 I learned something from you guys, something quite important actually, this is very good. All right, so okay, so then, so we had this, let's go back to Adelama, so first of all, so we computed the fundamental group of the Hilbert scheme, right, and we computed its homology, okay, all right, so let me also give you another little thing about the fundamental group, so we can prove that, so this is again, you know, it's a fact from algebraic topology, and it's a fact from algebraic topology if you like, and it tells you that pi1 of s to the r is equal to actually h1 of sz, okay, so, sorry, sorry, sorry, no, no, I have pi1 of s symmetric power, right, that's the fact from algebraic topology, and so then this, if I use the computation that I made of pi1 of the Hilbert scheme, right, I showed that, we showed that pi1 of the Hilbert scheme is isomorphic to pi1 of the symmetric power, so then knowing that pi1 of the symmetric power is h1 of s, this means that pi1 of the Hilbert scheme is, can be identified with h lower 1 of s as well, okay, so we've got pi1 of the Hilbert scheme, and we've got the homology of the Hilbert scheme, right, so as a corollary what do we get if s is a k3 surface, then pi1 of the Hilbert scheme is trivial, we know that and h2 of the Hilbert scheme can be identified with h2 of the k3 surface plus multiples of the exceptional device, and so in particular what you get, in particular you get that h2 zero of the Hilbert scheme, or actually let me, let me write that for h0 of omega 2 of the Hilbert scheme, right, which is h2 zero of sR is the same as h2 zero, it can be identified with h2 zero of s, which is isomorphic to h0 of omega 2 s, and this has dimension one, so what this tells you is that if s is a k3 surface, then your Hilbert scheme is irreducible, homomorphic syntactic, and all right, so that's that, and then but what happens if s is not a k3, if s is not a k3, then that's the complex torus case, right, so complex torus of dimension two, right, then we're going to take, actually let me also call it, let me call it a, so that I don't confuse the k3 case and then the the complex torus case, right, so then we're going to take the Hilbert, Hilbert scheme in one more degree, so we're going to add one more to the degree of our zero cycles, then this guy, this guy here, we already know that it's homomorphic syntactic, right, because all we needed for that was that the surface was homomorphic syntactic. What we don't know is whether it's irreducible, right, so so what is pi one now, pi one of a r plus one is, as we said, can be identified with h lower one of az, now this is no longer zero, this can be identified with z4, right, so this is not zero, this is not trivial, right, and h2, well this we said is h2 of a with with with rational coefficients, right, plus multiples of the diagonal, oh sorry, plus h2 of h1, plus multiples of the of the exceptional divisor, so again this is not, so again you see now that the the dimension of h to zero is bigger than one, right, you're going to get some two zero forms from here, so this tells you that this, in the in the complex tourist case, this is not homomorphic syntactic as you would have expected, okay, but we can actually, we can still extract and, oh sorry, it is homomorphic syntactic, it's not irreducible, so but we can actually still extract something irreducible from it, so how do we do that, so we consider the addition map, and let me call that sigma, from a r plus one, and this is my blow-up morphism row, right, and this is, sorry, no no no, so we have the addition map, let's say little s, from the symmetric power to the complex source itself, right, and then what we can do, we can compose it, so compose here, so we have the blow-up map, which we call the row to the symmetric power, and then we can put the addition map here, and I'm going to call this combination here sigma, okay, and so the key, the key is that, so you see what do we want, we want to produce something that's simply connected and irreducible, homomorphic syntactic, so if you look at the computation that we made, pi one is isomorphic to pi one of a, right, this is also pi one of a, and h two is what we want, except there's a wedge two h one of a here, but a is a complex torus, so wedge two h one of a is again a copy of h two of a, so if you look at this addition map, the extra stuff that you have is all occurring at the bottom in a, right, the pi one is the extra pi one, you don't want it, it's in a, and this wedge two of h one is also in a, so if you want to get rid of that stuff, the right thing to do is to get the, to take the fiber of sigma, you see if you take the fiber of sigma, that fiber is not going to have any pi one, because pi ones are all coming from a, and it's not going to have that wedge two of h one, because again that's coming from a, from the bottom here, so that's, that's going to be our irreducible, homomorphic syntactica, so put k r as sigma inverse of zero, so okay, so this is, this is what we call the generalized kuma, so this is definition, k r is the r plus, I'm not sure whether I should say r plus first or r, I think I will say if r is one, it's already, let me say it's the r plus first generalized kuma, okay, so now, okay, so now let's see, so we have again some nice properties of this k r and of this, this map which is induced by the addition map, right, which I'm going to summarize in another diagram, right, so, but first I need to introduce the action by translation, so the complex torus a acts on itself by translation, so I'm going to tell you, you know, I'm going to denote t sub a from a to a, the translation, translation by a, right, and then it also acts, the ta acts on the Hilbert scheme by pullback, so you can take an artinian sub scheme and you send it to ta upper star of c, okay, and then we have, we have a Cartesian diagram, again as follows, so I'm going to have a r plus one here, I'm going to map it down to a via my map s which was induced by, no sigma, which was induced by addition, right, then over, over here, I can write, I'm going to write the translation action, right, which is, which, which takes a pair little az and sends it to ta upper star of c, and then here I need to put something which will make the diagram Cartesian, so if you think about it a little bit, you will see that this has got to be a times a, and then to, to make this diagram Cartesian, so you have, you're going to take a pair little a little x, and of course you have to send it to some translator of x by a, but if you think about it a little bit, you need to, you need to take tr r plus one times a of x in order for this diagram to commute, okay, and you, you see that you have the Cartesian diagram like this, and then you can, you can restrict this diagram to kr to the kumar, which is the sigma inverse of zero inside a r plus one, right, so it induces another diagram, which I'm going to write, which induces the Cartesian diagram like this, so I'm going to put a times kr now, and this again, I cannot put kr here because once I do my translation, the fiber over zero can go anywhere, right, so I can't really do that, but I, I have the same map here, right, I can, I'm just restricting my map to the fiber over zero, here I have again the math sigma, and here now I'll just get a because I got rid of one copy, and here the map is oops, I think I made a mistake here, I mean there's something wrong, okay, well the a on the left, okay let me not put that, the a on the left is just the image of a times kr, so and this here will just be multiplication by r plus one, yeah, so basically your, if you look, if you look at this other diagram it takes a little bit of thought to see that when you restrict to kr over here on the on the upper left corner, then the image on the left is going to be just a copy of a, and this map that I've got on the bottom is going to become a multiplication by r plus one, okay, so all right, so now if we go back let's go back to our other diagram, so what you see now this is the Cartesian diagram, right, and what does this tell you, you can see here that you know the map on the left is is just basically a product, right, so it's just the projection to the first factor, so this as I said this kind of implies that this sigma is a smooth map and all the fibers are isomorphic, all of its fibers are isomorphic to the same right to kr, okay, and you can also prove is that kr is holomorphic symplectic and it's actually irreducible holomorphic, okay, so and also what you get is that and we have the homology h2 of kr is is isomorphic to h2 of a plus multiples of this of the exception of the bowser again, actually you have to take this, okay the accept it's this is the exceptional divisor of the the Hilbert scheme intersected with KR, right? So maybe I should write it that way, that would be less confusing. So you can prove all of this, it's not too difficult, but I think it's a little bit too much detail for our lectures. I will be putting more details in the notes and the lecture notes, which will be published later in the conference, in the proceedings. So are there any questions? This is a good place for me to stop a little bit. So I will take any questions if there are any. Okay, all right, so you see now, what are the dimensions of these guys? I mean, you see that to note that the dimension of SR is two R and it's also the dimension of KR, right? So we've got holomorphic symplatic manifolds of every dimension in this way. We've got two of each. And if you look at the Betti numbers, you see that here B2 of KR is one more than the second Betti number of A, which is six, this guy's seven. And if you look at the K3 case, you see that the Betti number is again, the second Betti number is again one more than the second Betti number of the K3 surface. And guys, I always get confused. Is that 20 or is it 19? I think it's 20, right? So I think I get 21. Is that right? Or is it 22? It's right. Is that correct? Is it 21 or is it 22? Because I always get confused about this. But anyway, we can fix it later anyway. So you see now this, so since these guys have different Betti numbers, they're not deformation equivalents. They have different B2s. Okay. Now, so these are all the examples of irreducible holomorphic symplectic manifolds that people know, except for two, with what people call sporadic examples. So there are two more examples, two more known examples due to O'Grady. One is, which are not deformation equivalent to these. One of them is 10 dimensional and the other one is six dimensional. So they are, they're not differential equivalent to these, but they were still obtained from K3 and K3 surfaces and complex to write up dimension two. So they were O'Grady constructed them by doing as modular spaces of certain type of sheaves on a K3 surface or certain type of sheaves on a complex source of dimension two. And then you showed that they're not deformation equivalent. There's a question in the chat. Oh, okay, somebody's saying it's 22. Yeah. Oh, so it's 23 then. Okay, I'll put 23, we can always change it later. All right. Thank you. There's another chat. Yeah, right, thanks. Okay, very good. Thank you guys. Okay, so these are, as I was saying, these are the only known examples. So it's a big open problem. Are there other examples? Are there other examples? Other families? Or let me put it in a different way. Are there irreducible, homomorphic, simplexed manifolds, which are not deformation equivalent to one of these? And Ham, can I ask a question? Yeah, sure. So the Kuma and the Hilbert scheme of points, are they never deformation equivalent? Or is there, how do you show that they are not deformation equivalent to each other? Oh, the B2s are different, right? Yeah, they can't be. That's right, okay, thank you. Yeah. Yeah, so this is a big open problem. You will probably get some prizes if you solve this problem. But okay, so the problem, I mean, the main thing is actually, we will get into that again, probably sometime tomorrow. I'm not sure how today we won't get to that, but the thing is that everything basically hinges on the second homology on H2 for these irreducible, homomorphic, simplexed manifolds. They are, to a large extent, determined by their H2. So it makes sense to think that they look a lot like surfaces. And the place to look for them is surfaces. So you should try and do more constructions with surfaces if you wanna construct more of these guys. But then maybe that's not true, I mean, who knows? And that's kind of like, that's a philosophical way of thinking, but it's possible that there are some very, some strange examples that have nothing to do with surfaces out there. But all the ones that we can actually construct have got something to do with surfaces. And that's the logical way of thinking about it. Anyway, so, all right. So now, okay, so exactly, so I'm getting to this H2 business, which has to do with Moduli and Torelli. So, and the period domain. So I'm going to now switch gears a little bit. And I wanna talk about Moduli of these things. So again, if there are no other questions at this point, okay, all right, so I will. And as I said, I would have to leave exactly at 111.30 today. So I'm not, I only have like 10 minutes now. So I will talk a little bit about that. So Moduli of hypercalors, the Boveal-Bogomotiform, the period domain and the period math. So let me first just talk about some generalities. So given a differentiable manifold, there can be many complex structures on X. And so we define, we have several, and we have various ways of classifying these different complex structures on X, right? So one way is the Tash motor space, which is defined as follows. So I'm going to denote it as Tash of X, which is by definition, the set of all complex structures on X, modulo and equivalence relation, which I will denote by tilde zero. And what is this tilde zero, where two complex structures, i and j are equivalent, which we write as i tilde zero j. If there exists a diffeomorphism phi from X to X, such that phi upper star of i is j, and phi is isotopic, or if you like homeotopic to the identity. All right, so this is the definition of the Tash motor space. And then the second space that, actually it's the second space that we would be really interested in, and that's the modular of complex structures. So the modular space of complex structures is again, by definition, I will call this one, and this guy is going to be the set of all complex structures again. Modulo though, a different equivalence relation, which I will just denote as tilde, where two complex structures, i and j are equivalent. If again, there exists a diffeomorphism, phi from X to X, such that phi upper star of i is j. So what I've removed here is the condition that phi should be homotopic or isotopic to the identity. I've removed that condition. So then these two modular spaces are of course related to each other, right? So this is how they're related to each other. So if we denote diffeoff X, the group of diffeomorphisms of X, dif zero, it's connected component of the identity, then I'm going to, I can call G, the quotient. This is the group of components of defects, and it's a discrete group, right? Because I killed the connected component of the identity. Then basically by definition, you have that you can get comp of X as the quotient of times of X by the action of G. So there's an action of G on times of X, and then the quotient by that action will give you comp. So as I said, a priority, what we're interested in is complex structures on X. But the problem is that comp is actually not a very nice space, right? So a priority we're interested in comp, it does not have so many nice properties. For instance, it can be not a non-housed orphan, it can have all kinds of awful things. So Tash is not, there's a lot nicer. And so what we do in practice, we will just work with small open sets of Tash. Which, what do these do? They describe small deformations of complex structures. All right, so now again, like I said, we wanna study these spaces. So we're going to, and studying them is equivalent to studying deformations of complex structures. So that's the next thing we wanna do. So let's talk about deformations. Let's start with a definition. So a family of complex manifolds is a smooth procomorphism of complex spaces like this. And then, okay, and then sorry, I'm gonna have to stop here and we will continue tomorrow if that's okay. Yeah, thank you very much everyone. Thank you. Bye-bye. Bye-bye, see you tomorrow. Yeah. Bye.