 Okay, so thank you very much. I mean, maybe I should start by correcting one thing Lothar said. He called himself the local organizer. Well, I would use the word surveying organizer. So if there, so I mean, it should at any point someone start to thank the organizers that Lothar deserves the main credit. The rest of us were just helping along. Okay, so the goal of my lecture series is really to as the title says is to explain how to use stability conditions on the right categories in order to answer somehow basic concrete questions in algebraic geometry. And I mean, as you can see, already from this long sentence, there's a lot coming in, there's stability, there's derived categories, stability conditions. And then of course, I also have to explain some context to the questions that I'm going to answer with the help of these techniques. But for today, I want to keep it completely elementary and just explain thoroughly the first building block of all that, which is the notion of slope stability in a building categories. I'm really trying to make this talk as accessible as possible. So please, please stop me if anything is unclear. All right, so let's go right into the definitions. So for today, this will always play within one abelian category. And the examples I have in mind are, so example A would be that A is the category of coherent chiefs on a smooth algebraic curve. Smooth projective algebraic curve over some field. I mean, of course, you can do variants of that. So you could ask X to be not necessarily smooth, perhaps reducible. Or you could also think of another example you could think of is where X is a non-complete surface. And A is the category of coherent chiefs with compact support. But today, I'll really just focus on this part A. But I mean, if what I'm doing today is to elementary for you, maybe you can think along the way how to generalize these things to cases A prime and B prime. Okay, in the other class of examples that will come from algebra. So let's start with the quiver. It's just a directed graph. So I could have, say, two vertices and then some errors from here to here. Or I could have maybe two errors like this and one error back. Then A is the category of representation Q. So what it is, I mean, if you want, these are just functors from this directed graph to vector spaces over your base field. So let's make this concrete. But that's, of course, just a fancy way of saying you take some v, which consists of a v i for each vertex in my quiver. I'll use the standard notation of Q0 is the set of vertices. And Q1, the set of errors. And then a map Fij from v i to v j for each error from i to j in Q1. So for example here, for this quiver, I would just specify two vector spaces and form for maps between them. And it's easy to see that this forms an abelian category. So I think of, so this is my notation for an error. So any error has a start point i and a start vertex i and an end vertex j. Yeah, so yeah, it's an abuse of notation. So really, this is a notation for this error, one of the errors with start point i and end point j. And I could also add some set of relations. It's of example, in this quiver here, let's call this vector y and these two x1 and x2. Then I could do something, by relation, I mean something like imposing, say, x1, y, equal x2, y, or y, sorry, I meant x1, y, equal 0, equal x2, y, x1, equal 0, equal y, x2. All right, this would be a possible set of relations. And then you can also look at the set of representation, where each such relation is sent to 0, right? I mean, for x1, there's a corresponding map between my vector spaces for y. And then the conditions as if I compose them, then I get the 0 map in my representation. Okay, and so now the key definition is that a central chart on A is a function C from objects of A to complex numbers. So for each object, I get a complex number such that, so first of all, it is additive on short exact sequences, right? So this is really equivalent to saying the factors via a group homomorphism that are also called Z by abuse of notation from the K group of A to C. I mean, if you don't know what the K group is, you can basically recover the definition of the K group from the statement, right? It's just a three-year building group generated by objects, modulate the relation whenever you have a short exact sequence, then the middle one is the sum of the left one and the right one. Okay, and the second condition is that if I take a non-zero object, then Z of E is in the following semi-closed upper half plane. So I take the upper half plane, union the negative real line, and I always denote this by this black balled edge, right? So in the complex plane, I allow negative real numbers and anything with positive imaginary point. And from what, once I have this, then I have, I can define the notion of phase. So phi of E is defined to be 1 over pi times the argument of this complex number Z of E. Of course, the argument is not well defined, and so, but it is well defined if I stipulate that it's in the semi-closed interval between 0 and 1, right? It's basically just the angle here if I have Z of E, then phi of E is up to rescaling just the angle over here. Okay, any questions so far? And so let me, let me give you an example. So when A is coherent sheaves on X, then I define my, I can define the central charge to be I times the rank of E, minus the degree of E, right? And so, I mean, I don't know whether all of you have seen the degree of an arbitrary coherent sheaves on curves. But I mean, some of one thing you learn in the first algebraic geometry class is how to get from a line bundle to a degree. And basically this degree function is the unique extension of this degree function on line bundle so that it's additive on short exact sequences. So it's really uniquely determined by being the degree of line bundles that you know and being additive on short exact sequences. And for B, here now I have a bit of choices to make. So I pick a Zi in this semi-closed upper half plane for every vertex. And then I said Z of V to be equal to the sum over all vertices of Zi times the dimension of Vi, right? And I mean, you should know that this is, these two cases are very different. Here, I have lots of choices to make. I mean here, I essentially made no choice. I mean, I could replace this i by any number in the upper, complex number in the upper half plane, but it would turn out that this doesn't actually change the notion of stability. And then you define an object of your abelian category is the semi-stable or stable. If for all sub-objects, and I should insist that they are proper sub-objects, their slope is less than or equal or less than the slope of E, or the phase of E, right? So here, I mean strictly less than, phi of A is strictly less than phi of E for stability and less than or equal for semi-stable. And I might sometimes also say I'm slope. Stable here. If it's clear what, what, what's the I'm referring to. Okay, any, any questions on the definition so far? I mean, let me just give you some cartoon sketches of motivations for this definition. There are, there are at least three and let me, let me mention each of them a little bit. And so the first one is for boundedness of materialized basis. Bay of coherencies. All right, so let's say we take a point x in my smooth projective curve. And now if you take E n to be the following vector bundle, you just take O of minus nx plus O of nx. And this has rank equal to and degree equal to zero. But I mean, the number of global sections is just the number of global sections of this one. And this will go to infinity, right? But eight zero of E n will go to plus infinity as n goes to in plus infinity. And what this means is that there is no bounded family, family, parameterizing all, all coherencies with these invariants. But even, even, even all vector bundles rank equal to and degree equals zero. Right? So, so what I mean by bounded family, I mean that there is a variety or a scheme of finite type that parameterizes all such, all such vector bundle with these two invariants. So why is that not possible? I mean it's an, it's an, it's an exercise to see that if you had a bounded family, then, then the space of global sections would, the, the dimension of the global sections is always bounded in a bounded family. Right? And then the next motivation is the connection to differential geometry. Right? So here I have to assume that X is smooth, projective over C. And E is a vector bundle. Then E is slope stable. If and only if it admits a projectively flat irreducible unit, unitary connection, unitary irreducible connection. Right? So that's a famous result from not already 50, more than 50 years old by Narazimhan and Sishadri. Right? So, so, so whatever this, whatever these, these words mean, that's really characterization of this purely algebraic notion in terms of the differential geometry of this vector model. So that alone is, is of course interesting by itself. And then, I mean the third motivation is purely categorical. Because it, it, it organizes your, once you have a notion like slope stability, in some sense it organizes your, your abelian category. And that's what I'll focus on today. So, I mean I won't say anything more about this one, but of course later in this week, the relation to motorized spaces will become quite important. But for today I'll ignore that completely. Okay, any, any questions? So, I mean here is a crucial lemma. If you have a short x-axis sequence, a into e onto b, when this is, if this is a short x-axis sequence in a, then why this is, you have to see some properties. So either the phases are increasing or they are decreasing. And then phi of a less than phi of e is equivalent to phi of e less than phi of b. And the same with less than or equal. And instead of a proof, just let me draw the corresponding pictures. Right, so what if I look at the corresponding center charges? Well, remember that z is additive on short x-axis sequences. So if I, if I look at the origin z of a, z of b, and z of e, then these form a parallelogram by this additivity. And either this parallelogram is oriented like this or it is oriented like this. Right, and so this corresponds to the situation where the phases are increasing. And this is corresponds to the situation they are decreasing. Right, and note that for this lemma, this assumption that everything plays in the semi-close upper half plane is quite crucial. Right, because only that way you really have a well-defined parallelogram in here. I mean, otherwise, if z of b would be allowed to move to the negative half plane, then these pictures wouldn't make sense anymore. Yeah, so here the a has bigger phase than e and here a has smaller phase than e. And if a has smaller phase than e, then e must also have smaller phase than b. Right, so this is this situation, which is the same as this situation, and this picture is when these two are not satisfied. Right, here I have phi of a less than phi of e less than phi of b. And here I have phi of e bigger than phi of e bigger than phi of b. Right, and the claim is that it's not possible to have phi of a bigger than e and here less. That's because there's no picture that would match this situation. I mean, here it's a destabilizing sub-object because it has, right here, e is not stable because phi of a is bigger than phi of e. The argument of a is bigger than the argument of e. And here it's not destabilizing. Right, so here e could possibly be stable. As long as the picture is the same for all other sub-objects and so on. Right, I mean, the point is this isn't just a picture in a two-dimensional vector space, it's a picture in c, in particular in the upper half plane of c. So there is an ordering like this, right? It's, yeah, yeah, these are all, yes, yeah, yeah, yes, okay. Okay, and then, I mean, basically as a corollary of the previous lemma, maybe let me state this as an exercise if you haven't seen this before. If e and f are z's in my signal, and if phi of e is bigger than phi of f, then there are no harms, right? So in this complex upper half plane, I mean, all the, if I look, if I fix a phase phi and look at all the semi-stable objects of that phase, they'll all have central charge on one of these rays. And right, my claim is somehow that this picture leads to a filtration of the Abelian category. And what this lemma says is that in this picture, harms can only go this way, right? There are no harms from anything here to anything here. Okay, so that's one part of the picture, and then the other half of that categorical picture is the existence of Hadron or Zeman filtrations, right? And the statement is in both example a and or a prime or a double prime, and b or b prime, the following property holds for every, for every object of my Abelian category, there exists a filtration. And let me add that this filtration is unique and functorious in a certain sense. So 0 equal e0, increasing filtration, that's that, so ei modulo, ei minus one is c semi-stable for all i, and the phases are decreasing. Phi of e1 modulo e0 is bigger than phi of e2 modulo e1, and so on. All right, so in other words, any object can really be written as an extension, as an ordered extension of objects sitting in one of these slices. All right, so let me draw a second area here, x1. So this, but this x1 means something different than this harm, right? So this x1 means any object can be written as an x1, some are ordered in this way. In the short exact sequence, the one, the object further on the right is always further on the right on this blackboard picture as well. Okay, so what I'll do for some of this is crucial for all the applications of stability condition and wall crossing and so on. And since it's so central, I'll use most of the remaining time to give a sketch of the proof of the statement. And I'll focus on case A since case B is actually a little bit simpler. And I'll start with the following definition. I'll define a convex subset of the complex plane. So HNZ called the Harron-Ziemann polygon. So what is this? I take all Z of A for A is sub-object of E. And here I should be clear that this maybe might well be a trivial sub-object, either zero or the object itself. And then I take the convex whole of the set, right? So let me sketch an example over here. Like you have Z of E somewhere here. Then if you twist Z of E by a negative line bundle, say O of minus Nx, then this will give you sub-objects with the same rank and smaller degrees. So they'll lie somewhere here. So all this is part of this convex whole. And then I may have some sub-objects with bigger face. And then I claim that this is a picture will look something like this. So this is called the Harron-Ziemann polygon. Of course a bit of a misnomer since it's not bounded. But I claim that it is a polygon on the left, right? And so there's a simple lemma, right? I claim that H and Z of E is bounded on the left. And so to prove this, I mean of course that's a equivalent. The claim is equivalent to saying that the degree of A for A is sub-object of E is bounded above. Okay, so I mean to some of you of course this statement may look obvious, but nevertheless let's think about what actually goes into this statement. Right, so what is the proof of this claim? Well it holds for when E has rank equal to 0 or rank equal to 1. Right, so for rank equal to 0 the degree is just the length of a 0 dimensional sheath. And the sub-sheath certainly has smaller length. And for rank equal to 1 again, I mean this is actually a non-trivial statement, right? I mean that's essentially the statement that the degree of line-bundles is well defined and if you have a sub-line-bundle then it has smaller degree. Right, and if you remember your first or second algebraic geometry course it actually takes takes a while to prove statements like this. Right, and then if you have a short exact sequence and if the claim holds for E1 and E2 then it also holds for E. It just follows from the fact that if you have a sub-object of E then you get a corresponding sub-object of E1 and of E2 that fits into short exact sequence. And so the degrees of the sub-objects here are bounded and by additivity on the short exact sequence it follows that the degree of the sub-object of E are bounded. Right, and then you can do induction on the rank. I don't mean in fact any E has a sub-object as a line-bundle, any torsion-free sheath has a sub-object that is given by line-bundle and so you have a short exact sequence like this where both this and this object have strictly smaller rank. Okay, so I mean so far this was just a justification of this picture over here, right? I mean h and z of E is bounded on the left and I mean also all the central charges lie in discrete letters z plus i times z, the Gaussian integers. And so the left-hand side of this convex head is really just given by this polygon. So let's look at these extrema points on the left. I let this be the extrema points of h and z, this Hiram-Ziemann polygon on the left and so over here I would have z0, z1, z2, z3 and z4 and then for each i let E i be a sub-object such that z of E i is equal to zi, right? And so I mean this exists exactly because the i is extrema, right? I defined h and z of E as a convex hole of certain number of points but since the i is extrema in this head, it really had to be in the original set of z of sub-objects. So there exists a sub-object with this central charge, right? And now my claim is that this is the hard-on-Ziemann filtration. Now I mean strictly extrema, yeah. Yeah, exactly, yeah. And I don't take anything here on this line segment, right? And so this was Lemma 1 that I had to prove, Lemma 2. So now I'm trying to prove this claim. So here Lemma 2 is that this is really filtration, right? So in fact I'm claiming that E i minus 1 is a sub-object of E i. How do you prove that? Now I mean the key thing is that in an abelian category, if you're given two sub-objects of a given object, you can always form the intersection and you can form their span, right? So here I don't mean the direct sum of these two objects, I just mean the span inside the given sub-object E and, right? And so then there is a short exact sequence, A goes into E i minus 1 direct sum, E i goes on to B. Right? In fact, I mean you have a natural map from this direct sum to E and B is the image of that map and A is the kernel of that map. That's how you see that this really has existed in any abelian category. And, right? And so from this you see that the midpoint of the line segment from Z of A to Z of B is equal to the midpoint of the line segment from Z i minus 1 to Z i. So let's, let me draw a picture over here. You have Z i minus 1. Here you see i. And the Harald-Ziemann polygon is somehow here to the right of this line segment because I chose two consecutive extremal points. Right? So you also have the Z of A and Z of B are in the Harald-Ziemann polygon just because they are sub-objects of E. Right? And so if you combine these two statements, this is only possible if, right? So in particular, they are always to the right of this straight line. And either they are strictly on the right or on this line segment. So if you combine this with the previous statement, this is only possible if Z of A and Z of B are on the line segment from the i minus 1 to the i. Right? But you also have that A is a sub-object of E i minus 1. Right? So this means that the rank of A or the imaginary part of, right, means that the imaginary part of Z of A is less than or equal to the imaginary part of Z of E i minus 1. I mean, why is that? Because you can form the quotient and the quotient is contained in the semi-closed upper half plate. Right? And so, all these statements together are only possible if Z of A is equal to Z of E i minus 1. And if A is, in fact, isomorphic to E i minus 1. But that's, of course, equivalent to, might have shown that the intersection of these two sub-objects isomorphic to E i minus 1. It's the same as this includes. And so for now, I'm just proving the existence. So I don't have to answer that question. But actually, you can use basically the same argument that I used for the proof of lemma 2 to show the uniqueness. Let me give that as an exercise. Right? Show uniqueness of E i and use that to show uniqueness of the Harald Zimmer filtration. Okay? So, although I still have to prove, I have to prove that E i moduli E i minus 1 is easy in my statement. I assume otherwise. I mean, any sub-object of E i moduli E i minus 1 is of the form, is of the following form, right? There exists an A that contains E i minus 1 and is contained in E i such that, right? So, from this, of course, I get an inclusion of A moduli E i minus 1 inside E i moduli E i minus 1 and phi of A moduli E i minus 1 is bigger than phi of E i minus 1. E i moduli E i minus 1. Right? So far, I've just written down what it would mean for this object not to be the same as stable. And let's draw a picture. All right? So, here we have 0. Here, let's say, here I have the i minus 1. Then, right here, I have the i. Here, again, I can form this parallelogram. But, right here, I get Z of E i moduli E i minus 1. Okay? And let's look at where A would lie. Well, I know, I don't write no, no, right away where Z of A would lie. But, I know that Z of A moduli E i minus 1. All right? This would have to lie somewhere here. It has bigger face than Z of E i moduli E i minus 1. So, it has to lie somewhere here on the left. And now, what does this mean for Z of A? Well, I mean, I have Z of A is equal to Z i minus 1 plus Z of A moduli E i minus 1. Right? That's, again, just coming from the additivity on short exact sequences. Right? So, Z of A has to lie somewhere, somewhere up here. Again, that's a contradiction to, that's a contradiction to Z of A, to the fact that Z of A lies in the Harald Zimmern polygon. And that Z i minus 1 and Z i are consecutive extremal points of this convex set. Right? Okay. So, I've shown, what have I shown? I've shown that I've gotten the filtration and I've shown that the quotients are, say, my stable. And finally, note that, I mean, right? So, of course, phi of E 1 moduli E 0, it's just phi of Z 1 minus Z 0, phi of E 2. Modular E 1 is equal to phi of Z 2 minus Z 1. And so on. And right? So, just by the convexity, these numbers are decreasing. And so now, I've proved, I've, I've proved at least the existence of Harald Zimmern filtration. I won't, I won't say anything about the uniqueness, you know. Any, any questions? Yeah, so here by phi, I mean, the argument, again, the argument of this complex number, normalized to be between 0 and 1. Yeah, I mean, strictly speaking, I've defined phi only for objects of A, but I can do the same thing just with these, with these complex numbers. And maybe I should make a remark of what, what did we actually use? Right? So, I used two things that I used that minus the real part of Z of A is bounded above for sub-objects of E. And I used that the, right? If you look at the Z of all objects of A, I used that this is a contained in a discrete set, right? In our case, because it was contained in Z plus IZ. But, but if you, right? And so whenever you have these two properties, then you automatically get Harald Zimmern filtration. Why these two together imply Harald Zimmern filtration? Because both of these imply, these two together imply that the Harald Zimmern polygon is, somehow this has a finer polyhedral property on the left. But in fact, if you think about it a little bit more, actually I need a little bit less. And sufficient, if it's sufficient, if I'm de-falling that, if I look at all the Z of A's for which the real part of Z of A is less than or equal to the maximum of M0 in the real part of Z of E, if this is a finite Z, then I'm good. Because then if I look at this Harald Zimmern polygon, it always contains zero. Sorry, here I mean for all M A sub-object of E. Right? If you take, if you're given an object, right? If you're given an object E that has the property that this set here is finite, for all sub-objects you look at Z of A, as long as it satisfies this property, right? So anything that's on the left to one of these two numbers, because that already means that the Harald Zimmern polygon is really a polygon on the left and then the same proof works. Okay. In particular it's easy to see that in this situation B that I started, that I mentioned earlier, Harald Zimmern filtration exists. Okay. And so I want to finish by discussing the formations a bit, right? And so recall in example B, low stability dependent on parameters. In other words, I could deform my notion of stability. And what I want to convince of there's an example A, I really couldn't deform it, right? And so my claim one is that this is interesting. I'll do this right now with an example and then the claim two which we'll probably take tomorrow and Wednesday to justify is that this can be generalized, this deformation property can be generalized if we use the derived category of A rather than only A. Okay. And so to justify my first claim that this is interesting, let me look at two examples. The quiver with two vertices and two arrows and B I take the same quiver, but then I add one arrow going backwards. Okay. So this is a quiver with, let me label the vertices as one and two and maybe let me label the errors here by x1 and x2, x1, x2 and y. And in both cases, right? So there are basically two situations for my choices. I mean either the one is on the left and the two is on the right. So remember for to choose the stability, I have to choose one complex number for each vertex. And I claim that there are just two situations. I'm going to consider namely whether the one is on the left or whether the two is on the left having bigger face, right? And so let's start with, and in both, in all of these cases, right? So there are now four cases. Let's look at, let's look at table representations of dimension vector 1, 1. I wanted v1 and v2 are both 1. Right? So you can think of, let's say we do all this over c. So they're both isomorphic to c, but of course not naturally isomorphic. And so maybe let's start with 2a. Right? So here what is my representation I have here c and c, both after rescaling. And I have two errors like this. But this representation always has as a sub-object the representation that has the zero vector space here and c over here. I mean, if you, if you look at the sub-representation, then this is clearly preserved by all the maps. And I mean, this one here has, this one here has center chart c1 plus c2. This one has the center chart equals to c2. Right? And so in this picture, you see that, right? So this is my v. And this shows that v is unstable. So here I have always, I have always just the simple representation just concentrated at the second vertex as a sub-object. And so when the configuration is like this, then this is destabilizing. But what, if instead we look at 1a. And so first of all, any questions on this example? So for 1a, again I have, I either have c on the left and c on the right. And I have two maps x1 and x. So for each error, I'll get the corresponding map from c to c. So when is this representation stable? Well, I mean, the only thing I have to verify is that the simple representation sitting on the left hand vertex is not a sub-representation. And so this is stable, if and only if x1 is not equal to 0 or x2 is not equal to 0. Right? And so if you look at the parameter space of all stable representations of this dimension vector, right? And this is isomorphic to c2 minus the origin. But then, I mean, you have to remember that this was, the two vector spaces we only see up to isomorphism. And they can rescale them. And rescaling them of course corresponds to rescaling x1 and x2 by a common scalar. Right? So I have to rescale by c star here. And so I get, I get p1. So on the one hand, I have p1. In one way, I have p1 as a modular space. In the other, in the other case, I have the empty set as a modular space. So here, the modular space of stable representation of this dimension vector is just the empty set. Okay. So what happens in this example? In this example, b, c and c, and I have two maps like this. And I have one map like this. And note that, I mean, both of these are c only up to isomorphism. But if you take the composition of two, two vectors so that you end up at the same vertex, then this is a well-defined endomorphism of a given one-dimensional vector space. So that's a complex number. Right? So in other words, you have a map from this modular space of representations of dimension vector 1, 1, 2, a2, just given by sending v to, say, y, x1 and y, x2. Okay. So that part we always have. And now let's look at the two stability conditions. Right? So for 2b, right? So again, here, this is the representation that is not allowed as a sub-reversent. Right? And so what does this mean? This just means that y is not equal to zero. Right? But if y is not equal to zero, then you can actually, I mean, after this rescaling, you can just presume that y is equal to 1. Right? And then this shows that from the image over here, x1 and x2 are determined. So this modular space of stable objects just becomes isomorphic to a2. And now what happens in example 2a? Right? Again, now it's the simple representation concentrated on the left-hand side that's not allowed to be a, to be a, to be a sub-representation. Right? So this means that x1 is not equal to zero or x2 is not equal to zero. Right? And so what does this mean? Let's look at what this means for this map to a2. So again, if the, if the image here is outside the origin, then there's a unique preimage, right? Because then the same argument as before works. I mean, if the image here is not the origin, then y cannot have been zero. And so x1 and x2 are determined by the image. So this is isomorphism outside the origin. And now what is the preimage of the origin? Well, right? If, if, if this map lands into the origin, then this, this means that either y0 or x1 and x2 are equal to zero. Well, x1 and x2 equal to zero is excluded. So in the other case, also over the origin, we get, I mean, y equal to zero. And so again, what do we remember? We have x1 not equal to zero, x2 not equal to zero. Sorry, we have now x1, x1 and x2 are not both zero. But now we haven't used the rescaling yet, right? So we haven't done any normalization to see stars. So the preimage here is, is p1. Right? If, so let me say that again, if I'm here, the image is the origin, then y must be equal to zero. And so the, the isomorphism type of the representation is again given just by the ratio of x1 and x2 up to risk. Right? And so in other words, this is the, the blow up of a 2 at the origin. Okay. And so I mean, I hope this example convinces you that something interesting happens when you, when you, when you vary stability and that it's worth doing the work of generalizing this, this deformation to other situation where we will, where we will have to go to the device category. Okay. Thank you.