 Okay, so welcome to the second apple today. And we'll start with the description of the proof of the Pico W conjecture. And as we did in the morning, we'll have a break in about an hour. And then we'll resume with something a bit more technical. Okay, so. So, let me start with this algebra, which is algebra. And the picture you should have in mind is that there is say, and we have it has a synthetic form. And that makes functions into a personal. And explicitly, you have the functions. Then you write. Okay, so this, so functions become. And more precisely. Okay, so we will get polynomial functions. So, this is. There is a little bit slightly different point of view, but equivalent. Alternatively, if you have a function, you can follow the corresponding function. Then the Hamiltonian is given by the same formula basically. Right, this is. And precisely how this function acts on another another function. If it is, you want to complete it was something. And then it turns out that. And the commutator. Also. This is basic standard. You can just plug it in and verify. Okay, so now. Okay, there is this geometric description of this algebra. But we're not going to see. We're just going to see, we will see it abstractly as an assignment. Right, so I'm right here definitions. But this was a picture on the right. On the left, I will write the definition. We have it's a, it's H two is a leo algebra. And this basis. And the commutator. And the commutator. And the commutator. Generators. Is given by them. So I need to write the determinant of the two vectors, which is. I write D. And I have M plus one. And plus M plus. That's that definition on the. With the convention that the zero. Is it this negative. Yeah. So this, I don't know some initial things about. So it's always like this in symmetric geometry, if you have some function, take some function of a function, then they will put some commutes. But one more explicitly, let's say that, so one x square and so on, which corresponds to d and zero, and this form a commutative. And one y, y square and so on, also form and this corresponds to d zero and also form commutative about. And another interesting thing can be worth pointing out at this moment already, that d two zero, d one one, d zero two form as a, d one one one, d zero two form as a absolute some scaling, maybe some, some way you have to take one half. So we can compute explicitly d one one, d zero, else the determinant here equals two, and we get d two zero, then d one one, d zero two is minus two, finally d zero two, d zero. Now here is where I see that I need some correction, this is part four times, d one one. If it was a, in SL2 it would be simply, there would be no four. So it means I just have to divide this two by one. But it's still less or true. It's the same number, it's just not the standard. Yes. Okay, so I like this algebra. The main, so expectation, so it should act sort of controls many interesting things. And here I don't even know, but this is a statement. What expectation means also? So I can say that I had this conversation with Razanski explains that some invariance in mathematics should be thought of as some kind of physical theory related on something times C2. That's why you could get the symmetries of C2. Okay, so I want to have some, there's some variations, so variations of this algebra. And I'm to, let's see. So we can also take C star times C. So replace C times C by C star times C, which is also the tangent model of the multiplicative rule. Or you can have two times the coefficient. But they are also symplectic. And therefore they produce algebra, the algebra in the same kind of way. And the difference is that similar. The C star times C star has the product of the magnitude of the sigma. And there are always a subset of the sigma. I used to think about this as having some canonicals. The X over X, which we have something on the data. Right, right, so this will have, exactly, so this will have, for instance, the X over X, which is on the Y. And that's why it will introduce some difference in some correction in the formula. So the commutator will look like the same determinant, but afterwards, you would have the n plus n prime, n minus one in the case of star times C. Or it will be simply the sum for the second case. Okay, so the difference is that you don't have, in the first case, we have minus one in both places. But then it can be just minus one in one place or not minus one. And this sounds, it can be little, looks like it's minor thing, but it's not. It's very important. So it makes them very different. But it's not about the attributes of names. I mean, Tonya and Victor are fields. No, just a letter, H2, I mean, very huge responses. Oh, maybe yes, maybe, but I don't know, I don't know. So there is this relation, there is a map between C and C star, which is given by exponential. And then it turns, makes them kind of isomorphic but only when that exponential can be kind of well-defined. What's the, the algebras are algebras? Well, they're not certainly isomorphic, they're different algebras, but sometimes you should think that, when for instance you have a representation of one algebra, one algebra, and you can define that exponential. So what does it mean you, sometimes it can't send any kind of, and what is X in quotes? It's not the same as X, it's always wrong to do that. All right, but you see it doesn't send polynomials into polynomials. And so, well, one example is about some, if you think about, let's first introduce some. So we call all the algebra C to rational, C star times C, trigonometric, C star times C star, and what I mean, the meaning is obviously, so for instance, if you have, let's try, let's denote it by X, the exponential, and then prepare it. So let's introduce this change of coordinates. And then we know that sort of, for instance, the trigonometric algebra has this operation corresponding to X to the K, and it corresponds to, and then what is it? It corresponds to the vector field, K X to the K DTY, right, trigonometric version, does again in the rational. So then you do the substitution, then you have to write it as a K times X, K X DTY, and so you can edit it in sum, K to the N, plus one X to the N over N factorial DTY. So if you can make sense of this infinite sum, for instance, maybe the operators. So, okay, then to complete the formula, sum K to the N plus one over N factorial is X of four, what's actually the context of zero. So imagine if you have, so if you have a rational representation, the representation of a rational algebra, and this infinite sum makes sense, then we can also define that. You should try to define the action of that, where we're using the formal rules. So this thing, not a lot of ideas. Okay, let's now turn to something more. So I'm going to, so I was saying that I want this algebra to act on interesting things. And for me, sort of the motivation came from the following. So we have to consider T and N plus one, and N plus one, that is a torus, torus not. For instance, take a T, C4, then you have, so how to draw it, you have to draw it. I have to do this wrapping around three strings four times. Then I connect the ends. This is, alternatively, you can also draw a torus and start drawing some line, which kind of wraps around three times, this way, four times, this way. So, and it's what is torus, so there is this family of knots, one knot for every integer, positive integer, and then we compute certain way, I've called a triply graded, for one of Brazanski's homology, and we restrict to certain parts. That's, no, that's just, so this is what I'm thinking, triply graded vector space model. Generally, I'm correctly not, it is a triply graded vector space. And so the part means you only take degrees, zero, five, percent, maybe. But it has this explicit description by Gorsuch, is given a source. So you start with writing polynomials in X1, Xn, Y1, Yn, and then you take, divide by the ideal of, ideal generated by symmetric, SM, symmetric, positive degree. So, you don't want to generate ideal by constant polynomial, but that's what we're going to make very interesting. So why take all symmetric polynomial? So, for instance, polynomials which go for this. Symmetric in X and in Y separately or? No, not separately, diagonally. Like for instance, this kind of expression is symmetric, but yeah, like this. Yeah, so well done. Transposition, any permutation produce X and Y simultaneously. So this is also called diagonal harmonics. Yeah, and the Heyman proved. Yes, here. Heyman studied this and they proved that dimension was n plus one to the n minus one. I together with Eric Carlson, put the shovel conjecture which gives the character of this. And send a presentation. Which is not. Somehow. Yeah, this. Right, which is all related to. But what is the duty? Yeah. Nice. Part there. So it's related to shuffle conjecture. But for this discussion, what is important is to notice that there is a structure which I think no one have used to study this vector space is nearly the algebraic. So claim is a sort of observation. The rational point. And then, so I tell you how it's generally direct. Like a. In X in the green X in the green. So it's put X by. So the end of some because. And then the URL. You can extend this. You can construct action on the tens of power of space. So you from this idea, you see that you should simply apply the differential operator in each coordinate. So you had some. And again, many mistakes, which I was doing that. I did not buy him. Two things. Let's now close. And then you're right. And then you'll write the corresponding operator and each variable. So you will have the X is coefficient. J X to the I J minus one. And then you can. Minus I J. It's J to the I minus one. Okay, so. Now, this is a certain differentiation. Oh, I see. There's not very visible. It's a J. Y K. I J at least. I Y K G. You see, I differentiate it. This one. And then you are divided. And I differentiate. I am. Yeah. And this is fine. Okay. Yeah, in my handwriting there. Okay, so. There's this highly non-figural vector space of this dimension. And, and we have this operator thinking. So it's easy to check that the sort of symmetry. Right, because it's. So the ideal is by. The plan is separated to any symmetric. Because. And that's why you can check that. So that's the instance. Of them. Action of H two. And together with. The worst key. Produce some. Operators for. That should be expected to be. Satisfying this out of relations, but we don't know. What is with me. You use their paper. This paper into all of us. Okay, three people. Okay. Since you've already mentioned. When we don't prove. But it's sort of some extension. Some. Before. Okay, so. This is the one example. So. Why what's the connection between this sort of. Comology of those spaces that I'm actually introduced. And there are some hints. Maybe you can think about this. This should be sort of a homology of certain. And this should be sort of with some kind of. Why. I think. So far I don't think. I think. So. I think. I think. I think. I think. So there are some examples. But it don't really become as I just. So. Some situations. Some situations. But. There is a story. For algebraic notes. Is expected. This is a conjecture. It's expected that. This is expected. Is expected. Because I only see. Right. There is. Yeah. The two gratings. You said something. There is sort of. This is. Yeah, it contains. I will read the. And one. And one. You get. Determinant is. Okay. Plus. Okay. And. Not the whole thing with this. And you're sure it contains so too. This also contains. It's different. Exactly. Thank you. Thank you. Thank you. It's the next. So here's another example. We consider. Let's consider a bit of writing. So I can ask you about that. Yes. So you said you produce some operator. I don't understand. I mean, it's necessary to. Yeah. So. So. For any more, you can just in the end, which is the strictly graded. And the. Construct some operators. I think. Well, we are motivated to do some other conjecture. But. But we can start some new operators and they expect them to have some new operators. Yeah. And. Yeah. So. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Oh, there it is. You can forget. You can the same thing. It's basically. Basically. You can. You can just put it back in your case. We should respond to your two variables. They satisfy the same. So. So. In the same. Two variables. One for the polynomial. The minutes. Term. As I can see, there in other case. The case of a. So it is a homology ring that is striking and it has two natural patches. So it is a ring. So as a ring it is actually just the exterior algebra generated by some generative like this exterior algebra in G generators if a bin and retic is dimension but also it is a bin and retic. But also as a bin and retic it has a group. So the homology has a common, so the homology has common amplification. And so the homology has common amplification because we have the addition of A times A to A and by pullback we can pull back the common amplification which for any group you get the structure of pop-up to practice. So what you see is that now, so if A has a sprint to recover the polarized then there is, so the homology of A is identified to the dual and you can then construct another multiplication and this is like a convolution. So you can think about functions on the right and multiply functions so you can take convolution. But then there is also, so you can write everything, so this induces some operators, so we have operators of multiplying by xi i, this will be denoted by n i. But then we will have these other convolutions with, so this gives us some other operators n i, so n i will be going from h i to h A to h 2 plus 1. But this, the convolution will give you operations, I don't know, I'll say n i will be decreasing and then you can also take the hard dashes, you give you some s o 2, three pull, and you can see that this s o 2 is part of, we get some s matrix in the chain, then you can also relate this to Yano-Kanitans one. Just to wrap up, I guess then I can, let's say, let's say from the beginning somewhere, so we have two causes of operators which form a super commutative algebra. So these are multiplication operators, they are related to some measures between some operators, and then you can think about this operator's coming from convolution as a k-key operators, like a kind of translation operators. And that's another commutative sub-algebra, and then there is s o 2 which permits one. So you get a similar algebraic structure, h 2. So each one is commutative, is it m i or something like that? They are super commutative, yeah. What is super commutative? Oh, they were in the other side, where this line came up. And what's the statement? So h 2, back someday? No, just I have again two commutative parts of the question, and I sort of, I think, what is it not to do with h 2? Yes. I thought this was another example. Yeah, I just, now I didn't study this one, but I have to say a precise statement here. No, but from what I've said, it's usually the top top things, and you give us two examples. And these are the steps you see often. This is, I mentioned it doesn't matter how it's done, but I was just trying to say it, okay, and then you should be able to see the truth. Okay. Now, finally, another example. Example of what? What's that for? Oh, I'm going to stop. So, in this, I want to consider the bedroom and the space, like in Tamar Shostok, but with modification. Maybe we want to introduce, so we have this curve C, and we pick a point, a point in the C node, and we consider the so-called parabolic character writing. Yeah, where are we? So, these are the representations of the fundamental group of the complement, G, L, and the complex numbers, such that pi of the same small group around G0 belongs to prescribed conjugacy class. It's that same conjugation tool, the diagonal matrix, this is A1 to AM, 0, 0, this, and this is all up to G1. Okay, that's a nice case to pick C0 and pick A1 in AM, complex numbers, 0, 0, 0, 0. Okay, now, this is our MD, and this variety has, so we take a homology, study this screen, so this thing depends now on C and C0, and that's the one second. Yeah, MD and Microsoft is a plenty of time. Yeah, and depends on all these A's, we actually want to put it in one place, because on C0 you fix the branch and throw, but A you can consider the parameter. You can think that after we introduce this special point, and in positive condition, we put two new structures on this point, namely, we can, now the diagonal, so this matrix cannot be 0, can be diagonalized, and the items they are one-dimensional pages, so they are lines, and they produce an N-line bundle on the model page, so that against page 4, 5, N was 0, produce N-line bundles, which we did not, L1, so multiplication by C1, L1, is the commuting sequence of operators, but then there is another structure, and which you can understand as follows. Why do people start moving A's with the entries of A and R, so we want to kind of, because this is a family of varieties, we will obtain the cart course differential equation. Well, now you said the y-throwing is just to come from the one-dimensional eigenspaces and one-dimensional values, so it seems to be the same with A. Right, they are independent, but this is now completely different thing you can do. Forget about the y-throwing, and simply consider this is the family of varieties, parent-righted by N-touple completion numbers, and then differentiate along that, along the parent-rights. So this is one piece of structure, which I call the one. The second is, when we vary the state, we use it on a drawing action, so we have a test of operators. Right, they don't know by one of the configuration space of these things. Do you want the AIs to remain independent? Yeah, I think they reset that. The A has to be different, you never said that. I did, I did. This thing, I didn't see. So, you know, then there's one of the operators, and then you check that they communicate operators. And then, so, wait, wait, what are they, what is that now? The one moving around A1, and the other one fixed. Right, so it turns out that it's sufficient to look at, sorry, the AIs going around 0. So, this, well, if you can move them, you will get a wild group action. But if you kind of don't, if you start with some table A, and then bring, go back to the same table, you'll see that you get a trigger action, just twist them twice. The statement is that you get the action of the final group, and you can see the letters on the back. So, sorry, but so, what are you, I think there's going to be AIs simultaneously moving around 0? No, you can, for instance, fix A1, A1, and move around A1. And move that one around 0. So, the point is that, how do you... The point is how do you exactly move around 0 is not just 1. Yeah, that's right. Because it doesn't make sense. So, although you said they're in C, it's C to the marked 0. It's not sure. C results in 0, is A to C star? But the AIs are in C, 1 is 0. Right, because the matrices are in JLN, they have to be in B1, I see that. Okay. Yeah, A, I said it's C star. But from what you said in C, maybe you said C star. Okay, so we have these two kind of commuting operators and you can start asking, can you play a similar game? Can we relate this one family term? But you can immediately see that it's not possible because the first kind of operators are important, but notice there's one, a greater one. So, L, these operators 1 and I are important. Namely, you start with some something in a homological degree I, you get that plus 2 and so on eventually go over the dimension of your variety and you get 0. But these MIs are important. Why are they important? Because yeah, they think of the one of the reactions, I don't have a group explanation. Yeah, but yeah, you can check that they are uniform. Somehow you found a way to see like in the model of registration and associated for the range of patients, the action should be added. So no quasi-uniform, they're actually uniform. So you cannot relate the two, but probably you can relate by the explanation. So you cannot hope that this should be like, so we think that so probably the algebra generated by both is like the trigonometric one version, right? And to get the other versions, you either have to explain the important operators, which means fast to the charm characters and then hope that gets nicely nice, it's symmetric. Or you should take logarithm of the uniform. And then we also hope to get rational or elliptic by exponential and all that. And let me say just one thing. Right, so another observation. Let's look at symmetries of c star times c star. This is g of 2z. But if you start with c star times c, yeah, let's say c is very difficult to get to. And then we have a whole test of 2z. Then c star times c, there's not so many things. But once you go to c times c, then you immediately see where it is the s of 2 star. And the expectation is that, so, so, why? So you say you see it, I don't understand that there is s of 2 r acting on this case. Which is s of 2 r, s of 2 c, okay, s of 2 c. So, and if you have something which will have action of elliptic algebra, then you will hope that by taking logarithmic coordinate, you get something of which you have action of that ratio. That is what we are going to do with this enrollment. So, then I want you to... Is it the sense of if like, that she had to cut out a parameter to know what she wanted to do? Right, yeah, so one with one correspondence should be KCOR and cohomology. But the other is more mysterious. So, then finally let me finish with another example. So, we started to think about the same case, Tamash. Take Tamash's... The relations in the ring of... Relations in the cohomology. So, here the cohomology ring is generated by... Here in this, there is some other ring which is maybe more natural. And there is this polynomial ring in alpha, theta and gamma. And there was one alpha, theta and sine. But basically, it's the whole thing. Sine, right? And there are some real ideologies. So, I guess that h2x... What do I do with that? So, I think it's still... It's getting taken to the following statement. So, this idea that this stuff, I think it's a very nice stuff, I think I wrote it. Yes, so what's the same? Okay, I'll say it again. But this is gamma, I think. That one here, degree 6 plus gamma. Of course. Those side-sides are very easy to process. So, this is gamma, I think. Okay, come on. Right. And then I wanted to write this operator again. So, I introduced the polynomial operator minus 2 beta, dv alpha, dv 2 beta. And it's not going to be gamma, dv 1 and 2, dv 2 and gamma. And then this... No, so this idea depends on d, which can be calculated as 3g minus 3. There is some version which incorporates another coefficient. Right. So, there is this differential operator, and then this l preserves jd. And the computer experiment shows that jd is generated by l and classes degree given. So, this is a homology of the space. The space is the dimension. And so, of course, everything about the dimension must be 0. But then you have, by then, you'll get all relations of the three dimensions. And what's interesting here for me was that he had infinitely many relations online. In that case, it was complicated. The accuracy, you know, he had exclusive relations for the heat region, whatever the word, you know, about. And then those relations are somehow connected to each other. All right. Yeah. So that point is that you have a single operator, and then you get the word. This is the operator. Come from this from the experiment. This is operator. It's d2z. 02 is one of the topological classes. So I will tell you more about what exactly the algebra. So this is, this is before the proof. It proves that there is action of this algebra. And the next. I'm going to talk to tell you how to construct action on this algebra, particularly obtaining proof, geometric of proof that this operator l reserves the other relations. Okay. So you want, this is. Yes. I want to start. Okay. So there are some questions. One of the things. Okay. Can you explain the last thing? I mean, so suddenly it was a d and then suddenly d was 3g minus 3 maybe plus some unspecified k. And the idea this is best. How big is it? So they are not dimensional. I mean, it's. So the, the idea is, is, is given as some, no, it's the same idea. Yes. No, no, the same idea. Like I said, yeah. Use your idea of the application. Our, our relations are explicitly given as those some, we also re-invade. Yes. And then recursively. But you only have three important relations and we have, we use your relations to prove the statements. This is the, I mean, we should. That's what I thought. But I thought it was another, because in computer you also have a computer. Yeah. That idea is contained among the, some, one of them, yours contains this. This is like on the side. But first, is this close to finite dimension? Yes. And what's the dimension? Well, you don't have to write the form. So, for instance, it depends on how much you wrote down the one correctly longer. Well, but the notation, which is the same itself. So, what is this? Oh, yeah, yeah. It's related to the one I showed you. So what is this? The most dear, what is this? This is treating my street. Right. But before it was that, what was the, so the D is in the, in the end? No. Yeah. D is a, is a half dimension, right? So, so GT is 1696. Oh, okay. So this is any is equal to two equals. Yeah. So it's not the previous D. It's not. Thomas has talked for any, it was two would be one. So this is what we did. So D is the dimension. Yeah. 3g minus 3 is the dimension of your 6g minus 6 is the dimension. And dimension is dimension, which is 2g. So you, on one side, you look at this paper and then you see that there is this idea is given by, for any IJ case, it is time set on equalities. You write some, some is very normal things. Some very normal some is this alpha, beta, gamma. And the other hand, we generate this idea by just saying, well, why is the operator L as many times as you want to any of, any element of degree greater than 2d. And you get the same idea of computer synthesis that we really get all the relations. But in some other cases, that's a conjecture, it's not a fact. The conjecture here is that GD is generated by L and classes. Yeah. So that's not the fact, it's only computer experiments. That is, because of GD. Computers, GD is generated by it. And it's the same we observed in rank 3. Can you say something more about the preamble transport? The, sorry. Could you say something more about the example of the a begin variety that you mentioned between questions? So you should think about this. So maybe, yeah, maybe I should say the following that if you, so you have this leaching system and expectation is that this construction, our construction of this action will work for all fibers system of the heating system. In particular, a begin variety, you should recover this, this structure, generally, precisely how related to H2. So you should recover the structure for the begin variety in fibers of the heating system. And the other thing maybe I should point out is for issue. How do I show the SFCH? So it constructs, well, not quite, but almost this H2 for family, for families of smooth curves, which you convert it to the family of the Corvians. Corvians. And then you have the two kinds of operators and the Chukari, Chukari and Chukari transform and evaluate them. So he has this similar structure, but I haven't studied this yet. That he is related to the hypercutter? Okay, then following our procedure, then we take a break. There's coffee there, the doggy there. Same place, and we come back, let's say, to 30. For a little bit more. Yeah, the humans. No, three 30. Three 30. Three 30. Yeah, just three 15. Oh, so we can't be back to two 30. No, that would be nice. But no. Awesome. He says, well, I think you're pretty good. I had applied my first part, first lecture. I mean, the second tomorrow, but the second now. Not tomorrow. Well, I told Anthony, he could make this as long as needed. That's a lot longer. And so there is going to be a similar spot. So we have those elements. It's not complicated. Okay. And then it's a little bit weird for you. It's not long ever seen when you start that joint. And it's just nine. This is I and KG and it goes through. So I KG is generated by. This is really correct. The C minus ice and numeric. The factorial. That's it. It's the formula correct and C minus I factorial. It's not a little bit longer. This should be correct. And I mean, every form I've seen in my life. In cognitorics, you can just put I goes to infinity. The coefficients stop. So if I is bigger than R, bigger than S. I should say, oh, if it's negative, if the C minus I, it wouldn't stop it to get bigger and bigger. So there's something very odd. Right. But see if it's horrible, completely thinking, right? It doesn't matter. It's still, it's balanced. So you have one factorial on top, three factorial on the bottom. But the sum of these factorials in the bottom is also something minus nine. Well, it's not, it's not balanced because C is not our process. Um, this, okay. But why do you stop it? It's the K. Where's the K going? Right. And this is generate I KJ. And then the statement is that the chronology, yeah, here. This means that you take this to a chain variant. But no, no, what is the signal? Oh, but N is zero, right? Otherwise it would reduce it, right? Yeah. And then, well, this, this part. So the model is generated by, so gamma is the sum of sine i, sine g. The ring is generated by size. Alpha, beta, 1, u star, sine 1, sine 2g, then h3, right? Alpha is in h2, and u is in h2, and v is in h4. And that's that. So that means you take a polynomial system, channel one, it's a symmetric polynomial system, side cross, degree k. Degree k. So you see that somehow dependence of sine is somehow just factor out. You have only, here you have only functions in alpha, beta, and gamma, which is that to generate the invariant ring under that sqlg action. Because there's a sqlg action. And if you go on websites, the gamma is in invariant and that, so this is the interesting thing. But I don't know since it's invariable, so. And then it's given in the paper. So, yeah, we have many questions, right? And I'm not thinking that. There are some more stuff. Namely, how to get the differential area. So the presumably it's compatible with these relations in some way. I'll be correct. And if you fly it out to this PRSG, see, you get something. Yeah, I don't know. So what we have is, I checked on the computer that radio is preserved and that it is generated by this help. But on the other hand, now we have a proof that this H2 algebra is from that you can extract the entry. So the proof doesn't go through what this example doesn't. It's not part of the. No, no, no, it's my application. So how did the help us the rest of the issue? Actually, this is only to zero. Right. Let's be 73. So what we have is the following that. And so statement that we could take. So H2 in the homology of C, this is a super. It's now a super reality and it acts. But not quite on the homology. I'm going to be spaced that some thing in is two x. So it's available for variables x and y. which one is it? So this is the. And this is range. So. And now we want to, okay, so we have for the number of these in two variables over the ophthalmology, and we have this one big page. So we write for psi in the homology of the curve, we write dmn psi for the corresponding tangent problem, then the relations become prime. This commutator is just n prime n minus n prime and d n plus n prime minus one, plus n prime minus one of the product. So there is a special thing we have that is d01. So we have d01 and d10. Right, but when I write it this way, it's just nice. But the commutator is actually not the commutator, it is the super commutator. So this is for any, so by psi I denote the elements in the homologation of the curve, but it's not very big, it stops at age two, so it has a zero, and you mean the cut product here, but you don't mean to stop it. Yeah, the cut product here, this is just a variation for the curve. So h star c has basis then point, which I really denote by omega, and then the class is one, gamma one, and so on and so forth. And this you can insert in this dmn. When you insert one, you get a copy of age two, but when you insert, then you insert other classes to get some kind of extra stuff, which is somewhat like an important addition. So kind of way, once you, for instance, if you have something not trivial here, then forming all commutators will give you again something positive degree. Eventually you will start getting zeros. And then another thing is, and what's super commutator? Well, super commutator of some a and b is a b plus minus plus a, minus, if you have a plus, you have both of them out. So kind of you start passing in this algebra from the smallest part from small degrees, and then you discover, so this commutator is d, this is already one again. Now I should give you this piece of information with d sign is, so it's a tautological class coming from character of the tautological bundle, and then you multiply it by, so you look at that. So e is on the space thanks to curve. Then you can multiply it by sign on the curve, and then you push it down to them. And this way you get one, one commutator. Then sign, you can ask, what is d? What is d zero zero? You see that I have to take zero from plus zero strong character, so zero strong character from my bundle, which is around. Then I multiply by a class of the point on the curve, so it gets from, from the strong times, the point on the curve, and then put it forward to m, because it's from forward. So I simply get the rank as a constant. So you see that you have a copy of the, of the y-algebra, because the commutator is close to it. Similarly you get, so we obtain d zero one, d one zero, and omega plus two, and then you can also put the other way, d one zero, d one. So our algebra has two, well, tensile by, as a module in the curve, as two, in the, in the representation, it will have this two y-algebra setting. Therefore, you see that you cannot have finite dimensional representations. On the other hand, every time you have some y-algebra sitting, someone can just come on into the factory. So you'll expect that y would be given y. So what's the statement of, implies what? It has d seven plus two y-algebra. And therefore no finite dimensions. And therefore no finite dimensional representations. But we are not asking it to have finite dimensional representation. Remember, we have this state star of, and two variables. That's why we didn't need these two variables. The variables identified with these operators in X's. So you can think that the differentiation with respect to y is, you can define it like minus one one, d one zero, the differentiation with respect to the X, d zero one. So now you have, suppose you have this, right, and all of this X on, on this star. Therefore, you can canonically write it as n. So using these two algebra actions, we decompose as a tensor product of this thingy, and we do not write it. Two variables. Where this is, by definition, identified with those elements which are killed by k. And then we also, inside the algebra, we find there is a way to, for this algebra, page two, and actually the homologation of the curve. Also it's written canonically as some smaller algebra for small. And then you have this x, y, the x, d, y. So this is, by definition, the centralizable of this x, y, the x, d, y. So this is, it's a general framework of why an algebra, it can stand away. Then, then, yeah, then there is also that kind of program expectation. Okay, so, and then, and then this smaller algebra is on, on this, but it is much, it is, it's more difficult to write down the regulations. So we prove that, so, the fact that it is generated, I forgot to mention one more thing. So also you can factor out that 0, 1, gamma i, sigma 0, gamma g. Also this is, so, but now this is all the element, and this is all the element, and there is a non-anti-permitative commutator, which, so this basically tells you that you can also factor out all these, these elements, which actually come from this, the map from the, and b to the G-cognom. Sorry, we are looking at the general two-model space in the original framework, but if you want, you can also, so you can also factor out the odd biology, just missing it away. And then, so, and you get even smaller part. So then it is generated by, by some like H, this sign, these are this, these tautological masses, and the, and the operator D to 0. So this unit H2 is smaller. Yeah, there's a little smaller one, yes. And then, so, so now we have obtained that this following statement, that this becomes the following. So this means that, so on, so on this data gamma, model as an ideal, we have action of, of these operators, multiplication by, multiplication by R, multiplication by beta, multiplication by gamma, some operator L, which corresponds to this D to 0, 1. Okay. So, and we want to say that we want to extract some information. And then you compute and double, so first you see that certain, you compute this kind of commutators. This is some computation in the algebra one, and so on. And you see that also that, and this again, each time there's certain polynomials. So the triple commutators will range. So L has, has this differential operator, as you check, because you see that you start with D to 0, commuted with any of this, you get something with 1 in the first position. You commute that second time, you get 0 in the first position. And then, if you commute again, you get 0. So it's a degree or degree, degree 2. And you commute explicitly all the commutators, all the double commutators. Okay. And then you write therefore, you have the L minus. So if you write this for instance equals C alpha, alpha, alpha. And you subtract C alpha, alpha over 2, D to the alpha, D to the alpha. And so on. So you, you want to, I see you put this, you obtain, take this, you obtain a operator now which puts double commutators. So you can compute first commutators. And proceed. So you after this, you obtain that L minus certain explicitly differential operator commutes and alpha will be done. And vanishes. And therefore, it must be identically zero. Because the pathology is generated by the logic of path. Therefore, L is given by the specific differential. And in particular, since L of 0 is zero, you must send the yield to the next specific differential operator, presents some equations. Some of our relations is PRSGC, you gave us. Right. So you can't use the fly out circuit checklist combination. That's what would be possible, but that wouldn't generate. But this, here we have a procedure that can be applied for everything. That's on for rank two. Okay. And you quick question before we go. Now we're about to clap. So let's thank Anton and also Thomas. We're soon tomorrow at 10. Thank you, everybody online. See you tomorrow.