 I want to thank the organizers for the invitation. So the object I'm going to talk about are quite classical. So these are the surreal writings. So what's the surreal writing? Well, it's just you taking a big surface and fix a line bundle and looking at the curves in this line bundle, which I would quite think could be irreducible, reduced. And with the fixed genius, fix the geometric genius. So I put this everything inside this with the fixed genius into this variety. It's called the surreal writing, the very classical object. So for the surreal writing, we do have some kind of the expect behavior. It's a very naive consideration. So if you take a general member in the surreal writing, we're expecting, well, almost, and this actually not necessarily true, but we're expecting this thing is nodal. So we're expecting it has exactly the edge of this linear system minus the geometric genius. So expecting we have the delta nodes. So that's one thing. The other thing is we're expecting the dimension of the thing is the dimension of the new system minus delta. So it's like if you each node should impose exactly one condition on this new system. So we're expecting you have this dimension. Obviously, this actually will fail in many cases, especially when it's x of general type. So that's the very classical object called the surreal writing. So now I'm most just interesting things which over the KC surface. So look at the surreal writing on the KC surface. You fix a line bundle and look at the curves with a fixed genius in this line bundle. So for a KC surface, so a KC surface for the surface with a trivial cognitive model and with h1, 0x, 0. So for KC surface, if you look at the KC surface, if you fix a big net line bundle on x and if you look at the dimension of the linear system, it's actually exactly given by the asymmetric genius, the PAL. So if you look at the expert dimension, the workhorse exactly could be cheap because the dimension of the linear system is exactly the S-net genius. We get the dimension, the expert dimension is the genius geometry genius of the curve. So if there's no guarantee that this writing will be non-empty, but if it is non-empty, the some deformation theory of curves will tell us that every irregular component should have the exact dimension. So that's the form of the deformation theory that will tell us that the expert dimension will hold for every component if the variety is non-empty. So let me talk a little bit about the non-emptiness of the submit writers. So there's no guarantee that actually you fix a line bundle. You can find a curve with these fixed genius. If you pick any number of G between 0 and the asymmetric genius of L, there's no guarantee you can find such a curve. So that's called the existence of the genius G curve on K3. So let me say something about these non-emptiness of the rewrite. So in the end, I think everything will come down to find the rational curves because if you have a rational curve, that's a genius 0. You can deform this rational curve. Basically, you can deform the smooth side of the singularity to get a curve of high genius. So for general case 3, this is not that hard. I mean, you can find a nodal curve, a nodal rational curve in every linear system and a very general case 3 surface. So that's, well, so from that nodal rational curve, you can deform those nodes and make sure you get any genius curve of any genius in this way. So that's for very general, you can always the serivarity is always non-empty for any linear system. Now, if you go to the question for arbitrary K3, it's become very difficult. So this is a collection of, well, the whole program of our existence is actually a collection of these works from various people. First is if you have a K3 with some extra structure, if you have a K3 or a K3 with infinite automorphism. So for this K3, you can be this kind of, well, for a different K3, you have this self-rational map from x to the self. So in this way, you can produce infinitely many rational curves in this way. So this is by Bogumov-Chinko, then Hazard, and Paiyue very recently for over a carousel P. So for such K3, indeed, you can find infinitely many rational curves, okay, infinitely many rational curves. And then by deform those rational curves, you will get a high genius curve as well. So that's the four K3 which are elliptic and the infinite automorphism. Now for K3, if you take a K3 which is, well, arbitrary, so there's without any structure, especially if you just assume the car rank is one, there's not much she can play around this. So the big breakthrough is by Bogumov-Chinko. So the most important about this result is the technique they introduced, it's by carousel P reduction. So what they prove is you have many rational curves on K3, with the car rank one, with genius of the K3, is two. So the most important thing about their work is actually the introduction of this method, it's called the carousel P reduction. So it's roughly, you can reduce the statement to K3 over Q bar. And then you can do the reduction to carousel P and then using deformation to find infinitely rational curves. So their result only works when X has genius two. So when this is the double cover of the static curve. So the next breakthrough is by Dringley and Lidica. They can prove this using the same method, but with a lot of refinement. And they can prove that this is true actually for rank, car rank is odd. So it's with odd car rank. So if you look at the combining the previous statements, you see that the fact here is if the car rank is big than to five, X becomes elliptic. So it's covered by the previous Bogomorov-Chinko result. And then so the remaining case, well, if you take the Dringley, well, Lidica's result with Bogomorov-Chinko, and the remaining case is car rank two and four. And also in car rank four, there's only two cases which does not have infirmary automorphism or does not have any inhibitory vibration. So the remaining case is basically rank two case. So the rank two case, we eventually just finished this rank two case. We show that not only for rational curves, it's for any curves of gingers G. And you can find infirmary such component of this severe writing. So, well, there's still a lot of questions you can ask here because we have no control of this L. So we don't even know the things where this is infirmary of L. We don't even know this L actually generate the ample comb. So whether you can say I can find this L, which VxLg is now empty, and we this L actually generate the ample comb, we don't know how to answer that question. So, in particular, if I can formulate this conjecture, is that for any case-free surface over C and every very ample line bundle, then this, well, the severity is now empty. So we don't know how to prove this. So the existence is only for infirmary, but we don't know what these L are. So it's construct by the, well, we don't even know that these L actually generate ample comb. Okay, so these are the existence. So I think the next larger question is what these variety look like, right? So it's now empty. So what does it look like? So the first property is whether it's irreducible. So that's the conjecture about the severity. Again, we can only, well, this is definitely going to fail for some special K3. So that's only for the general K3. So you can ask this for any, well, you can take this for any very ample divisor, but let me just stick this to the previous polarized K3 and in the previous class. So the conjecture is for a general polarized K3 by that, I mean, it's X with polarized by ample and bundle. Required this L to be ample and the primitive means it's not indivisible in the in the group. And so I expecting, of course, you cannot, you're going to ask the question when G is bigger than one, because when G equal to zero, the dimension is zero. So it's going to be, never going to be irreducible. So the conjecture here is if G is at least one, then this variety is expected to be irreducible, okay? Expect to be irreducible. So this actually obviously true if you take this, the number of these nodes, okay? If you take these P, A, L, mind G, it's expect number of nodes of curving this variety. If this is small, this is true because if you delta is small, the delta nodes impose independent condition on the linear system. So that's when delta is small, that this is true. So that's the conjecture. So we're trying to prove some sort of variety is irreducible. So where should we start? Okay, so the proof of this kind of things like some variety or more just space, well, usually we look at the first such result which is the Harris proof of plain curves. So we look at the, his proof, okay? So that's usually the starting point of these kind of statements. We look at what kind of things we get technically follow from the proof. So your heads prove that if you take the survey variety of plain curves of degree D engines G and this variety is irreducible, is irreducible. So it's a classical result. And we look at when, whether we can just carry his proof to KC surface, okay? Whether we can carry his proof to KC surface. So Harris proof consists of two parts. I will call this one is easy. Well, it's the easy part, the other is hard part. So the easy part is that first, well, first is if you're taking rational curves on P2. So it's obviously irreducible because a rational curve is given by just three polynomials of degree D. Obviously it's irreducible. And we need a little bit more that if you look at the monogamy group X on the nodes of rational curve, that's monogamy is the fourth symmetric group. Okay, fourth symmetric group. So that's the easy part. The easy part is basically have a base case, the base case VD0, it's irreducible. So that base case. Now the hard part is show if you take closure of VDG, okay, taking G bigger than to zero, taking closure, it always contain these components VD0. So that's the hard part that most of the proof is actually to justify it by degeneration. So that's the hard part. So the idea key is, I hope you can see this. So I try to avoid to write here. So the idea here is, for example, if you take V40, so that's the quarter curve, rational curve, which has three nodes, okay? So the general member is like this, three nodes. Now, if I want to show like the four V41, that's the curve with two nodes, so that's V41. So when this deform to a curve of two nodes, basically it will smooth out one node, okay? So the local geometry looks like if you take, say this is V, this whole thing is V41, and the locus of V40, it's like a triple node, okay, a triple point. So this, I think this is V41 and this point is a V40. So that's the local picture of this V40 system type V41. So the thing is, the second statement, first of all, V40 is irreducible and the nodes at three nodes of V40 actually, it's at the monogamous group acts as a four-symmetric group on these three nodes. So they change these three nodes. So that actually change all these three sheets. So it shows that this variety, if they closure, is locally irreducible along these V40s. So that's how he act to prove that. So every component of this one contain a copy of this, well, contain this thing, and it's locally irreducible along this. So this V41 is going to be irreducible, okay? So that's the strategy. So let's see, how do we actually carry this over to KC systems, okay? So we carry this over to K3 surface. So we basically follow the same strategy. We need a base case. So the, well, we take base case, it's VD1, okay? So that's the unit curve on K3. So show this thing is irreducible. And then the hard part, well, then the second part is show, if you take energy is bigger than one, the closure will contain this component of VD1. So that's the strategy of K3. So if you want to carry this argument over to K3, what we find actually the situation will be reversed. So the first step actually hard part. So the base case for unit curve on K3, actually the severe variety is this case at the hardest among all these irreducibility, all these severe variety. So the situation is reversed. So the first one is hard. So this one is hard. So to show that the severe variety of unit curves on K3 are this irreducible, it's actually, it's a difficult part. So this one is difficult part. The second part is not that bad. So that's what we know. So every, if you take the closure of any VX, VXLG, it will contain at least one unit curve. So the argument is something like this. So it's based on the following. You take, so the way to show that you can actually, well, any closure contain at least one unit curve. You look at this, the modular map to the modular space of curves, okay? So this is edge. And this map actually is a generic finite. So the image had dimension G. So the dimension of this edge of VXLG, okay? So then there is a classical result of D8 which said that there's no complete variety inside this modular space of dimension G. So edge of VXLG, this is not complete. In MG bar, okay? So it's not complete variety. So it will hit the boundary somewhere. So that means in boundary of these, the severity, you'll pick up some curves with the lower G. So that's the, that they can prove that, that the second statement. So the second statement now is easy, but the first thing, it's become hard. It's like a reverse situation with Harris, the plane curves. So what's the, so what, what, what we can do? Okay. So basically to prove this, we don't have an easy base case. We don't have an easy base case. So the base case with the curve actually is the hardest. Okay. So it's not complete variety. So basically, the base case with the curve actually is the hardest. Okay. So it's not like in the case of a plane curves, you have easy case of rational curves, which obviously is irreducible. So what, what do we can do? So there's a recent paper by Bruno and the Lillian kids are, and they approve that they claim this. So the, if you take a very general case, the surface applied K3, general K3, then this, this is connected for all G. And it's irreducible for a large number. We're actually almost they solve everything for almost everything. Well, the main case is G is less than three. Okay. So, I think there's some gaps in there too, but I want to say, there's still a lot of new technique that introduced is quite useful. So this, they introduced this two new technique. One is like for this backward induction. So for the river, actually G could be one is the hardest case. So, but when G is very large, close to the advantageous. So when the delta of the nodes are very small, actually it's easy. Okay. So actually it's known, I mean, that's the doubt is one or two, then it is irreducible. So that's known. So what they do is try to do induction backwards. So, starting with high G and the reduced to the lowest. So from the irreducibility or connected for G is large and reduces to G is small. So that go backwards. So that's one thing. The other thing is, they, they try to derive this, you reduce ability from connected. So that's actually the thing I want to talk about. So, so the, so the message is these derive this, you reduce ability from connected. So that's that's active piece on the top of. So the, so, so let's ask a general question. So if you, if you know a variety is connected, how do you show that is it reducible? You need something else extra. So let V to be a variety. Then, if we know it's connected, what's the extra condition we need to make it, make it reducible. Okay. So I was adding two conditions. So that's the way that actually what they, as they suggest is you need to show two extra things. First, you need it to be coconut colleague. Okay. So actually can replace this by a S5S2. So you need coconut colleague. And the other thing is you need to be a smooth in connection one. So if you have these three conditions, then it goes from connectedness to your duty fit. So a connect is usually it's easy to handle them. You reduce ability, basically because when we prove this statement, we'll want to do the generation. So when you degenerate, the, the survival become reducible. Okay. But a connect is still preserved in generation. So it's usually easy to prove something is connected. And then you reduce it. So if you know the connectedness, how do you actually divide this to the, this V is irreducible. So that's, you need to addition condition. One is it's coconut colleague. And the second one is V is moving for dimension one. So if you have these three things, then we say V is connected with irreducible. That's from the pressure on connected principle. So the argument is, is following. I mean, V has a lot of components, right? A lot of components. You can look at the intersection between two of them. So, because these V is a smooth in connection one. Well, V is a colleague. So it's obviously a few of pure dimension. So you don't have any funny things like you get the component. So any two components will insect. Well, if they insect, it must be co-dimension less as co-dimension bigger than two, because assuming the V is a smooth in connection one, so they cannot meet along some variety of dimension, co-dimension one. So the co-dimension of the section has to be at least two. So now if you remove this, this intersection, you remove a bunch of things with co-dimension two. So there is this harsh on connectedness, which says if you're coconut colleague variety, if you remove a subset of co-dimension at least two, and you still get connectedness. Well, that's the kind of job actually intuition is if you remove a point from C2, you still do these things connected. So that's the... So then if you remove this intersection between these components, you get something connected. But if this is reducible, you get contradiction because if you remove this intersection, obviously if you have more than one components, obviously this becomes disconnected. So that's how you derive from the connectedness to irreducibility. So we need two things. One is for co-dimension one. So roughly we want to do this for the variety. So assuming we already know it's connected, assuming that the real result for the connectedness is right. Then how do you get this irreducibility is you want to show first it's co-common colleague. The other thing is smoothing co-dimension one. But there's one thing. One thing is, so the variety is actually not the good object to study because it's... Well, it's not smoothing co-dimension one. So it's too singular. So we look at the more just the more space of map. So we look at the more just space of state maps of G to X, which had the class of the image is in this L. So I call it mxlg. And look at the open part is the one dominated the survey variety. So look at this mxlg, which is the open part which consists of mobilization of curves in the survey variety. So that's the more just space. Now you look at this map from this mxlg to Vxlg. So this one, you can easily see that it's one to one and on two. And these two actually are bi-rational. Okay, bi-rational morphin. So for the one side, it's not an isomorphin. Okay, so actually I made the mistake myself. I thought it's isomorphins. I say everything in terms of the survey variety. Actually it's not. This one is not isomorphin. So the situation is something like the normalization of a cascaded curve. So you still get these homomorphins, but they are not isomorphic as teams. So these two are not safe. So actually this is exactly the normalization map. The exact normalization map. So you'll see that this mxlg is normal. And that's the normalization map. But these two are not safe. But as far as connectedness and irreducibility is concerned, it really matters because these two are homomorphic. Okay, so that as a couple of just space, these two one to one and on two, so one thing is irreducible the other one. So this Vxlg is connected or irreducible if and only if this modus space is. So these two are as far as the reducibility is concerned. These two things are safe. So we look at this modus stack of maps. And there's some, let me just say something we should know about everyone, which is what's the 10-year space for these 10-year space to these modus stable maps. It's given by the normal bound. So that's the 10-year space. And then you have the objection H1 and F, okay. So the difference of them is the virtual dimension but for cases of if there are some twists for virtual dimension something we'll mention later. So that's the 10-year space. Now, what's the dimension? Well, dimension you can estimate using the H0 and F. So the normal bundle of the map and this is given by if the map is we call this unrampified then this nothing with G. If it's not, then there is a standard method by this abarita and conabas lemma, which is on the general member of this modus space. The portion part that now contributes to the 10-year space. So you can model these torsion, you still get estimate. So the dimension is always the best with G. So that's the estimate of dimension. And also this is a local company section. I kind of try to find the located reference for this. It's kind of a little bit hard to find. So I just formulate the argument myself. So the argument is using a twist family and lifting everything to the help scheme. So you can look at basically I taking the portion of the help scheme get this stack. So you can embed a curve. Well, first you need to get the right dimension. You have to using a twist family of K2 surface. So the twist family is the central fiber is the K2 surface. The general fiber is not projected. It's not projected. So there you can do the estimate. So once you get the right dimension, so by bound this into a two direction. If you have this as the right dimension, then it's automatically it's a local company section. So that's from the give me if you have this, the virtual dimension same as the cable scheme, and then it's automatically a local company. So these mx, LG is a local company section. And well, then because it's smooth over the stack. So the low at this mx LG is also a local company section. Okay, so that's the. Yeah. Well, so you put this into your general deformation. It will get definitely get something which is with it. Yeah, I just abstract with family. It's not, I mean, you have some, I can be in the job to have a different job to have different way to specifically with contract between family. But for me, it's that they can get general deformation Compact deformation. No, no. Well, general compact deformation not definitely not. Not, not that. As long as you have the property, then everything works. Okay, so, so, so it's a local company section. The other thing is, well, we have to, well, so that's fixed the one piece of the puzzle, which we need to be a local company section. So it's, it's a common colleague. So I'm guessing it's one, one, this thing has smooth in commission one. So what's the smooth locus of these more just base. Well, if, if the F is unrefined. Okay, so it's automatically identified estimate I did before, then this normal bundle of this is going to be same as the canonical bundle of the sea. Okay. So then it's automatically smooth at the point. But you can do a little bit more. It's just unrefined. So you can allow to have at most one double point. So if you're one double point, what happened here is the normal bundle will have a portion. Okay. So but this portion is only have at one point. So it's only have one portion point one point. So if you look at this normal bundle is given by these, the kind of Kc twist by this neck and P and plus this portion support at T. So that's the normal bundle. Okay. So you can compute this. You can still get dimension G. So the service companion space still have the right dimension G. So such F is also a smooth point. There's also some fun. So allow the in the more just space of the map. You allow the curve has a double point. Okay. A double point is okay. If you treat up once it's kind of tricky. One double point is to guarantee system. So that's the, the observation. So you allow this curve. Well, the simple point can consist either this F is unmerified, like all other security are nodes. And all it has, it can most almost all of them are nodes, but we won't cause that's fine. Okay. That's fine. So that's what that's how we actually do this. Okay. So what is this is still a walking in progress. Is we clear, we prove that that's more less expected. If you take a, well, a general subfamily of subfamily of the celebrity. Okay. So the, we expecting every member in the subfamily has, has at most one cost. So, so that the way to do formalities is, I can fix G minus one general points and look at the, the curve of G passing through the G minus one points. So that's going to be one family. So in this one family, every member is no, though, outside a cost. Okay. At most costs. So we have, we say, we say, it has at least, at least P a L minus G minus one notes. Okay. So because you're expecting the, the, the, the, the, the number of nodes is P a L minus G. So if you have these many nodes, the remaining singularity has dealt with one. So that means neither node or cost. So the other, the other way to say this is every member of this one family is, is a node. Well, has at most one cuts at most one time. Okay. So the case for genius good to zero actually is no, that's every, every ratio curve is actually no, though. So some, some consequence, okay, some consequence is possible, because I, my comment is, if you have one cost, you still get this more just space is smooth. Okay. So that means the more just space is smooth in connection one. So what I've faced you, let's move to find everyone. Now, you have, you can, well, you have this CS criterion. If the, if the variety is called McCauley, and if it's moving from one, it's normal. So MG XL is XL G is, is normal, a local company section. Okay. So that's the support of all that space. So if you look at the structure on the singular on the server variety. The thing is, if you, this curve has a cost. The, the local geometry at this cost cost is also cost. It's all like, it's a tricky, it's a kind of interesting, interesting. If you look at the, the single locus of XL G, one locus, which has, which is custody though. Okay. The global about if you cut this by high planes, you'll get a cost of, well, if it will cut out, cut by G minus one high plane, you get curve with a cost, with cost. So, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the issue of these irreducibility is same as connected. If you can't prove connected, you could be really big. So if you're assuming the Bruno and the keys are the paper is right. Then, then this will finish the whole program. So that connects will inspire you to be really good. Okay. Connect will probably, will, will, will actually, prove the, you will be really good. Okay, so I think. So, so let me just say a little bit about the proof. So the proof is by the generation and it's very close to the proof body, which we specialize to this. This is the Brian Young K3, which is the case three, with a minus two section. So the advantage of this case three is, it's very useful to study the previous cost because the previous costs will completely become reducible breaks into the section, of course, these and fibers. And this is the case three is a C part the end fibers of the vibration. So, so the way to show this is by taking such a family, then I take a family of corresponding curves in this case we, and on the general fiber, it becomes this reducible thing. So to study the singularity, I had to blow up this family along these fibers F1 to FG and G1 to G24. Okay, so, so that's, you have to study the local deformation of those singularities. So I always keep the rest of them basically, there are a lot of this is involved some local study of deformation of singularities so that is one of these here, which is about tech note. And the usually, usually, people actually to capital Harris but actually on this one you do run. The, it's about the, you have a tech note of all the m, then the deformation of this all the m tech note will becomes M minus one note. Okay, and once we know. So, our situation have a request dealing with a little bit differently, which has at the rational double points at that point, it's just changing situation a little bit. And the deformation of those things will becomes one cast with and M minus two nodes and one cast. So that's where it comes from. So I think I will stop. Questions for the speaker. Are there any questions on zoom. Yeah, so there was one part you said it was in progress. So, so what was that, what was that step your store. And then five outside. So that will guarantee the more just place. Are there any games you can play where where you map between components of the service of very variety, as you vary the L. That's what the Bruno paper about I mean they they have the reduction from hiking. Okay, I was just thinking you could take advantage of the structure I mean since you're showing that they're not empty. Yeah, and then you can take advantage of the structure of play some game one bundles. Well, I'm just saying, you could go from one component to another and try to construct a map, relating the severity of violence. Oh, yeah, we'll, we're seeing that there's, well, effectively the people so they're the ones. No, I mean, I mean for each L. And so, so then you could play games with actual maps between. One to one. It's, it's not your problem, but, but it just went once you showed the non emptiness. Okay. Any other questions. All right, well, let's thank you.