 Okay, welcome back to the second half of Lisbon, which of each talk, so please take it away, Lis. Thanks. Thanks for returning. Okay, switching gears seemingly completely for the second half. I'll tell you about Schubert polynomials and a particular model for their monomials, which will then be equipped with a Demeser crystal structure, hopefully by the end. So, maybe more familiar to this particular audience, the Schubert polynomials, and I'll focus on type a so I have in indexed by permutations in SN represent a basis for the comology of the flag variety, the complete flag variety in CN. And let's go and shoot some Berger presented a recursive definition using divided difference operators for Schubert polynomials, but that's not the perspective that I will take here. The monomials of Schubert are generated by many different combinatorial objects by this point so here's a subset of the history of those. So the RC graphs of Billy Jokushin Stanley, the planar histories, or pseudo line diagrams of Fomin and Kirolov, and the reduced pipe dreams of Knudsen and Miller. So, these models are combinatorially equivalent to one another so they sort of come naturally in a cluster, but I will focus on the just to pick something the specificity of the reduced pipe dream model to present the crystal structure to you. So here's an example where here's how we build a pipe dream. So first of all you have to fix and, and then you draw kind of part of the end by end grid just this upper sort of triangular portion. And for each of these little locations in that upper triangular portion I cover with one of two tiles either across or an elbow oriented in exactly these directions. And so here I've started building a pipe dream by covering with four crosses, and that then determines the rest of the diagram, because I fill everything else with that one remaining tile, which is elbows, following these particular and you'll easily see by how wide I've tried to draw everything that all of these bits of the figure connect up well, they actually connect uniquely, and that you can then track the pathway of a pipe where we'll call the direction of flow sort of like this way. And flowing from spot three out column five from row three out column five. And the results of this building such a diagram is called a pipe dream. So, if further each pipe crosses at most one time. So for example we see that this pipe is only crossing this pipe once at that spot. The pipe dream is called reduced. Okay, so don't cross pipes extra than necessary. You can also read off a permutation corresponding to a pipe dream just by watching. So where, where you enter and where you exit, and then recording. So, let's see the three. Whoops, that's not the right thing to highlight three goes to five is actually this location that that location, they happened anyway. Sorry, so watch where this number ends up. And that's what you record in the window of the permutation. Okay, so I think this is a repeat. So, as now, what I can do is fix the permutation and then ask for all possible reduced pipe dreams that have that corresponding permutation. And I have a notation. Well, this is the common notation or P reduced pipe dream, a corresponding to w is the set of all such. And here's a fact from Knudsen and Miller that the length of the permutation, or if you'd prefer to think about the length is preferred to think about the number of inversions so the number of pairs I less than j that actually get such that the corresponding spot in the window is put out of order is. So that number is consistent with the number of crosses that you have in the pipe dream. This is in the reduced case. So let's just noticing that in this example that I have four crosses. I built that example in the first place. And then if I look at the window and each of the entries in the window and I look at all the possible pairs I less than j. I see that one and two are out of order, and three and four out of order, three and five are out of order and four and five are out of order so I also have four inversions. Okay, so that's the, how the way that this fact plays out in this example. Okay. I can also calculate a weight for each pipe dream so I just look at the number of crosses in row I and record it as a vector. So in this particular pipe dream D. I note that there are three crosses in row one and one cross in row two, and then no crosses in the rest of the rows. Okay. So here's a theorem due to Billy Joker Stanley initially with a slick proof by foaming and Stanley a bit later. And this is the version of that result that appears in Knudsen Miller's paper on pipe dreams says that you can write down a sugar polynomial, just by looking at all the reduced pipe dreams for that given permutation. And checking out their weights, and then recording a monomial with that weight as a sex pointed vector. So in our example, this example I've been following through now we've calculated that weight 310000 this so far I just know one monomial in the corresponding sugar polynomial, namely x one cubed x to to the first. Okay. So 13 other monomials, not all distinct. And we'll, we'll see them shortly constructing them in a different way. All right, so that's the story of Schubert polynomials. And now, what I'd like to do is blend these two stories together so the first entire half of the talk was focused on crystals and further demos or crystals as a special family of crystal structures. And to place a crystal structure on pipe dreams. Okay, so much like the Tableau case, which I showed an example of in detail, we will start with a pairing process. Okay. And I also have to decide what row I'm focusing on. So, look at row I. Let's start with the right most cross and row I. Okay, so I'll follow the same little pipe dream example that we've been using over and over, and I've highlighted here the right most crossing row one so I'm fixing I equals one, why not. Okay, so that cross is going to try to be paired. It's going to try to be paired with something below and to the end to the right so something southeast of it. It has to be paired with another cross. So in this case this cross has nothing, no crosses southeast, none available and so it will just remain unpaired. Okay, then you proceed to the next right most cross in row I so in row one the next right most crosses here at this point we do see a cross southeast in row two of that of that cross that next cross upstairs. And so if there would be more of them we pick the closest to it, but but in this case there's only one to pick. So these two get paired. And then we continue. There's further one more cross in row one. But I have already used all of the crosses in row two at this point. A cross in row two can't be paired with more than one cross in row one so once something is paired off it's just paired off. So I have yet another unpaired cross remaining. And you just repeat these steps as many times until you make decisions about everything in row I being paired or not. Okay, so the summary is this picture greens are paired and reds are unpaired. Okay. Now we are moving towards defining a family of operators. These operators will eventually respect the pairing process, but these operators also have their own life and had their own life long before my collaborators and I ever started to look at them. And they were defined by Bernie and Billy in the early 90s. And they called them shoot moves and they were working on RC graphs but here's the version of a shoot move and how it looks on a pipe dream a piece of a pipe dream, which is combinatorially equivalent. So, so just kind of in spirit, what a shoot move does on a pipe dream this is just a local picture of a piece of a pipe dream. So, we'll have a cross that's kind of in the upper right corner of a rectangle. And within that rectangle, all the other corners are elbows. And in the interior of that rectangle, everything is crosses. So that configuration, which is summarized in purple here is the the scenario in which you can perform a shoot move in the spirit of Bertrand Billy. And the shoot move will just take this cross and it will jump it into the lower the opposite corner kind of pops across the diagonal down here and you replace what it came from with an elbow. Okay, so this is the spirit of a shoot move general. And this is so this is just Leo what kind of I had in my one thing that I had in my mind as we were talking during the break. So this is a theorem of Bertrand Billy that you can actually generate the entire Schubert polynomial from one reduced pipe dream, they call it D top or one reduce, you know one RC graph they call it D top by applying all possible shoot moves. So, all of the menomials of a of a Schubert are connected by these operators these shoot move operators by a theorem of Bertrand Billy. And that gives you a lot of good combinatorial information about the shoot the Schubert polynomial, but it doesn't give you the combinatorial information that that captures the underlying representation theory perfectly. Okay. But nevertheless, the crystal shoot moves as their name suggests are built on the backs of the shoot moves. And that you do shoot moves with respect to the pairing process. Okay. So, just a reminder up top of the general shape of a shoot move, but now I'm going to explain the crystal shoot moves that we've defined. And it will require checking some further information that comes out of the pairing process. So let's start applying a crystal shoot move. First of all, you have to fix what, what I you're actually looking at, as always in a crystal operator. So fix what I you're looking at and then you first run the pairing process on that row I. So here's the outcome of the pairing process on the same little pipe dream that I've been abusing. So here I equals one, again, I've run the row one pairing process. If all the crosses in row I, at this point are paired, you define that the operator's value is zero on that pipe dream. This is not our scenario in this example, I mean that shows an example such that you'd be able to do something but it does happen that you might have everything be paired after the pairing process. So we have two unpaired upstairs so we get to proceed. So in this case, we're going to take now the left most unpaired cross in row I so that's this cross up here, and we do a shoot move. But in this case, we can do a shoot move because the scenario of the shoot move just doesn't exist we have no space over here. So actually, we do define F1 to be zero on this example, but for a different reason. So there could be two sources of why a lowering operator could be zero. Okay, so here's a one that won't be zero. So here's now a different, just a different pipe dream and I've done a row two pairing, and I have here an unpaired. So that's my left most unpaired and read. And I do have the opportunity across this little rectangle of crosses to jump and do a shoot move. And so I shall. Here's the outcome of applying that lowering operator. Okay, so the shoot moves are still the underlying operators it's just that they're restricted now with respect to a pairing process. And that thus they could be zero. A lot of times, you know, that billion version on would maybe not have a zero operator. By the way, it's really by the construction across this moving from row I into row I plus one that this weight condition is satisfied that you're just subtracting an E1, sorry, I and adding an EI plus one. Okay. So let's see. Yeah. Also now, we have to go backwards, if we're going to have a crystal operator. And that's hopefully intuitive by the visual, but I actually want to say a little bit about the detail. So if you're, you're preparing to do a raising operator so now I'm reversing the direction and across will move up to the row above it. Then you still do the pairing, you do the pairing kind of with respect to the row where you're planning to move. Do that pairing process, but look in the row below it for an unpaired cross. I should probably say, I'll write most unpaired cross, maybe I'll say that in a second. And then if you have such an unpaired cross in the row below in row I plus one, then you reverse the shoot move. So now this cross has jumped back. Yeah, it's clear at least by the visual that these, these operators are the reverse of each other there's some, some details to work out about like, you know, well defined in this and when things can and cannot be applied, and being zero together and things like that. But here's the proposition that tells you exactly the recipe for when you have a raising operator. So you have to have an unpaired cross in row I plus one after the running the pairing process on row I, and that's always going to guarantee you that you came from a lowering operator, and that therefore it can be reversed. So this is a little proposition that you have to believe along the way in order to define the inverse e eyes. Okay. So this is just a summary. Again, of all of the, the details for now just applying an AI. If you've got everything in row I plus one paired after doing the pair process, the pairing process and row I, that's the situation in which you declare that EI acts as zero and otherwise you can just do an inverse shoot move. And because you just did everything in reverse, the weights are moving as they should in order to obtain a crystal structure. Okay. And the EIs and the FIs are mutually inverse. Also, as they should be to obtain an actual on a crystal structure. So here's the theorem. In one, one way of stating the theorem. Again, this is joint work with Sarah Gold and new Johnson. This is so if you choose a permutation. So the operators that I've just defined for you on reduced pipe dreams, define an honest type a and minus one demos or crystal structure, and more precisely the set of all of those pipe dreams will decompose into a union of crystals and that union is indexed by all of those pipe dreams such that EI is zero all the time. And then there will be a permutation associated to such a pipe dream and you can pick out the weight of such a pipe dream. So, regarding this condition, that's indexing the demos or crystals. So if EI is access zero for all I this is a common language and representation theory that we call, we call such a D highest weight pipe dream, if I always access zero on it so you cannot raise it anymore. And so it's weight is like as high as possible. And so, so we have these highest weight pipe dreams which want to index the components of a collection of reduced pipe dreams. And I'll just remark that if you have a highest weight pipe dream, then this weight vector will actually be an honest partition so we're in the scenario of the special cases that we looked at before where you start with a partition. As your, your input to build the crystal and all you need to know how to do is figure out this permutation in order to truncate the operators according to that demos or permutation. Yeah, so this is all this permutation however is also uniquely determined by that highest weight pipe dream. We have an algorithm for for picking this out. I don't have time to share that algorithm in this particular talk but it's, we're finishing writing that section up now. It's kind of cute. Um, so at this point maybe also I should say explicitly that this crystal structure on the monomials of Schubert polynomials has been worked out before by I'm sure more than just these three people that are listing here, but in a very explicit way in terms of the model of reduced factorizations with cutoff so kind of looking at reduced words meeting certain criteria, and that was work of a safe and chilling. And also maybe 20 years ago Christian one are wrote a paper that included. I think in parts, it included this, the ideas behind this theorem or maybe the spirit of the operators, but the details and the explanations and proofs there were all carried out in the models by words so kind of more attack more of a monominal and using the co-plastic operators on by words to make any, any similar combinatorial claims and so yeah there is definitely related work using different combinatorics. So, then maybe the way that I'll leave you is to actually finish the example that I've been picking on this permutation, which swaps one and two and also swaps three and five. So here is a picture of all of the 14 reduced pipe dreams for that particular permutation. In the arrows. Yeah, in the arrows. We've indicated the results of applying the f eyes so our arrows in this picture just happened to be directed downwards. And we've color coded so once f ones are blue arrows and f threes are green arrows and f twos are red arrows. So we'll get into this picture a little more. So here's the Schubert polynomial just recorded all of the all of the weight vectors and then put those weights as exponent vectors of monomials and here's the Schubert for this particular permutation. Here's how it breaks up in terms of the crystal structure. So there are three highest weight pipe dreams here here and here. You'll note each of those is a permutation sorry partition shape weight, although that doesn't characterize highest weight pipe dreams just to say. And so we have like three different chunks to this graph. And there are three different crystal structures with three different truncating permutations. So for example, the truncating permutation once you see the graph like this you can read off the truncating permutation. You read sort of backwards to see okay what what things what operators am I allowed to fly first. And what operators am I allowed to fly second and so forth. And then yeah third is all of the reds. So the key polynomial can then be obtained by reading off that permutation and applying it to the highest weight vector the weight of that highest weight reduce pipe dream and recording the composition so here's just like the sum of this particular Schubert polynomial as the sum of three key polynomials and those three key polynomials are characters of three different demos or crystals. And yeah so as a corollary of this exact same demos or crystal structure, you can write any Schubert as the sum of keys, and here the key, the key polynomial are indexed by that permutation, which is obtained by knowing that pair of data, the weight the highest weight vector of your highest weight pipe dream and the truncating permutation. And I think. Oh right. Yeah, so I think I'll just end here by adding that. I've been a lot of work, I don't know, decades ago, 80s 90s, I think, by people like less goon shoots and purge themselves and maybe in the 90s by writer and Shimizzono various ways to take a Schubert and pick it apart as a sum of keys. And I guess it must be the same underline common torques but it's definitely not there sort of phrased as a crystal structure. And I guess some, some sort of like other kinds of common tutorial models for doing this decomposition have also been around for quite a while. And yeah, I think I'll stop there. Right. Well, thank you very much for the very nice talk.