 Okay, maybe let me start. So what I'll talk about today is to describe this space of stability conditions, right? So let me define this. This is the, so recall last time I fixed this finite rank lattice that's a quotient of the K group. And now I define stop lambda of X to be the topological space. I also sketched the topology, the topological space of stability conditions, where the central charge factors via lambda and that satisfies this property of pre-stability conditions, satisfying support property. And I'll call this a stability condition, it's a pre-stability condition that satisfies support property. Okay, and I mean the goal that I'll prove today that this isn't just a complex space, it's actually a finite dimensional complex manifold. Right, so go, this is a complex manifold and I'll make this statement more precisely. Okay, but before I go into these, and I mean somehow I'll first do more of the abstract machinery and then do some, hopefully some examples to date towards the end and then more tomorrow. So tomorrow I'll describe this for algebraic surfaces. Okay, but before I do this, I want to make something explicit that I already touched upon implicitly. Right, so you can, I mean we want to study the formations of the stability condition as we deform Z. And of course, there are sort of two kinds of deformations, one that leave the kernel of Z unchanged and the others that do change the kernel. Right, so in other words, GL2, let's say plus R, acts on arm from lambda to Z. Right, via the identification C isomorphic to R squared. And I want to lift this to an action on the space here. And so for this, you should note that it acts on, right since it acts on C, it also acts on S1. I just taken a vector S1, apply your element of GL2R and then rescale it to be in S1. And this means that the universal cover GL2 plus R acts on the universal cover of S1. And some are in, to keep in line with my convention so far, this, I'll explicitly think of this cover as given by phi maps to e to the i pi phi. Right, so this matches the relation between central charges and phases. Okay, and so the proposition is that this universal cover acts on star-blumter of X-wire. So G dot, if I'm given the pair Z comma P, then this gets sent to G composed with Z and P prime, where P prime of G dot phi is equal to P of pi, right? So we're here, this is this action on R that I described up here, right? And so to make this as explicit, right? So you have the, of course you have Z star inside GL2 plus R, the universal cover of that is C, right? And if you take P in here, and then psi to be an argument of a choice of one over pi times the choice of argument of C, right, which corresponds to lifting this element to the universal cover, then in this case you just get, right, in this case, P comma psi dot P comma P, is C times C and P prime, where now P prime of phi plus psi is just equal to P of phi, right? And now we had this, we had this other description of white stability conditions given as a pair of a central charge and A, and an abelian category A, right, and so how can we see this in this other description in the other definition, right? So here you have this upper half plane corresponding to A and then, I mean, some of what this definition is saying is that, right, I'm saying my complex number is something like of the form one plus I in that direction, right, and this is rotating objects in this upper half plane to some more tilted upper half plane, right? So this is C times, right? And why this still corresponds to A with respect to the new stability condition, but what corresponds to the upper half plane with respect to this new stability condition? Well, for this we have to cut off our original heart into two pieces, right? If you look at the negative real axis in this tilted half plane, this corresponds to this ray, I guess blue is difficult to stick. So maybe let me draw a separate picture over here, right? So I have this ray over here and I can use this to construct my torsion pair. Here's my F and here's my T, right? And it turns out then after you rotate, the T is below the real axis, but if I shift it back by minus one, then I get, right? So in the new picture, it looks like this. I have here F and here I have T shifted by minus one, right? So in other words, the new heart, A prime, this is the tilt that the torsion pair shifted by minus one, right where A sharp is the tilt of A at the torsion pair, torsion pair T comma F, right? If you remember, A sharp contains M T and it contains F shifted by one. So if you shift everything by minus one, you get F and T shifted by minus one. And so it is, you already see that somehow this action which in one description does something completely trivial, does something slightly less trivial in the other description. Sorry? So the blue line is somehow the pre-image of the negative real axis under the, multiplication. Any more questions? Okay, so now let me just state a precise theorem. And, all right, and so this is, so let me here use a big letter, a spread line, some of that's the main result of the theory. But the way I'll state it, it's actually, so we made this a little bit more precise in joint work by myself and Emanuele Macri and Celari. And, the proof I'll explain today is due to myself. So I posted this as a pre-print a few months ago if you're interested in reading it up afterwards. Okay, but let me state the results. So, the statement is that stop lambda of T is a complex manifold, the finite dimensional complex manifold and the dimension is equal to the rank of lambda as you'll see in a second. Namely, why does a natural forgetful map from the space of stability condition to this vector space come from lambda to Z? That's then Z comma P to Z, right? So this is a local homeomorphism, right? So for example, it could be a covering of an open subset, but it could also be more complicated, right? So it just says locally on the source it's an isomorphism. And, let me state a slightly more effective version. That appeared in this joint paper with Manu and Maghri and Paolo Stilari. Right, so given such a stability condition, let's say that it satisfies the support property with respect to a given quadratic form from lambda R to R. Okay, so if you came late as email this person, if you lost or found anything. Okay, so let P of Q inside arm of lambda to see the open subset, see the set of Z prime, such that Q restricted to the kernel of Z prime is negative definite, right? And then typically this will have two components and I write P of Q for the component containing Z. Then what I'll claim is that as long as I deform Z within this open subset, then I can freely deform the corresponding stability conditions or freely and uniquely, right? So then there is an open neighborhood that's that the restriction of this curly Z from U sigma to PZ of Q is occurring, okay? And moreover, all sigma prime in U sigma satisfied as a property with respect to the same Q, right? And I mean, some of this last statement maybe will become more interesting tomorrow because you'll see that this, right, what is Q, it's a quadratic inequality for chairing classes of same as table objects. And it turns out that in geometric situation this is really something deep and meaningful Bokeh-Mollaf-Geserke inequality for surfaces and something we don't quite understand yet for three-folds. Right, and so, I mean, maybe let me rephrase this, right? So in other words, for any deformation, right, ZT of Z equals Z naught, right here, this is for T in the unit interval. Whenever I'm giving a deformation, a path in the space of central charges, such that the kernel Q is always negative definite in the kernel, then I can uniquely lift us to a path of stability conditions as a unique sigma T equal ZT comma PT, right? So this, maybe let me make one more comment on this, right? So if you think about the second definition of stability, it's kind of very unsatisfying because I had to explicitly give you the list of same as table objects, right? The P of phi, the list of same as table objects was part of the data and there's usually when you try to define stability, you define a condition of stability and then work out what the stable objects are. Well, it turns out that once we fix this at the beginning, at least, this problem is solved once and forever, right? It says once I've stopped with some notion of stability, then when I deform it, I'm told uniquely how to adopt a set of stable objects along the way. So once I've chosen the starting point, a set of stable objects along this path has uniquely determined the kernel ZT, right? So that's my short end notation for negative definite. Okay, so basically most of what I'll do today is just a sketch of the proof of the statement. I mean, the proof is not really simplified enough that it makes sense to present it in a lecture like this. And so let me first make a simplifying assumption. I will assume that q has signature m2 comma rank lambda minus two. But in particular, it doesn't associate that non-degenerate symmetric form. Particular associated symmetric form is non-degenerate. Right, and that's only a smaller assumption. I mean, the kernel of the, has at least rank equal to dimension equal to rank of lambda minus two. So we know the negative dimension, the negative part of the signature must be at least this. Right, and this can be, I mean, in non-degenerate cases, this will always be one or two, so it's only a milder assumption that this is two. Okay, and so let me write k equal to for the kernel of Z and then k orthogonal for the orthogonal complement with respect to the symmetric bilinear form. And note that under this assumption, Z from the orthogonal complement to Z is an isomorphism. Right, so, quite just because by construction, k is, as Z is injective and restricted to the orthogonal complement and k perp has rank two. And now we can always use the GL two plus our action to make this an isometry. Isometry, right with respect to, I mean, of course, here I have the standard norm and here I have a positive quadratic form which also gives me a norm, right, with respect to norm on k perp associated to Q and the standard norm on C. Okay, and then if you let this be the norm on the kernel associated to minus Q, minus Q and P from lambda R to k, the orthogonal projection, or projection, then you get that the quadratic form is just of the form Z of v squared minus norm of P of v. Where it's basically just a slick way to choose coordinates, right? And so, in particular, I mean, maybe this is worth rephrasing. So, Q of v, Q of v of v greater equal to zero, this now becomes equivalent to saying that the norm of z of v is greater equal than the norm of v of v. Right, so another way to think of the support property, it is saying that, I mean, if you look at the norm of z of v for semi-stable objects and then it cannot be too small compared to the norm of v inside the lambda arm. And that's some of the version of support property that you would find in temperature papers. Okay, now, here's another easy lemma. We up to the MGL2 plus R action, every z prime in PZ of Q is of the form Z prime equals Z plus a function just defined on the kernel. The norm of U is less than one. Right, so some are saying this condition that Q is also negative on the kernel of Z prime, just not corresponds very explicitly to the fact that Z prime does not differ too much from Z. We just write part of the proof, right? I mean, Z prime restricted to the orthogonal complement. This is always an isomorphism because Q is positive definite here and Q is negative definite on the kernel of Z prime. Right, and so you can always use, you can use the PL2 plus R action to achieve the prime restricted to K orthogonal is equal to Z restricted to K orthogonal. And then the claim easily follows. Okay, and now some of the first key idea is that we can now break this up, right? This is some variation of Z. We can break this up into one first one deformation where U is completely real and then a second one where U is completely imaginary, right? So write U equal real part of U plus I times imaginary part of U, right? And then first you deform V to Z plus the real part of U. And why is this useful? Well, the reason is that these kinds of deformations look very nice with respect to the first definition of stability condition that I gave, right? When, in other words, I mean, what is the saying? It is saying that the imaginary part of the central chart is constant. And so if you think about this upper half plane picture, then the objects, all the objects are somehow wandering on horizontal lines, right? And so as long as we can control what happens with objects here on the negative real line, we don't actually have to change the heart of the T structure at all, right? So here we can keep A equal P of one constant, right? And then the second step is to deform V plus real part of U to Z plus real part of U plus I times the imaginary part of U. And I mean either you say this is analogous when you replace all these arguments using the upper half plane by some are using the right half plane or it's just equivalent to the previous one by the gl2 to rx, right? That's equivalent to first step via gl2. That's all. Okay, so from now on I'll assume that I'm just in the first step. So I assume that U from the kernel is just a real function. It still satisfies this property that the norm is at most more. Yeah, so it doesn't, I mean, what I'll show is that this assumption actually will imply that I never hit zero here, right? So in fact, that's exactly my first claim. So the claim is that, right? So let me write Z1 for Z plus U under this assumption. I claim that this is a stability function for A equal P of zero comma one, right? And so for the proof, there are two cases. If I take E and A, then if the imaginary part of Z of A is positive, then the imaginary part of Z1 of A is positive, so that's okay. And what if the imaginary part of Z of A is equal to zero? Right, then this means that E is Z, say my stable. Yeah, if E is in an object in the heart, that's imaginary part of Z of, yeah, thanks. Right, so now the point is that this means that, right, this means that Q of E is greater or equal to zero, or in other words, that E of E is less than or equal to the absolute value of C of E, which is equal to minus the real part of C of E. So this means if you take Z1 of E, this is less than or equal to, right, which is the same as minus C of E here, assumed to be real, right? And Z1 of E is less than or equal to Z of E plus Q times P of E, just by the definition of the operator norm, but this is strictly less than one and this is at most this, so this is less than zero. So in other words, I mean really where does it come from? It comes from the fact that the support property guarantees that if I deform my Z a little bit like this, then the central charge of stable objects only varies by a controlled amount, right? This is exactly what this argument is showing. Of course, this is the key idea that we'll use more. Here, well, I'm assuming now that U is real. My time throughout the, I'll assume throughout now that U is real. And now, so what's the next thing we have to prove? We have to prove that Z1 satisfies the, I mean we want to construct a stability condition from the heart A and from Z1. So the next thing is we, is that we have to show that Z1 has the hardware element property, right? And so the key, the second key idea is that right for each E and A, we know that that it has a Harrow-Ziemann polygon with respect to Z, right, this H and Z of E, the Harrow-Ziemann polygon of E with respect to the start in Z is find a polyhedral on the left because I mean maybe I didn't make this explicit on Monday, right? So what I showed on Monday that if the Harrow-Ziemann polygon is find a polyhedral on the left, then we have Harrow-Ziemann filtration, but it's also easy to show that the converse holds, right? And now we'll show that it also has, that also H and Z1 of E is find a polyhedral on the left. Okay, all right, and so we'll start with this Harrow-Ziemann polygon of E. And somehow we have to show that, right? Some of we have to show that there aren't infinitely many objects of subclass that can come here to the left and make this Harrow-Ziemann polygon suddenly not discreet on the left with respect to Z1, right? So now we have to show that if you have any object, if you have any Z1 of A that some are here on the left to, I mean, right, so T will deform a little bit, let's say to T Z1 of E, then if you look at a center of charges of sub-objects of E that have some center charge with respect to Z1 on the left, then we want that there are only finitely many such objects or more precisely only finitely many B of A possible, right? In some of, of course, part of the idea will be that, I mean, if Z1 of A here is on the left, then we'll show that Z of A is can't be too far away. And I mean, if A was where some are forced to be Z stable, then this wouldn't be too difficult because as we've already seen, the center charge of stable objects doesn't differ too much between Z and Z1. But we also, I mean, A doesn't necessarily have to be Z stable, so we have to make the argument a little bit more general. Okay, and some of the key ingredient that makes this possible is that, right, A is just a sub-object, so that Z1 of A is somewhat to the left of Z1 of E, right? So that it could possibly contribute to a, you know, infinite set of extremal points of this Harrowisman polygon of, right. So this, I mean, this is the goal, that this is finite polyhedral on the left and some of it have to show that it's almost spanned by finitely many points on the left. So if E is in A, then the mass Mz of E is defined to be the length in quotation marks of the Harrowisman polygon. And so here by length, I just mean the length of the path on the left, right? So here, just this part over here. And I could also write it as the sum over the absolute values of the central charges of the Harrowisman filtration factors, right, if EI is the HN filtration. Okay, and so, I mean, basically, this is the quantity that will control everything. And the first lemma is that if E is in A, then the norm of the projection of E for the kernel of Z is at most, is bounded by the mass of E, right? And so, and the proof just follows from the fact that if you take the norm of the projection of a Harrowisman filtration quotient, then this is bounded above by its central charge. Why that's exactly this condition that for thermostable objects that the support property gives. It's a by, this is by the support property. And then if you sum this up, on the right-hand side, you get the mass. And on the left-hand side, you get something bigger than P of E by the triangle inequality, right? And this is already good, because I mean, somehow we've seen that if I compare Z of E and Z1 of E, then some of the differences controlled by the norm of P of E, and now it's controlled by the mass of E, which is something we can control geometrically. The following lemma is completely trivial, but let me nevertheless state it, make this explicit. If A is a sub-object of E, then of course the Harrowisman polygon is a subset of the Harrowisman polygon of E. It does nothing to prove here. That's because every sub-object of A is a sub-object of E. And this now implies the following. If you have, if A is a sub-object of E, then I have, right? I mean, I would, what I would really like to do, I would like to bound the mass of A in terms of the mass of E. But that's clearly not possible, right? Because I might have a sub-object somewhere over here to the right. And then clearly the mass of that sub-object could be bigger than the mass of E. In some of what this lemma is saying, that this is really the only thing that can happen. So more precisely, if you control by the real part of the central chart, then for every sub-object, the mass is less than the mass of E. Okay, and so the proof is really by picture. Let me write, let's say this is the, this is the Harrowisman polygon of E. And then our Harrowisman polygon of A will somehow be contained in here. And now I compare the following two paths, right? I take a, I take a number X somewhere sufficiently big. And I look at this path that follows the Harrowisman polygon of E up to this point, up to this point which real value equal to X. And then I do the same thing with A. Then the gamma is the path that follows the Harrowisman polygon up to Z of A. And then just continues horizontally until the real path reaches this value X. But it's, I'm constructing two paths here. They first follow the boundary of the Harrowisman polygon. And then follow the, just go. It stops right, I've chosen just the real number X. That's sufficiently big. And I just choose like, right? And then it's, I mean, from the picture, it's sort of clear that the length of gamma A is less than the length of gamma E. But I mean, the length of this one is equal to the mass of A plus X minus the real part of Z of A and the same thing over here plus X minus the real, right? And that's how you get, how you get this boundary. Okay, I mean, that kind of argument hopefully explains why I like drawing these pictures so much. But of course, I mean, the definition of mass already appears in the physics literature. But I mean, to make statements like that, I mean, just in terms of this formula, but I mean, to make statements like that, it's really helpful to have these pictures. Okay, and now the final lemma is that I claim that given C in R, there exists a prime in R, that's that if A is a sub-object of E, that is somewhat to the left. With respect to Z1 of this real number, then the real part of Z of A is bounded by C prime, right? And I mean, really, I mean, intuitively, I hope you already have an idea that this may follow from these objects because some of these sort of bounded the mass of sub-objects, we know that the projection to the kernel is then bounded in terms of the mass. And we know that the variation of central charges is bounded in terms of this projection, right? And so, yeah, so let's make this explicit. So if C is bigger than the real part of Z1 of A, then this is, of course, greater or equal to the real part of MZ of A minus U times the norm of P of A. But here I'm just using the definition of operator norm. Okay, but I mean this, of course, I can replace just by the mass by this lemma up here. And then if I rewrite this, I get the following. I can write this as one minus U times the real part of Z of A minus U times the mass of A minus the real part of Z of A. And so far I've just, I haven't done anything. But the point now is that right this I can, this I can bound. First I can bound U by the operator norm by one. And I know I can bound this one by the corresponding expression for E, right? And so now you see that this one is independent of A. This term is independent of A. And so if you solve this inequality, you get an inequality for the real part of Z of A. So I mean I'm replacing this one by this one. Yeah, so this, yeah, yeah, using, using this lemma here. Okay, and so now the conclusion goes as follows. I take, if you're given an A of a sub-object of E with real part the one of A less than or even less than equal to the maximum of zero and real part of C one of E. Right, then from the previous lemma we see that the real part of Z of A is bounded above. But of course we also know that Z of A is contained in the Hardware-Ziemann polygon, right? So this really means that Z of A is lies in a compact region. So or in other words the absolute value of Z of A is bounded above. But then this also means that the, right using this lemma here, this also means that the mass is bounded above. And then using the lemma before that this means that the operator norm, that the norm of the projection to the kernel is bounded above, right? And so this means both the projection to the kernel and the Z of A lie in a compact region. So this means they're really, they're only finitely many possible. Only finitely many possibilities for V of A, right? On the one hand it's contained in the lattice, on the other hand by these two conditions it's contained in a compact subset of lambda R. And so, right and so by what is said on Monday this means that the Hardware-Ziemann polygon is finite polyhedral on the left. And so C one has the Hardware-Ziemann property. And by the way note that, I mean, at all times these inequalities or these estimates that are used to work quite sharp, right? As soon as the norm of view becomes strictly equal to one everything goes down, everything collapses, right? And I no longer get the bound over here. Okay. But then I mean, then the second step is really just the same argument with, I mean are really the same up to the GL two plus R. All right, so, but I mean, what remains to prove? No, is that Z one comma A still satisfied as a property with respect to Q. Okay, and so I mean, I'll be a bit more sketchy here. So I mean similar arguments. So that they're really only finitely many classes. I mean, if I said ZT to be Z plus T times U composed with P or T in zero one, then I claim that they're only finitely many classes of sub-objects, many classes, V of A of sub-objects A into E that destabilize E for any possible for some T that destabilize E destabilize E for some T, right? You can easily imagine that this follows with similar arguments. And particularly, I mean, you have something like wall crossing, right? They're only finitely many possible classes that I have to take into account when I look at where could the given objects become unstable. You know, I assume that E is element A is C1 stable and it violates the sub-property, right? Then it follows that it is C0 unstable. And so it follows that it is strictly semi-stable at some point in between, T1 semi-stable at some point in between. And it also follows that E has a Jordan, what's called the Jordan Heller filtration. So what is this? It's a filtration where all the filtration quotient are stable and of the same phase which factors CT1 stable of the same phase, right? And now the observation is simply that, right? So somewhere here you have CT1 of E and there is a filtration E1, E3 and so on whose center charges all lie on the same ray so that the quotients are CT1 stable. And so it follows that Z of E on the one hand that's equal to the sum of the absolute values of the central charges of the filtration quotients, right? And now this is by assumptions this is less than the norm of P of E, right? That's equivalent, I mean that's equivalent to Q of E less than zero. And of course by the triangle inequality this is less than or equal to the sum of the norms of the P of the projection of these filtration quotients, right? And so if you look at this chain of inequalities it follows that one EI, modular EI minus one also violates the subproperty, right? Okay, but now I can play the same game. I can play the same game. This will be, again this will be unstable with respect to Z zero. So there's a T2 in between that becomes strictly semi-stable. And then I mean that part I'll really sketch, right? I won't say much about, I mean the T2 semi-stable and so on. And I mean really quite similar finiteness arguments as I already sketched showed that this can't go on infinitely, right? This process terminates. Terminates and so I do have, which is a contradiction because we know that no stable, no zero stable object violates the subproperty, right? And so this shows that Z1, A satisfies the subproperty with respect to the same Q. And in particular now I can apply the same argument for the imaginary part, right? This of course only works if for the second deformation I start with the same estimate for the semi-stable objects. Okay, so I wanted to do some examples, but maybe let me write a step two minutes early since I wouldn't fit this into two minutes. Thank you.