finish line, it gets a prize, it gets augmented by what. Right. So, so you got these runners that are running and the one that runs past the, this, this blue guy. The one that runs. What did it do. How do I get rid of that. It's down here. The one that runs past this blue guy, someone know what I did. I think you did some kind of like, like, focusing in on a little region so there's probably like a little axe at the bottom somewhere to get out of this. Oh, fantastic. You're genius wonderful. Okay, so that. So, when you run past that little blue guy you get the rotation, and that describes which basis elements appear in this factor here. When you expand. Now, what happens in the towel I need to tell you how CJ check is. And how CJ check is the complement of C. So it's going to be these guys. And it's the power. So it's going to be tau six, tau four, and I this tau CJ check started. So I'm taking tau six, tau four, three, one. Tau pie check, because when I go past this starting point I get a tau pie check, and then tau 11 minus one check and tau eight minus one check so. So this is the sequence of operators that contribute to this rope. So the way that we got this wrote to me, the rotate me is that we applied these towels that are appearing up here, and this sequence of towels that produces the road. Okay, so the next way that I'll tell you about is. Is this f CJ. So in this case we're looking at FC seven of why. That's this factor in this box over here, and that is why why seven inverse minus why. To why five inverse. So the seven and the five are this seven, which is my J and the one that comes before it. And the two is the first red one. It's the first one in my sequence of those. And that when you convert that f C seven J C seven why you get F mu C seven so this factor in the corollary is going to be Q to the mu seven minus mu. To minus times T to the V mu seven minus V mu to minus Q to the mu five minus mu to T to the V mu five minus V mu. So, this four down here, and this tells me how the wise act, and the cues pick up the parts of me and the tease pick up this V mu factor. And that's what we're seeing here is that the wise have turned into you with the part of me and T with V mu factor. So those conversions are easy once you once you prove the operator formula, which is top one, then getting this FCJ for the bottom one is easy. Okay, now I want to. Let me copy this road see because there's one more I'm not going to tell you all the details but there's one more. Which I think is really cool. And so I want to take that. And there's still time right. So, so this some extra factor weight mu of C. So that's the last factor in the corollary. And I tell you what that looks like. So that is product of several, several pieces. So it's first piece is T to the minus the number of eyes, such that mu I is bigger than mu I plus one. So that you can just compute that's like a descent statistic. Take your composition and just count how many. And then there's a factor that depends on the elements of C. Remember I have my blue spokes and my red spokes. And so now I'm going to look at the, the red spokes. And I get to there are two cases that naturally appear. There's the case where K is not equal to a one. There's an extra factor that comes from K equal to a one. This is the K equal to a one factor. Those are the a one is the first. This is a mu a one. It's the first read the first element of that subset C. And after it looks like one minus T over one minus Q to some garbage T to some garbage one minus Q to some garbage plus one T to some garbage plus one. And that the garbage always looks like this kind of exponents. I'm not going to write it out because it's not instructive, but then there's one more factor. There's the product over the case not in C. And this one I find really cool. So this one is, is built from factors I call weight C mu Q, weight mu C, C K. So let's think about what is, what is a K that's not in C, a K that's not in C is one of the red spokes. No, okay, this not in C is one of the black spokes. So these are for me the, the spectators, the red spokes are the runners who run around. And then there are the spectators. And what happens if you're a spectator in one iteration is that you see a runner running by. A runner has some value, because that's mu seven or mu nine or something, and you have some value because you're mu six or mu four. Right. And so what happens is there are several cases here. But if that runner running by let me put it in red so it looks like a runner. So the value of the runner mu K that's the runner that runs by the spectator at position K. If, if that runner has exactly the same value as you do, then your blood brothers or something and you say oh this is great to see you I haven't seen you for years, and the whole race stops and everything is zero. Right. The busts. You just get a factor of zero. If the runner who runs by is bigger than you. Then they just run by you get a factor of one. But if the runner that runs by is smaller than you. Well then they have to pay for that. Because here you're going to extract a total you're bigger than they are, and you can get the toll. And the toll factor is looks like T. So it's got a one minus Q to the something T to the something plus one one minus Q to the something T to the something minus one divided by one minus Q to the something T to the something squared. So it's a it's a messy factor which you can look up in the paper for the exact way to compute it. But somehow I found this, this business when we were trying to trying to figure out these weights, and we finally got these runners running past the spectators, then we could actually sort of compute them by hand. And so I think that's, that's what I have to say is hopefully it gives you a picture of what these monkeys look like and how we got them where we came. Nice. Let's thank the speakers.