 Zdaj zelo, da je to kontekst. Zelo, da je nr. 1 kontekst. Zelo, da je nr. 2. Zelo, da je to zelo hrv, kako je x zelo smuzim projektev kaj je zelo zelo c in omega v h2 je code in klast for hyperplane atrban so ease for other people it is C1 of the ampoule class ampoule line bundle then there is a operation of multiplying biomega and it goes if you look at it minus k, to hd plus k, je enizomorphizem, da d je dimension d. D je dimension d. V medelu je kohomoložija, v zelo sem taj semetri. In vse obzervacijo, Is it something similar exists in the different world of symplectic holomorphic affined varieties. So this is a vague kind of thing and I want to show some examples just to kind of bring you into the context of what I mean. So up to which level is it okay? Is it everywhere fine? So example one will be here. Example one take T will be even dimensional algebraic torus. This is just the baby example. And omega is a form dTi over Ti dTj over Tj. This kind of form is called log canonical. So it's called log canonical. And suppose aij is minus aji as matrix non degenerate. Then it is a linear algebra exercise to check that if you take operator of clapping with omega, k times from hd minus k of T to hd plus k of T is isomorphic. The cohomology is just the exterior algebra over the character lattice and this is just property in the exterior algebra. But there is something very different between these two examples because here d is the dimension and here d is half of the dimension. So it's not the same. Example two. So we take c star times c star with the form dx over x dy over y. And we look at this as a coordinate line. This is one c and this is another c, so x and y. And then we take a point here and we blow it up. So blow up one zero. And this is in the situation of affine varieties, it is natural to pick a point at infinity and blow it up. It's different from usual blow up where you take a point inside and blow it up. So in the equation, it means, well strictly speaking, I have to first add this line, then I blow it up and then I remove it back. But in the level of formulas, it's very easy to compute. So you just write this equation, x minus one equals yz. So x is still in c star, but y can be arbitrary now. So this blow up is given by this equation in variable x, yz. Now you can eliminate x, so this is equivalent to complement to just yz plus one is not zero. And so you get the complement of the set, yz plus one equals zero in c2. I was happy to see it in Heracov's talk. So what is important about this set? First you check that this form extends to this, so this is my x, x is a blow up in extends to this x. And second I want to say that as I said, x is a union of the algebraic torus in c. So blowing up adds one copy of c. You can see it in the equation. If y equals to zero, then x must be one and then z is anything. So if y is non-zero, then it just coordinates on c star. And then we want to compute cohomology of x. So this is a picture. How do you compute this cohomology? There is some spectral sequence, but so you draw this b called t. So you have h0 of t. So x is an open set t and complement is c. You use a standard technique to compute the cohomology from this data. So I put here cohomology of t. And then there is a residue map from h1 of t to h0 of c, which is z. This is z two times and this is z. And there is a residue map. And after you're supposed to do, you're supposed to take cohomology with respect to the arrow. And this will cancel one copy of z here, with one copy of z there. So the result is to obtain h0 of x is z, then there is h1 of x. Also zh2 of x equals z. Ok, that's how the cohomology looks like, one dimensional in these degrees. And then there is this form. So what does it do? You see that it's kind of, the picture is symmetric and the form provides anisomorphism from upper part to the lower part. So here this is isomorphism. So up to this point, comparing to the torus, nothing changed. We just removed one z here. But now in the next example, maybe before going to the next example, how do you raise? And is there also some? So I just write on this. And of course, after you saw that you can blow up one point, you can try blowing up more points. And so example three is the same, but also blow up some other point, 0, 1. So now you're blowing up two points. Now you can compute equations, you get x minus 1 is yz1, y minus 1 xz2. You eliminate x, you get yz1z2 plus z2 minus y plus 1 equals to 0. And I don't know for people who are in the, who knows this. So this is so-called, so-called augmentation variety, variety associated to the trifoil knot. So to the trifoil knot. And if you, so there is this triply graded, graded Havana homology. And it produces some polynomial a plus q plus t. And at a equals 0, you get q plus t. And, but this computation of cohomology gives you h0 of xz, h1 of x now is 0. Right, because after you do the same, it will kill the other z in h1. And z, and omega does this. And, and still nothing interesting happened. But I just wanted to give you this example, because this is geometry behind this formula. Is it at a equals 0 is q plus t, it's two dimensional. The cohomology of this is two dimensional. And there is a way to match this. Anyway, so, so you see that in this world already some kind of knots appear. And then finally, finally example 4. I'm going to blow up 6 points. And so, example 4, blow up 6 points. Let's see. Now, remember that if you blow up 6 points on p2, you get a cubic. Recall, blow up 6 points on p2, produces, produces cubic. And it turns out that our blow up will, our blow up will reduce affine cubic. And this is, this way you can obtain a character variety of, so this will get character variety, 80 of p1 minus 4 points rank 2. And this is computed by Freake. The famous, you have three variables, one cubic equation. But, so again there will be symplektic form. But let's do the cohomology computation. Here something will have, something will be slightly different. So we have the torus. This is z2, zz. And then there will be six copies of, so it's h0 of c, c, c6 times. And this is z to the sixth. And this map residue. Now, now the residue map will kill all here. But there will be z to the fourth here. So results of the computation is, we have here h0, z, sorry, x, z, h2 of x, z. There is zero here. And now the question is, there is some z to the fourth here. So where does it go? You know the spectral sequence business you should read your result by diagonals. So this will be, sorry, this is not what I meant. I meant there is z here, there is z4 to the here. And this diagonal is h2. So h2 of x got z from the torus, one z from the torus, and four z from these attached lines. Here it was z2, z6, z2, and z2 here canceled, I got z4. So what happened? So the omega still gives you this azimorphism here. But the symmetry does not respect homologi indices. Because this is h2. If you wanted to have curious hard left shits as before, then this h2 must be azimorphic to h0. But there is no h0 paired with this h2. So the explanation is that picture is not nice, is not symmetric from the point of view of kohomology index. But it turns out, but it's symmetric is from the point of view of weight index. So what is the weight? Weight is when you do this calculation, you should keep track of number of points you get over finite field when you do this. So, for instance, here this is like h0 of t is a generic point of the torus. So over finite field you get q square elements here. So this has weight, so weight is q square. And here it is q and here it is 1. And here, so I mean 2 copies of z would mean 2q. And the number of points on the torus is q square minus 2q plus 1. Here we have 6 copies of the affine line. Over finite field we have 6q points here. And when you keep track of the weight filtration, then this 2q kills 2q from here. So if you put weights on this picture, this will be q square, this will be 4q and this will be 1. So in the picture degree in q will be always the height. So in general the picture will be symmetric with respect to the q degree. So here are some notations, so you always have q to the weight over 2 where w is the weight. So this is standard. But I mean I could spend a lot of time explaining precisely what it means. What is the Hodge theory, what is the mixed Hodge theory and so on. But for this talk I hope this kind of picture is sufficient. And now I will formulate first some precise statement. No, no, no. So we have like first theorem. So this theorem is just naive generalization of what has been going on. Nothing really new here. So fix D. Supose x is filtered by closed subvarieties i, goes inside xi plus 1 and so on. And xi minus xi minus 1 is isomorphic to some algebraic torus and some affine space which are related by, once every time you decrease dimension of your torus by 2, you increase dimension of xi by 1. So suppose x has a closed holomorphic form 2 form omega. Saj set, that restriction of omega 2, such c star to each piece is a pullback via projection of log canonical non-degenerate form on the torus part. And then we have the curious hard layers holes. Which means that, so basically I explained what it means on the picture. Sorry. So you take some, you start somewhere to the left of D. We divide by half. So this corresponds to q to the D minus this weight. And then hj. And then you take product with omega i times. It is increasing the weight by 4i. And then h. But it messes up the homological index as it should. So j plus 2i. The symmetry is respect to weight index, but the homological index is not controlled in this. So this is an isomorphism. So basically in meaning of this is that the same thing happened as in the examples 1, 2, 4. Questions for the first part. OK. Now I will give you. Yes. So this is beyond my reach. There is some construction of the lean that everybody is using as a black box. So roughly whenever you do all computations with cohomology, you should know that all operations preserve the weight filtration. And from this you are supposed to deduce the weight filtration on the result. And then you take associated graded. Grr is associated graded. So now we go to context level 2. So go sort of more real technical stuff. So let me draw the picture. The usual picture character variety. Yeah. So non-canonically you get the direct sound decomposition. And this is the pieces in the direct sound decomposition. So I assume, so parabolic. This is a picture. It's not precise. But this is always a fine variety. Well, in most cases naturally. Like the cubic from Freakia. Yeah. This is usually a fine variety. And then they have this model space. Same stable of stable. Stable Higgs bundles. This, which comes with the heating system looks like. Your model is space and goes to the ID. So, and then there is a non-nabilian hodge correspondence between. Non-nabilian. And what do we know about this picture? So. Observation one. So howzel. Letalia. And Rodriguez. Gave. Made a conjecture. And in the conjecture. Is explicit. Explicit formula. For. For the refined. What is punkare polynomial. So refined here means. That. Means that the weight filtration is taking into account. The usual punkare polynomial has one variable. But if you had this, you get two variables polynomial. So this is a polynomial. In the Q and T. Two variables. Of character variety. So I will take a long time to explain exactly. What is the level of generality they make this conjecture. And what kind of character variety. So. Let's keep it here. So there's some character varieties. Pretty general conjecture. But one thing that. They noticed is it. After making the conjecture that there is mysterious. Mysterious. QT symmetry. This is one observation. So if you saw this conjecture then it has some McDonald polynomials. Some arms and legs. And. And it's symmetric in Q and T. Now if you. Translate to what it means in the. Co homology. Then it's exactly this kind of symmetry that I'm talking about. And they. Conjecture that so. Now another observation maybe deeper one. So maybe also inspired by this observation. With the cattle don't make any. Is that P equals W. So the quest here is that we have this weight filtration here. But if you pass to here there is some other. It's not clear what it is. What filtration it is. And P equals W conjecture tells you what it should be. That. This weight filtration is. Coincide with the perverse. Filtration. Is the perverse liraj filtration. For liraj filtration you need the map. So. So it is this induced by the hitching. By the hitching X to C. To the D. OK. So. They checked it for some case with gel to. And then the question is. Yeah, so what happens now for the perverse filtration on that side. And there we have some sort of relative. Relative hard left sheds. Usual hard left sheds for. Because the fibers of the hitching map are compact. So relative hard left sheds translated. P. W. Implies. Would. Impli. So first. So somehow the picture is. Complete once we assume this conjectures. And then I just want to make some. Remark. So it turns out this curious hard left sheds. Like you can deduce it from P equals W. But then you can ask can you do maybe the other way around. Can you. Deduce P equals W from. Curious hard left sheds. And it turns out that. That. Curious. Hard left sheds plus. Plus P equals W in one direction. Implies. The full statement. So this is probably the way to. We will use in the future to prove P equals W. There is a paper last week. An archive by. Maulik. And the cataldo. And do this for some special case. Genus to. Curve. The first is. Yeah, just somehow observation. I'm going to take. This was. If you want to. Here. So. We prove it for character varieties. By. Using the observation. About the decomposition in the Torah. And then some more work. And then so here. So. There will be a sequence of. Things. I want to say so. I start with one. So there exists. Such picture X. Tilda. And then there is Jill in parabolic. Character. So this is X. Variety. Is at least. One parabolic point. So I require that. In at least one point. There is some monodrame. And the monodrame has distinct eigen values. That's what it means. Or. There is slightly modified definition. When you are allowed to have the same monodrame eigen values. But then you have. Need to. Give me a choice over complete flag in that. Over that point. So for those who know what is parabolic. This is. Then. So first thing says that there is this. And choose to. See to the interest to battle. X tilde. X tilde. Admit. A stratification. So that. Each. Piece. Is. A vector bundle. Over. Certain. Variety. Why beta. Associated. A braid. Beta. So. This is. I'm not telling you what is why beta. I'm just saying that there is certain construction of some class of varieties associated to braids. And then I can stratify. This X tilde by. This varieties. Up to this thing that you might pass to a vector bundle. But which. Passing to vector bundle doesn't change kohomology too much. Just the weights changed. So. Now. Each of these. Why beta. Admit. Stratification. So that. Each piece. Piece. Is of the form. C star. 2d minus. 2i. Times. Ci. For 60. For same. So. All. In in the end you get some stratification of X tilde. And the point here is that. All these these are the same. So we have this nice decomposition. If you. You want to know where it comes from. It is very brutal construction. I have this one point with this parabolic point. And I have some loop which comes back to the point. After I follow the loop I have two flags. One original flag. But after manroma I have another flag. I compare them together. I get some permutation. So once you fix all permutations appearing in the picture. And you ask what is the corresponding part of the character variety. You get this variety associated to braid. And braid is just given by these permutations. You know sequence of permutations give you a braid. Each permutation can be lifted to. Sort of smallest permutation. But then when you compose them it becomes non-trivial braid. So for instance I know. This will be braid corresponding to transposition. But then you compose them. You get a longer braid. So. So here we have a permutation. And then composing. In this. Pi. Gives a braid. So now. Ok, so we have this cell decomposition. And we have. And remember it's still in the special case. When we have one parabolic point. The second thing we do. We have to keep track of the form. So we. So restriction of. Of goldmans on the. Character variety. To. To why beta. Gives certain canonical form. Gives. How did I call it? Certain natural form. Form on why beta. And then we have to deal with this form on why beta. In this bit painful to compute. But. Further restriction. To the. To the component. Gives canonical. Loc canonical. On. C star to the 2g minus 2i. And the whole point is to prove. That this form is non degenerate. So we need to. This. Is non. Degenerate. And then I don't know following. I followed. Suggestion. Of. There is a construction. Construction. Of a surface. Whose. Whose intersection form. So it's a topological surface. And there is intersection form on. The one. And this coincides with. This form written in coordinates. Aj j omega. So. This is. So. There is certain combinatorial construction. And. And then. Calculation that tells you that. This. Agree. So this. Surface is like. Covering of the original surface. The strange happens. Around singularities. There is this braid. If you are interested, I can. Tell you. More about this surface. OK. So. So from here we deduce the curious hard left sheets for. So this implies. Curious. Hard left sheets for. For the character variety. With at least. One. Parabolic. So this is somehow first. Milestone. Maybe I have time to make one. So one comment here. So if you are interested in this. If. At least two parabolic points. Exist. Then. Character variety. Contains. A dense. Algebraik torv. So. I don't know how well this is. This is known or not. I talked to Simpson and he told me that. He has a conjecture. Like this. And. I can. Show it under this assumption. But I think it should have been. It should be known to. To some people. So. Is some. There is this cluster. Coordinated some character varieties. And this should be the open. Open cluster. One open chart. That's. Okay. Now we want to prove it for general character varieties. Not assuming parabolic point. So here is what I can say that moving around. So it's. In trying to beam. Kind of. The standable. The eigenvalues. Moving around. Around a good point. Just. Kofomology doesn't jump. So. So it means that. Forms. Local system. On. Over. Neighborhood. Of a good point of. So specifying. To. The eigenvalues. Good means that. The character variety is non singular there. So. Or if you read. If you know the notations of how the. Rodriguez Vegas, then. Good point means. Means generic. In the sense. Of how the. Generic. So somehow this is some. Technical point, which is. So I was so that it's conjectured. By. Let me. But somehow it didn't follow from general theory. Because. Character variety is not compact. So if it was a proper map. Then smoothness. Guarantees that kofomology doesn't change. But if it's not proper, then. Doesn't follow. But you want to deform not only stability condition, but also eigenvalues on the other side. There is some. Almost. It's almost there, but somehow. Not just. Okay. So, but using the cell decomposition, you just see that each cell is just constant. Over that. In this vibration. So it's clear. So this kind of observation. Plus. Springer theory. Theorie. Implice. Six, which is. Six is here. And six will imply. Seven. So six is that. The kofomology of X. Is non. Parabolic point. No parabolic points. Is kofomology of X. Was parabolic point. In that permutation. So you can. Find the kofomology of. Your. Character variety for complete curve. Inside the kofomology of character variety. Is parabolic point. By just taking invariance. And then numbers. So. And this implies that curious hard left sheets. For general. Without the assumption that. Parabolic point. Parabolic. So it's not quite true that for instance the class of omega. Of the form is constant. But if you pass to the associated graded. In the weight filtration it becomes constant. Basically the picture is just constant. And for the associated graded. This is just action of a sandwich. Is also constant. And there is invariance. You cut it out. And that's nothing happens. So. Right. The last thing I wanted to say. Maybe if you have a question. Or maybe I remember some comment. Sorry. I want to. Show you the favorite. Thing. That. This surface. That. Looks like this. So for instance if you. If you want. Trifoil case. It was example. Three. The surface looks like this. So I have this building blocks. And then I have to do it again. The trifoil. Is that you have three intersections here. So I have the surface. Now this is. Is has one boundary component. Right. If you follow. The boundary component. So if you attach a disk. And intersection on this. So this. Intersection form here. Form encodes. Dx over x dy. Over y. From example to. From example. Three. But in general you have this funny. How do you say this? When the for Christmas you do the. Cut some stuff out. And you get such pictures. More complicated pictures like this. But they just. Generally you attach some disks. And you get some close surface. And there's some intersection form. That's what controls the simplicity form. Yes. That's it.