 Okay, so I'll continue with some applications and I'll be readily speaking, speeding up a bit. So what I'll start with is with discuss applications to pre-neutotype statements. And so let me first give you the setting. So access to smooth projective K3 surface now. And let's say that the PECAR, let's say for simplicity that the PECAR group is rank one. H is the class of an ample divisor. And maybe as a, let me already point out that it's, one should change the set up here a little bit, namely, so as lambda again, you see algebraic homology, which is here, just for C3. And, but as this vector V, now I used the mokai vector. So that's the churn character of E multiplied with the square root of the top class, but really that's just a very small change. So it's churn zero, churn one, and then churn two replaced by churn two plus churn zero. That's just to make this map compatible with the, to make the Euler pairing compatible with the natural pairing on this homology group. And I mean, Bernoulli is about the following question. So if you take a curve in this linear system, what are the possible embeddings? Embeddings C2, sorry, morphisms, C2PR of given degree, right? And of course, if you have one such morphism, then you can always compose it with linear projections or with automorphisms of PR. And so we do all this up to such projections, also automorphisms. And so then this is, of course, classified by globally generated line models. Globally generated line models under L of degree D and eight zero of L equal R plus one, right? And then here, this is just, of course, L is just a pullback of a waffle. And then, I mean, what is the question that Bernoulli's statements answer? It's about the existence of such line models and when they exist, the dimension of those line models. What is, when is the following? So let me write V, R, D of C. These are all the L's in peak D of X with eight zero of L equal to R plus one. Right, so when is this non-empty? And if so, what is its dimension? Okay, and some of the invariant here is the Bernoulli's number row equal G minus R plus one times G minus D plus R, which you should think of as G minus eight zero times H one, right? And this is the expected damage. And you can see this from the deformation theory that this is the expected damage. Okay, and so let me also introduce a variant for the complete linear system here. So here, I just take the union over all C and H over V, R, D of C, but now I also allow, also allow torsion-free sheaves, a pure torsion-free sheaves when C is single. Okay, and so what does this have to do with what we looked at previously? Well, if you now look at the following mochi vector, V equals zero H D plus one minus three, then there is this materialized space M H of V of G-seqor stable torsion-sheaves. So what is this? This is really, this is a compactification of the relative Picar variety of degree line models on curves in the linear system, right? So this has a map to the linear system, isomorphic to PG. And here, if you take a smooth curve in here, then the pre-image will be the Picar variety. Okay, so here, yeah, when C is smooth, then yeah, yeah, yeah, when C is smooth, these are just line models and when C is single, then I'm allowing arbitrary torsion-free sheaves supported on C. Right, and so we've already, some of it is problem into a modular space of the type that we can do, that we can study. And right in particular note that M sigma alpha beta of V, so the stability conditions that I introduced yesterday, it's the same as M H of V, four B equals zero and alpha sufficiently big. Okay, now the goal is to study this question here using wall crossing. And so the first, the first lemma is that there's actually no wall as long as beta is equal to zero. M sigma alpha beta of V is equal to M H of V, alpha bigger than zero. And here, the reason is really very simple. Right, so here I'm still using this formula for Z alpha beta from yesterday, right? Minus, minus turn two beta plus I alpha H, germ one beta plus H squared, alpha squared over two, zero beta. And sorry, here I mean for beta equal to zero. Right, so for beta equal to zero, when here I'm just using the normal charm character, then I claim that I'm not crossing any walls, right? So for alpha large, I know this statement holds and I claim that this holds for all alpha. And the proof is really as simple as you could imagine. So I mean the imaginary part of either the alpha beta for any is in Z times alpha H squared, right just from this formula. And imaginary part of Z alpha beta of V is equal to alpha H squared, so it's as small as possible, right? And so there's no way this objects of class V could be, say, my stable because the central charges of the Jordan-Herter factors would have to lie somewhere here. And so that's impossible. Or in other words, this is, I mean, this is basically you should think of these objects as having rank one in co beta for beta equal to zero. Right, and so there is no wall for alpha bigger than zero and beta equal to zero, and so it follows that. And so the claim follows. Okay, so I mean in this picture yesterday, that I drew yesterday, this means that here when you have zero, zero, one, which corresponds to alpha going to plus infinity, then here you can draw an entire line segment. But this is beta equal to zero. That does not cross anymore, right? And so V is somewhere over here. Means really the first wall that could possibly appear is this one here. Like going through here, which is one, zero, zero, corresponding to alpha going to zero. So let's take a sigma naught on this wall. And the claim, the next claim is that this is actually a wall. And we let, the first of all, I claim that OX is sigma naught stable. And so if you have an L in MH of V, then this is automatically sigma naught semi-stable, right? It's stable all over here, and here it diversely could happen, it's semi-stable. And so the claim is not that this is strictly semi-stable, if and only if it has global sections. And in this case, I claim over, I can tell you the Jordan-Herter filtration, right? So in this case, I have automatically a map like this. And since OX is stable somehow, right? So, I mean, of course, maybe I should have said the obvious point is that OX and L have the same phase along this wall. So that's just because the kernel is contained in this band of, right? This here is churn of OX. So the kernel is contained in this line. So they have the same phase. And so this is a simple object in the category of semi-stables of this phase. And so this map is automatic in injection. And I claim that the quotient is stable, which is the same thing as saying that this short exact sequence is the Jordan-Herter filtration of L, right? And I mean, I won't say anything about the proof, except to say that this is really, it's not so hard because I mean, the basically, the Jordan-Herter factor, the, I mean, if this was strictly semi-stable, then all the Jordan-Herter factors would have to be automatically contained, have churn corrector contained in the rank two letters spanned by these three vectors here. And one can show that there's simply no room for this to decompose further. So it's a purely numerical computation using the letters inside the homology class. Okay, and so, okay, now the idea is we can, we can use this wall to describe VRD of H precisely. Right, and the first observation is that if you let WR to be equal to V of W, right, which is just V minus R plus one times V of OX, then it's an easier computation to see that WR squared is equal to two times rho minus two. Right, and so in particular, this means that if such a line bundle exists, then since this is, okay, sorry, I was going too fast. So, right, so I mean, let me for a note from here, just that if VRD of H is non-empty, then it follows that M sigma naught of WR is non-empty and moreover it actually is the stable locus, right? There aren't just semi-stables of this class, there are actually stable objects of this class. Okay, and so I claim that this is basically enough to describe this entire locus VRD of H. And so, but for that, I need a few more ingredients, right? And the most important one is really the description of which modular spaces precisely are non-empty. Right, so this, I mean for coherent sheaves, this is due to a number of people and then it was completed by Yoshioca and then it generalizes to bridge unstable objects due to results by Yukinobo. So let's say that V is primitive and that takes sigma in stop lambda of X. So maybe let's say in the component of geometric stability conditions that I described yesterday. First of all, M sigma of V isn't just a set, it exists as a stack or cos-modulized space. And the important thing is now that if sigma is generic, then we can say precisely when it's non-empty then M sigma of V is non-empty if and only if V squared is greater or equal to minus two. Right, so this is similar to the Bogomol of Gizek inequality that I had yesterday. So there is, I said if a modular space, right, if you have a Gizek stable sheave, then you have this delta H greater or equal to zero. But now here this is a much stronger statement. You have an if and only if you know precisely which modular spaces are non-empty. And then when non-empty, I mean maybe this is really Mokai's argument or maybe that we had Mokai here. It is a hypercalor variety of dimension V squared plus two. In other words, it has a holomorphic symplactic form that generates H zero of omega two. Okay, so the first corollary is already that just directly from this theorem, you see that if we are the of H's non-empty, then rho is greater or equal to zero. And the proof is just that if this is non-empty then M sigma naught of WR is non-empty. And so WR squared is greater or equal to minus two. Right, and so combined with this formula, which is a completely trivial computation that disproves the claim, right? And somehow this is, I mean this is due to a laser sphere. But somehow, I mean just looking at this wall crossing we are somehow forced to re-proof laser sphere results. Right, but now let's assume that we are in this situation then we can say a little bit more namely we can say the following. So let M, so the first lemma is right so let's assume that rho greater or equal to zero. Then I claim that M sigma naught of stable objects of, right so we know this modulate space is non-empty for generic stability condition. So this set of semi-stable objects on the wall is also non-empty, but I also claim that there are strictly semi-stable objects. And again that's a similar argument as before basically there's just no room to, there's no room to destabilize all objects of this cloud. And the next lemma is that once you have such an object right if you take W in M sigma naught stable of WR and you're given a short exact sequence O X goes to R plus O X R plus one goes into E on to W by just any short exact sequence like this given by an N dimension R plus one dimensional subspace of X one from W to O X. Then E is sigma plus stable. Yeah by sigma plus I just mean a stability condition slightly off the wall. So here I have my stability condition sigma naught and then you can imagine if you look at the stability condition nearby then if E was unstable then it would have to be destabilized by objects that are somehow semi-stable of the same phase with respect to sigma naught as all the other objects. And it's not so difficult to show that such objects cannot exist. Right but then this actually, but this means since there's no other wall between that and the large volume limit this really means that E is in M H of V, right? In other words E is a line battle or a torsion free sheet supported on one of these curves. In other words E is in B or D, right? And so as a corollary you get that this V or D of H is a Grasmanian bundles of R plus one dimensional subspaces from the X one from your stable objects to O X over this open subset of M sigma naught of W R given by strictly stable objects, right? And now the point is that W and O X are stable of the same phase so they have no homomorphisms either way. So by ser duality they have no X twos either way so you can compute the dimension of the X one, right? And then so you can compute the dimension of this vector bundle over M sigma naught and so you can compute the dimension of this Grasmanian and as a corollary you get that dimension of V R D of H is exactly what you expect, namely it's equal to the Brill-Neuter number plus the dimension of the linear system, right, Ro? Ro plus D. And with a bit more work, right now you can look at this map from V R D of H inside M H of V. This has this projection to H and with a bit more work one can actually show that this map here has equidimensional fibers, one can show that. And that's actually, I mean that's actually new result, a slightly improvement on these also results so you can really show that V R D of C has precisely expected dimension, right? I mean a priori this has expected dimension but the map to here could still have fibers where the dimension jumps, right? So these are the fibers of this map over here but the statement here is that this map really has expected dimension everywhere. And I mean the arguments here, I won't have time to explain this today. No, this works very easy. Again if you use the definition, so here if you allow, again allow torsion free sheets then it works well. And I mean the arguments for that some are used some tactic geometry, I mean some hypercaler geometry type arguments but I won't go into that. Any questions so far? Yes, so I mean this in, this is a short exact sequence in co-beta effects. Or if you want more precisely in, right? I mean in fact in coherent sheets this is typically subjective if your line model is globally generated. If the line model is globally generated then this map is actually typically subjective in the category of coherent sheets, yeah. So sigma plus stable where sigma plus is the stability condition here slightly off the wall but then it means it's also stable at a large volume limit so it's really it's a geyser stable sheet. I mean it's exactly one of these line models or torsion free sheets that we wanted to study. Let me also mention a variation on this of this theme. Given X is an abelian surface. Now there's a crucial difference. Let's again look at this short exact sequence in co-beta. Now we might have a zero of W not equal to zero. And the reason is just that, right? If you look at the long exact sequence in co-mology for global sections of the short exact sequence then by construction we have an isomorphism from H zero here to H zero here. And so H zero of this one goes into H one of O X. And of course on the abelian surface H one of O X is not equal to zero. And in particular and moreover, I mean for many numerical choices of R and D we are actually forced to choose a W with global sections because otherwise the X one space from W to O X would not be of sufficiently high dimension. It's just a computation that the numerics have to work a little bit differently, right? And so this means we actually we have to iterate this process. Again for the multilay space of objects of the class of W we again have to analyze which are, how big is the set of objects with zones or many global sections. And again you look at this by considering the short exact sequence like this. And of course somehow one way that stability conditions help is that somehow you know what you're doing here is just part of studying the Jordan-Helder filtration of L. And so in particular the methods are really simple to other wall crossing computation that people have already done. And so when you do this and you iterate this procedure then you get the following result. So I mean quite similar results were obtained with other methods by Knudsen, Lely-Kieter, and Mongaudi. And some of it is small crossing methods. We can prove a little bit more. So that's joint work by myself and Chunni Lee. So here let me now assume that P is less than G minus one. All right, so this means that the Euler characteristic D plus one minus G is negative. Then the statement is that VRD of H always has expected dimension when non-empty. But the non-emptiness now suddenly becomes a little bit of an arithmetic condition. So it's non-empty if and only if the expected dimension which is rho plus H is greater equal to D times minus chi minus D squared, where D is the remainder of division of R plus one modulo of minus chi. All right, so in other words, when R plus one is divisible modulo of chi then the statement is this variety is not empty if and only if it has non-negative expected dimension. But in general the condition is more subtle. Right, and so in particular this means the behavior is very different than for key three surfaces. So for key three surfaces if this variety VRD of C is non-empty for one curve then it's not empty for all. It's whenever the Bernoulli number is non-negative in this generic situation. There's here this result means right when you have, when rho is negative but rho plus H satisfies this inequality then this means some of the curves in a linear system will have such line models even though the Bernoulli number says the expected dimension for the individual curve is negative. So in other words the individual curves might be Bernoulli's special D times minus chi minus D squared, where chi is the Euler curve. Yes, yeah, yeah, it's all for primitive curves and peccaring one, I mean, I would precise something slightly more general than peccaring one, but basically. Yeah, yeah, any curve in the system is reducible, yeah. Here, so Giovanni Mungali, right? And I mean where does this arithmetic condition come from? It really comes from the fact that it's some of the right inequality that controls this inductive process here. I mean that's really all that I can say when it comes. Okay, and so the next application concerns by rational geometry of modular spaces. And so again, let me start by describing the question we are answering. So given a modular space, let me stress that this is, that this is really interesting already in the simple case when you consider the Hilbert scheme of n points on your surface x, then you can ask the following questions, what are all morphisms from HMH of v to another variety. And now to make it even possible for this question to have someone sufficiently fine at answer, let's say we ask this to be subjective, we ask y to be normal on projective, and that f is connected, has connected fibers. Then we might ask a similar question for rational maps. For rational maps, let's make it simpler and just ask for birational maps. Let's also assume that again, to make it possible for this to have a finite answer, that wait, so this is defined on an open subset and let's say that g of u is a large open subset, so it has a complement of co-dimension two, at least two. So in other words, the inverse map is not contracting any devices, or to say differently, I mean it can't just take any map and follow it up by arbitrary blobs, right? And then of course, one is essentially is given by studying pullbacks of ample classes which are NAF devices on MH of v, so devices that intersect every curve non-negatively. And similarly, so for two, here you study the upper star of ample, right? So in the class of mobile devices, right? So maybe let me just explain the second just a little bit more, right? So if y is projected, then it's equal to the product of the section ring of an ample line bundle, but then I can just the same because of this assumption that I made of eight zero of u of the upper star L to the u, right? So for the global sections, it doesn't make a difference when I restrict it to u and g restricted using isomorphism, so I can pull it back. And so that's the same as project of, again, because this can be chosen to have complement of very small complement. This is same as this one, right? So you can, I mean sometimes you can go the inverse and go, I mean this describes how to go from such a biracial model to mobile devices and sometimes you can go the other way and that's in some sense what the minimum model program does. All right, and so I mean answering this questions one and two really become the following questions of one prime, what is the, what is the F cone? What is the, what is the movable cone? And it's chamber decomposition corresponding to biracial models, right? So assuming such a nice that MMP works for all mobile devices. Okay, and somehow the key result that makes it possible to answer these questions using wall crossing is the following positivity lemma, right? So the idea is we first, I mean we know we can reinterpret Mh of v as a modular space of stable objects in the device category. And now the following lemma will show that this such a modular space automatically always comes from the, with the family of, of NAF devices. So for statement is that for any family of, so let me first state a slightly more technical statement and then give you the corollary for modular spaces. So whenever you're given a family of sigma, I say my stable objects over S, right? So what this means is you have an E in the derived category of the product of S and say yourself as X. So that for all S and S, ES by which I mean the derived restriction of E to S cross X. That is a sigma same as stable. Okay, so the positivity lemma says whenever you're in this situation, there is a divisor class and want to be precise in numerical r-card here, divisor class L sigma on S. That somehow, it does not depend on any choices. So you have to, you have to, you have to that somehow it does not depend on any choices. So it really it's completely determined on the, by the stability condition, right? And so I mean this numerical r-card here divisor this can be defined for arbitrary singularities. There's no problem at all, right? And so this means, right? So it's a class that you can pair with any curve and it gives way to equal to zero for all curves C and S and moreover, I can say it tell you precisely for which curves it is zero. L sigma at C is equal to zero. If and only if for all MC, C prime in C, the two objects EC and EC prime are what's called S equivalent. Right, so what this means is that they have their Jordan-Herter filtrations the same, have the same factors, right? I mean earlier you saw how I constructed stable objects so right, so if I come here in this situation, assume that W is stable and that I'm wearing the shortage of sequence just by wearing the extension class, then all these objects would become semi-stable, would be S equivalent, right? And so as a corollary, if you're given a chamber C inside star-plumptor of X, all right? Then I have, I can look at this motorized space MC of V, then I get a map L from C to the left cone of MC of V because here this class depends on the stability condition and so as I vary sigma here, the class of, the divisor class of L sigma varies but the motorized space remains constant and I mean just by continuity, you can of course extend this. Right, and moreover, right? You have this additional statement about when does L sigma dot C become zero, right? For, so I mean on the interior of C, I really get into strictly left classes that are positive on every curve and moreover, I mean for the, if you have a sigma in a mall, sigma over here, then we can in principle determine curves L sigma dot C is equal to zero, right? So if we can understand are there S equivalent curves, curves of S equivalent objects, then we can really find such curve to pair a zero with this divisor, so in other words, we would have found a class in the boundary of the nefco. So you're really constructing nef devices and dual extremal curves at the same time. Okay, and then I mean moreover often, it's when you have a ball like this, you have your ability condition on either side, sigma east and sigma west. I mean in many examples, one can often show that M sigma of west, sigma west of we is M by rational to M sigma east of we, right? So things don't actually look all that different on the other side of the wall. And I mean, this is automatic when some of these modular spaces have a dense open subset of objects that remain stable on the wall, but I mean, that's not always the case. And for K3 surface, there are actually many examples where this is not true, where all objects become same as stable, but still the modular space, between the modular spaces, there's an induced by rational map. Okay, and so when are these methods really effective? Well, they're effective when you can control wall crossing. When can you control wall crossing? That's when you know precisely which modular spaces of stable objects are non-empty. So let me phrase this a little bit imprecisely as a theorem. So whenever when X is equal to P2, so in this case, that's due to Chunilie and Chao Lhazong at building on earlier work by Akhara, Gertram, Vascoon, Huizhengar and Wolf. So being on papers by several papers by subsets of these five people. And for when X is an abelian surface, this is Minamina here, Nagida and Yosheoka. And for X equal K3, so in that case, it's in my work with Manila Makri, is that in all these cases, you can use these ideas to completely determine the vibrational geometry. It's okay to use these ideas completely describe the vibrational geometry. Yeah, abelian surface then. So this is always for primitive we. And there's also one case of we non-primitive that's due to Kieran Mikan and Zheozong. But all the other cases are for we primitive. That's Kieran Mikan and Zheozong. Yeah, maybe I've lived too long in Scotland to remember that this name was difficult to pronounce when it fell. Right, and so maybe let me make this a little bit. More explicit for K3 abelian surfaces. So what you can do is you can piece together these maps L from the individual chambers using these vibrational maps to get a map L from star lambda of X to the movable cone. That is, that's, I mean, it's almost objective. But the image contains, in particular the image contains all classes corresponding to birational maps. It just misses those corresponding to Lagrangian vibrations. And it's compatible in the following sense. If you have sigma in star lambda, then M sigma of we is exactly the birational model corresponding to the movable divisor class L of sigma. Right, and this holds, this actually holds both in chambers and if you interpret everything correctly also on walls. So for walls and for M sigma of we, you really have to take the course modulate space that identifies two same as stable objects with data as a good one. And yeah, it's piecewise essentially linear, yeah. So on every chamber it's essentially linear and at some walls you have to reflect the map. So when you hit the divisor contraction, then you have to reflect the map with the corresponding reflection to stay within the movable cone. Right, and I mean some of the key idea is, right, I mean, right, so you have, I mean, why can you use this map to precisely describe the chamber decomposition over here? So it's really that the point is that walls here on the left are easier to describe. The key idea is just that sigma in stable length of X is on a wall for a class V. If and only if you can write this V as A plus B, A squared and B squared greater equal to minus two so that Z of V is parallel to Z of A and Z of B. You're all pointing in the same direction. Right, and so where does this come from? Well, at least naively you would expect that under this assumption, there are same as stable objects of this class. Well, you know the same as stable object of this class if you can also find stable ones and you can find corresponding extensions. Then the argument is really very similar to the first lemma that I showed how to construct in this Bernoulli argument, right? So if you have A in M sigma naught of A stable, let's say this is your sigma naught and B in M sigma naught of B stable. Then one can show there are always extensions of the form like this. Inversely extensions like this so that E is in M sigma east of V and E prime is in M sigma west of E, right? And so what's the actual work is that as I said, sometimes all objects become strictly same as stable so there might not be any objects of class A that remain stable on the wall. And then you have to somehow find a different decomposition so that the first part of the wall and the second part of the wall is in order really to reduce the chamber decomposition over here, you not only have to decide is there a wall on here, you also have to decide are there curves of S equivalent objects. Okay, but then I mean the end result is really completely explicit description of the movable cone together with its chamber decomposition just in terms of the lattice of the KT surface. Okay, let me stop here, thank you.