 There's a quantum product defined that this way. Roughly speaking, it cancels the number of genus zero stable maps to G1P passing through the two inputs and then the point guide deal of outputs. In current quantum sugar calculus studies this quantum product in terms of sugar bases. And today we are going to discuss this quantum graph module ring. And we will focus on three results, including quantum equals RFI. It doesn't variety presentation and the module mirror conjecture. So RFI, which is defined like this here K is the formal loop. Ring of formal, I mean, ring of the ring of low on series. And always formal power series ring. And there's something similar to the G1P, like we have RFI sugar bases, which is defined to be the fundamental class of orbit closure or orbit closure. But here the B is the other end of a program. And T here is a torus fixed point of the other grass money and in general it is indexed by the code with letters. I mean, if G is simply connected. And we will assume in our talk. And but there's another set indexing this. The set of torus fixed one, namely the, the set of minimum length, who said representatives in the. W it is more convenient for us to, if we want to discuss the foreign results. And we have a fun general product on this homo, you can run homology range given by this. Very simple formula, but it is simple respect to this base that's the fixed point. I mean, well, let's also form a basis of the homology, but both after localized, the equivalent parameters are the simple respect to this basis. But it's not with respect to the alpha and super basis. So we could run our fine sugar carats studies for the onion product. In terms of alpha and super basis. So here's the first result. Discover by Peterson and proof by lamb and she was a little more. Namely, the alpha, so we have quantum equals alpha, but that's the ball the case when P equals borrow. In general, the alpha should work as determined the quantum in the following way. Namely, we have a explicit ring map. From our fine. At the homology of alpha grass on gas money and to the column called more G. Explicitly, because it sends a basis element. Of a super basis to the, to another basis of the quantum super basis. And for peace also explicit, but. Because it's a communicator so roughly speaking just send a basis element to another basis or zero. So we may go to zero. But what I mean by equals and determines means that in general this assignment of the basis elements are not bijective, even for people to be based not so if we in the following sense. Both, but this is injective. And it is a jacket if we multiplied by suitable quantum parameters. And here for general P is, it's also the same. If we take away all the basis element going to zero, then the assignment will be injected as well. And subject it in the same sense. And the public spoof by lemon she must always lies on some algebraic and computer identities. But I can. But for example, but quantum shape shape formula and Peter's and Woodward's comparison for me. These are formulas for the quantum. And there are some for the. I'm not knowledgeable enough to tell you more here. So the question is, can we prove the result. And more geometrically, but it is already, I mean, their proof is geometric because it depends on this formula, which is geometric, which are geometric. So let's look at something from sympathetic geometry. We can define a map and a pure another map with the same source and target. In the following way. First, let me introduce them. A more practical definition of that. Namely, the guy find out my cross money and represent the fungal, sending each scheme to the set of isomorphism classes of principle g bundles over P one cropped times gamma with civilizations over. I mean, away from the zero. And this result is proved by both view at last row. And by algebraically, and by precisely, analytically. So, this map costs surface side of map is defined in this way. So, so we call the domain is the homology. So it's represented by cycle going to Africa's money and so by the above result, it is represented by principle g bundle. Over P one times gamma with some civilization. So we can use this space to define comorbidant theory. Yes. Yeah, so this is the space. So we have a principle G bundle, and we can form the associate fiber bundle. So it's now a G more P vibration over P one times gamma. And we can consider the moderate space of stable map. Going to this space. With certain degree. The degree is not arbitrary, but the stable mess is required to contain a component. Which projects isomorphically downstairs to this section. But since we are, we are going to compare the five so we are allowed to some bubbles, but the bubbles will be projected to point downstairs so they are vertical. Vertical line inside some fibers, which are G more P. So these are the, so these are the requirement of the degree. So after and also we are sorry one more point. There's a one more point and it's required to hit the cycle over infinity. And because we have a civilization, we can spread it like this trivially using this civilization. And this output. So the, the one will be what we really mean. Is that the map is defined to be like this. Something like this. In the standard way how we, for some of this is, that's the way how we define for some of the common cop product. Yeah. And there's some issue about how to record a degree using some library, but I'm going to omit it. So we can define such a map. Actually it's not just for G more people any for for any G variety. And this is history. So side of first defined map. Where the input consists of zero dimensional cycles. In the larger group, namely the base group of the Hamiltonian group. For anything better manifold and surface extended it to the to include our cycles of arbitrary dimension. So here's a lemma where we appear. We don't know how to compute, but we know it is a ring map. And it's proof by surface, which extend extending side of this argument. For the long equivalent case. So it originally they, they prove it only covariately. And you can vary the we can use localization to reduce to the situation where the inputs are consist of point or six point. But this equation is proof by Italy tiny. And it turns out that the map I've just defined CEO to PLS. So this proof gives an alternative proof of the results. Because we can pull it by theory. I don't know theory that it is a ring map, the right hand side is a ring back. And now we are able to compute this map. It turns out to be equal to PLS map. The idea of proof is that the more we can show that the more dry stacks are smooth and expected dimension. If the input cycles are both semaphones. Resolutions of the. I'm actually but varieties. So after doing that we can compute like how you don't work with compute or two point of common rhythm invariance. Here they used the fact that all the be always and be minus all this in the session personally to the final stack. Meaning to have an for some for their cases something like this. The final step here means we take a fiber product. And we do with some cycle representing, representing, representing the insertions. So this is the final step. And because of the transversality with the turns out that the final steps are smooth as on expected dimension, which is zero to the invariant will be to be computer by finding the number of elements in this deck. In the second of the borrow and opposite is still a continuous group. The stats will consist of. He invariant course. So that we can determine the stack computer really. And this is the same way how we compute. And this approach can take us further, because we can consider inputs different from the alpha and super classes. It turns out we can prove to Peter's and Verity presentation. And also we can consider an extra action on the alpha class money, which is loop rotation. And we can put it in the definition of software side those maps. And it turns out we can obtain the demo your mirror conjecture. So Peter's a variety presentation. I think it's also announced by Peter in MIT lectures. So roughly speaking is that we have a presentation of the quantum core module rain. We are why P is a scheme. And moreover, we know we can find expressive functions on the scheme which go to the quantum parameter under this isomorphism. So here's the definition of why P the pism variety. So first we start with. The land and use group of G. And later I will recall a practical definition of the land and do group. And then we have a principle important E and. Isomorphism between the. The cut down and do cut down. Then we can define the. Compatify, but I mean, I mean, I think this is what we call pism variety. So which is a positive variety defined this way. And our YP would be the. A stratum of the previous and variety. That means why cutting with some. And this result is proved for the foreign cases. Including all G mob B. That is complete flat varieties and all fair varieties of type A and some projects. Easy D. Skill varieties. So here's a practical definition of the land and do group. And because of it below and proof. Geometrics attack a conference. Which is a equivalent between. The representative representation category of land and do and. A category of parish chiefs on Lafeng Tasmanium. So in the course of their proof, they introduce. What we now call envy cycles of type lambda and new and wait, wait. New. Here. Let's say. Line the code letters, which is dominant. And meal also line the code letters. And it defines to be a closure of it's usable components of. Some of the set. It's like the. Attracting set. I mean, we, we. Consider flow on the Africa's money and then we consider the. Like stable manifold and stable manifold. An example is that. When the weight is the. W zero lambda W zero is the longest element. Then MV cycle. Would be the largest one. The closure of the spherical. I mean, the geo orbit. And if the weight is land itself, then the smallest. I mean, then the cycle would be the smallest one. Consisting of the point. And one important name is that. If we do loads. As. Lambda. By the highest rates. Representation. And we decompose it into. The rate spaces. It turns out that the rate each rate space. As a basis in that spider and recycle. Of type lambda and wait. Of up off. Wait meal. Here, I use as because he loves the sure module. Because everything was over integers and. In that case, I find sure model more. Convenient. And you choose isomorphism. Well, they use the geometric. To construct a ring isomorphism. From. The coordinate ring. Of the centralized group scheme. Of this element where we have seen this element. When we define piece of variety. So, so, I mean, I should give an example. So he is the. Is the matrix, which are all one in the off diagonal. An effect will be. Correspond to the. We correspond to the diagonal entries. And so B E T is a group centralize a group scheme of. Matrix of this form. So we have such a ring isomorphism. And this isomorphism has already been constructed by. And. There's a cutting off. Finger burn. But earlier, but without geometric satellite care. And one advantage of using geometric satellite case that we can. Compute it. In fact, Bowman can is a cruisin. Find that. And you use map. Sensor the matrix coefficient. To the fundamental class of the every cycle. Namely, so we. We have a S lambda, which is a representation. So we can take a consider the matrix. Like this, so that, so that the entries. Are regular functions on G. But we can restrict to B E T. And we look at the particular role. Corresponding to the smallest and recycle. So that is. The highest wage matter. And they show that the map send this. Regular functions to the fundamental class. So because. We have a, you know, we. We have a basic, right? So here's this matrix is taken with respect to this basis. So the input is, let's say, is that every cycle. And we can see that this, the entry sitting in this row. So that it sends the. Make this confident to. The fundamental class of set. So, so I've introduced. Another family of cycles. Other than super cycles. So we can ask when we compute. I mean, what, what, what if we input. The fundamental class of every cycle into a PRS map. It turns out we can do it for. For. If way. For this way. Well, so in this picture, they are. They are calling, right? Well, you see that it's not as it's not a large class. But turns out we can. We have some surprising application. So first of all, we observe that. If the weight of the enemy cycle. I mean. For example, let's consider all the ways. A kind red, except the, the vertex. Then we see that the dimension is quite large. It's larger than this number. And it's equal to this number is. The way I mean. The exact correspond to this word as. So we call this because this is an extremo weight. This is one dimension, the way space is one dimensional. So that is generated by one every cycle. Let us below this cycle by this. Set Wp W zero. Wp is the longest element. Of our fine. I mean the logo of P. We're going to play with this cycle. We give it a notation. Okay. So before the break, I would like to. Go for a few slides. So let's, let's go back to the surface side of the map. So we have seen this picture before. So we. We define some notation. We consider the moderate state of. The stable maps I told you before. And now the. The cycle output cycle will be denoted by C. And the gamma will be some resolution of the. Geo, I mean the closure of geo orbit. And there's a fact that. Well, if we consider M zero. You want P. We know that the moderate space is not. And if. When the degree. Light in the cone, right? In certain conspan by some code. For our case. It is. The set of degrees such that the. The. Moderate stack of sections. And it will be. The same phone, but with some shift. So it's no longer effective. But. But the effective. Shift. So this original effective. But now we. The cone will be shifted down to us going to the anti-dominant chamber. And by some computation. I mean some dimension argument we can show that. If we. With X. Oh, I forgot to say. For any X. Lying in the. The geo orbit. Closure. Lambda. Then. I'll met. When. Would be zero. If the decrease. Is sufficiently large. And when. Because in this case. All the moderate sets have no contribution. But there's some exception. It turns out that. There's only one. Moderates that. Contributing to. To. The computer invariant. So it turns out that this number would be some intersecting numbers. Intersection number. Of X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. X. Y. Y. Y. X. Y. W. W. turn the PRS map will be zero. This is what we can see from the corollary. But I have to say that it's not very difficult to prove that from the component tolerance of the PRS map. We call PRS map is defined explicitly and we can actually use it to prove these results. Almost from definition. But what is press non-tribal is the computation of this interception number. We probably can't prove it easily from the definition of PRS map. So let's take a bit after I explain this slide. So what the the foreign puts Northbury's records but I find this is a good way to see why this is true. So first we determine this stack. It turns out it is first we don't concede for simplicity we don't consider resolution. Don't consider resolution but but the geo-orbit project itself it turns out that the stack is asomorphic to the some sub some orbit by some subgroup here error is the level of p to we have we inform the aphron-grasmaniac error which is a sub in-screen scheme of the aphron-grasmaniac g and we can take the orbit closure like this. We have such a isomorphism and and roughly speaking is proved by and why is this true because so recall we have a principal g bundle over this geo-orbit closure and if we restrict this principal g bundle over this aerobic closure the structure group will reduce from g to l and because e p is fixed by l so we have this map air equilibrium map and we can take the isociate fiber bundle and the left hand side for the left from the left hand side we can construct a section I mean for every point of this aerobic we can construct a section and then we map it into the side of space so that we get a map from right hand side to left hand side and we can use the regularity and dimension argument to prove that they are equal and so the next step is to show that the intersection number is one and we can consider the c-star action on a geo-orbit closure so let's just assume we have a bb decomposition but it is not true because it's not smooth but we can show that we can show that the largest one largest bb that exists so this is a a fiber bundle over the fixed point component and it turns out that the aerobic closure is is the fixed point component and our enemy cycle is a closure of one of the fibers of this fiber bundle so that the intersection is one yeah so here's a proof of yeah this equality so um I think I should pause it here for a break okay yeah all right yeah how long will the break take yeah we'll take uh maybe maybe three minutes break until five or eight okay yes