 just wanted to check. Okay. Okay, so, yeah, let me make one more adjustment. Yeah, okay, so should we start? If that's okay. Okay, all right. So, yeah, this, so this, this lecture is, there's nothing, absolutely nothing original about it. In fact, it's a material that's, it's, it's, it's about a subject that was developed for 30, 40 years ago. So, but, but it's inspiring for some, some things that one can do new things that one can do in a different direction. So it's sort of, it's some sort of perfect. It's just an example where everything works perfectly. So then it's worth kind of keeping in mind that's why I want to go over it. So, so let me start and let me, I'll share. That's okay. I just, I checked before, hopefully this, this is okay because, yeah, okay, so yeah, so we'll do the following. We'll talk a little bit about, you know, just some basic geometry, basic tools that we need from the geometry of a billion varieties. Sorry, sorry. And the speakers initially with the Alina and Lodar, and this is the second talk. Also, I put in the chat a link where you would be able to find the recordings of the talks that happened in December and you will also find the next one there. So, sorry for the interruption and go on. We are happy to hear. Thank you, Julia. Thanks a lot for organizing and for, for putting everything together. So, yeah, sorry for the slightly disorganized start here on my part. So, but we'll, we'll follow the following. We'll go through the following. So first of some geometric preliminaries, then we're going to talk about the Fourier-Mukai transform. Then we're going to discuss the Chowdy composition. And we're going to talk about the SL2Z action. And finally, hopefully we're going to get to this in finally the SL2Q action. Okay, so that's, this would be the plan. And we're here, you know, the main, I guess, the main actors developing this subject where, you know, it starts with Mumford. And then also, you know, goes for Mukai, Boville. More recently, the Dutch school, so Moonan, Polish Cuk, etc. And in this talk, we'll essentially follow Mukai and Boville in terms of just the basic exposition. So let's, so let's start and please feel free to interrupt it at any time. Okay, it's not, it's good to clarify things as we go. So, yeah, so let's, so A will be a principally polarized, a billion variety of dimension G. Okay, and we're thinking about this over the complex numbers. And so, L is the principle polarization. So it's an ample line bundle on A. Okay, and so here, principle, the space of sections is one dimensional. And this is the same as the homomorphic color characteristic. Okay. So, yeah, so actually it's not necessary to restrict to the case of a principle polarization, I do it just because it's just a little bit simpler technically. And but in a kind of a trivial way in a way. So what I, so I will just, it's not in any way essential. Okay, so let me talk about some natural morphisms in this context we're going to use. So we have the, the addition map is important. So here X, Y, we call it M goes to X plus Y. And the fiber wise version of it, which is the translation by X, let's say. So this Y here, Y goes to X plus Y. Okay, so that's one. And then for any integer N, we have a multiplication map. Okay, so here X goes to NX. And this N is of course an isogenic. And it has degree N to the 2G. Okay. Okay, so I should say also that's just basic identity, of course, is it's a, if I, if you take a cycle, so any class alpha, and you pull back and push forward by N, you get the degree of the map times alpha. So for, for any alpha, let's say in the Chao class, so a cycle on A. Okay. So, so these two operations up to this multiplication by the, by the degree are inverses to one amount of pullback and the push forward by N. So I should also, well, I will also have the assumption that actually I can phrase all that is L, the polarization is symmetric. This is not that you can always do this. So in some sense. So in other words, it's pullback under minus one is itself. Yeah. So, so it's a symmetric line bond. Okay. And so under these conditions, then if I pull back L under, under N, then this is in fact L to the N squared. So this follows, this is, this is a formula of Mumford essentially. And an excellent, I mean, I, you know, depending on your familiarity with a billion varieties, but in any case, just an excellent reference to learn the subject is still Mumford's book on a billion varieties. So, so, so this is important. And if I, this, this will use this formula because we'll be interested very much in how either sheaves or, or, or Chao classes behave under this morphism, the pullback under N under the isogenic N. So, so this is important. And let's say that we set the first term class of L, we're going to call it theta. So this means that if I, if we pull it back, this is N squared times theta. Okay. Okay, something that we will use. Okay. So now, so this is as far as the essential maps, but so the second main point here, as we said, the stage is the Poincaré line bundle. Okay. So, so we have P on A cross ahead. This is the dual a billion variety. Of course, this has the, this Poincaré, of course, it's a universal object. So if you restrict to any A times a, so this is a dual a billion variety, this is peak zero. So they can view it as the group of line bundles with C one, zero on A. Okay. So if you restrict to A times a line bundle, so point which parameterizes a line bundle, this is actually isomorphic to the line bundle. This is the universal property. And also there's a normalization here that we use if you have this property and then if you restrict to zero to the origin on A, this is, we want this to be the trivial shift on A hat. So this is a normalization. You can normalize this anyway we want. But with these, it's sort of, it's a unique with these properties. It's a unique line bundle with these properties. Okay. So what's immediate to see in terms of the interaction of the maps with the Poincaré is that if we pull back the Poincaré under the map which is multiplication by N on the second factor, this is P to the N. Okay. So the Poincaré line bundle pulls back like this. And the way you check it is really, it's really fiber wise. So you use this seesaw lemma. So it's using the seesaw lemma which is extensively used by Mumford. So, you know, the two restrict similarly on the, I mean, it restricts, I'm sorry, identically on the fiber, the pullback and the P to the N. So then it's easy to see. Okay. But yeah, so for our purposes, we're going to simplify things. We're going to always identify the dual abelian variety with A. So this is, I guess, the last thing we're going to mention before we move on to start discussing the Fourier mochi. So we'll identify a hat with A using the polarization using L. So this is Mumford's map. So this is the hat and here we send X to, right. So this is the identification. This is a line bundle of degree zero here over first-gen class zero. And this is a group homomorphism. And moreover, yeah, so this is this fiel. So this is basically structural results that go back to Mumford. So fiel is an isogenic and only fiel is ample. So it has a finite dimensional kernel. And if L is in principle polarization, in fact, this is an isomorphism. Okay. And so under this, we're going to work with, so we want to work with the Poincaré on the product A times A, not A times A hat. So then we're going to use, in fact, for us the Poincaré, so we identify and you know, you're going to excuse the abuse of notation here. So we we're going to, the Poincaré on A times A will be simply the pullback. So we pull back under the Mumford map and of course the identity on the first factor. Okay. Now we're going to call this P as well. So we don't have a million names for it. Okay. And then important point which is sort of useful in calculations is that again, it's easy to check. And again, you proceed here by the same idea with the CISO, that this Poincaré can be explicitly written then in terms of the polarization as follows. So it's pulled back under the multiplication map. It's very symmetric. So this really, it's immediate to check. So I leave it to you to check, but this is the expression for the Poincaré and we're going to use this every, yeah, so this will be used for us. Okay. So this kind of ends the preliminaries. These are the basic objects. We're going to work with the Poincaré line bundle and we're going to work with these natural maps or multiplication by end, the addition map and so on. Okay. So let's get started with the Fourier, the Fourier mucay font. And here we're following mucay 1981 paper. So this is 40 years ago now, yeah. So this is his classic paper duality between the A and the A hat, et cetera. And the applications would be to pick our sheets. Okay. So 1981. Yeah. So we're just, you know, we, let's define the functor. It's just a functor with Kernel, the Poincaré line bundles. So let's say that we fix projections from the product A times A. So then f of alpha is two star f tensor, I'm sorry, p here, Poincaré. Yeah. And here everything is derived. Yeah. I mean, it's all, it's all, it's, it's, yeah. So you think of derive push forward and so on. So derive tensor product. So, so then the mucay's main theorem is calculates the self composition of f. Yeah. So it's kind of foundational theorem in the subject. So yeah. So up to this shift, this is the theorem. So up to this, well, some multiplication by minus one, there's just the composition is the identity up to a shift by G places in the derive category. Okay. So, so the main, so the main ingredient, so there's one, the theorem is, is not hard. I mean, it's just a great way of organizing this. But there is the, the main ingredient with the geometric ingredient goes, it goes in, it's this calculation of Montford. So well, in fact, I think it goes before, actually before Montford. So anyway, so the main ingredient is to understand how to calculate the push forward of the Poincaré line bundle. Yeah. So, so the point is that, so I'll just say, I mean, I won't, I won't do the proof of this theorem, but I can tell you that once you have this ingredient, it's, it's easy. So this is the skyscraper shift at zero shifted by minus G. Okay. So if you, and so what does this mean? Okay. So this means all direct image shifts are zero, except in the top dimension. Okay. And of course, in top dimension, you just have the structure shift for the origin, which in turn is based on the following fact that is just a classic fact that the cosmology of a line bundle with C one zero. So, so topological trivial, topological trivial amount is trivial, zero for all I now, because that's what that's to calculate the push forward except in the case when, in the case of the trivial sheet, yeah, when it's obviously not zero. Yeah. So for all line bundles, which are non-trivial, but topologically trivial on a, all the cosmology vanishes for all, obviously this is not the case. And so that's why you get this, this, this formula that that's exactly what the push forward of the Poincaré line bundle calculates. And this is the main ingredient in Mokai's theorem. Okay. Yeah. So now I should say this, we're going to talk about it in a few minutes a lot more, but I should say that using the churn character, you can view this f in chow as well. So, or, or cosmology for that matter. So we can consider this f acting on sending the chowing itself. Okay. So, so then on this level, of course, this f of alpha, it's exactly as before here, instead of the Poincaré, I have the churn character, it's a line bundle. So let's say that the first churn class is lambda, where here lambda is the first churn class of the Poincaré. Okay. So this is, this is the transformation on chow level. And then of course, if I write this, if I write Mokai's theorem in chow, this is all this involution stays the multiplication by minus one. And the shift becomes a sign. So it becomes minus one to the G on the chow level is that's okay. So let me also state, and I won't prove these, but again, they're not hard. So the beauty of this result of Mokai is this whole paper that got the, got the subject started is that in fact, it's just a very elegant way of organizing geometric information that sort of already exists in some sense, it was floating around them as this result about the Poincaré about the the homology of line bundles was was was known, right? So it, but it gets organized very beautifully under the Fourier Mokai. Okay, so let me say that so this is two important properties here of f that we have there. Well, and again here, the main reference is really Mokai's paper is wonderful. This is wonderfully explained. So if you take a product of two objects, this is and the Fourier Mokai of that is the so called Pontriagin product of their images shifted as a shift by G here, I'll explain immediately what this product is and then conversely, if you take the Pontriagin product and you apply the functor f, this is the product of their individual images. Okay, so here, this is the Pontriagin product is defined as the push forward under them, the addition map of the tensor product of the two shifts on of the two objects on the two factors. Yeah, so here and so here you see you have a cross a you have a one on the first two on the second and this is M. Okay, so this is called the Pontriagin product. Okay, so the Fourier Mokai exchange is tensor product and Pontriagin product. Yeah, and the second property that we will use is just it commutes in a sense with involution. So let's let me and this is what I want. Yeah, so let I'm sorry with them. So let's say that we have an isogen tau, then applying f and then pulling back by the isogen is the same as pushing forward by the isogen and then applying f and vice versa. So in particular, applying f commutes with pullback under the involution minus one. So it's something that we'll use so we don't have to worry about the which order you apply this in. Okay, so so this is the second so these are the foundational we're going to get to the action but we need these foundational results. First thing that I said is these two properties of the Fourier Mokai functor are actually not difficult again it's well some calculation but it's not a long calculation and there are some tricks just slight tricks involved and we'll do one calculation with one trick so you get a sense if you have never seen this type of manipulations before and we'll do it in a second when we talk about the SO2 action. Okay, so let me now move on and discuss the the Chao decomposition now we're in a position to do this and then we'll finally get to these actions. Okay, yeah so this starts with Mokai's identity so but we're going to use it in in Chao so we're going to write so we're going to write the following we're going to write that we're going to write a formula for the identity means the diagonal right as on the identity as an operation on Chao corresponds to the diagonal is a correspondence so um so here is what we how we write it so the self-composition of F up to these two things rather trivial multiplications here is the identity and we're we're viewing this on Chao now on Chao level okay so that so that gives a formula then for the diagonal I can write the diagonal explicitly with diagonal this is the identity diagonal this is the identity operator yeah so this is so this is a formula then in ch star of eight times a so viewing so viewing these um functions as correspondence so this is a composition of correspondences and we're we have a way to write the identity this is sort of comes out of Mokai so here let me make sure I have space well so there's nothing I have to I have to carry these trivial factors here and in fact I'm sorry yeah so this is just on the second the minus one comes after you apply the functor so in fact if I write it as a corresponding correspondence form I have multiplication by one on the so it's the identity on the first factor minus one on the second and now this is applied to all the composition of the two functors of f with f so here is how I write it on Chao level you just have multiplication of the the two-chern characters and you see here this is pulled back this lambda one two is pulled back so this is a how do I you read this formula well this is a push forward from three copies of a yeah so this lambda one two is a class on the first two lambda two three this is how you compose correspondences and then of course I push forward to one and three yeah so very simple okay but so now I'm just going to write this explicitly because what I want to understand is how so you can split this up this is a mid-dimensional class obviously the diagonal so I can I can just write it explicitly and if you want a little bit an entry kind of it's sort of a rather dumb thing but it has a consequence so here is what we're going to do we're just gonna write these exponentials out so that you see how this plays so this is a sum oh there's a push forward here let me okay so I forgot there's a p1 3 here push forward sorry this I should write this a bit nicer it's terrible this sum right from i 0 to 2g yeah so these these are uh yeah so okay so but so now well let's just write this a little bit one more time because we know what so this minus one will just act on on the second on this guy yeah so this this will just act here because it acts in fact on the third copy of a if we move this inside okay so so in fact then um yeah so here we have so let's write it like this so there's a sign that gets picked up here at the expense of just getting rid of this involution so this is a combinatorial factor and then you've got this okay so we haven't done really anything but the point is that now so I will give a name so this this term will just call this this is a class in a mid-dimensional class in in in the chow of a times a and we're going to call it delta i so what we've written so then we've written this delta as a sum of these delta i's and what's the reason we were interested in this is that their eigen values these delta i is the way we wrote them their eigen values with of the multiplication by of the pullback by multiplication by n so in fact we have that if we pull back under one n this is n to the i delta i and this goes back to you see this end and acts in here acts on this copy you see acts on on this one and as we talked about how the Poincare is an behaves nicely under under pullback by under one n star right so this is exactly what this translates into so um okay um so so then it's also easy to see that these correspondences delta are actually orthogonal so if you compose two you get zero for i not equal to j and this is the important so okay so but then the conclusion is rather nice so this is a it's a very simple conclusion out of mukai's main theorem namely this means that because we wrote the diagonal in this manner we broke it up into uh into eigen correspondences for the operator which is pullback under one one n this means that every um chow class can is written in fact as a linear combination of of eigen vectors for for for n star yeah so any this is a conclusion and this is the chow decomposition and here of course we have q coefficients in fact whenever we talk about the chow class in it's always with q coefficients because you see we have all these denominators floating around so there's no one there's no other way to do this um is written as a linear combination of eigen vectors for n star and namely in the in the following way of course we we just write using the diagonal being very it's very elementary now i'm in here so we intersect with the diagonal we push for so so then in fact this is this alpha is written as delta i okay zero two g let's call this alpha i this and because of the property of uh this is an i because of the property of delta i of the correspondence itself when acting with one n and one m star well alpha is inert under this action because it's pulled back from the first factor but overall this this pullback then this push forward this alpha i then is an eigen vector with eigen value n to the i okay so so then so then we have obtained that the chow the chow groups decompose as a sum of eigen spaces for n star and any n gives the same decomposition it's completely clear from this uh so so this is the result of boville sort of classic result of boville and then refined on refined by so one one so this also plays a role for our s l too so which is why i'm insisting come together in but so let me then also note so it's just a basic remark here that we're going to use uh is that um as a correspondence we can write the multiplication by n star then correspondence we have that n star is equal to the sum of these um eigen eigen correspondences delta i weighed by n to the i okay so that's the first and another way we could write this is also it's n star what it's the transpose of the graph of multiplication by n so here just incidentally this is just multiplication by n so it's just the graph is a mid-dimensional cycle in in in a times a and and the transpose means that you're going uh your your um pulling back from by q and pushing forward by a and if i'm looking at the correspondence itself that's the push forward this is standard so gamma i need to push forward and the transposes is the pullback okay and just the way that's the way me stick to the way boville actually writes this in terms of the decomposition itself let's say on on cycles of co-dimension p or by checking what's possible with with these with these delta i's you will see that in fact not all values not all eigen values are actually allowed well well actually yeah okay so um so this so the summands are um so the eigen vector eigen spaces so this is alpha such that this again boville's notation which is actually yeah it's pretty good so we it's just a slightly different way of labeling yeah so so here is the the co-dimension also enters yeah so so the interest in this chow decomposition comes i should say so so so it's an immediate consequence of of the free mochi as we saw um but it's it's a it's a it's a testing the the chow when a billion varieties over testing ground for these uh sort of sort of overarching uh uh block balancer idea circle of ideas and conjectures yeah so you want to check them on a billion varieties and this is this was boville's interest in it as well okay so yeah okay so now i'd like to uh so this was the third part of what we want to discuss and now i actually want to move on and talk about these these actions finally so we've sort of prepared the ground i think we're in a good position now um so let's talk about the so to z action first so so we go back to this the basic to the basic identity in mochi's theorem that if we self compose f we get this okay so somehow with s l to z we're gonna focus on just we want this to represent an element of s l to z this f and which element would it would it be so let me write the basic basic general sort of a standard presentation here for s l to z okay so these two these two matrices generate as is well known and so of course we have this is the identity so in fact s squared is minus the identity and of course s d cubed is also minus the identity yeah so you see um we'd like to essentially map s and we will to f okay as a first to to start defining the action yeah so then we're gonna worry about this t in a second um and and that's that's promising you see you but it's up to a shift right i mean if i yeah well so let's let's go with it a little bit okay so so let's say that we aim to i should say that i mean so we aim to produce a representation s l to z so morphism into the um group of auto equivalences of the derived category and in fact on the level of chow and the automorphisms of the chow of the chow ring and that's actually going to be simpler you know so yes we want to send s to f and here is what we're going to attempt with t t it's going to go simply to tensoring by the principle by the ample line model the principle polarization okay and so then we do have that f to the fourth if we this is a shift by minus 2g so in the derived category this is just up to a shift and what i know this is these are ways that you have an s l to action up to a shift this can be made precise i mean you can actually save more in in more rigorous mathematical terms but certainly in chow since this becomes a sign it's okay actually all right so this is the identity okay because it's minus 1 to the 2g so it's not no problem there but let's um so the one calculation i wanted to do is to actually check this relation so i want to check that um now if we compose f with the operation of tensoring by l and we cube this this is all that there is to check in fact you get this okay so i propose to you we check this completely explicitly and then we know we have this s l to z action okay so uh yeah so here is what how we're gonna uh we're gonna start so so we get started here so we're gonna start by observing this is a crucial thing we're gonna work a little bit on the Pontriagin product so bear with me so e is a random element in the derived category and we're gonna take the Pontriagin product with the polarization l okay so this by definition this allows us to work a little bit actually in an abelian variety which is kind of nice to do so so this is what this is okay by definition but then what mokai notices so this is entirely mokai's argument for this it's a beautiful argument so this can be viewed in fact in the following way well i'll explain in a second let me write it so this m becomes a projection on the second factor and this p2 becomes well we're gonna call it delta and this delta is a subtraction map x y goes to y minus x okay so how do we understand this well here is the trick mokai trick so we map so we have the addition map here that we want to transform into a projection and how do we do it we use a map which sends x y to x and x plus y so this is an isomorphism let's call this m tilde and now this we project on the second factor this is a yeah so so then this m is actually the composition of p2 with this m tilde this is the addition map and you see under this um but upstairs you actually do have an isomorphism so so in fact you you you you obtain this you obtain this this second line here yeah and the reason why you want this pursue the second line because this looks a little bit like it starts to look like a furie mokai right i mean you pull back from one it's um it's it starts to resemble the furie mokai a little bit if you pull back e from one um uh from from one factor in in the end you push forward you twist with something and we're gonna see what this is in a second and you push forward via p p2 oh actually i mean i i mean i yeah i'm a little bit inconsistent i used p and q before so but it's okay so it's not a big deal um okay so now let's let's continue a little bit here so this this delta in turn i don't know this it's not a hard calculation as you'll see in a second now well this delta in turn is the composition of the addition map was just flipping the signs right i mean changing the sign on the first factor it's a subtraction map so i'm sorry so this i'm sorry so so then that's what i wanted to say so then of course if i'm looking at the pullback i can write it like this okay so so then we're we're gonna work a little so we have now this pontriagin product we managed to write it like this and we're gonna continue to work a little bit on it so well i'm gonna kind of artificially pull this pull this out here and then i have so we're gonna apply a minus one star to you and we're going to undo it here um and instead of delta because of this formula i'm just gonna have a pullback under multiply k under multiplication map okay and this disappears now because we project on the second factor this is uh this is irrelevant so i can this this pullback by the involution is does not matter for us does not make a difference and this is what we have but now okay now we know though that we can write the pullback of the polarization there are the multiplication in terms of the poincare bundle this is the explicit formula for the poincare okay so we we we replace it here and so then well so one factor of l just comes out because we're pulling back and push it forward via p2 and here is what remains we have the poincare appearing explicitly which means which this will be an free mokai and from the first factor we have minus one star e tensor with l again which means if i so this maybe looks like you make make it very complicated but remember we have to produce an sd cubed so um so so if so we conclude this is a final point of the calculation that um e pontriagin product with l is in fact the following is the composition of the following operations tensor with l composed with the free mokai functor applied to what applied to again tensor with l and um minus one star but yeah so you take this test object e and you first apply minus one star then you tensor without uh then you apply f then you tensor without that's what we're talking about okay so you see already we have like a nice string here um so now we're almost done this was actually the um maybe the harder right so now we apply f on both sides and this produces a longer string so you have f applied to the pontriagin product this gets transformed into a tensor product okay and then the string here becomes longer we compose at the end we add one composition with f at the end of our string so it this is almost so if you think about is this is sd st we have here now so then we have the right hand side so we're almost done um so let me and the the main point is what is this f of l and um and again this is easy to calculate that in fact f of l is l inverse okay so if you run the polarization through the free mokai through the functor itself you get the inverse and this is just really by definition so you know this is p2 star of um well so this is p2 lower star of i wrote this terribly could you do a word's job i'm not sure okay so you just write definition is um the pullbacks under p1 give you nothing so so f of l is in fact the pull pullback under p2 comes in front and then you have p2 lower star of m upper star l and then it's easy to see by using the same trick that we used before this is in fact oh okay so you're really left with l inverse here okay is it just to say that this is an easy calculation okay so where does this leave us so we have um yeah so we're we're we're we're done in in a sense so here we so we conclude we're here without with our calculation we're at the following stage that if you tensor with l inverse because this is now uh the Fourier mokai of l and you apply f to e this is the same as applying f everything on the right hand side that we have tensoring with applying f tensoring with l minus one star okay so now we just move to the other side and you see that you we have obtained therefore we obtained precisely the cube right i mean we we have that f composed with tensorization by l cubed um well um we so the it's actually given exactly by what you have when you um yeah so you're left with a minus g okay i think that's correct yeah if you all right so you move this to the other side you have another factor of f and you have a minus one coming from f squared but you also have this shift by uh you have a minus one star coming from f square but you also have this shift by minus g so this is almost right yeah um in fact you would need so this is what we deduced and this is the correct identity but we actually need we actually need also minus one star there yeah so we actually would need to be to be equal to minus one star minus g okay and the fix to this is to change this f assignment a little bit so instead of assigning to s f which this is a little bit off you have to assign it f tilde which is minus one star f because then out of you know since this is a cubic in f you will actually get a minus one star yeah I mean so this is the fix so it's a it's a slight fix so that the first star first guess was almost almost right okay and and this is uh and this is you check that you have an s l to z action yeah I mean so and it's a it's a very simple right I mean the s is is actually the furie mochi up to this very small twist and um and t is just multiplication by the polarization and yeah I'm sorry maybe this calculation seemed dull to you because it's this relation you you have to check but it's kind of a nice to check something because otherwise you know it's just sort of a this dry thing where does this appear you know this s t cubed how do you check it geometrically and in fact it's uh it's it's it's as you as you saw it's actually simple to check now so I think it's good to do to to see one thing so um yeah but of course you know here it's up to shift but in chow let me point out and which is the level we're interested in we this this calculation was in the derived category but you in chow in fact um um uh of course this shift goes to minus one to the g and everything is everything is okay so you actually have phi an action of s l to z on the automorphism of ch a where s goes to minus one star and the furie mocha in chow um and t but t is multiplication by the line bundle but so which means that in in chow this is multiplication by the churn character so e to the theta we said the first churn class is multiplication by e to the theta okay and but I want to point out something else which is nice since we are right and then we're gonna well we'll see then what we do we have a decision to make them but um if I write this matrix now this I'm interested very much in this matrix and this is um I guess this is t s t okay it's the product t s t well then the the calculation we did also shows that shows us how this acts on the chow yeah because we saw so in other words phi of this by our previous calculation this is tensorization with l composed with f tensorization with l and in fact precisely composed with minus one star because in fact s is not quite f yeah it's this composition okay so but we calculated this and this was in fact the first calculation we didn't we said that this is actually taking pontriagin product with l yeah this is taking pontriagin product well so it's it's kind of a beautiful picture here right so I mean this 1 1 0 1 means that you multiply by take a tensor product with l is sent to tensoring by l and 1 1 0 1 is sent to taking a pontriagin product now so it's just kind of an amazing thing they were beautifully symmetric okay um yeah so now I mean okay the hour is out but uh so I have and I thought this might be a little bit hard to do in exactly one hour but so the so we we have to we can make a choice here the reason why we'd like to extend this to sl to q is because then we see also the le algebra I mean we want to understand also the sl2 triple and so on the algebra level so far we're on a group level so I don't know if it's not much but I don't know uh you know if you want to we have two choices either to uh stop now and it's not a bad place to stop because we have the sl2z action and then we can just sort of complete it and we can do this uh in a in a future meeting or I can I can finish this is going to be I don't know maybe 15 minutes maybe less in fact and so it's it's entirely up to uh up to the audience basically you know I can do either yes yes exactly exactly so that that would work very well in fact you know because we're in a good place and we're sort of uh kind of in the setup and in the setting and I can definitely do and then it kind of makes sense because I wanted to talk about these left sheds sl2z anyways we haven't yeah yeah yeah so let me stop sharing okay so and it's good to keep it to a certain you know yeah not not for this to stretch infinitely um but um yeah you know so uh I think we're you know so if you have any questions uh just let me know but I mean we so far we have an sl2z action on the uh uh on on the chow of of this principally polarized the billion providing which is given very simply as you saw the uh one main player is the free amokai and then we simply have the tensorization by the polarization here and this is all in mukai's paper yeah so it's 40 years ago he talks only he's not interested in the chow level in that paper specifically so he talks only on the derive level while you have this shift so it's not quite an action because of the shift but basically you know this calculation is in which I presented the sd cubed is is in uh is is in his paper so it's a beautiful paper to read and still the best paper to read I think to understand this you know um yeah so I as you saw it's just not nothing yeah I'm just sort of presenting mukai as well but we're gonna get to this uh to to where it connects to something that's sort of current yeah I mean when we're looking at the left shits sl2 that comes out of this action so this this this naturally accepts this extension to sl2q and then you're talking about the algebra as well and so so so it would be nice but um okay so um then I sorry thank you for the very nice talk oh okay I am sorry it was uh yeah and maybe uh if people also online have questions yeah if there are any questions I'd be happy to yeah I don't know so yeah yes we'll we'll have to we'll have to confer on the exact schedule but uh yeah that's uh yeah but Wednesdays this time is is is a good time basically so yeah okay thanks a lot for for coming to listening to a very old uh yeah I'm an old old topic yeah it's not nothing super original in fact there's nothing original but it's a beautiful beautiful picture anyway so okay all right so thanks very much and uh I'll uh I'll see you see you before long one okay ciao