 And we will, yeah, our next speaker is Portia Anderson. I hope I'm saying that right. Who will tell us about it's warping symmetries of Knudsen-Tau puzzles with hexagonal and parallelogram shaped boundary. Thank you. Oh, and I guess so we should, yeah. Wait until you have your slides here. Yeah. Good. Okay. So you can see that. Yes. Please go ahead. Okay. Okay. It's actually Portia, but that's okay. Sorry. Yes, no problem. Okay. So this is going to be about this interplay between this combinatorial puzzle method of doing Schubert-Catholus and the geometric perspective. So Schubert-Catholus is about computing structure constants and the cohomology ring of the grass monion and true to the origins of this area of math. One way is to look at transverse intersections, triple intersections and count the points. So that's one way. And then we also have puzzles, which compute the same constants. And so sometimes an observation or symmetry that's clear on one side, like the puzzle side can show us something we didn't expect on the geometry side and vice versa. And so we'll see some good examples of that with this project I've been working on. Okay. So here is our space that we'll be working in, of course, is the grass monion of T-planes in Cn. And so that is a smooth complex manifold and projective variety. And we have these Schubert varieties. We have these Schubert varieties and we index them, especially for puzzles with binary strings with K ones and N minus K zeros. And we can think of those interchangeably with partitions that fit in a K by N minus K rectangle. So, and then have opposite Schubert varieties which give us a dual basis. Oh, sorry. I didn't even say, okay. Schubert varieties give us an integral basis for the cohomology of the grass monion. And then the opposite Schubert varieties give us a dual basis for that. And we have with this perfect pairing, which is taking the integral of product of two of them, which I will say what that integral is talking about more soon. So, okay, let's define a puzzle. A, okay. So let's let this triangle lambda mu nu denote this boundary shown here on the right. So, we take this equilateral triangle and then we label the edges with strings of ones and zeros. And they go, they're right in the orientation shown. And, okay. So then the first one is going to be the first one. And then the second one is going to be the first one. Okay. So then those are lambda mu nu. And then a puzzle is a filling of this boundary with pieces in this allowed set here. And the way you fill them in is just as you would expect like a jigsaw puzzle where you have to align the any touching edges of the pieces such that they have the same number. Okay. And then at the bottom here is just some random examples of what these things look like. Okay. So then we have this theorem by my advisor, Alan Knudsen and Terry Tao. And it says that puzzles do Schuber calculus basically. So puzzles compute the structure constants for the homology of the cross-monion in the Schuber variety basis. So what that means is for these product expansions of two Schuber varieties, you expand that out in the Schuber variety basis and the coefficients you get are these constants. And so the C lambda mu nu equals the number of C lambda, or sorry, of lambda mu nu puzzles. And this, and then here is the like geometry version of this. And if that's unclear what this integral is, which it was to me for a long time, the integral, you can think of it as pushed forward to a point. And then how we turn this into like an intersection theory kind of problem is you can, if your varieties intersect transversely, the product of their classes is the class of their intersection. And so we can like squish them into one class with their intersection in there. And we wanna turn it into a triple transverse intersection. We have to perturb at least one of these with something in GLN to make it transverse. And then the integral counts the intersection points if it's a finite intersection and it returns zero otherwise. Okay, and then here's just an example of puzzles computing some structure constants. So what if we want to expand out the product X0101 squared. So then we put those 0101 strings or lambda and mu on the sides of a boundary. And then we fill in many puzzles that we can find. And then what ends up coming out is just two. And then we have one with the new equals 011 0 and then new equals 1 0 0 1. So then the C lambda mu news for those news is one and then everything else is zero. So then that tells us that this expands out to this sum here. All right, now we're moving on to think about some puzzles with different boundary shapes which is pretty unorthodox I guess because normally they're always triangles. So we're gonna do parallelograms and the next slide we're going to talk about how to take this and complete it to a triangle so we can think about it as a normal Schubert calculus problem. Okay, so let's take a parallelogram boundary shape and we're gonna label it with strings of zeros and ones going clockwise. And we demand that the two opposite labels have to have the same content which means the same number of zeros and ones and same for alpha and beta. And that's what this sort notation is saying. So sort means take all the zeros in the string and put them ahead of all the ones. So this is just saying alpha and beta have the same content and same with lambda and mu. And then what we do is we can complete this thing to a triangle in a trivial way. So we glue on these gray triangles where, so with this, these labels turns out that if you have a string of zeros and ones and then you have another label on the other side say beta in this case then that forces this other edge to match beta. So gluing these on just forces these to be the same and it preserves this inner parallelogram so we can just glue those on and chop them off as much as we want. And so that gives a bijection between these puzzles and ones like these. So, and then we have a geometric way of thinking of this as a ordinary Shubert-Taco's problem. Okay, and then the next thing to note is that if you rotate this parallelogram puzzle 180 degrees then you get another one of the same sort. So that says that this rotation gives us a bijection from parallelogram puzzles with labels going around this way and just the rotated version. So that is silly from the puzzle perspective but we wanna ask about the geometric meaning. So let's put those two bijections we just saw together. So we have this bijection with gluing on these triangles so we follow that up. Here we can rotate the parallelogram 180 degrees and then we come back down and complete that again. So now we have this triangle here with alpha and beta on there and then here they're swapped and same with lambda and mu. And so this is making this equality is claiming something about something geometric. And so the first question I wanted to answer was why is that? So how can we understand this geometrically or prove it? So when we did that, we ended up finding a stronger symmetry actually which is a combinatorial statement which we don't understand combinatorially. So that says that we can just swap two of the opposite edge labels instead of both pairs at the same time which would be the rotation. So if we just swap alpha and beta then we get the same number of puzzles still. And then this picture kind of is suggesting how this was done which was that we had to sneak through this triangle version in order to prove that equality and we don't know how to just directly combinatorially prove this. And then by the way, this also is true for if we allow some of these different special pieces which compute different homology theories. Okay, so we can also swap lambda and mu. So here's just a glimpse of a cute little puzzle-based way to do this which is where you commute sides of the triangles because puzzles have commutativity because the homology classes commute. So you commute these sides triangle, it turns into this and then you can fix this edge down here and then again you can commute the sides of this triangle so you get the same number and then you just follow it around. That's, yeah, so that's just a brief glimpse of that and it doesn't work for equivariant because that piece has no rotations. So, okay, and then I'm just gonna probably place through this really fast. So it turns out that this yellow bit, this yellow parallelogram can only have at most one filling and then it does have a filling. If and only if these fixed labels down here, eta and theta fulfill a certain condition in relation to each other. So we sum and that is, you know, the theta choice has to be unique to eta so we call them eta and eta prime and we sum over choices of eta and then it turns out this number of these is sum of products of Littlewood Richardson numbers where one of them is the number of puzzles in this green triangle and the other is the number of puzzles in this. So it's kind of nice that we can write as a product of Littlewood Richardson numbers and then here's the just a glance of the geometric version of that. Okay, and then we have this stronger thing coming up where, so we have these puzzles as I mentioned that with this special piece that computes the structure of constants and the equivariant homology, T equivariant homology of the cross monion. And so those constants live in this polynomial ring in n variables over z and it's just, it's a richer version of ordinary homology. So if you prove something, we prove this here then it's also proved with the ordinary. And these compute these polynomial structure constants while each of these special pieces contributes a weight, yj minus yi, which is given by where it's located in the puzzle. And then to get the weight of the whole puzzle you take the product of all the weights of the pieces and then the structure constant is the sum of all the weights of all the puzzles that you can get with those edge labels. Okay, and then here is this result about swapping edges of equivariant parallelograms. So if we take the same exact setup and now we're allowing the special equivariant on this pieces, then what happens to the structure constant if we swap alpha and beta. So it turns out that... I've minutes left. Okay, it turns out that what that does is you, it's a permutation of the variables in the structure constant. And so for, let's look at the alpha and beta case. So if you are swapping alpha and beta you look at, here's this permutation that will act on the structure constant. So what it does is reverses the first A variables, the first Y1 through YA and leaves the others alone. And then, yeah, so that is this here is saying that if you swap alpha and beta you effect on the structure constant is just you're reversing Y1 through YA and then leaving the rest of them alone. And then swapping lambda and U is something very similar but you're reversing the last, quarter of the last variables. Okay, and then this is just a very brief flash of what the gene of the group is like where it's you pull the problem back to this product of smaller gross monions. So, and then we also have this take it a little bit further and we find that the actually the number of these equivalent parallelogram puzzles is the same as when you swap. And that is not automatic. So when these equivalent puzzles sometimes you can have even the same structure constant but you will have a different number of puzzles say for a commuted version and then they just add up to the same polynomial. So this is not a given. So the proof of that is simple though. And anyway, so this bijection is definitely not understood combinatorially at this point. And then this is just a brief flash of like an example. So this is some of these puzzles, the set of all of them and then you swap lambda and then you get these. So still glimpse and finally hexagons. So we can ask, we can generalize some of this stuff to questions about hexagons because a parallelogram is a degenerate hexagon for one thing. So we can do this thing where we complete that to a triangle in a very similar way. And then we can, we also asked this question about 180 degree rotation yielding a bijection between these. And that was a weirder problem because this is, it can be a bigger triangle than this. So a different cross-monion and everything. So I did prove that for ordinary homology at least. And then we also find the similar edge swapping result where hexagons with like symmetric edge content. So there was these two cases that are nice where the first one is the opposite edges have the same number of ones and zeros. The other one is you have this like three ways, symmetry of the same content. And then for each of those, you can swap between those symmetric edges and end up with the same number of puzzles. So... Two minutes left. Okay. So yeah, this is the last slide. So this is just further questions that I'm interested in. So it's, we want to extend these results potentially to other homology theories. So do these, can we swap edges and get the same number of puzzles? Or what does that even mean? What is the right question to ask for some of these other ones? So like there's the K theory version and we have these puzzles with an upside, the right side up 10, 10, 10 that computes the structure sheaf basis and the upside down 10, 10, 10s that compute the ideal sheaf basis and the, yeah, the parallelogram symmetry kind of implies something interesting with that. I don't think I would like to understand the geometry of this better. And then, you know, echo variant hexagons, I haven't really felt much into that. I'm not sure what the right question to really ask would be right now. And then lastly, there's these segue shorts McPherson classes which have both the 10 right side up and upside down 10, 10, 10s. And so I'm, yeah, I'm interested in seeing what is the deal is with that. They have any symmetries and what that might mean. Yeah, so that's it. Thanks very, very nice talk. Thank you. And I guess there might be time for questions after all the talk.