 Okay, good. I think my computer is getting old sorry guys it keeps overheating so I'm going to have to turn off the camera again sorry about that. All right. Okay, so, so let's see so. So we have this. So we had Burgess theorem right about whole and only groups. And, and so Barbara asked me about this case five right case number five at the bottom of the page you know what is this group you know SPR times SP one. All right, so let me just explain that a little bit better here. Yeah, here we go. And I think I need a new page so let me go and explain this in a new page so what is the. So this is just an aside. So, right, so SP so what did we have so we have that SPR we said it's contained in the automorphisms of H to the R, right so that as we said these are basically the, the R by R matrices with with entries in the commutes with a given you know her mission, paternity permission matrix. And what is so then what is SP one. SP one we said can be also identified with SU two and also as three, this naturally lives inside age itself right so this is the group of unit length, unit length, what turns. And so then this guy. So with the way the way that it works is that SP one. acts on on H to the R via scalar matrix multiplication. Okay, so so that's still that's then so then when we write SPR times SP one. This is this is really the subgroup, this is the group generated. So the subgroup of the automorphism groups right of HR generated by these two actions so generated by the actions of SPR on the what on the on the one hand and then the units on the other hand. So, so they do not I mean these two SP one and SPR do not commute right, but you just take the subgroup that their actions generate inside the automorphism group and I guess the notation is, is, is not very good but it's the usual notation that people know but now it's much more clear thank you. Yeah, yeah, right. Okay, great thanks. Yeah, so it's not, you know like I said they don't commute but they don't they both they both act and then you take the subgroup that the actions generate. All right. And yeah this is standard notation so you know I didn't feel like I could change the notation. It's not very good, but okay. So, um, let's see so now let's go back to where we were. We were talking. So I wanted to give you guys examples of hypercalamide falls right, and we started with Kate with the case of dimension to that's the smallest case right because the dimension of the hypercalamide fold has got to be at least it has to have at least four real dimensions always right. So, the smallest ones are two dimensional ones and for surfaces, as I said, as we said before SP one is equal to SU two, and SP one is the hypercalamide case but SU two is the Columbia case so what you get for surfaces is that being Columbia or being hypercalamide is the same thing. And the, the only surfaces that you get our K three surfaces and complex Torah. And I gave you the definition of the K three surface. K three surface is we kind of we're kind of kind of going here with a minimalistic definition that this is the minimum that you need to define a K three surface some people define a K three surface as being simply connected. But I didn't do that they're just at each one of x o x is zero. And it's a compact complex surface with trivial canonical bundle. So omega two x is isomorphic to o x, and each one of x o x is zero, and then everything else you can deduce from this you can prove that each one of x o x equal to zero actually implies that it's simply connected you can also prove that the integral of the K three surface is totally free. And you can also prove that they're all caler in fact we didn't assume that that this thing is caler, but you can prove that it is caler that is a hard theorem. And they have they have unique caler metrics. So, for complex tour I well complex tour I think there's no what they are because of, at least because of Angela's lectures. So, it's, you know, C two modular lattice. And so they are, they are not simply connected obviously, but they are hypercalor right they have a hypercalor structure, and the omega two is also trivial. Okay, and then I gave I gave on the next page we had the simplest examples of K three surfaces at the top of the page here. So you can take double covers of P two branch along smooth sex sticks. These are actually algebraic K three surfaces right so non algebra K three surfaces could be projective varieties or not they could be just compact canada manifolds without having any kind of embedding into a projective space. So, but, but there are a lot of them that are algebraic you know you get the co dimension one families of algebraic k three's and these examples that I've got here are algebraic ones. Double covers of P two branch the long smooth sex sticks smooth cortex in P three to three complete intersections in P four to two complete intersections in P five. And then the others as I said are a little bit more complicated so you can you know you can describe more complicated ones but I'm not going to get into that here. All right. So, um, and then there are other examples. So what about higher dimensions. So higher dimensional hypercalors compact hypercalors. What are these higher dimensional ones. Actually, sorry, let me my computer is still overheating let me try to do something about that. Allow it to cool down a bit. So, okay, so what about higher dimensional compact hypercalor manifolds so let me also explain why I'm so I'm insisting so much on these examples. The thing is that it's actually hard to write down examples of compact hypercalors, what people know a lot of examples of non compact hypercalors, but with compact ones it's much, much harder. So the examples that I'm going to, I can basically today basically I can sort of tell you all of the examples of hypercalor manifolds that people know about. And it's a big open problem actually to try and write down new examples, or maybe to show that there aren't any other examples maybe this is all there is maybe, or maybe maybe it isn't you nobody knows right. I'm not sure anybody is even conjecturing anything about it at this point. People don't feel that they have enough information for to even make a conjecture. Okay, so. So let me so then what are what are the examples that we already know in in in dimension to we know all of them, right, but in higher dimensions. We basically can construct them using the two dimensional cases so there are two infinite series of examples, which which just use the surface case to produce higher dimensional compact hypercalors. So, and the first construction involves taking Hilbert schemes of points. So that's number one, actually, maybe I won't remember it. So how does that work. We start with a compact complex manifold of dimension to meaning a compact complex surface, and we, I will denote as to the R the art Cartesian power. And let me denote as to the R in parentheses the our symmetric power. Actually, let me just define it first so this is going to be as to the R modular the action of the symmetric group SR. So, this is the our symmetric power. The action of the symmetric group is, is by permuting the factors of S to the R right so this. So you can naturally think of, think of this, the symmetric power as effective zero cycles on your surface so as to the R. And we can also think of it as effective zero cycles. Right. So a priority, you know, a point of S to the R of the Cartesian power will be will have coordinates like x one. X R. X INS. And if I'm looking at P bar, which is so let me denote pie, the quotient map from S to the R onto the onto the symmetric power, if P bar is high of P, you can write P bar as X one plus plus. Now you see that if, if some of the exercise coincide, for instance, right, then you will get multiple degrees inside P bar. All right. Okay, and I'm going to also introduce some more some more notation that we call delta ij the diagonal of the Cartesian power where the ice and the J coordinates are equal right. And we know that the action of the symmetric group is not free on the diagonals right and. Okay, and then we can also describe the stabilizer right so the stabilizer of a generic point of delta ij it's just a transposition. Well of course you get the identity I can take and I'm describing the subgroup right so I get the identity and then I have the transposition that in that permutes I and J. Okay, so. All right, and this this quotient morphism right so this pie is et al away from diagonals from the union of all the diagonals. So now basically so now this already tells you that the symmetric power cannot be smooth right so because the diagonals are co dimension to right so the co dimension of delta ij in in in SR is to right so so then. This map you know this map pie pie is ramified along ramified exactly along these diagonals. So then this means that SR cannot be smooth because because whenever you have a finite map of. Of manifolds the there we have the purity theorem which tells us that the ramification locals has to have has to be put mentioned one so if it's not if the ramification locals is not put mentioned one that means that we don't have manifold. Now we know that s to the hour is a manifold because s is smooth so the cartoon power is smooth, but then this tells you that the symmetric power is not smooth. All right. Okay, so this symmetric power is not smooth and what we're going to do we're going to resolve it so this SR has a natural singleization. So this is the Hilbert scheme. So this is the Hilbert scheme, which I will denote as between square brackets. Of length are our team and sub schemes, our team and you know, analytic of course sub schemes, we're in the analytic setting here right of s. All right, and you have a you have a natural map from this Hilbert scheme to the symmetric power, which sends a sub scheme see to the underlying cycle, which is going to be a zero cycle of degree, degree little on right because the length of the sub scheme is our so then the degree of the zero cycle will also be our. Okay. So then this, this Hilbert scheme is, you know, is a little bit complicated in general but it has a nice description in in co dimension one, or maybe I should say it's co dimension two over the symmetric power but co dimension one on the on the other side. So, so basically so let me. So, let me do a geometric description. So let, so let s star are, and or, respectively. Let me just introduce this in a different way. So I'm going to have s star are contained in SR and then I'm going to have SR, the symmetric version inside SR, and I'm going to also have the Hilbert scheme version. And in all of these cases. So this one, the first one. So this is going to be the open set where at most two coordinates are equal. And then the same thing. So the others will be the image so this one will be the image right of that one. And then this is our star this will be the inverse image inside the Hilbert scheme. So what if you would have what what's going on here so if you have, if you're given a cycle. And where you know you have multiplicity two at one point. And so this one is as in, I want this one to be in SR lower star right. I have a cycle like this. How do I get an artinian subscene supported on the cycle right so the data of an artinian subscene of like our supported on this on this on this cycle is equivalent to the sum of a tangent direction to us at x one right so if I want to specify a subscene, while the subscene is going to be reduced everywhere except that x one, because that the multiplicities are one elsewhere right so the only place where you have a multiplicity to is x one, and then to the to this to you know to specify the scheme structure you just have to specify a tangent direction, right. So then there's the set of our team and sub schemes. So, so what does this means that the set of our team and sub schemes supported on to x one plus XR is can be identified with the projectivization of the tangent space of s at x one. Okay, so if I specify tangent direction that's the same as specifying a point of the projectivization of the tangent space. Okay, so then, you know, so then from this, you will see that, you know, so the theorem that I'm going to state now will make sense I'm not going to prove the theorem but it's not very hard to prove if you keep this this statement in mind right. So, the complex analytic pair. So, the sum of s are actually I need one more piece of notation sorry about that so denote. So let's do let the inside the symmetric power, the, the image of the diagonals. So this is just the diagonal right the symmetric power has but only only one diagonal, right. And then the lower star is going to be the intersected with SR lower star so the lower star is going to be exactly the places where two of the points coincide right. So let me put have that notation there. And then for my theorem the complex analytic pair, SR lower star. The lower star is locally an isomorphic to the pair, which I will call big times C and B times zero, where B is a complex ball and C is a cone with vertex. Oh, over a smooth conic. So this is this is a complete analytics description of the symmetric power right but away from the very bad points right these are at some bad points. And the Hilbert scheme you can describe the Hilbert scheme. So this is a complex manifold this is the blow up along the lower star of the symmetric power. So if I remove the points where more than two coordinates coincide, then I can describe the Hilbert scheme as just a blow up of the symmetric power, which is which is very nice right. So next, you can also then you also have a commutative diagram right, actually a pullback diagram, the Cartesian diagram like this so you can. So you have the Cartesian power mapping down to the symmetric power. And then here we are. And then you can put the Hilbert scheme, which we said is the blow up along the lower star of the symmetric power. And then here you can put the blow up along the diagonal, the union of the diagonals of SR lower star maybe I should also call that the delta lower star because I am removing all the points where more than two coordinates coincide. Okay, so, so basically what are you saying here you're saying so this guy here was a quotient by the action of the symmetric group right so what are what we're saying here is that the Hilbert scheme. It's the blow up of the symmetric power but it's also the quotient of the blow up of the Cartesian power. So the action of the symmetric group. So we're also saying that the action of the symmetric group actually lifts to the blow up, right. So the action. SR lifts to the delta lower star of SR lower star, okay. So you have a very nice picture, right. And then, you know, once you have, once you have this picture like this, then you can, you can in fact you know we want to show, what do we want to show we want what in the end what we want to do we want to produce examples of hypercalium manifolds, or if you like examples of the holomorphic symplectic manifolds which we saw are the same thing, right. So, here we're going to produce an example using the holomorphic symplectic language rather than the hypercalium language so if we assume that s has got a symplectic form right and everywhere non degenerate holomorphic to form, then we can use this description this diagram to produce a holomorphic form a symplectic form on the Hilbert scheme. Okay, so here's how we're going to do that. So, proposition, if KS, which is omega two s is trivial, SR the Hilbert scheme admits a holomorphic symplectic form. And let me give you a sketch of proof, ideal proof. All right, so how do we how do we do it so so we're going to choose a generator fee of, or actually let me call it omega. I'm used to calling it omega I shouldn't change my notation omega of h zero of s KS right now this this is holomorphic symplectic because KS is trivial so this guy doesn't vanish anyway. It doesn't vanish anyway right so our form doesn't vanish anyway it's a holomorphic symplectic form. Then we're going to pull it back so see is, let's pull back via the first projection. We're going to put it back to the Cartesian power first okay so I'm pulling back via the first projection to the Cartesian power, and then doing it for all the other projections PRR upper star of omega. So this is where we are I from SR to S is the eye projection. Right. Um, okay so then so now this is now a holomorphic to form on the Cartesian power right, and we were going to pull it back so if I if I go back to my diagram here let me give some names to this morphisms. I'm going to call a to this blow up morphism here. Epsilon is this other blow up morphism. My quotient map was pie. And then let me call the other quotient map from the blow up of the Cartesian power to the blow up of the symmetric power to the blow. Okay, so I took my form on S, I, I, I took an average right I pulled it back by the various projections to the Cartesian power, and then I took an average right now I'm going to pull it back to the blow up so. to BL delta of SR star so what do we do so we take a to upper star C restricted to SR lower star right. And so this is this is invariant under the action of SR right. And so this is a to upper star of C itself right so they're both been variant right so this means that there exists a holomorphic form. S star lower S lower star little R such that a to upper star of C is equal to row upper star of fee again remember row is the quotient by the action of the symmetric group. So we're saying that a to upper star of C is invariant under the action of the symmetric group so then it's a pullback from the quotient by the action of the symmetric group right, but this guy a priori this is only well defined on the. S R lower star. Okay, now because the co dimension so because SR minus. SR lower star has co dimension to. Then this this fee is actually. The extends to all of the Hilbert scheme. Alright, so you've got we've got our form it's a it's a two form right it's holomorphic. Now the only question is whether it is non degenerate. Okay, so I'm not going to get too much into the details of why it's not non degenerate so need to show that this fee is not everywhere non degenerate. Sorry. Yes. What are it and the raw in the previous page. Oh yeah yeah right right. Let me show you that. Oh, there it is. Okay, so yeah so it's the blow up diagram right so. So, you know, Ata was the quotient by the action. Oh sorry sorry no Ata. Yeah, Ata is the blow up map right for on the Cartesian power and row is the quotient by the action of the symmetric group on the two blow ups. That's what it is. Okay, so you you use the same notation. Okay. Yeah, sorry yeah I don't I didn't want to change notation because it's. You know if I if I just it would be too much notation so I'm abusing the notation a little bit right I mean I'm kind of. So, so here you know when I, so I really should write see again restricted to SR lower star right. And the fee again is not a priority fee is only well defined on SR lower star but I'm just saying I'm going to extend it, I can I can extend it to all of us are and then I'm going to denote it with the same letter. I'm not going to use a different letter you know to denote the extension, if that's okay. Thank you. Okay, so. All right. And so we've got okay so we've got this holomorphic. It is a holomorphic form it is a holomorphic to form right, and we need to show that is everywhere non degenerate the idea for doing that is to show that. We need to show that it has a posterior power, this which are a fee we have to show that this guy does not vanish anywhere. And in order to show that it doesn't vanish anywhere. You need to do a little bit of analysis. I'm not going to really get into that but just let me give you the idea the idea is show that it's divisor. The line bundle now right I mean, which are a fee is a section of the line bundle. And to show that it doesn't vanish anywhere so is a section of the canonical right because it's holomorphic to form. So, we show that it's divisor of zeros is just the zero divisor zero that doesn't vanish anywhere that's exactly it right so, and this you can do it's not it's not that difficult you just you go but you can you go back over the definition of fee right fee was obtained. We took the holomorphic simply form on S, we put it back by the various projections and then we, you know we, we took the average and then we said that this that the pullback of fee was discussed if you look at the definition you can sort of work out what the divisor should be, and you can show that divided that there's no divisor. So, I mean if I have time I can go back and do a little bit more details on this maybe today or something but for now let me let me go on with the other stuff we need to do. Okay, so we have a holomorphic syntactic form. But, but it's not. I mean, of course, SR itself you know the Cartesian power itself has got lots of holomorphic forms you know you just take your homomorphic form on S, you can put it back by at the various projections you can take combinations of those that's why we've got our team to one such right. But what's interesting about the Hilbert scheme is actually that it's going to be irreducible holomorphic sympathy right so in fact, this guy is irreducible holomorphic syntactic. And remember what that means it means that there is up to up to a constant multiple, there is only one holomorphic to form on this manifold. So, so that's. Okay, so that's and you can do that actually by basically showing by basically can computing, you can compute the co homology and the fundamental group of the Hilbert scheme okay. So let me just maybe explain a little bit how that how that works, right. Okay, so we compute the fundamental group and co homology of SR. Oh actually it's irreducible if, if s is K3. Okay, if s is, if s is a complex torus of dimension to, it's actually not going to be irreducible so we're going to have to do a little bit more work to get our holomorphic syntactic manifold, I mean our irreducible homomorphic syntactic Okay, so but anyway we can compute the fundamental group and the co homology right. Either way just assuming that s is holomorphic symplectic itself right, whether it's K3 or a complex torus. So to do that we're going to write we're going to go back to our blow up diagram again right so. And to our quotient by finite groups right this diagram here is, you know the diagram that I had before this Cartesian diagram is very important right this is how we're going to compute everything basically the co homology the the fundamental group and everything. So, the fact that we have quotients by by the symmetric group right gives us. These are quotients by symmetric so they're Galois covers right so you get exact sequences of fundamental groups right so so we have that SR to SR and. Also the blow up map. The delta of SR to the Hilbert skin actually lower star here are Galois. And both of them have Galois group SR is metric group right. So that means that we have exact sequences of fundamental groups as follows, I can ever go one. So the star goes to pi one of the delta of SR star goes to pi one of the Hilbert scheme. And then here, I have a map, I can, I can map this to pi one of the symmetric power. Here I can put by one of the Cartesian power. And here I can also put SR again. Right. And now these two are of course equal. So what have I got once in the middle here what have I got well, the pi one of the blow up. Well, the, it's a pi one of the blow up along a co dimension to space right so this is equal to pi one of SR lower star. And then again because we're just removing the co dimension to space this this is also equal to that. All right, so you have, you have a commutative diagram of exact sequences you have isomorphisms in the, in the middle and on the left. So this means that, you know, if you do a little diagram chase what you get is that pi one of the Hilbert scheme lower star is isomorphic to pi one. And then once again you do some, you do the similar argument that we had before this pi one here is also isomorphic to pi one of the Hilbert scheme. Again, because you're you're always doing things up to a code mentioned to raucous, so you don't have any problems. And then also that that means I can remove the star here. Okay. All right so this is the pi one. And then as I said we can compute the comology going to be a little lemma here. The comology of the symmetric power. I will do everything with rational coefficients can be identified with the comology of the Cartesian power. You have to take the invariance on the rest or this is pretty standard right whenever you take a portion by a national defined finite group you can do this. Number two, you can compute. Now, now we're going to compute the Hilbert scheme okay this was just for the symmetric power, the common the age two of the Hilbert scheme with rational coefficients is isomorphic to age two of the symmetric power. Multiple of the class of the exceptional divisor. So I mean again this makes sense right because, well, we're talking about age two. So we can again because we're talking about age two we can replace SR by SR lower star right remove the remove the bad loci. And then it just it's just a blow up. So if it's just a blow up along the diagonal, then you get that, you know this is just a blow up formula the comology of upstairs is the comology downstairs plus multiples of the exceptional divisor. So this is also standard then. And then number three, then we can also write down the comology of the symmetric power in terms of the comology of s. Direct sum wedge to age one of us. Again this is this is this is relatively standard you just do. You use number one which tells you that the comology of the symmetric power is you have to take the invariance of the comologies of the Cartesian power and then you use the cunec formula. So the cunec formula will give you the comology of the Cartesian power and then you take invariance and you get you end up with this one. So let me just briefly write down some reason how we prove these things right so how remember you should take at some point the five minutes break. Oh, sorry, yes. Yes, you're actually it's exactly time for the five minute break. Yeah. But maybe let me just write down the. Yeah, I was just the first one. The first one I guess I'm not telling you anything I'm just telling it standard. Number two is the blow up formula and replace. SR. With. And number three. Use one with the cunec formula and the same thing and replace SR with SR. And that's it. That's all I need to say. Thank you. So I will okay so we're stopping now for five minutes. Thank you. All right, thanks. Yeah, sorry for interrupting. I was just, I thought it was a good moment for a break so. Yeah, no thanks I mean I was actually about to forget. And actually I should say I mean I need, I need to leave exactly at 1130 today if that's okay so. Yeah, okay.