 Alright, so welcome back everyone to the second part of the talk by Chang-Zeng Li on Pluker coordinates and mirror symmetry. Okay, yeah, I'm afraid the first part may not be so acceptable for the audience, but the second part should be better. Yeah, I hope. Yeah. So for the second part, I focus more on quantum super calculus and then we just focus on the type A flag variety. So for type A flag variety, let me just remind you of some simple factor. And so the flag variety, okay, so the flag variety that parameterizes the patch flag in here. Also, I would say like, yeah. And it has a basic superclasses and learn the combatorium of the subset is a subset of the permutations such that it is like W1 less than W2 less than W1. And Wn1 plus one less than Wn1 plus two less than Wn2. Yeah, that means W has descents at most L position, Wn1, Wn1, N2 until Wn. Okay, so in this case, the quantum core module as a vector space is as much to the classical core module tensor with valuables. And the quantum product is like this, at least in non-zero. And in this case, so many things are known for flag varieties, the representation were known in particular for the complete flag. The idea is X1, Xn minus one minus one zero, this is Q1, Qn minus one. So this is the determinant is equal to TTIn minus determinant is equal to Tn plus EI Tn minus I. Okay, and the idea is given by E1 to En, it's known. And there are many people for meta contributions to here as I'm sorry, I just list one. And for quantum super polynomial also in the complete flag variety case, it was constructed by forming a polyethylene polyethylene core for patch flag. We can find it in the paper by Schrodinger name. And in this case, even for classical core module, so much less is known for the positive formula for the stretch constant. It will involve the quantum module part as far as I know, the only one case, the only known case is just for Grassmanian. Only known for Grassmanian in Taipei. Of course, for some special case like FL1n minus 1n, this is a joint type, this is a bit special. But besides that, the Grassmanian was the only case that we know. This is the same to the quantum to classical principle proved by Anders and Andrew Kress and Harry Tanwakis. And in our paper, so we proved some new identities in quantum cohomology in this form. So, we take the notation m is 1 to m and AB just means integer is from A to B. Once we are given elements i between two. So that means once we are given some i under here, we denote at least C to be i minus m plus m j plus 1. And then we define a control set C in this form. Also, we consider the weight to be the sum of all numbers in this set. Then we can prove in the quantum cohomology of the patch flag, the quantum product of this, the summation equals 0, this kind of identity. So, let us explain the notation a bit more and see one example. So, given each j, we can define the Wj explicitly. This is some precise definition, but we can ignore it at the moment. We just see one example, FL247. So, in this case, this is equal to FLn1n2n and n is equal to n3 by convention. Then n1 is equal to 4n2 and then in this case nj1n minus n2 is equal to 3 less than 4 equal to i less than n minus n1. This is equal to 5. Yeah, so i does satisfy the assumptions. And d is equal to 1. So, if we will show you just from the definition, we can see C is the subset that just consists of three elements. And also in this case, nj is larger than d. So, this is the first case. Then, once we write it as j1, then here is equal to here. We can write down the definition of Wj is given as follows. Precisely, for the three cases, for j equal to 1, then the presentation is given by 1, 5, 2, 6, 3, 4, 5, 7. You can see this is for FL. This is for gr47. Oh, I'm sorry. This is really for FL 2, 4, 7. Because this is increasing, this is increasing. Yeah, this is increasing, this is increasing. Yeah, indeed, we could find this. And then the other three is defined easily in this way, for instance, for the first one. For instance, for this one, this is W, this is the sigma nj plus d run here, plus 1n plus 1n. And so nj equal to, for the first one, this is 2 plus 1n. And 1 equal to, I want to say, let me say in this way. So, oh, sorry, here. So, this is 2 plus 1 minus j. I'm sorry. So, this is 3 minus 1. This is, so this is why it's 2, 3. Okay. And therefore, in, so our theorem in this special case tells us the identity of this form. The question is that how do these identity come out? Yeah, apparently we could have a lot of identity among the product of quantum converging. Why we care about this kind of identities? But how do we find some meaningful identities? So our expression is in this way. So first, we study the mere symmetry, the quantum converging of lag as a mark to jacqueline. And mere symmetry is packed that the image, if this is the right super potential, then the image of f minus 1, I mean, the class of f minus 1 should be equal to the image of force class. This is one part. And the second part is less. So once we have a pre-coordinated, for example, like 1, like 2, 3, 1, 4, 5, 6, 7, something like this. So then this is a Guass-Mannian permutation for gr27. And so we have some partition, like 1, 1. Yeah, there are some correspondence. And in this way, we could have some expected identities. And to learn, but this is something we don't know, we want to prove. And once we have expected identities, so we check the corresponding one in complete flag variety, then we really consider some super polynomial for some spatial superclasses, and then we can find some identities. In this way, we can prove the spatial case. And then from here, the complete flag passing through the patch flag. So this is like GOB, this is like GOP. That is, we have some identity holds for GOP, then we also hold holds for GOP. This is less obvious, but it's something more obvious in quantum case theory. In quantum case theory, we have something like this. We have some ring homomorphism. So if some structures with the image keep the same, then the product in quantum k maps to the product of quantum k of the patch flag. But this level in quantum commons is less obvious, but it turns out to be true. So to express the identity, so let us say a bit more how we did for the break coordinate expression of super potential. So originally, Rich defined the super potential of GOP, which are here. And we modify it a little bit. So here is WPW0 inverse, and we modify it a little bit. So almost all the essential statement keep the same up to some sign. And then this is Rich, the super potential, this is a radically contractor. So then there are some isomorphism to this open neutralized variety. Again, there is some isomorphism from GP to this is the sub variety inside the patch flag variety. And this is the sub variety in the patch flag variety. This just means lower triangular. This just means the upper triangular. So in this way, so we can define to super potential f minus and f plus. And in the special case of glass meaning minus and plus are the same. They all both give the super potential by mass and Rich. In general, f plus generalize, this generalize, but this not generalize f plus. And in precoordinated, so this is some precise expression. And this is precise expression. And we just see two things. First, so f minus is of this form, f plus is of this form. So you just need to be careful that this means p13 p12 times p1 p3. And this is p13 p23. So even for this simple example, even for this example, so we can see that this is a simple poll. Along one component of integral divisor, but this is not true. So this is a simple poll along two. So you can see this is more complicated in this sense. In general, the expression for patch flag of f plus is much more complicated. And that's really just means we take the derivative along the x dilution, not along the q dilution. So pt as moism tells us there is a moism between Jacobi rain to quantum commodity. And we show that the image of the class of the function is indeed equal to the first class. Now let me explain how we expect some identity. So we want to say, so this is some C1. And this is some C1, but we collect C1 is equal to C1 is equal to 4 sigma s2 plus 3, 5 sigma s4, something like this. So here I'm lying, sadly, because we can see the mere symmetry. Some landline's dual would be involved, but in type A case, so the dual landline's group, dual landline's flagrity and the flagrity are as a multiple each other. So if we modulate as moism, so we use the same notation, but in our paper, we distinguish them. So here there are one, two, three, four, five, six, seven, eight, that's eight ten. And eight ten is actually read as like one, one, one, this looks like one plus one, and then this is one, one, one, one. Yeah, so you may like this roughly, roughly like a sigma s2 plus sigma s4. And then naively, naively. So as we mentioned, so once we, because this is a pass, this is how we write it. So one, four, five, seven means the first pass is one, two, three, four, five, seven, like this. So this is a partition that's three, three, one. Yeah, so for p one, five, seven, like this. Or you may naively, we may naively just say this naively, we just naively consider the superclasses. So if we, if c1 equal to f, then this would equal to this. And so it doesn't make sense to divide superclasses by superclasses, but it doesn't make sense if we can, if we consider the denominator times this equal to this. That is, this is equal to this. The good thing is that this is a super device class, and this is super device class. Then we have the connoisseur formula. And from connoisseur formula, we can do some reduction. So even this holds, this is equivalent to this. This is the first time when we discovered the identities. Okay, like this. Now if, so let's assume, assume, assume this identity plus connoisseur formula, then we obtain this. But now we want to prove this. And after, our mathematical way is that first we prove this. But this is not using connoisseur formula, this is not using connoisseur formula. But using some geometric interpretation of this super potential. Okay. And for this, to prove this identity, there are some key observations. First, WG is always 321 awarding permutations. By 321 awarding permutation, that means that does not exist. IGK is actually that this holds. And for 321 awarding permutations, there is a, there was a formula. So by Billy and Juxie and Stanley, in terms of the determinant of complete homogenous symmetry functions, sigma w, luckily in this way. And along the same time as FGP, we also had some quantum super polynomial for certain permutations. And in her notes, in her paper, in his paper. And so he had constructed that the quantum super polynomial for 321 awarding permutation has the same form. Has the same form as the classical one. And just by simply replacing the complete homogenous symmetry polynomial by quantize one H. H is in some queue here. Okay. Then using, but this is for complete flag. This is for complete flag. Okay. And then in FL247, the good thing is that. So, first, we can write down all the WG by using this. And this part is, you can see the second one. Got many permutations. So he has the common sure functions. So we can also write it here. Then what we want to prove is like is a product. This time it's like this times this times this times this minus the plus equal to zero. And this means this should be equal to zero. Then we can find the same thing. Right. That is, that is, this is equal to, so this identity is equal to the determinant is equal to zero. And this determinant is indeed equal to zero just because this color and this color are the same. So this equal to zero. So this is one key idea. So let me as we can flown some trivial identity. This is something trivial. To your identity by taking the last expansion and then using the determinant formula for it turns out that after we painted it will be written as the determinant formula of two superclasses. And from here, then we can get the plus sum. In the presentation, then we can prove the first and cloud equal to super potential. In like, not, not done yet. At the moment, we just released these two in complete flag. We still need to prove these two in partial flag. We, we want to show the same thing. If we replace this by P is by P, then this should be equal to zero in quantum module of G over P. This is the G over B. So as we can see later, so WG and the both the both permutations. All these permutations are in W other pieces. So this is what I mentioned right now. And so you will have you will treat it as classes in complete flag and we have some identity in here. Then we can also have identity in here. This is the less obvious in quantum module. So there could be two ways to obtain this. So one way is later. So for the latest some kind of function property in terms of functions. The supercarsis could be read functions. And this is a function for functions in the variety functions in the variety. And this function is a variety under the shear is a function in YP. And YP is plus is inside Y. This is Peterson. And I'll not take this is. Yeah, Peterson. So we will obtain a function here. Then we can do the analytic continuation to the whole space to become a metamorphic function and then restrict to restrict to give the metamorphic function on the whole space. And then restrict to YP plus. So in this way, you obtain a function of sigma. It gives a function of sigma WP. So in particular, you will have this fun at least the identity limits as functions that it holds it. And then this is a general function on the total space. It means it must be a general function over here. So in other words, the list of holes for the YP. And also there is another technical issue. So we in the previous proof we need to assume here. So in this case, when we're using the duality reduce to here. In this way, we are done. Okay. So this is how we find how we find the nightly find some identity and then we try to play it and then learn how to play it. And then we again, we're using some kind of mismatch phenomena and together then we pull some mismatch statement. Yeah. Okay, I'm done. All right. Well, thank you.