 Okay, yeah. So welcome everybody. Today we are delighted to have Sami Asaf from USC speaking about an insertion algorithm for multiplying super polynomials by super polynomials. Please go ahead. Thanks. Thanks for the invitation to speak. The slides, I think the link to the slides will be posted in the chat. So if you, if you like spoilers, then you can go ahead and look ahead at all of that. Most of what I'm talking about today, oh, hang on, I'll put the chat up right in front of me. So maybe I'll see if you have questions in the chat. Most of what I'm talking about today is joint work with my amazing collaborator, Nance Alabair-Jaron. So here's the plan of my talk. We'll have a, we'll have a break right here, which is actually the halfway point. So the first thing I'll tell you about is what I did during COVID. I played around and found a rule for multiplying a super polynomial times a sure polynomial. And then in the second part of my talk, I'll tell you about super polynomials that most of you here just judging by the names, no more about super polynomials than I do, but I'll tell you my way of thinking about them because it leads to the second half of the talk where I'll tell you the insertion algorithm that proves the conjecture up here. And this part here is all joint with Nance Alabair-Jaron. So that's the plan. We'll get started. The first part of this talk should be very easily accessible. So if there are any questions, then feel free to chime in. It should be pretty understandable. That's sort of the goal. So I'll start by telling you the problem. I'm not defining a super polynomial yet. Somehow I don't need to. There are these polynomials that are a basis for the polynomial ring. They're a basis. So I can multiply them and I can span back in the basis and I can look at the structure constants that we get. So the fundamental problem that I'm trying to solve is I want to have a combinatorial formula for these numbers. Okay. So, okay, here's a combinatorial formula for these numbers. These numbers count flags in a triple intersection of super varieties. That's a combinatorial formula. I can't use that because literally if you give me U, V, and W, I can't compute that number. So I don't count that as what I'm looking for. I'm looking for something simple. I'm looking for something I can explain to a child or a computer and have either of them do it correctly. So the rule that I'm looking for, which I mean, I think this is open, but I've been told that it's been intrinsically solved and I don't really know what that means, but I don't think this exists. So I'm looking for a simple, positive combinatorial rule for these numbers. And the gold standard here is the Littlewood Richardson rule for shore polynomials. It's skew tableau that satisfies simple conditions. And it's something that you can just write down and check and you can work with. And that's sort of the standard that I'm aiming for. So this is the problem. So maybe we can debate whether it's open, but if you put this word in here, maybe we can. So I'm going to be doing a special case today because everyone does special cases. So here's my special case. I'm looking at this guy and I'm going to let the indexing permutation there be k gross monion. So what's a, what's a gross monion permutation? So it means that the permutation has a unique descent. So here's my example here. One, three, six, seven. Those numbers are increasing. And then at position four, where the seven is, I go down after that. This is a number two. Seven is bigger than two. And then the numbers increase again forever after two, four, five, eight, nine, 10, 11, 12. How are you like? Okay. This is for gross monion because there's this one descent right here. So k gross monion permutations are in bijection with k partition. What's a k partition? So I will be reversing a lot of the conventions that maybe you're familiar with in this talk. There are reasons that I do this, partly because it all makes more sense to me this way, but partly because it's actually easier to do all the constructions later if you allow me this reversal. So here is a k partition. This is a four partition. Why is it a four partition? Because it's got four numbers. They weekly increase. Okay. And there's a simple bijection between these. So what's the bijection? Well, here, what you do is you subtract one from the first thing, two from the second thing, three from the third thing and four from that thing. And when you do that, you get zero, three minus two is one, six minus three is three, seven minus three is four. And that's this thing. Obviously you go back in the reverse way, plus one, plus two, plus three, plus four. And then you add all the numbers that you haven't used yet in increasing order thereafter. So these are in bijection. So instead of giving you a generic permutation for this one, I'm going to give you a k graph monion permutation. But instead of doing that, I'm going to tell you a k partition. I'm going to say, hey, k is four. And here's my four partitions that I'm doing. Okay. So the simplest way to state the rule was actually stated without the details. So in this generality by Bertrand and Sateel in their 1998 paper, they said that they expected the answer to look like this. Saturated chains in brew hot order from U to W, from here to here, with some condition depending on V. And now I know what that condition is. It's this thing that I'm going to call lattice weight V and you have to use k brew hot order. So in order for this rule to qualify under my definition of what I'm looking for, in terms of simple, I have to be able to explain it to you as if you were a child. And that's what I'll do next. So I'll explain it to you in the simplest way that I can on the next slide. But that's what I'm aiming for. And this is the problem that I'm solving. So we know what these numbers are when V has this special form. And here, there are no restrictions on what new can be. And there are no restrictions on what k is it. Or U or V or W. Okay, so brew hot order. So again, I think you guys all know this better than I do brew hot order. It's just you take a permutation permutation just the numbers one to 80 written in some order. Okay, run one to eight. Okay, and you switch a smaller number followed by a big number, you switch their order. Okay. Now you want to do it in such a way that of all the numbers in between them, none of them has value in between. So if I'm switching the two and the five, the three and the four can't be in between them while they're not. So that's a good switch. When I do it across k, I put a line here. So in this example, k equals four. In this example. So when I do it here, I put a line at k equals four. And so I'm switching something from the left of the line to the right of the line. The smaller number is over here and the bigger numbers over here. That's k brew hot order. So that's pretty simple. So I can do this switch and then I could do another switch. I could switch here the six and the seven and switch them to switch the six and the seven. And I can keep going and just keep switching things and take a chain. This is a chain. This is a chain in k brew hot order where k is four. Okay, so I told you one of the things. So Frank and Intel decorated these chains with numbers. So they kept track of this chain. Some information that's important is what's the big number? So that I'll call that a decoration on the chain. So every step on the chain, I write down the number that corresponds to the big number that I switched. So here too was the big number is the one that was past k. These are the decorations. Now from this, I'm going to derive more information, but it's all derived from the chain. So it's sort of information carried by the chain. So that I'm going to call these the D's and I'm going to tell you how to get an E. So you start with k. k is four in this example. k will be four pretty much for the whole talk, I think, mostly. So k is four. So I start with a four. Okay, these are these are my decorations up here. So you see five, seven, eight, two, four, seven, five. I just write down my decorations for my chain. And then from the decorations, I drive these E's. I so from five to seven, did I go up or down? I went up. So keep the four. Seven to eight, you go up, keep the four. Eight to two, you went down. So you need to decrement that four and it becomes a three. Okay, two to four, I went up. So I keep the three. Four to seven is up. I keep the three. Seven to five is down. So I decrement the three. That's it. That's how I get the E's. In general, nothing guarantees that the E's couldn't go to zero or negative. That's fine. I'm not interested in the ones that do not right now, but actually, Frank and Natel were, and they used it to define a really awesome quasi-symmetric function that we're also proving to sure positive with the same structures. But never mind that. What's the weight? I kind of don't really care about the weight, but just for fun, it's the multiplicity to get down here among the E's. There are zero, one. So the first position, zero. There's one, two. There are three, threes and three, four. So the weight for this particular thing is zero, one, three, three. Okay. So I have told you what K Bruhaut order is, what a chain is. I've told you what a weight is. So the last thing I need to do is tell you the lattice condition, but actually Littlewood and Richardson told us that. It's the same thing. So you can go with this. The computer likes this one, but kids like the following one. So you take these numbers, okay? And you're going to keep the pairs. So you're not going to split up the D's and the E's, okay? And you're going to sort the top row. What does that mean? That means the two comes to the front, the four comes to the front. And then when you're moving the seven, the seven is going to go right here. It's not going to go in front of the other seven. You don't want to over sort, right? Okay. And then you're going to get this sequence. The E's are inherited. And now from the right, you look and you always want to have, as you read the numbers, you always want to have seen at least as many fours as three, at least as many threes as two, at least as many twos as one. And here we see a four, three, but the same number, four, we're ahead on the fours. Four, three, three. So we always saw at least as many fours as threes. In particular, the number of fours is at least as great as the number of threes. Same thing for the threes and twos. Here's a three. Here's a two, bunch of three. And the twos and the ones, here's a two. There are no ones. So we win. So that guarantees that the weight will be a K partition, and that's the rule. Okay. So I conjectured this back in 2021, but Nantel and I have now proven it. So this is what these numbers are. So that's the rule. So if there's no other takeaway, and if you're really short on time, that's it here. You've got everything I have to say. I'm going to tell you another way to think of this, which is, so this is the way that maybe is the easiest to explain it. This is the way that the computer, well, this one in particular is the way the computer likes to do it. But I'm going to tell you a graphical way that I like to think of it next. How do I think of this? Well, remember, my gold standard is the Littlewood Richardson rule. That has to do with skew tableau. Technically, those are saturated chains, but we don't think of them that way, and we don't draw them that way. We draw them with boxes. But really, we should have drawn them with bubbles. There's a reason for the bubble. So here's a bubble diagram. So this is a Roth diagram and an inversion diagram. What you do is like here in when you have a six. So I've written a permutation up this way. So U3 is six. So that means in row three, I put a dot in column six. And then what I think of doing is I tie all the whole plane with bubbles, and then I pop all the bubbles in the row and column of every one of these bullets that I get. I just pop all the bubbles, and then whoever survives, that's my diagram. The survivors, the bubbles that didn't get popped. That's it. That's a familiar thing. Okay, so I'm going to do something with that. I've written a definition on a slide. The key takeaways for this are I'm going to move bubbles down, I'm going to add a box, and I'm going to move bubbles left. Okay, that's what I'm doing in the classical case of little enriching rule. All they did was add a box. But I have to get a little more excited. I have to move things down and left. How do I do that? Well, I'm going to go with that example. The same chain that I had before switching the two and the five. Okay, I start with the Roth diagram, which is literally this one up here. Okay, and I look at the dots that I'm switching. So the columns, the dots correspond to the columns that I'm switching. So here is the two dot in column two and the three dot in column, I'm sorry, five dot in column five. I look and say, is there a bubble in the top row, in the row of the top dot? Yes. Okay, then I'm going to move that bubble down to the row of the bottom dot, which I've done here. It's highlighted red. And I always add a box at the bottom dot. And in the box, I put my E. Okay, that we didn't use this one. It didn't apply, but it will in a second. Notice this is the Roth diagram for the shape I get by switching the five and the two. That's one way I can think of chains. Here, remember the next thing I switch was the six and the seven. Let's see that six and the seven. It's this dot and this dot column six and column seven. Now we don't have this case, but we do have this case. There's a bubble here, sort of, I think about the bubble, let's see, I'll draw it here, as in the rectangle between these dots and sort of the convex hole of those dots, there's a bubble down there in between. Okay, so the bubbles in the top row are going to move down, but the bubbles here, we need to move them left. And the rule for moving them left generically is ignore all the columns that have a bullet below and just move it to the next column to the left. So that will maybe come up, but I think this example might be too small to see the intricacy there, but it's not so important. The idea is that bubble moves left, and we add our new box. And here we are, here's our new box. There's a difference between the bubbles and the boxes. Okay, that's why I'm using them differently. Okay, and let's keep going. So we can also switch the seven and the eight. And here there's just no excitement. And in fact, this is what always happens in the Littlewood Richardson rule case, the classical case, is there's never any excitement. So is it easy stuff? So we'll switch the army out of the box. And now what happens with our next one with the one and the two, notice two is smaller than eight. I've gone to the left with my column. So I'm going to put a smaller number in my box, but all these guys are going to move left, including this, this box, right, the box is now a bubble for all moves that come after. So I'm going to move that left. I added the three, keep going. Moved a lot of stuff left here is another move something down rule because this bubbles in the row of the top dots that move down. The last step was the four to the five. I shifted over. And this is what I'm going to call a lattice permutation tableau. This carries with it all the information I need of the chain. Why? Well, to any Roth diagram, you can associate a canonical reduced word, which is pretty easy. I guess I can do it for this guy. This is in row six. So I'm going to do this six, seven. This is in row five. This is in row three, four, five, two. So if I read these numbers, like a book, that's a reduced word for you. If I do it over here, that'll give me a reduced word for W, the shape that I have. If I read only the numbers that are in the bubbles, I get a reduced word for you. That's a subword property of Brewhart order. So for example, there's another chain with the same you and the same W and the same lattice weight. So this is the other lattice chain because this Littlewood Richardson coefficient is two. So this is the other one. And notice that the configuration of where the bubbles are is different because they're two different subwords of the canonical reduced word for this guy that give me a reduced word for you. So this is really about subwords. So it doesn't matter that the boxes don't line up just so, okay? It still counts. So this is two. The Littlewood Richardson number here is two. So that's the rule. That's how I think about it. And I'm going to point out what I'm going to talk about next is I'm going to talk about down and left, which is how we prove the rule, but that's a little bit backwards from how I came up with the rule. I came up with the rule because of Ponert's rule that moves things down and my way of decomposing subwords into keys which moves things left. So these are the two main ideas that why I think of the rule this way is because things move down and left. Okay. So that's the rule. That's the conjecture theorem now. And that's the first part of my talk. So now I'll talk about super polynomial so that maybe the problem I'm trying to solve is well defined. Okay. Conert's rule. I love this one. I don't know why people don't. Conert's rule is pretty simple. You start with any diagram you like. This happens to be the rock diagram of a permutation but whatever. You start with anything you like and you pick a row. Okay. So here I'm picking this row and I take the last bubble. So I could have picked this one or I could have picked this one but you take the last bubble in any row you like and you push down. That's Conert's rule. It's really simple and you just keep doing. Kids love this one especially if you have like flattened marbles on a checkerboard here. What do you do? If you want to push this bubble down you just jump over a bubble if you have to. You don't go below the x-axis but you can jump bubbles. That's fine and you just keep pushing it down and I'll just keep pushing down. I push down any bubbles I like as long as I'm pushing down the end of a row I can push a bubble down. Here I'll jump it down. That's enough. So the set if I start with a diagram the set of everything I can get by doing Conert moves is the set of Conert diagrams for D. So in this case this is D and this is T. T is a Conert diagram for D. Okay. Conert proved many things. One is that this generates Shure Polynomials. He actually did it more generally. He did it for Dimeasure characters but for Shure Polynomials you need to associate a monomial to each diagram. That's pretty easy. Here it's x1 cubed because there are three bubbles in the first row. It's x3 because there's one bubble in the third row and there's no x5 because there are no bubbles in the fifth row. So the number of bubbles per row gives you the monomial. Add up all the monomials you get by starting with a young diagram which is a diagram of a K-Groth monion permutation and you get a Shure Polynomial. And actually the reason I like this is because there's a super simple bijection with Tableau. Again I promise you I would reverse everything. I'm using reverse Tableau. Again, there are reasons and one of the main reasons is because this makes this bijection really simple. I will talk more about this bijection in the second half of my talk. But for now just know it exists and it's super simple and straightforward and that's why I like Conert's rule. So Conert's rule gives you Shure Polynomials. Now it seems to be debate about this. So maybe we should be debating peer review and how we should do that because there are four published papers that prove Conert's rule. So personally I tried to read these two and I could not. I couldn't get through them. I didn't necessarily find anything wrong with them except that I didn't understand it but maybe that was something wrong with me. I don't know. So I wrote my own proof of Conert's rule. And then because I so love Conert's rule I worked with Sam Arman, Grant Boling, and Henry Erhard who were a fantastic graduate students that I have. And we actually proved Conert's rule holds for any Flag Shure module of the Southwest shape. So that was proven at Conjecture Dominic Thurl's and I had that Conert polynomials in general are characters of modules and we found the modules and Conert's rule works in a really beautiful way with MagYar's recurrence. It sort of falls out very nicely. So if you don't like these two and you don't like the one I'm going to show you in a second then you can go look in this paper too. And maybe this is actually my favorite proof but I'm going to give you a different one because it's easy to give on one slide. So if you don't mind Conert's rule then you can skip what I say next. If you if you don't like Conert's rule maybe you like Pipe Dream and if you don't know what a Pipe Dream is you can also skip it. It's a Pipe Dream. So you write a permutation up the y-axis like I've done here one five two eight six nine three four seven. You tile the plane with crosses and elbows one per box. So you tile the grid with either cross or an elbow. It's a Pipe Dream that when you follow the pipe that starts in row five it ends up going column five. Okay and the pipe that starts in row three or sorry starts with the three goes to the three. Okay we've written the identity here and it's reduced if the pipe don't cross more than one. Okay that's what a reduced Pipe Dream is. So this is a really nice model for Schubert polynomials. I'm going to give you the bijection between this and Conert diagrams. Okay so first get rid of the clutter. Okay so we we don't we don't need those elbows. The crosses carry all the information of the polynomial and you know that anything that's not a Christ crosses an elbow so you might as well just give it an elbow. Okay and then I actually want to do this with reduced compatible sequences so the pipe dreams are kind of wrong so I'm just going to shift everything diagonally. So there's no shift here and here I just you notice that I just shifted one there are three here and there were only two here so I just shifted everything one column to the right and then two columns to the right three columns to the right and so on. Okay so I have a bunch of crosses I'm going to turn the crosses into bubbles right crosses for pipe dreams bubbles for Conert's rule and I'm going to do it in a way that moves things left by process that I call rectification. I'll justify this name of rectification later but it's the first case is really simple but here you look above the bubble the cross you want to turn into a bubble and you look in the column to its right you look also above and you try to pair bubbles that are in those two columns and the rule is that if you have a bubble down here you pair it with something weekly above it that's a good pairing that's what we're looking for but here there's nothing to pair there's nothing up here so I don't do anything I just turn that cross into a bubble and go to the next one and then here there's nothing above it right there's nothing up here so okay I just keep going here oh this is exciting we have a bubble so this is a bubble that's in the column here but it doesn't pair to anything so if I'm in the second column and I don't pair to anything then I move left that's what rectification does from the second column and I don't pair I move left this is literally a crystal on the transpose diagram it's a crystal action on the transpose diagram so everything that crystal crystal moves has all the properties they have this also has okay and then this one there's there's nothing above here there's nothing above so we'll go to the next one the next one's exciting um oh sorry this one's boring um the next one's exciting this one there's just one thing to move so it moves now this one's cool because these guys pair they don't move but this top bubble that's red it doesn't have anything to pair with so it's gonna have to move left and then when I get here these two bubbles they pair and it would pair even if the this upper one was a little higher it would still pair you pair up and left and so that that red bubble moves left and then here this bubble to move left nothing happens there are three bubbles that move left there are three bubbles that move left and if you were paying a whole lot of attention or if you want to cheat and look back at the slides this is the corner diagram that we got before for the Roth diagram for this permutation that's the bijection so when I say it's simple I'm not lying the proof has some limits it's eight pages I welcome you to look at that um but from here on out I assume conor conor's rule holds for Schuber polynomial um so well we're good with that and of course the weight I guess the weight here the the weight for pipe drains is is the rows of the crosses and you can see that I haven't changed the row of anything I was on the plane with the column so that works and you can see this move left thing is important so back to back to the point the gold standard for the proof of the Littlewood Richardson rule what is that it's rsk in my mind I mean there are many many nice proof um but I mean rsk is particularly elegant um you take a tableau and another tableau you insert this one into that one you get an insertion tableau and a recording tableau and the recording tableau gives you your multiplicity it's just beautiful and all the properties it has it's just awesome so that's what nintel and I did we took conor's rule for Schuber polynomial and we look at the conor diagram and we take tableau because you know whatever tableau of k gross monion shape new here's my k we insert this tableau into this diagram we get an insertion diagram which is a conor diagram for w and we record it by the lattice permutation tableau by a saturated chain and taking generating functions that's how we prove the rule so this is a good point to stop pause for questions and this is a good point to take that promise five minute break um so if anyone has any questions but I will explain in the next half of the talk I'll get into the nitty gritty of this why I put numbers in the cells the colors are just there because I don't like looking at number three I'd rather look at the color green so if you like colors better than numbers that's what the colors are for they really don't mean anything but the numbers mean something I'll stop there see if anybody has any questions yeah thanks very much for a very nice first half and yeah time for questions if there any quick ones