 Yeah, so. Yes. Okay. So let's start with the finding the logical process. I cannot already explain. I can't part yesterday. So let's. So let, let M be. The model is space. So, in fact, for this part, I don't even need to be anything. So let M be some family of kids bundles. Then. On C. So that means that he has. So there is a, there is a bundle key on. This is a bundle on M times C. Right. And then you have this. Tita. He's going to be 10. Seen. And then I want to construct some classes in the co-homology of M. And then, and we do the following just to fix the fix everything we, we write the total charm character of. And then times. I want to. The top of C. I will write it as a sound. Okay. So it's a little bit. By definition, this, the right hand side is this. And so, and then I, but this is this class now needs an H. Okay. Then one thing you can do, you can decompose according to current formula, or you do this. It's better side to side in each IFC. Then you write. Okay. Side. And then you push forward along the curve. So. Project to M. And this, and then it isn't. Each. Okay. Plus I minus. Okay. So this is a construction. And this we call. Sign. Sign. So this is a construction of the political classes. And so you see that you can do it on every, but every family of kids bundles. But in particular, you can do it for, let's say for, for stack of stable kids bundles for some versions of this. And in any case, the suffering generated by the side. So age, the pathological of M by definition is. The suffering of. Ecophomology generated by this. So this is one thing we want to play with. There is another. So, so I need to, I don't need some concept of. Maybe. So this is, that doesn't have anything to do with six bundles. It's about X. Let's play algebraic variety. Smooth. Then we choose if you choose a smooth compactification. And then we look at the image. So the image of the restriction map from. Homology of X bar to homology of X. We call it the pure part of the co homology H. I should be more consistent with my indices. Should probably put out the bottom. So this is a pure part of X and, and you can easily check that it does not depend on. Chose of the compactification doesn't depend it. Well, it can easily check depending on the resolution of singularities. So this is then. Purity and then you can see that the total logical ring is. So it is a fact that operations of push forward. And pull back and multiplying by classes coming from, from multiplying by pure classes. They will preserve, preserve the purity. So this means we find that in general, the total logical ring is a submarine of pure pure. And you kind of play with these two. We are going to play with these two notions. So. And it's there as before. Yes. This kind of family of heat models, or the stack of whatever you put the conditions on the balance. Okay, so this is general. Right, so now I'm going to proceed with construction of. And then I'm going to construct them for a particular particular choice of M and here it's a little bit of technical thing. So we have to fix. And this point on the curve. And we will consider consider parabolic his bundles. And it means that you have a T is a bundle on on C, one on C. Then you'll take a flag, like this is the flag, complete flag in the fire. And then T time. So this is precisely twisted for work. Tita is a map from E to B answered by Omega one and then also twisted by and condition is that Tita, the residue of Tita P preserves the back. This is our M. And then we choose that an open set. And now I'm going to say what is this open set condition is that first you have for each such thing you consider the spectral curve. This is the curve. The pictures as well as you have a, you have your model is based on panels with each man. And fixing a fiber of the kitchen map. And you fix the eigenvalues of all the fibers of detail of fibers. Then the second way is traced out certain curve in the total space of certain line bundle of them on C. And then the line bundle is precisely this, this final. This curve inside this total space of the final. And we asked the spectral curve is. Yeah, it's reduced until usable usable. So this is called the most. Okay. And then we want to put some other conditions, and also we force the residue of Tita P is distinct. Is regular. So this is the. This is the space of one of we will look at. And this is. This open space can be sort of some open subset inside the some space of stable. Parabolic twisted parabolic bundles and but the compactification depends on the choice of some stability condition you can choose once you have parabolic structure, you can, you have a choice of stability conditions. And this is now partial competition and depends on this. But the original set and open turns out that all kids bundles in an open. So, oh, because this spectral curve is irreducible. In his bundle was his spectral curve does not have any proper sub bundle. Because the sub bundle would have spectral curve and some, some component of the spectrum. That's why the piece opens that completely belongs to the set of stable to the monthly space of stable bundles for arbitrage of stability condition. This is our and open. And we are going to deal with, we will consider, consider the optical ring logical of the same open. So, we have the cure. We have the whole going to construct the leak operators. So on this open model space. There is no problem in constructing the operators because, because everything is, is automatically stable. Defining a hacker practice on the usual model space was that that the modification can fail to be stable. But here is everything is everything is fine. So we have this sort of a diagram. And here we need to fix the degree, so the extent for the degree. So we have this one. So let me look at something's correspondence. So this now will parameterize things as before. So it is a point E flag, Teta. And yeah, and E prime, which is inside E, such that one dollar the quotient is a skyscraper shift. And this one was a skyscraper. E prime is colonized one. He's a colonized one search. And then this map is EF, Teta, this arrow and this arrow is the prime. All right, and Teta has to preserve, I forgot, Teta restricted to a prime. And then it's to go E prime. Yeah, so this is there. And therefore you can also restrict everything to the prime and then you get the prime. You get Teta, of course, and then you can ask what is the flag, but the flag is completely determined because, well, if you think you, the flag is just a choice of the ordering of the egg and Teta. And again, very, very restricted to this prime and the same, because the spectrum curve is actually the same. So, so you have this diagram. And then I can define the map, which I call TK of side. The XI was a class in HI of the curve. And this is a map from the pathology of the open to how it is defined. You first pull back along the first arrow. So, so it's given by sending alpha to pull back on the first arrow. Now you have co homology class on Z, which can be singular. It doesn't matter if it is singular or not. Then you multiply by the by the fundamental class of Z. And here I have to use certain virtual fundamental class which I will just say a few words, but then I can multiply by. So what else can we do. There is a map from Z. We can multiply Z to the curve, which is given by measuring reading off the point where we are doing the modification where the skyscrapers leaves. This is. So we can multiply by. By three. Of this side. And furthermore, furthermore we can find the fact that we have So on every point of Z, we have this skyscraper ship. And it's corresponds to one dimensional subs, one dimensional space sitting over some point of the curve. The reading of that one dimensional space gives you a line bundle. And we will multiply by C one or that line bundle to the power key. Okay, so this is general. Okay, the response to the power of this line bundle. And side enters as you can use this class in the curve. The line bundle was on Z. This was, you, this is reading off the one dimensional. That's fine. This fiber of the skyscraper sheet. Or you can say reading. Even the prime is a one dimensional space. So we can see is this correspondence. This trickles for thoughtful. This is a very standard way to take a correspondence is a little bit maybe more general usually you don't have to start this. See one. Yeah. So it doesn't lie. I is here. So I can be one, it can be gamma, gamma, one of the gamma classes or it can be the only one. But what is K is this is here is the power of the line bundle. Okay, then. So these are well defined operators. And they often act on the whole homology of, of the, of the, of the, of the space. So the virtual class, you can see it. Quite easily by looking thinking about the Z is being a zero set of certain section over vector bundle on something nice. So, first, you can look at all the friends. This is some projective bundle over M open times C. And this is, it's really nice it just is projectivization of a vector bundle. And then on it, once you have chosen to find, you have to fix this, you have to use this condition that guitar is strictly defined is preserved. And failure of this is measured by, by certain section by map from from from the fiber of and you're trying to. So if I'm responsible subset of a fiber of E, and you're restricted to the subspace, and you measure if you plan again in the prime or not. So this is, this is some section of a vector bundle. And generally, if you have some variety to describe as a zero set of section of a section of a vector bundle, it has some natural, virtual fundamental class, which does not have to be the fundamental class. If, if the dimensions are nice. If everything is transverse, then it is exactly that. But sometimes it can be something else. Kind of measures this multiplicities or measures some excess intersection. Right. So, you have the condition on this spectrum closing. You have to check that. The spectrum curve does not change. And both the crime and the degree, the degree because I can really don't change. The degree. Yeah, the degree. But I also saw, and this is the, but I thought you said, these bundles were simple. They are not simple. I meant, they don't have proper sub bundles. But it's not the, it's a sub sheet, but it's not as. But it's not as strict. It's not the same. Yeah, it's. What's the best. So a sub bundle means in every point you have a subspace and it was the same dimension. But here you have in one point subspace of smaller dimension. It's not the same that we believe. There is the inclusion that's not going to get the way of the sheet. No, but for instance, if you have a line bundle. The sub bundle is for a right. Right. Yeah, I don't want to say. Yeah, yeah. And another way to say it doesn't have sub bundles of lower. Okay. So, this is, this was the definition of cake operators. And now let's formulate the result. So serum is it. This operators. So first is to this things preserve reserve the total logical summary. And second is that on. So this is the sum of all the logical rings of this MD, when you now change the degree from minus infinity. So we have an action of this trigonometric one. Okay, so where, where D zero M side is given by side. And the one hand side is given by T. So you will think, let's draw the picture. Yeah. Okay, this is the first is just that they preserve the total logical ring and second is complete determination of relations. So we can think about this like they're sitting here side zero side one. And then here is it's T zero T one T two. And then sort of this pictures. This quadrant is populated by other operators. So this is maybe something like, something like D. The relations of trigonometric of this trigonometric algebra. Show you that there is a state forward to the tool to write every Dean's every day in terms of size and T's. So the second row, and it's bigger than zero and things independent of them. So D and M. Right in general, it's populated by DMS. So for three. You have the definition D zero and something D and something that is the one, the one of the one right. And then. So how do you prove this theorem. It's not surprising. So, oops. The idea is, I will say that. So, you can do things one thing is the commutation relation between T's and pathological classes. So first you prove that that. So, that I am. So what will be the relation in the. So this one is the zero and this is the one and so you're supposed to take the determination. Then to the team. And then it is. And this one. Right. So, this is one of the relations we, we have in the trigonometric algebra. And what does this relation mean geometrically. So let's go to the future. So this is the thing here. And imagine what happens if we first multiply by some chart class of some tautological bundle here. This is some chart class, depending on the crime. Right. Some kind of polynomial. So we put it back here. And we get this polynomial in prime. And we know that the prime is he minus the skyscrapers in case you. So you can simply write this character is the difference of John characters of the, and this is extra steam. Now, the job character comes from here. But by projection formula, you can instead, first push it forward and then multiply by by corresponding cycle. And then, and the extra stuff comes from the web tools to the curve and to, and this line bundle which we have before. So, besides we calculated that you get the formula. So this is basically just projection formula. So use that E prime is E minus the skyscraper. Save this and projection. Okay. So this is. Now, second step. Now you imagine you want to apply T and to arbitrary polynomial side classes. You already have a combination relation, you can sort of move them out one way one. And then you will have to apply it to to one. So, second thing is the conflict. T and side of one. Let's go back to our. So we have one here. And before that, we get one. And then we have multiplied by this virtual fundamental cost. For one, this range is this, you don't need to take. You simply have to take the virtual fundamental cost multiply by something and push it forward. So this is a complete completely completed in the series, the framework of cigarette classes. This is the general thing like you have a some projective bundle, you have some section of some vector bundle on it. And you have you multiply by some power of the ontological line bundle and you push it forward. So, using serious process. And what we get is explicit polynomial site. I don't want to write it down, but it's actually code writing some kind of sound or depending on. Okay, so once we have these two computations. We in principle know how to apply T to any polynomial of side classes. And so this gives us a way to write. So any. So now there are several ways I mean, for instance, you can define the separators acting on the folk space. And like, you can define this tease some vertex operators on the space of polynomials on the polynomials in this variables side. And then the non-trivial type would be the reserve of relations. Homology. But so compatibility with this action on on the topology and then check all the relations there for instance explicitly. So, so you see that this theorem now allows us to compute them. So we have everything on the topology. Okay, but there is also pure part of the technology which can be bigger and then the next. So, so it turns out that by Markman's argument theory. It will tell us that that the pure. So I don't know that I think this is somehow not noticed by Markman and by others that he in his argument has to construct some competition that he kind of there is some strange thing. But what he in then proves is true on this. Exactly for the pure, basically what his argument proves that is a pure part equals to the pathological class. The pure part of the space is generated by pathological class. Even if the modern space is open and maybe some more complicated conditions, but this is all true. And here we have one catch that is one thing. One problem. Because we are dealing with the parabolic modern space. And the original argument is about. For the space. So, and then resolving the problem tell you that you have to extend the class, the set of topological classes by something we did it before. In large, by adding these classes that come from the flag. Yesterday you were mentioning the non-parabolic case. So we prove everything on the parabolic case first. Only afterwards we will. So you use the parabolic? Yes. And this is also true in the case of Molly and chance. They work on this exactly the same. Open space. This is related to. So we can when we add this one so we can actually somehow modify. We can replace the place and replace the co homology ring of the curve by some kind of something slightly larger where we so we have ordered this thing. But then we add some classes. So I don't know exactly know it's probably the co homology of some kind of or default curve of the inertial stack over the or default curve or something like this. So kind of to, to keep the same notations you, you just enlarge the string by this mess here. Then the whole complex set of the set of relations everything works. And also T. So then T. I think side K of this, I will be something like C one of this quotient. Maybe after some constant. Yeah, and then TK. You have to then define what is the key guy. And this is something which is certain there is this case. And this is given by doing a modification in the over the points of. The Eigen spaces for the rest of the time. Here you use this one dimensional subspaces where you can do modification. So there are, and there are like, so that this identifies this for this model spaces for different degrees. But there are there are key choices to make. No, what is this? And then times followed by some something. So it's a game something's missing. And then so algebra extends, extend to action of this trigonometric H2. Now tensed by this, by this new thing, which the extended fits. See, which means that's what it's. So now we have that. So conclusion is that this algebra. This algebra acts on. Of the co homology. Open. And there is one. I forgot one thing that you have one extra variable. Here, which you can just think that instead of taking this variety. Here you take into account the C star, trivial six direction by. Yeah, so it's a parameterize it's not just. Yeah, so the actual vector one, but he has a C star. So in principle, it will get a stack representing the corresponding factor. It has this component. Yeah. T, I, TK is the general kind of version of the stopper logic. I mean, it seems to me that inside some kind of version of double, double, double, double. Oh, yes, right. Yes, usually it's called w. Sometimes it's called w. This is in these. Yeah, they are. It's not really. No, just so. Just some part of the co homology coming from complication. And they're all the same right there is a mothic because they always a mothic model space is what you can do. But the, I mean, in the sound each term, it's has no connection with the. You know that the psi, they preserve the degree but the T's, they change degree by one. That's why I need to take this time. And if some of the operators need a line, why do you divide it? Well, as I'm saying it to be, if you're really careful, then you will see that's everything. There is this one. Why is it something like psi of one? Because on the model space, there is no canonical choice of pathological bundles. You have kind of choice up to tendering with some arbitrary line bundle. And once you take into account dependence on that line bundle, you'll get some why some parameter. Okay, now, right when I am now. Okay, so now we have action of something algebra on this pure part. The next step we do, we do the generation from H2 trigonometric to the rational one. So H2 is not really a curve. It is something who's, we think of with cohomology as the cohomology of C and then this extra class is in H2. So you see that the point B becomes N of a point, but they are obviously the sum is again. So the generation to rational, this is some form of procedure. And it uses the fact that if you look at the sequence of psi N of psi, it basically depends polynomially on N, roughly depends. So this you can write it as a y N times polynomial of N, y to the N times polynomial. Operator valid polynomial. And then instead of you see what it does to the whole. So then you implies that D. The DMS side is. So you have this DMS, which is from M D to M D plus M. But then you. So first thing you have to do you can have to identify them somehow and you can use the operator T zero. So T zero omega is turned out to be vertical and take this sequence. And this now depends polynomial on M. And then taking the coefficient of that polynomial, you will get some other operator. So this. Now I have to distinguish I will call this operators. So, once you take coefficients. And this is the right thing to do if you remember last of yesterday how the operators in trigonometric generation should be related by exponential. So, we get some operators D and rational, and then it acts. So theorem is then, then that age to rational. And then there is the homology of C tilde. Now acts on the co homology of over fixed module space. But now it's too extra variables. And, and in this notification D zero M is the same with the econometric. So this is our cycle and pure part. So that's that. So now we have some algebra acting on on the pure part of some model space, and we want to relate it to some other. One proposition is that the pure part equals to the image. So we consider co homology over stable model space. This is asked. And look at the. So we'll consider this map. And of course, because right so so this is known to have pure homology. Because of this direction and everything comes to the zero fiber. On the other hand. So, so this map has to preserve your part. So we have this, and the claim is that this is restriction. Is that this algebra. Maybe I can call this is not just notation. On the corresponding. And now we look at that. So I am supposed to prove equals W furniture. Here's the position. Identifies the penetration. So, because I have to show that there is a perverse iteration on the pure quality. This is actually not so difficult to deduce. So, remember that. Tell us. Explain how to get the perverse filtration in the compact setting. By restricting to the. So the college of the comfort space. So Jacks. There is things to the quality of any open sunset. So you get some filtration on the pure part. And then you can check that that also satisfies hard left. There is some theory. So, and then there's so there is this. We have perverse filtration on the people. So, of the, of the open one space. And it matches. Then it matches office. And moreover, so we have this operator. Okay, this identified as a span of. Okay, I'm going to place it for the operator. 111. I can talk about some details. But how to see this. Basically, there is a cell to the definition of the new status. Yeah, here it means as you stable. But in the end of the table. So in open. It was a little bit empty for the rest of my office. Yes. Yes. But it was automatic. By definition. In the. So you mean this empty. Yes. Now, all, all, all everything in this open part is stable for. I mean, Yes. You have the definition of the public. So the same thing here is twisted and parabolic parabolic. So, which one of those letters is I can see D what. Yeah. Yeah. D 111. This letters are all one. They're all one. And this is, if you remember, it corresponded to that age generator of a system. So. What's the eye. I'm listening. Okay. Yeah. I am in space means what. This is a greater than some I didn't expect. There were greater than some I didn't expect. For some I didn't notice. It's fine. I didn't. I didn't. I didn't. So yeah. You have an order. They're all integers. Yeah. All right. So I use the idea. I think. Ah, okay. Yes. So. Okay. And then let's see. And then you can, in particular, he's a polynomial. So. Okay. And then let's see. And then you can, in particular, he's a polynomial. Since. Since multiplication by side classes, change the eigenvalue. By just changing. Well, controllably, then you see that. This implies. People's W for. For MD stable. We split my parabolic. Then you have to. And then you can, in particular, he's a polynomial. Since. Then you have to. Restrict to the stable parabolic. Don't we sit with this means that. The condition given by the condition. So close subset. Determined by the condition that. So. So it turns out the restriction map on the co-homology is an isomorphism by a lemma from. From the paper. For that. So, because of the C star action. This particular case. The co-homology of the. The stable twisted parabolic. Stable parabolic. And then we prove that we checked that. Pre filtration agree. It's the next step. And then finally. We are in the position to relate that. You look at the co-homology. Look at that. Non parabolic, but just stable. And we have a stable as parabolic. And then there is some space where you have. So this is the spaces are related in the following way. Here you have. It has an important. This is a subset where it is zero. So this condition makes it. Zero. And this, but then it is just. And then you analyze the. Who pushing this diagram. And you prove that. So with this way. The map is given by multiplying anything. By a product of. I was in J. And this map is given by. Anti-symmetrization. With respect to. Action. And sort of this. Let's you identify this. The homology. Direct salmon here. And then we. Some more work. About the perverse equation. So. So. The homology. As a direct salmon. Compatibly. We get. On the. On the. That's. And. Like us. Sorry. Sorry. I don't actually. No, that's, I might be. That's it. Stop. All right. I don't know if there's going to be a little bit after a break or. Yes. Right. It's up to you. Yeah. Yeah. Okay. So then. Let's take a break. And. Maybe. And there'll be. A little bit more. So. So we say. 1130. When there should be coffee. On the same kind. So. In the same place. The same is. Let's start the second part. I already. Formulation. First. Exercise. Exercise. So. So. So. So. This is a vector space. And. And suppose it satisfies this. Property. That's missing. This is precisely like. The much talk. Then this category is a billion. No. Generally, when you have category when something like. The. Vector space and filtration. It's usually not a big. Because. The filtration induced on the image from. One side. It can be different from the filtration images. For any map images as a core is a portion. And at the same time as a sub. And when you induce filtration from the portion can be different from the one used as a sub. In some situations, for instance, this is a category of mixed host structures. Another example of this. And then this is also. Like this. I. Me. The kind of a kind of small miracle. Describe. The possible. There will be. Let me give him some. No. The problem is that. If you take any omega like this and then add arbitrary. Map. Which sense if I to have. I plus one. You have generic map like this and then I don't see how you can. Probably generic one will be decomposable. Right. Associated graded for F. So it sends this triple to. To. So this is generally holds for. In the context of a map. Of algebraic varieties like in Tamash. And you have relative. Projective morphism. Relative ample class. And the perverse rotation. Gives you such an option. Such. Sorry. And by, by my proof. Yes. Yes. Okay. But there is a little bit more. It's. I don't see how you can be used from. People that would be from this. And. I was thinking about just speaking, but. Okay. I will talk about. After this. Yeah. Oh, okay. Maybe then you should. So the corollary of this is that. That if you have. Some. Relative context. Then that. A pure part of the homology. You going mapping to a. Projective. Star. You. Pure. And then. Perverse filtration. Intersected with the pure part. And. Some. Relative ample class. They satisfy. Is a left shift structure. Just. Probably can deduce it from. From theory perverse chiefs. Using the way. Am I. I don't understand correctly that way. The composition. Is clear. So. Your part would just correspond to some. Part of the decomposition. So. Yeah. Okay. So this is where this was used to. Define. Perverse chiefs. Using the way. Am I. I don't understand correctly. The way. Retraction interacts with. With this. The composition. Is clear. So. Your part. Is used to define the. Perverse filtration on the. On the. The open space. Another thing. Now that's. That's answered. That question that. There is another lemma. Suppose. Suppose. So to. Acts. On. Homology. Some. This. So that. Is. E. F. H. That's. The three generators. With. E. With. F. Send acting as follows. H. Of degree. Zero. E of degree. And suppose E is. Is a multiplication by. E is. M. Class. P. K. Is a. Spun. Of. I can value. For H. Where I. Is less than equal K. You have this. Left structure. And suppose you can in some way. To extend the vessel to action. Compatibly with the. Chemicals. Yes. It may be completely determined by. By this. Yeah, so for instance. It is a. I mean, maybe you know. If you have an operator. There exists a unique iteration. Satisfying some problems, even an important operator on the vector. Space defiltration. That's the rush. Row down. Is actually. which is satisfied certain properties in this one and symmetry properties that it gives isomorphism between complementary and graded pieces. That completely determines the interpretation. But I mean, maybe it's better to prove. Proof is that on the associated graded associated graded, we have now two SL2 actions. So in particular, we have, and moreover, yeah, I was mixing up the notation here. Some associated graded, you have two SL2 actions. One coming from, because of the hard left-hand theory, coming from E. Another one coming from this SL2 action. And then you check that the difference, so you have some H and E prime, so E is the same, E prime, H prime. Now take the difference and then notice that the trace, okay, now I have to remember that. Look in the paper, I guess. There was some clever. So the difference H, H prime now commutes with E with E and with H, because H pK is in pK, therefore H preserves on each graded component. So it commutes with the other grading operator. So this, and so it implies that H minus H prime commutes with both SL2 actions, because SL2 action is completely determined by just E and H commutes with both actions. And then somehow I was using that H is the commutator of E and F, H prime is a commutator of H prime is a commutator of E and F prime. And it concludes that the trace of H equals trace of H prime equals zero. And then somehow on the other hand, you can see that the trace of, there is an equality between eigenvalues of H and H prime. Well, anyway, sorry, that's the lemma. So it's not sufficient to have SL2 action for some kind of class, but if that class is an MPL class, then it's okay. Then sufficient to completely determine that the rest of the action. So you have a sense story on the W side and what makes you on the other side of the eigenvalue of H? There it is more kind of just more much easier situation. On the W side, we have W side. We have the splitting of the weight filtration by the Hodge filtration. So because everything is state, state implies that the Hodge filtration splits is opposed to the weight filtration. And therefore there is a canonical decomposition. So the cohomology, of course, there is also conjugated to the Hodge filtration that gives you another splitting, but then you can also say that this filtration is used to split things or equivalently filtration and certain unipotent operator which sends one to another. So, well, so this becomes a direct sum of this some graded pieces like FK would be. So two I's plus some shift. For instance, if it was pure FK to K, oh, that's fine. Okay, that seems correct. And then if you modify the class, so we have this class side two of one in my notation in H2, in F and W4 and H2. And we can choose it to be in that. So this in general, so this is just H2 and this has the following parts. It has F2 intersect W4, F1 intersect W2. So this, you choose a representative which belongs here. We choose modified by the project here. So that W be the projection two of this guy. Now you have a graded vector space. You have a degree two operator and it satisfies hard lectures property. Therefore there is a unique extension to SL2. So we have now grading plus degree two creator and knowing that so by my result satisfies, we have W plus two I to be minus one is an isomorphism that there exists unique SL2. And so I have a part of the dimension of the sub-conscience. It might carry itself on both sides. The dimension of the sub-conscience but this is like a few... I have to actually say two of them. You have to a theory of the same state. You have to this question. One from P and one from W. Why is this sub-conscience the same future? Well, it's not quite... Wait, wait, wait. One from P, one from W. I don't... Are you asking about... We already have the P equals W, right? The P equals W means that the sub-conscience is the same on both sides. Yes. So how do you deduce what is that to action that the sub-conscience is the same on both sides? So wait, wait, wait. Don't you need the sender result? The sender result is the same more than that. Well, this sender result is the same. What does it say more than that? Well, this sender result is just saying that it's just... It doesn't say anything non-privileged. They're already at this decomposition. So there's this decomposition. It is one to be petty. Every psi plus... Every plus psi belongs to some part here plus lower... Well, this is decomposition. I said it's kind of automatic that... Ah, okay. Right, so you write something like this. You write psi, K of psi. And you know that it is in... It has an F, in W, 2K, right? But this has W, 2K intersected FK plus W, 2K minus 2 intersected F. It's minus one and so on. Okay, so I think... So Shem, they're saying that this part... Right, in the original... Okay, in the original Shem, he says that this class belongs to here. It proves that it's just pure waste. In general, the point is that you have to show that you can construct this class by using kind of K, polymorphic K forms. And that's... Right, this is anything that you generate like some political classes. Ah, PHS3. And they start from the same... You just have to have something for the name. And you can generate some by classes of pure waste. Yes, but the errors are buses. Yeah, well... Something could go wrong if this class has something here and that's something new here and this something new cannot be killed by other than this class. But as in general, it's completely... Now imagine if it has some component here and some other component here and you cannot kill this component. Then you're completely... Construct this class exclusively and then there's construction to see that you're only using polymorphic K forms. Well, that's wrong. And how is it done? You have... You have model space of local systems. This H-local system is given by some gluing matrices. And you kind of... You have your surface. You have a local system. And each triangle... You kind of triangulate it and then each triangle has some co-cycle. This way, this is the group. Then you have... Sorry, you have some co-cycles on the group corresponding to topological... corresponding to characteristic classes. Characteristic classes are classes in H2K of BG. And BG can be completed... I'll tell you that it is a simplicial scheme. So you have a point, you have G, you have G times G. And you apply the DRAM complex. You have DRAM complex of the point, DRAM complex of the group. So I have a picture like this. M-G, G times G, and we get two. Well, that kind of goes this way. Right. And then here, there is a simplicial scheme structure. So you have two maps here, three maps here, given by the projection, two projections and the composition. And then there's four maps here. And then by pullback, you'll get maps like this. This is double complex. So these elements here give you some terms in the DRAM complex. And you choose those representatives in case of specific data problems. And then you can see how to cook out some form on the space of work. For instance, a four is given by form fitting here. It is a two form on G times G. More details. Yes. And then in the parabolic case, you should see what changes. Yeah. Individual consultation. Yeah, okay. Well, I, I don't know what, what kind of other steps I skipped. Yeah, okay. Maybe stop here. Okay. Thank you. Okay.