 Okay, so let's start with the usual question I asked at the beginning of the exercises. So are there any questions or any comments to what we've done yesterday? I know that most of the people are a bit shy, but there are no stupid questions, and I guess that's, you know, if you want to know something, it's good to ask simply and then try to learn. I have a stupid question actually. No, no stupid question, I just said it. Okay, well, no, it's, so when we yesterday, Angela defined AG as the upper half space, the seagull upper half space mod out by the action of the symplectic group. So I have two questions actually, maybe very naive. So the first question is why do you have to take the symplectic group on the rational numbers? Like, is there any reason why or, okay, this is the first one. And then maybe the second one is also, I didn't understand the computation of the dimension, because so why does modding out doesn't lower the dimension? That's what I didn't understand. Okay, so let's give me a second. Yeah, or I mean, also if someone else has other questions, please go ahead. I don't want to. We have time. Let's, okay. So just to tell you, now I will just hide the videos. So I may, okay. So we have AG, yes. And we said that AG is AG divided by SP of something. Yes. And, okay. So the first question is, the first answer is somewhat, is a good question. So usually we want to write here Z, yes. And if you have a principal polarization, you really can write Z here and everything is okay. But then if you start with a non-principle polarization, you have in some sense a problem, or you have two different groups and two different actions that gives you the same thing in the end. So either you do SPQ, so you allow rational coefficients and then the action is the same. So we have a group acting on AG or you take SPZ, the thing that you like, but then the action will be twisted by this D, which is the polarization type. So in some sense, first of all, if you write SPQ and you write this equation that it has to be, and it has to satisfy the equation with the polarization, you will, you know, for principal polarization, you will get SPZ anyway. So that's why it's not that stupid to think about Q, yes. So Angela was having in mind more general construction. That's why she wrote SPQ. So this is the first thing. And the second thing is that, again, it's similar to, you know, for G equal 1 case, yes, that you have, yes, you have something like that, yes. So in order to show the, okay, so in order to show that the dimension does not go down, you really need to show that SP acts, and actually, you know, you have to show it. And I think in the Birkenhake-Lange, wait a second, chapter eight, I will find it in a second, and I will show you that SP, you know, SP needs to act, it acts probably what? Locally free, yes, and actually not, wait a second. So that you just, you just need to compute the, you need to find the action, not to find, but to find the properties of the action, and you will see that it does not change the dimension. Wait a second, I will just find the correct statement. I think this would be like the easiest way, okay. Yeah, the analytic model is space. So now I probably need to share screen with Birkenhake-Lange, okay. So as I, wait a second, I need to probably make it a bit bigger. So as I said, here is the argument about your first questions, yes. So sometimes you want to have integer coefficients, and then the action is a bit different, or you want the action to be the usual action, and then you need to take the different group. Yes, so this is what we said before, and this is just how it is. And in terms of your second question, you just, yeah, okay. So you have to show that the action is proper and discontinuous, and then the dimension will stay the same. So that's the... Okay, so it's just the way the action works. Yes, so in some sense you can try to find, yeah. This is how I see it, at least. So yeah, so this is the answer to your questions. So basically this means that if they are in the same equivalence class, they are also connected by an action, that's... Yes, yes, yes. And vice versa, I mean. Yes, yes, yes, yeah. So the idea is that actually, so if you want to, the HG really, yes, the HG, it's really like the moduli of a billion varieties with a variety with a symplectic basis. Yes, with a symplectic basis. Yes, because this is exactly, if you see Z here, yes, and ZD, usually we write like that, yes. Then that this H equals the inverse image of the imaginary part of Z, yes. And this is exactly this H, this is the matrix. Okay, so maybe not, we will not. I don't want to write it like that, because it may be confusing. Imzet is the matrix of H in a basis E1 up to E, ED, E, EG, yes. So and D, if you remember, this is the diagonal. Imzet is the matrix of H, yes. And then H restricted to this lattice, you can just, you know, you can compute it. Z times Imzet inverse times D, the imaginary part of those will be exactly D. So what you can do in order to have the same, the isomorphic abelian variety, you can change the matrix by the, you can change the basis by the symplectic action, yes. Okay, okay, okay, I see. And then Z will still be in the upper half space, yes. So this is more or less the same thing as you're doing in the genus one, yes. So you can, you have one and tau and you can change it by some SL action, yes. And you will have one and another tau, yes, that's the idea more or less. So that's, okay, I think this is the answer. Okay, are there any more questions? Yes, so I just came up like 15 minutes ago with another exercise or question, just for some of you maybe to think about, because you may have a feeling that, you know, we are working with prims and we get all the Jacobians that are not simple, yes, they have abelian subvarieties. And, you know, you may think that this prim theory will solve any problem like that, in some sense, so that having subvarieties is more or less the same as having some kind of prim of something, yes. So, because there are lots of, lots of those. So there is an exercise to show you that the prims construction is very nice, but it doesn't solve the whole world within the Jacobians and abelian varieties. So what's the question? You should find the curve C, such that the Jacobian has some abelian subvariety, but this abelian subvariety doesn't come from the prim construction. So there is no covering such that this X is the image of the smaller Jacobian or X is the, there should be, this is the prim of this, this covering, yes. So this kind of, this kind of subvarieties can be called as a generalized prim or generalized prim Turin varieties, and they do exist. So if, if you have an idea, maybe I give you like a few more, a few seconds, minutes to, to think if you can find such a curve, especially for those who are more familiar with the, with the prim and with the prim theory. Yes, for, for G equals three, there is no such a curve. Yes. Because as exercise, we said that one of exercises says that it always comes from the prim construction. So you need to look for genus four curves and that's the, yes. So maybe I will leave this exercise for you and now we can go with, with another, with other exercises. Okay. So, so let's go back to our exercises. Does anyone want to comment on any of the exercises? I would really encourage you to say a few words. I'm pretty much sure that for example, Iran, who was talking very nicely about the prims know some solutions and I don't know many of you. You shouldn't, you should just, you know, go for it. It's my point. Yes. Like I will not be mean to anyone. No, it's obviously we're, we're, we're learning. If not, I, I will, yeah, okay. I will go through, through the solutions, through some of the solutions. Okay. So I wanted my, my aim for today would be to talk about this exercise number five. So injectivity of pullback map five and, and then nine, because as you said, as, as, as we have seen yesterday, Angela said, what are, what are those cases? And actually the proof is not that hard. So maybe we will be able to date to, to, I will be able to, to persuade you that this is like not, not that hard, assuming you know, the, some background. But before that, okay. So before that, let's start with exercise number eight, because there is a, wait a second, let's start with putting some power into my computer, into my laptop. Hopefully that, yeah. Okay. So now I will stay. Okay. So, yeah, obviously the, so after what we've said yesterday, you may, you may find that the exercise number eight is strongly stated, because having embedding is, is too much to ask for. So it's not an embedding, but it's just the birational map. And we talked with, we talked about this, wait a second, I'll try to, by the way, I will try to put those, to put those notes, notes online at some point. So, yeah. Okay. So let's go here. So, yeah. So, okay. No, maybe I can write it once again, it will be, it will be a bit easier. So exercise number eight, this one I will leave. Okay. So exercise number eight. Okay. So assume that we have, so what we have, we have C, which is embedded into X. And now assume that it's not an embedding, but just a map. Non-constant map, X abelian surface. Okay. Then by universal property of Jacobian, we have a map from JC to X. Yes. And since C is of genus three. So C is of genus three. So X is abelian surface, hence C generates, okay. So, because it has to generate some abelian sub-variety and by definition of generating a group or a sub-group and it cannot generate the elliptic curve because it's already of higher genus. So it has to generate X. So therefore, since C generates X, the universal property of Jacobian, this map will be surjective. So that's why, so by the universal property, you have this, you have the surjection here. Yes. You have the, you have the map and you have a surjection. And therefore, you can think about the kernel, which will be some elliptic curve, yes, by computing of dimension. JC is of dimension three, X is of dimension two. So the kernel will be, will be of, okay, maybe kernel, yes. And then you can dual, dualize, dualize, and then you will have here X JC dual, which is dual, which is isomorphic to JC, and then you will have here E. Okay, maybe, maybe this is not here. Maybe it will not be an embedding. Maybe there will be some connect, some more connected components. But anyway, you have here a map. And then again, if you have such a map, then restricting to C inside JC, yes, you get a map from C to an elliptic curve. Yes, so if you have really, and here, okay, so just to write it down. Here is rational embedding could be not a good word. Okay, so generic one-to-one, okay, I assume. So, yeah, and that's, and that's, and this is what we want. And the other way around is, sorry, I have a question. Yeah, okay. At the last morphism that we construct, could it be constant? Which one? This one? No, the map to the, yeah, this one, to the elliptic curve. Okay, so this is again, so this is, this is Apple Jacobi map, yes. Okay, okay, so first of all, this was a good question. Maybe I didn't comment on it enough. So you have a map, you dualize it, and then you have a suggestion here also. So there is a map from JC to E, which is surjective. And then again, the Apple Jacobi map, the image of C in JC generates everything. So the restriction to C will be a surjection to E, okay? Okay, thanks. So you're right, that's, that was a good point. I didn't comment on it, you know, this is, as usual, sometimes needs this one more line to, to persuade others. So yeah, that's, that was a good question actually. I forgot about saying that. Okay, so the other way around is this, you know, the, the philosophy of this exercise is to show that three is just small, that three is one plus two, yes? The other way around, if you have C that goes to E, then again, you have a map from JC to E, zero, because this is surjective JC to E is also surjective. And then you find the, the kernel K, which will be an abelian surface. And then you dualize and you get here E dual, JC dual, which is isomorphic. And then you have here K dual. And again, here, here you have, here you have a surjection, yes? So again, restrict, restrict this map, call it, let's call it F, no, F is now, let's say alpha, alpha to C, no, alpha is for, usually for double Jacobi map. Okay, let's go with F. Okay. And now the question is, so what can F, what can F be? Yes, so the question is, is F generically one-to-one? So why F is, is F restricted to, so again, let's make it a little bit, because I'm, I'm, I usually, I'm too fast in such a, so consider C as embedded into JC via other Jacobi map, and then restrict F to C. Starting from C to K dual. And obviously I need to show that it is generically one-to-one. Okay, so, but what, what it can be? Okay, so now we know that C is of genus three, yes? So if F is not one-to-one generically, then the genus of C, not C, but F, F of C is less than three, yes? So now you, so, you know, the, what you can do, you can actually compute what's the genus of, of, of, of, of, of this image is, yes? Note that, again, genus of F of C has to be bigger than one, because there are no rational, rational curves in the Aynabilian surfaces, actually in all Abyian varieties, there are no, no, no rational curves, and it cannot be an elliptic curve, again, by the same, by the same argument, because if it was an elliptic curve, it wouldn't generate the whole K dual, okay? So if, so hence it has to, it, it had to be, yes, the only option, let's say like that, the only option would be G of F of C being two, and F from C to F of C, et al, double cover, and this is not the case, yes? Because for et al, double coverings, we know, we know that, we know how it, how it looks for et al, double coverings that you can find the K dual, and it is, there exists, I will say, yeah, by exercise, by another exercise, this is not the case, because for et al, double coverings, you have an embedding into one, two polarised, you can write by barf, barf results, okay? So we solved exercise, exercise eight, but it's not an embedding, it's just an embedding with singularities, yes? So you, you assume a singularities, and as I said yesterday to Barbara, if I remember correctly, you really have those, you really, in general, you really have those singularities, because you can compute the arithmetic genius of polarising client model, yes? Because O of C will become a polarisation on K, yes? And you will compute, you can compute the arithmetic genius that will be d plus one, one, if the polarisation is of type 1d. So the geometric genius is free, arithmetic genius is bigger, so therefore you need to assume that there are some singularities, yes? So that was exercise number eight, and yeah, and the embedding, as I said, it's not, it's not a good word, it is embedding up to finite number of singularities. Okay, are there any questions in terms of this? Yes, then there is this second part, no? Of the exercise eight, which is really intriguing, somehow, yes? So you can, so you can have all types of, of all degree of, of maps, yes? How can you construct this from the planar model? I have no idea, this is what I've told you yesterday, I have no idea. It's, you know, the thing is that I thought at some point that, you know, that this is something like the problem of finding roots of polynomials of degree bigger than five, yes? So that you cannot do it, you know that they exist, but you cannot do it. And here, I don't know, I don't know if this would be the case. The thing is that really the, this covering will be non-Galewa, yes? This will be non-Galewa covering, so it's really, Yes, but there is a theory of covers, it's well developed with, there is a series of papers of Kasnati and Ekadal, and so you can, every single one understood, but I was trying to do some computation, but it's not clear to me because this is this phenomenon that you have that the ramification divisor, the degree of the ramification divisor is, in any case, always equal to four. Yes. And then whenever you have a covering between curves, you can embed the recovering curve in a projective bundle of rank, okay, is the scroll has the nation n minus one, if n is the cover degree. And also you can say that the topological divisor of this projective bundle restricts to the ramification divisor of C. So the fact that this ramification is fixed and independent on n, it's, it's very strange for me, I don't know, so I mean. Yeah, you know, I completely agree that it's, you know, that this, you know, this is from my point of view, obviously, you know, the question is somehow strange, yes, who would who would like to look for such curves. But, you know, by as I've told yesterday, yes, by by the construction of on the level. Okay. So, you know, usually the philosophy is as Al-Hella said, yes, that that we have lots of constructions on the curve side, yes, on the theory of curves, and we want to using those constructions to understand the model of a billion varieties. And this question, why this question is strange, because the question goes the opposite way. Yes, we started with a theory of a billion varieties. And in theory of a billion varieties, we have this non-simpleness, yes. So we have this, this fact. And then what we want to do, we want to understand how this non-simpleness, what it gives you on the level of the curves. And, you know, usually the easiest way would be to just go back with Torelli. But since Torelli is, is far from being explicit, you cannot do it. Yes. So that's why, yes, because, you know, I can write explicitly, I can write explicitly the period matrix of a dimensional three, three dimensional, a billion variety. And I will be sure by what Al-Hella said, that it is a Jacobian of some, of some curve, but I don't know how to find this curve. Yes, this is the, this is the thing. So you can, you can, you can certainly write as similar to Humbert surfaces in, in, in dimension two, some low side of non-simple three-folds, like having an elliptic curve with a restricted polarization of 2021. So I can do it really explicitly on the level of a billion three-folds, but then there is this big question mark what happens on the level of curves. And I don't know how to do it. I don't know if anyone actually tries to do it. And, you know, and there is certainly no, or maybe not certainly, but I don't know any, any paper, any research going on in this direction. But I think that the question is nice. This is, yes, I have one last question on this. So do you know, so the, the, the Teta divisors on some quartic curves is given by, by tangents, right? So do you know if these genus three curves, which admit a birational map to a nobillion surface, have some condition on the, on the bi tangents, so that there are no, in general, 28 bi tangents to a quartic curve. So, so classically, there were some projective quartics studied in terms of these bi tangents. For instance, there is the, the class of Lurot quartics, which have a format divisor in the space of quartics curves. So do you know if there is some genetic characterization of these curves? Okay. Again, okay. Again, if, if this degree is small, namely two, for example, then you can do quite a lot, I think, because then you can, you know, you can just play a game with, play some games with automorphisms. You know, the thing is that, so maybe you can find for, for some specific degree, you may try to find something, you know, the, the, the idea of the problem is that you can choose any degree. Yes. In, in, in next January, I will take 2010-22. And so, so the thing is that, you know, there is no general, you know, it should be something which will be like for, for any, for any degree. And I don't think there is a way to do it. Yes. There may be some specific low side that you can, you know, you can try to describe and see if they, if they have this kind of, this kind of covering, but, but it is certainly not general. And that's the, and that's why I think this question is quite interesting. But yeah, you know, I, I decided that if we are on the conference and not on the, you know, on the proper lecture, let's have some exercise which has no solution. Maybe some students who don't know that there is no solutions yet, we'll discover some. Yeah, that's the. Yes. No, this is was an excellent idea from my point of view. Yeah. Okay. So going back to, to our exercises, I would like to talk a bit about exercise number five and then number nine. So we have a finite covering between smooth cares and show that the pullback is not injective if and only if F factorizes via a cyclic ethyl coverings F, F twiddle of degree two. So again, the statement can be find in the, in the Bukken hakelange. So I will, at some point, I will at some point go and comment on the proof there. But before that, I would like to make some, some other comments, because in some sense, you can think that, that this, that the, this condition is easy to, to check. And it's like, you know, you just by looking you will see what's going on. And I want to persuade you that it is, it may not be that easy. And you have to actually be a bit more careful. So wait a second. Okay, so let's go back to where we, we are in the, we are in the bamboo paper. Yes. No, we are in the exercises. Okay, so let's go to bamboo. Okay. So why, why the pullback might not be injective. So, you know, so there is this, the nice thing is that so isogenes are maybe not necessary cyclic, but are ethyl galua. Yes. And if you, so, and, so, and cyclic coverings are defined by curve and, and this, and torsion point. So they have a lot of, a lot in common. So how, how to see, see till the given ethyl cyclic given by C eta. What do you do? What do you do? You take C. Okay. You can embed it into JC, the Jacobian, as usual. Okay. You can divide this by, by eta. So you take JC divided by eta. Okay. And now I also, okay. And then, so there is some map, whatever, however you call it. Let's call it F. Actually, F dual. Let's call it F dual. Maybe F is not, okay. Let's, so let's call this map pi. So let's call it F dual because then, again, JC is the principally polarized abelian variety. So it is isomorphic to its dual. So now I will just make this identification. Then you have a map F from JC divided by eta dual. And this map is exactly, is ethyl cyclic map between, isogenic between abelian varieties. And now, if you have this map F, JC here and JC divided by eta, then you have here S, C. Here you can take the preimage of C and this will be pi. So actually, any pi, any ethyl cyclic covering of curves can be seen as a restriction of some isogenic of abelian varieties. Jacobian and Jacobian divided by eta. Obviously, you can, you can see that if, if the Jacobian is of the type 111111, then here you have one d d d d d the polarization. So the polarization. So here is the polarization 111111. And here you have polarization 111111 and then the, which is in the end degree d. And this is a polarized isogenic between those two, those two things. Yes, it is exactly of degree d. And then if you restrict it to, to C, you will have, you will have the, your, your map pi. And this is how I see the, the cyclic coverings, which is quite nice because then you, you have and then, and then if you see it like that, then it is easy to, to see that if you have a map from so this then is isomorphic to C tilde. And then if you have a map from JC tilde, yeah, the norm map norm pi JC tilde to JC it by what we see it norm. What I want to say, I want that it factorizes via JC tilde. You go to here to JC divided by eta dual. And here you have JC. Yes. So this map actually factorizes via this, this special variety, obviously for, for surfaces, if you assume that C is of genius too, it is very, it is very nice that you really can see it because then all those, those guys are divisors here. So you really can compute it. But then so if you have such a such such a factorization, then obviously you can do here JC divided by eta. And then only then here you have an embedding. Yes, because here on the level of curves, if you will have, you will have embedding because C tilde is embedded here. So, and you see that here you just divided by eta. So JC, so if here this is pi, then the image of JC by the pullback is actually the image of this, of this, of this Arbidian variety. So this, this shows you that if you have an eta covering, then, then the pullback is known, cannot be injected. Yes, this is, this, this construction shows you the first part. If you have a, if you have a, yes, if you have an eta covering, then the induced, the induced pullback map is not injected. But obviously the other way around is the similar, is the similar construction. Because if you, if it's not an embedding, you can find the quotient at some point, because there is something in the kernel, you can just find a point in the kernel. And then you can, you can do the lower part of the diagram. And then from the lower part by the utilization maps. Okay, I haven't said it, but it's again, like a lemma to say that the pullback is, is the dual to the norm. Okay, let's go maybe now to, to Birkenhake-Lange and I will show you the details. So this is exactly what you, what you want. Okay, so the, the proposition 11.43, the homomorphism is not injected. So yeah, so, so that's the, that's the statement. If f factorizes by a cyclic et al covering. Okay, the, the important part is that this et al covering has to be the, you know, if you have two coverings, it has to be the lower one. Okay, so you don't, you don't care if fb is cyclic et al or not, but the thing is you have to find some f prime. And as I said, the proof is, the idea of a proof is the construction that I've told you. Yes. So if you find the f that factorizes like that, then the homomorphism of the pullback, yes, factorizes via two pullbacks and it's enough to show that the first pullback is not injective. And this is exactly what I said the construction. f prime is given by a line bundle of some order n and the C double prime is unit section under the nth power map. So, yeah, and you have a natural projection. So the, since the tautological bundle p star l is trivial. So the f prime pullback is not injective. This is exactly how I see it, as I've told you, is that you can, you can find this, this line bundle. Yeah, because again, if I write a point in the Jacobian, this is exactly the line bundle. Yes. And, and it's an torsion if and only if some, some tensor, tensor power is trivial. So that's the, that's the one direction. And then the other direction is again, as I said, yeah, you take, you find a non-trivial line bundle in the kernel. And then obviously it has to be of some finite order, because you just, you know, you just compute that it has to be an n torsion, because if not, it would be like the not finite covering. And then you, you find the diagram to see that you, that your section factorizes somehow. So, yes. And then, and then you can, you can factorize it more and more. So that, this is the proof of, of this, of this exercise. And now I want, I would like to comment on the, on this, on the statement because you may think that, that somehow, you know, it's, it's, it's not hard to check whether, whether it is injective or not. So I would like to show you the construction that we encounter in the, in, in our paper with, with Angela and, and show you that it's not that easy. So, okay, so let's start with h, g of h is two. Okay. So here we have h. And then take a two to one et al cover. And we have c. So c h is two to one et al. So g of c is free. Okay. And then we know that by results of Manfort, for example, that the genus, that the c is already hyperliptic. So c is hyperliptic because, because it's a double cover. Second. And then consider c tilde to c. Again, two to one et al. So what we have, we have here h. Here we have c. Here we have c tilde. And because c is hyperliptic, here you have a double cover of h. And double cover of genus two curve is hyperliptic. If you know if it's b elliptic. So there is here the two to one cover to the elliptic curve. So here you have, everything is two to one, two to one. And here you have two to one, as I say, plus four. So we have four branching. Okay. So this is, and here you have two to one. Here you have two to one et al. And we were interested with with Angela with, and we call those coverings client coverings of h into c tilde. And now we are interested in this map. Okay, let's take some other color. This map f. It is obviously four to one branched. And as you see, in the construction, it factorizes. Yes, it factorizes, but the curve, the map downstairs, it's not cyclic et al. So it's plus four. So it's not, at the moment, it is not going within the proposition. So we may think that the that's, okay, so we may think f pullback is injective at first. Okay. But actually, depending, the, why the second I'm writing to, depending on, let's call it, let's call it alpha, depending on alpha, f, f star may be injective or not. So it's really, it really depends on the, there are two cases, we call them isotropic and non isotropic. And in one case, the pullback f the pullback was injective in the non isotropic. And in the isotropic case, it wasn't injective. So references, you can find it in the Bo19 paper. So it's really, you know, if you have a, and why is it so? The thing is that the construction, this f, because we have all two to one. So there are lots of involutions playing around. So we have H, which is a hyperliptic involution here. You have hyperliptic involution here, you have elliptic involution here also. And then you can, because to do one, you can you can have, you have a lot, you have a lift of involution on the hyperliptic involution here. Okay. And then you can think about what kind of, what kind of lifts can you have here. And in one case, in one case, in the, in one case, the lift of the hyperliptic involution was hyperliptic. And then it was in some sense, the only f was, this was the only f was what I want to say, I want to say that this was the only factorization of f, but for nonisotropic case, we had a different, we had a different, we have a different factorization, what I can say so. So again, B019 for nonisotropic case, we had something like C tilde C E, but there was, so this was two to one, two to one plus four, but there was something like here, two to one plus, this is a genus five, so plus eight, and here it was two to one et al. So you actually have, you actually have had that, you know, you had this, this diagram. So F star E is E prime in JC tilde. So it's really, you know, so this exercise, although, you know, you have to really look for, for your, if you have your morphism, if you have your covering of curves, you should really be careful, because sometimes if you have one way of factorizing it, maybe there is some other way of factorizing it and, but you don't, you know, you don't see it at first. And in the end, there exists and your pullback is not injected. So yeah, so that's the that's the outcome of, you know, that's the punchline of this exercise. So although you think it's, you know, it's not that hard in precise statements, you have to be a little bit more careful. Okay. So, so this was the exercise. Yeah, the exercise number five. Okay, I see it's 1230. So probably I should finish now actually. No, which is not that bad. Okay, so maybe, yeah, are there any questions? Okay, so maybe next time I will say a few words about this, about this frame being principally polarized, but it's really like a easy dimension count, and knowing the restricted, how the restricted polarization looks like for the complementary abelian varieties. So yeah, so it will hopefully be a bit shorter than today's. Okay. And obviously, I really encourage you still to ask questions about lectures and so on, because our next meeting is on Friday, so there will be like two more lectures. So hopefully there will be more, more questions about what's going on in the lecture. Sure. Okay, so thank you very much. Thank you. So I guess, thank you again, Pavel. And I guess we can have lunch now and meet at two.