 Without further ado, Alex Fortress is going to tell us about Bruja orders on 01 matrices. Please take it away, Alex. Howdy, everyone. Thanks for having me on. I'm feeling a little bit sick today. So pardon my raspy voice, but we'll, we'll, we'll, you know, struggle through it and we'll be good. So, yes, um, first I'm sorry, one minute and so not want to change the next slide. Here, let me try resharing it real quick for the technical issue. Okay, I'm sorry. One minute. That's nice. Don't move. Yeah, I couldn't get him to move. But what if I try. Aha, okay. I think that would be good. That's perfect. Okay. So first, consider the space across money in two four. And consider the action on it by a tourist of rank 4. Of course, this acts naturally on C4 by scaling each coordinate. This action sends planes to planes. So it gives me a nice action on this cross money and as well. And so, then for this action, the set of fixed points is the set of coordinate planes in C4. And so, you know, on this picture here, when I write 1, 2, I really mean, you know, the planes spanned by E1 and E2. Okay. And then, if I have 2 planes that intersect along the coordinate axis, then in that 3 dimensional subspace of C4, I have 1 degree of freedom to rotate 1 of the planes on top of the up. And in Grossmont 2, 4, that looks like an invariant curve. It's a 1 dimensional orbit whose closure contains those 2 fixed points. For example, take this fixed curve between the fixed point 1, 2, and 2, 3. And really, this fixed curve is a copy of P1. So it really looks like the Riemann sphere. And when the tourist acts on this sub variety, it really looks like rotating the sphere by multiplying by some character. And if you're seeing on the top of the sphere, it looks like you're rotating clockwise from the bottom, it looks like you're rotating counterclockwise. And so, the tangent weights that 1 fixed point sees is the reciprocal of the 1, the other 1 sees. So here, I make a choice. The fixed point on top here sees, here, A is the coordinates of the tourists. And if you see in the numerator and A with a bigger subscript than on the denominator, we say that that fixed point is here. And we use this to define a partial order. So, I do this with all of the invariant curves that I can draw for my space. And in this case, this is the partial order that I get. And so, for now on, I will call this the geometric order or the geometric Brouhot order. I will note that it does agree with maybe the normal Schubert calculus notion of the Brouhot order. So that's good. So this is maybe the main geometric object of study for me today. Okay. But I want to make another point about these fixed points. In this case, with Grossmann 2.4, the fixed points are in bijection with two by four matrices that only contain zeros and ones. And such that every column sums to one and every row sums to two. For example, maybe if I was talking about the fixed point one, two, I really mean, I could describe that as this matrix where I'll put a one in the first and second entry. And then the second row is then forced by the row and column sum properties. And so really, I could describe fixed points in this sort of, I don't know, almost combinatorial, you know, more discrete way. And from this, what I want to do, I'll generalize now, if the case of flag varieties, provide a general, you know, partial pipe A flag variety. Here, I still want to consider fixed points in this new way, where I will consider a fixed point as if my ambient space is Cn, I'm going to consider k by n matrices that only have zeros and ones entries, such that the column sums are always one. And the row sums are the pairwise differences of the dimensions in a flag. For example, in this flag, when I see my first vector space is span of E1, I'm going to put a one in the first column of the first row. And then my next piece of my flag, I'm adding in vectors E2 and E3. Again, fixed points, you are, you know, flags with coordinates vectors as my spanning my pieces. So, you know, here, if I'm adding E2 and E3, I'll put a one in the second and third entries of this matrix. Finally, for the fourth piece of my flag, I'll have E4 as well. Great. And so from this, I can really describe each of my fixed points as zero one matrices. But then also note that in the case of a full flag variety, the differences of these dimensions in my flag are all one. And so the matrices that get out are permutation matrices. And permutation matrices, well, permutations have also a notion of Brouhot order, the combinatorial Brouhot order. And so maybe now I can consider combinatorial codes for Brouhot orders on these zero one matrices. That's the goal. And I'm going to see how they align with this geometric order that we just that we defined earlier. So for me, whenever I say BCT, that stands for binary contingency table, that's one of these zero one matrices with fixed row and column sums. Here are is the vector of all the row sums and C is the vector of all the column sums. And so now I'm going to define a combinatorial order on these matrices. I'll call it the combinatorial Brouhot order. And the idea is for a given matrix, M, I'll build a matrix called sigmas of M, such that zero is the sum of all of the entries above and to the left of it, which is just zero. This one is the sum of all entries above and left of it, which is this one, so on. I'll get these and so now if I have two such matrices M and M prime that are in my that are BCTs, I'll say that M prime is less than or equal to M if sigma M prime is greater than equal to sigma M entry wide. I know it's a little bit flipped, but I claim that that's the right thing to consider. So this is some nice sort of simple partial order one could define on these things. Here's another one. I'm going to call this one the secondary Brouhot order. And here's the idea. If I in one of my BCTs see a sub matrix, it looks like zero one one zero. If I replace that sub matrix with one zero zero one, I call that an L2 to I2 interchange. And if I, and so here's an example in this matrix, I see a zero one one zero. I could do an interchange to change that into a one zero and a zero. If I can do a sequence of these moves to get from M to M prime with L2 to I2 interchanges, we say M prime is less than or equal to M in the secondary Brouhot order. Okay. So that's the those are the partial orders that I've just defined on these BCTs. And so of course, the good news is for the case of flag varieties, they're all the same. That's the big important thing. So if you want to get your hand on this geometric order, we have a nice combinatorial code via these two other partial orders that we've defined. They're all the same thing for partial flag varieties. But do recall that there was one restriction that happened, which is, you know, when I was describing the fixed points of partial flag varieties, I was only allowed to use BCTs where the column sums are one. So what if I remove that restriction? These two, you know, combinatorial orders make as much sense as before, they're perfectly fine. Now I'm not really doing any geometry anymore, it seems, because what space am I working in? It turns out I am still doing geometry. There's a good space to still consider. And they're called bow varieties. Maybe I'll tell you a little bit about that. And so first, this is just some sort of combinatorial picture called a brain diagram. It has some, you know, slashes that go with positive slope, some with negative slope in between each, any two of them is a segment and it's given some sort of decoration. That's a non-negative integer. And for every picture of this kind, I'm going to associate a smooth variety. If this picture is called D, I'll call that variety C of D. C stands for churcus. This is called a churcus bow variety. And maybe some other names. These segments in between are called B3 brains. These slashes that have positive slope are called NS5 brains. And these ones with negative slope are called B5 brain. Okay. I'll tell you just a little bit about what this space is. But whenever you see an NS5 brain like this, if you could look at the two decorations on either side of it, for each of those numbers, consider a vector space of that dimension. And consider the set of all linear maps going this way and the set of all linear maps going this way and take the direct sum of those two HOM spaces. And whenever you see a D5 brain like this, again, each of these decorations associate a vector space of those dimensions, but also associate an extra one-dimensional vector space and consider the set of all linear maps go this way, this way, and this way, and also these two endomorphism sets as well. Okay. But then also modulo by one relation. So take the direct sum of these HOM sets mod by this relation and we'll do that along your whole brain diagram. Direct sum all those together. Then from there, you'll play some sort of Hamiltonian reduction game where you'll find some sort of moment map. You'll take the preimage of zero. You'll find some nice open subset of that called the set of stable points. And then you'll do a quotient by some sort of group called the gauge group. This is GLN acting on each of these vector spaces that are sitting over your D3 brains. I know that's a bit much, but that's maybe a quick blitz through what the space is. And I claim these are the spaces to think about. So one thing is that they're quite good spaces. They're smooth, complex varieties, and they come with a torus action naturally that comes from acting all those extra copies of C from those D5 brains. Also, it's going to come up with one other little action whose coordinate I'm going to call H bar. Maybe this comes from fifth. Okay. So here's the good news. For any partial flag variety, its cotangent bundle is an example of a bow variety. So all those previous stuff we've already done are already captured in this case as well, which is good. You could say, oh, well, it's actually the cotangent bundle. It's a little different claim. In this case, the H bar will act by scaling the cotangent fibers, and you won't get any new interesting invariant curves if you don't bound it once. Furthermore, any type A quiver variety is also an example of one of these bow varieties. So here's an example of cotangent bundle of flag variety. It's a type A quiver variety. It's also a bow variety for some brain diagram that I can input. Okay. Great. So I would like to have some indication that these things are related these BCT problems that I was working on before. To do that, I need one other little gadget. Every five brain has something called a charge, which is just some sort of number you can assign to it. You can read it directly off of the brain diagram with these formulas here. And I claim that these are very natural objects to consider for bow varieties, these numbers, these charges. Because in fact, the fixed point set for a bow variety is in bijection with some BCT set where the row sums are given by the NS5 charges and the column sums are given by the D5 charges. And so now suddenly, we have BCTs that are telling me what my fixed points are. I have a quick question from chat asked whether cotangent bundle of GMOD P is always a bow variety. And the answer is in type A, yes, these are type A bow varieties and all partial flag varieties of type A. Examples, yeah. So now I have a great thing. I have a nice geometric notion here of a partial order that can build from my bow varieties. And now I want to check again, does it match up with these combinatorial orders? And before I can be sure, oh, here we go. Some bad news turns out that the two combinatorial orders in general are not the same. They happen to be the same if the column sums are one, but otherwise they're not. This was first proven by Brualdy in 2006. And to prove this, here's a counter example. Here's a pair of weird six by six matrices that are related in one partial order, but not any other. As far as we know, this is the smallest counter example that anyone's found. This set of BCTs has 89 elements. So this is starting to get a little bit big. So the fact that these two orders are not the same was not obvious for quite some time. So this is relatively... But now the big question is if these two combinatorial things are not the same, then are either of them the same as the geometric order? And so that's the big question. And so our recent theorem, it's in a paper with me, Richard Ramani and Tomasa Bota, is that for a bow variety, this geometric order is actually the same as the secondary Brouhot order. This is the one that had those L2 to I2 interchanges. That's actually the correct description for bow varieties. And I don't have a whole lot of time left, but I'll just tell you a little bit about this result. So one might suspect, oh, these L2 to I2 interchanges, are they just the same as invariant curves? Maybe these two objects are actually very naturally the same thing. It turns out not really. Here's an example. So in this set of VCTs, here the blue lines are invariant curves that are not interchanges, and the red lines are L2 to I2 interchanges that are not invariant curves, and purple are things that are both. And so you can see the blue things and the red things are really quite different. They just happen to generate the same partial order. And so that's sort of interesting why a heck to these things do that. Also, if you're wondering about what the heck this is about, really between these two VCTs tree as fixed points, there's a one parameter family of invariant curves between them. We call that a pencil of curves. That can happen in bow varieties. That's a common occurrence. Okay. And so now if it's not just a perfect matching, how do you start doing a proof of this? I'll tell you a little bit about the proof. I don't have a lot of time. So I'll just tell you a tiny bit. So we actually have two different proofs in the paper for it. One idea is that maybe we can just learn what the invariant curves in bow varieties are. And we actually have a full classification and a recent paper with me and my collaborator Ian Shu. And so I'll give you one example of what an invariant curve looks like in a bow variety. Here's an example of there's a curve between these two VCT trees fixed points. And any kind of little move that looks like this is one. And what do I mean by this move? Well, pick two sort of columns, two columns, and find some two sub blocks of these columns such that line up, they have to be contiguous blocks such that the sum of this column is equal to the sum of this column. When that happens, we say they're matched. And if you see that, swap those two, gives you another VCT on the other end. And I claim whenever you can do this sort of move, you get an invariant curve. And if you can do multiple them at the same time, you'll actually get a pencil invariant curves if they have the same tan. And so maybe this is a nice very combinatorial thing to describe these. And from here, maybe there's a good proof to be done of just checking that these are, how these are related to the L2 dye interchanges. This is your three minute warning. Excellent. Okay, we have one other idea for a proof. This is some other new technology in the world of bow varieties, what we call brain resolutions or geometric fusion. The idea is, imagine in my brain diagrams, whenever I have a D5 brain with some charge C, I split it into two other D5 brains, whose charges add up to C. So I'm making my charges smaller and splitting into more D5 brains. And the idea is that whenever you do this, you actually will end up getting some sort of nice embedding of the bow variety for this picture into the bow variety of this picture. So then the nice thing is, if what if I just do this as much as possible until I split all the charges to just be one, when that happens, the space you're looking at over here really should be the cotangent bundle of a fly variety. And those we understand quite well. So this new embedding comes from new work between Richard Ramani and Tomasa Bota in their most recent paper. And so that maybe gives you some idea of how to prove something if we know how to put things in here that might simplify our problem. And that's what I want to talk about today. Go. Thank you very much. I guess we have a minute for questions. Yes. So some quick questions. Well, I have a question. So the tangent spaces at these fixed points, do you know are the weights distinct? Not necessarily. In fact, when they're not distinct, when we could expect the pencil to happen. For example, this, this, the same tangent weight. And because of that, all curves in this pencil also do. So, so then if the weights are not distinct, how do you embed this into the cotangent bundle? Ah, right. And so there's a little bit of a lie, which is that it's not quite a perfectly invariant embedding. Because this is actually just some sub tourists of the tour. It omits that H bar action. I see. Yeah. Right. All right. Any other questions? Well, if not, let's thank Alex again for a wonderful talk. So thank you, Alex. Thanks for having me. I appreciate it.