 All right, so welcome, everyone, to the Schubert seminar. Today, we're very happy to have Liz Milicevic from Harvard Fort telling us about crystal shoot moves on pipe dreams. So please take it away, Liz. Thanks. Thanks to the organizers so much for the invitation to speak. I'm a regular attendee of this seminar and have, if I find it to be one of the few lovely things that came out of the COVID-19 pandemic, and I'm happy that it's still going. So I'm glad to be able to make my own small contribution to what is a rich and lively seminar. So when I get to new results in the second half, everything I say there will be joint work with Sarah Gold who were Haverford graduates of class of 2020, whatever, this past spring, 2023. So just recently, graduated. All right. So I will start with what I hope is a friendly and somewhat helpful introduction to crystals. So I think that that is not necessarily a common piece of background for everyone who regularly attends the seminar. So that's the perspective that I'm adopting. So I'll first start with a brief and extremely incomplete history of crystals focusing on the origins and the kind of work that was happening around the beginnings of the formalization of the notion of a crystal. But crystals in broad strokes are combinatorial structures which are defined to encode representations of Lee groups and Lee algebras in particular. And so work starting around the 1990s was happening, I would say, strong origins are due to Kashiwara, who was working on representations of quantum groups, really around 1990. And then at the same time, but independently, Lustig was doing work on his canonical bases, which turned out to be related to crystals. A bit around the same time, maybe a bit earlier, work of Drinfeld and separately work of Drimbo on quantized enveloping algebras was coming out. And it was not being framed in the language of crystals, but was, again, underpinning the origins of that theory. And then by the time we get to the mid-90s, people like Littleman were using path models to formalize standard monomial theory and the likes of Lakshmi, Bayan, Sashadri and putting crystal structures on paths. And yeah, I'll have a picture later. So I'll maybe say a little bit more about what I mean about that model later. And then just to get the word Demazur in, so thinking about building modules via line bottles on flag varieties. This is the perspective of folks like Demazur and that representation theory is also encoded in a similar way through the language of crystals. If you've seen the definition of a crystal, then perhaps your impression is much like your impression of this slide is that there's a set that satisfies a bunch of axioms. I will give an example, which is very concrete, where I tried to sort of highlight the flavor of this typical flavor of each of the pieces or the components of a crystal. But at the bottom, at the heart, you have a set. That set comes equipped with some kind of weight function. And I will make that concrete. So each element in this set is typically represented as a vertex in a graph. And you can evaluate a weight function on those vertices, which will spit out a vector, an integer-valued vector. And the real heart of crystal theory is a pair of operators, EI and Fi. EI are usually the raising operators and Fi, the lowering, but they are mutually inverse. And so they kind of come as a pair. And these operators have a predictable action on the weights. So if you calculate the weight of a vertex and then you raise, then this will add a bit of weight. And it adds the vector alpha i, where alpha i here is the difference of the two standard basis vectors, EI and EI plus 1. So that definition doesn't necessarily help you hang any meat on the bones of it. So let me give you an example, an example that uses some of the combinatorics that many of us use in Schubert various aspects of Schubert theory. So fix a partition lambda. So a partition will have here, for this context, n minus 1 many parts. By the way, that n minus 1 matches the I didn't say it explicitly, but I wrote it that I'm talking here about type a n minus 1 crystals, or n minus 1 crystal structures. There are crystals in all the various types, but I'll focus on type a in this talk. So OK, so we have a partition and it has a specified number of parts. So here's an example of a vertex set for a crystal. So semi-standard young tableau of shape given by lambda. And we're filling those semi-standard young tableau with the alphabet 1 up to n. And n is linked to the length of that partition. So here's an example. If lambda is 321, so I've represented that lambda in a young diagram as the number of boxes goes, well, yeah, I guess I'm using English style. So three boxes, then two boxes, then one box flowing down. And this filling is such that I have to strictly increase in columns, but I weakly increase across rows. So there's an example of a filling provided. So semi-standard young tableau of a fixed shape with a specified alphabet are vertices of a crystal. So now let me build the other aspects of the crystal for you. So we need a weight function. And in this example, the weight function just counts how many of each letter in the alphabet do you have. And so in this example here, this semi-standard young tableau, the way that I filled it, I used two ones. And I used two twos. And then I used one each of three and four. So that's the weight function in this case. All right, now the most complicated thing is to get your operators defined. And this usually comes in kind of a three-step process. So I'm sort of breaking down the definition of a lowering operator for you here in three parts. So one, I'll grab a reading word corresponding to a tableau. Second, I'll do some kind of pairing process. And third, I'll give a rule which allows me to switch one i to an i plus one and leaving all else the same. So still picking on the same example. My reading word, I'm going to read up columns. Sorry, up rows. There we go. And I'm going to read left to right within columns. Oh, sorry. Gosh, up rows. I'll read rows going up. And within a row, I'll read left to right. There we go. So I would get four, and then two, three, and then one, one, two, creating the following reading word. OK. Operators are defined for every single i between one and n minus one inclusive. So you have to fix an i. And then I can tell you what the operator is. So in this example, I will fix i minus i equals one because whatever. So for every i, I will convert the reading word. And i will be converted to this kind of parentheses. And an i plus one will be converted to this kind of parentheses. So once I do that conversion, the four and the three, they stay untouched. But the two, the twos here and here have become these parentheses. And the ones here and here have become these parentheses. OK. OK. And so then we look at those parentheses and we look at which ones have been paired and which ones are left unpaired. So I've drawn in this example, it's sort of clear by inspection that these two pair off together and that these other two are just hanging out unmatched. And the rule is that the right most unmatched one of these parentheses, so closed parentheses. So that would be this one here, the right most unmatched one of those closed parentheses will switch. So this one becomes that one. And I've left everything else in the reading word the same. And then I just read back the new reading word at the end. So all that's quote unquote, all that's happened in the end is that I had a one here, which became a two here. But OK. I think that the first time I saw this definition, I heard it from Anna Schilling. She did probably a similar example. And I felt like I could not absorb it. If that's also happening to you, I empathize. But yeah, I want to advocate for crystals also in the context of Schuber theory today. So if this example doesn't do that advocacy, hopefully something later will, we'll find out. OK. So I just want to finally note to sum up this particular example, the weight change or the effect on the weights given that particular low-wearing operator that I just did. So let's record this is the pre-F1 and the post-F1 semi-standard young tableau. Oh, by the way, there's a lot of things to believe here. Well, if my goodness, you get a semi-standard young tableau back in the end, blah, blah, blah, blah. There's a lot of things to check. I'm just claiming they all work. And so let's see, where's this one has become that two. And if I pull out the new weight, I have changed a single one to a two. And so whereas I had two, two, one, one before, I now have one, three, one, one before as a weight factor, which is exactly the required difference. Take what you have before and subtract off a difference of alpha one. So I'm just saying that this equation holds in this example. OK. If you do that exact thing to this fixed partition with an alphabet of one, two, and three, then you get exactly these eight semi-standard young tableau. So this example is the exact same crystal structure that I just gave you. It's just on a smaller partition so that we could actually easily draw out and write down all the semi-standard young tableau and their relationships. So I've drawn this here such that ones, like using lowering or raising one operators are all happening in the same direction of flow and same with twos in the other direction. But in any case, I get eight semi-standard young tableau altogether. And perhaps it's less mysterious to observe that if you record the weights of all of those eight semi-standard young tableau, so for example here, and you just make monomials with those corresponding exponent vectors as the weight vectors and you record what you have for all of these eight semi-standard young tableau, you actually build the Sherpa polynomial for that shape lambda 2, 1. So Sherpa polynomials. Oh, sorry. So how can it be that there are two lines from the ones in the middle, the lines away from them have the same color? Could you not apply the one to those in the middle, those that have twos going from them? I couldn't, actually. Yeah, so the way to see that, I guess that's not good. So the way to see that I couldn't apply, whichever one you were just asking about would be in this pairing process to see that, oops, everything was paired, and no, I couldn't. I don't have a remaining unmatched parenthesis of the kind that I would be flipping. Yeah, so it's kind of, I'm not claiming that it's obvious, even in this small example, how to work things out. But yes, technically, this one, this one, and this one are actually distinct. They're not like, they're not touching these ones. There's, yeah. I draw them in this way because, well, you'll see in a second why I like to draw them this way. But I do get that one in the middle twice. I get the weight. They're two different semi-standard young Tableau. The two's and the three's are in different places, but the weight vector is the same in each case, 1, 1, 1. And so I do get two copies of that guy in the middle. So they're kind of sitting on the same node, but they're not connected in the graph in the same way. So yeah. And just to remind ourselves, although we love sure polynomials in this seminar because of how they represent homology classes in the Grismanian, if you're looking at, yeah, if you're doing Schubert calculus, that's probably why you like them. They are characters as well. And thus, that's the way that sure polynomials are arising here in the context of crystals is that they're characters of irreducible representations of GLF. OK. So I have more stupid questions. Oh. I thought the E's and the F was supposed to be inverse to each other. So how does that work in the middle? Well, I can only say from here to here I could apply an F1 to get that way, or I could apply an E1 to get that way. I see. OK. Yeah, thanks. Good. Yeah. Yeah, sorry. I'm abbreviating on the edges. I'm abbreviating just with the subscript I because the E, the EI and the F are mutually inverse. So if you can apply one, you can always apply the other. Yeah, so it's enough to decorate it with I. Great. Great. OK, just in case for anyone who knows my usual tricks, in the audience who thinks that I've gone astray, just a couple of slides to explain to you how I came to crystals, which is this picture on the right. And that's via alcove walks. This is the same exact crystal as the one on the left. So these two crystals are exactly the same. They have the same vertices, the same edges, and the same labels on all of the edges. But one is packaged geometrically in terms of the model of alcove walks. This is a little bit different than the model of Littleman from the earlier slide, but sort of more similar in spirit. And for me, the reason that I started to love crystals a lot is because I learned, and this is a picture from a project from like 10 years ago, that they give me a way to construct an entire family of combinatorial objects starting just from one top element. So this is a picture of a construction that I needed. And somehow this gray walk that goes all the way up, quote, unquote, to the top is the top most element. And then I'm applying a bunch of crystal operators to get initially to this one and then eventually to everywhere, everywhere else here in this picture. So they give a nice, yeah, they give a nice construction for, you get a lot for one. Maybe that's the moral somehow. And yeah, OK. OK, so maybe that intro was less comfortable for this audience. But oh, actually, sorry. No, I have to tell you about Demiser crystals. And then we'll get into the other piece. OK, so Demiser crystals. OK, then we'll take a break. That'll be perfect, actually. So Demiser crystals now are actually what's going to appear in Schubert polynomials, so the pipe dreams that I'll tell you about in the second half. And the spirit of Demiser crystals for the purpose of this talk, I want you to understand them as truncations of the entire crystal where we're restricting the operators that we're allowed to apply in a prescribed way. So here's the, here's one key definition to construct in this way to construct a Demiser crystal. So if I have any subset of vertices in my crystal, given a fixed I, I can define a set of a bunch of other vertices in my crystal, which come about by taking something from x and lowering it some number of times. OK, so for example, if the green set that I circled over here is x, so the green set would just be the top vertex and this one additional vertex circled in green. And then I took i equals 2. I would look at all the other vertices that I can get by applying F2 as many times as I want. So this k can be any positive, non-negative, non-negative integer, and I just see where I get from there. So including that k equals 0 works, so I would still have these two original vertices in x. But I could also now apply this to and get to here, or I could apply this to and get to here, and then I could further apply, or I could further apply like 2 twice, F2 twice and get to there. So I have a subset of the vertices, which are all reachable by applying powers of F2 in this example. And now you could define a similar sort of sequence of operations where you start with any permutation pi in Sn. You write it as a product of simple transpositions in a reduced way. And then the demazur crystal associated to that same top element lambda, or that think partition lambda, and this new data pi of permutation is given by just starting with the highest weight element and then applying in sequence all of these demazur operators. So let's just compare. Let's look at an example and see which demazur crystal we actually built, how it's indexed in the previous example when we were just getting used to these curly d's and what they were. So in this example, ui, the highest element, is this top semi-standard young tableau with the highest weight vector. And then if I apply d1, curly d1, that's when I get this collection, which was my green circled bunch before. And then after that, I applied all possible operators to whatever I had before. And I have created the demazur crystal associated to this permutation S2S1 with the same partition, shape 2 comma 1. OK. There are also characters. So characters, they also provide. Demazur crystals also provide or yield characters. So the character of the demazur crystal indexed now by two parameters, so a pi and a lambda, is the key polynomial indexed by the composition, which would take that partition lambda and then put it out of order by the permutation pi. So here's the key polynomial corresponding to this demazur crystal that we just built. And I've just recorded, for the heck of it, the composition. I took a partition 2, 1, 0, and now I've made it out of order. And I get the composition 1, 0, 2. And I just record the monomials that I get along the way. And there are five of them because there are five vertices there. And they all happen to be distinct in this example. Yeah, doesn't have to be the case. OK. And so just as a remark to tie that to the big full crystal picture together to the demazur crystal, the sure polynomial is actually the character of a demazur crystal as well. It's just when you take pi to be the longest element, w0, which doesn't provide any restrictions on your operators in a way that will, yeah. It doesn't provide any restrictions that will affect your overall vertex set that you obtain. And so yeah, you can get the sure polynomials as a special case of this construction as well, where you sort of take the biggest possible permutation. OK. So maybe now this is actually a perfect time to take a break from the crystal theory, and then we'll do more Schubert stuff after the break. Sounds good. Thank you very much. So it's 4.56. Let's take five minutes break. So I'll see you at 5.01.