 Thanks, so I mean I'll start with two post-cryptome to yesterday's talk and so the first one is post-cryptome to to Sasha Kuznetsov's talk on introduction to derived categories so namely I want to say a few words on the octahedral axiom Which is sort of not so I mean most of the time when you work with derived categories is not so important But when you do deal with T structures, it's really crucial And I mean any of the acts I don't think you'll be able to do any of the exercises I'll mention today without the other octahedral axiom and so I might have this reputation as being something fairly complicated, which is partly due to Due to the name octahedral axiom, which makes it really hard to draw because really it Answers a very simple question. So if you have the composition of two morphisms in your trial Angulated category, then what can you say about the corresponding cones and The statement is just that these cones live into an exact in an exact triangle There is an exact triangle like this and and and I mean then there are many situations where you've already seen this say These are morphisms and I'm dealing category that the inclusions then you just get one of the isomorphism films and they The second post-cryptome is to to my own talk Right, so say you have this quiver with Two vertices and some arrows Going like this and maybe let me start to introduce this notation that Right as one is the simple representation concentrated at the first vertex and As two It's a simple representation corresponding at the second vertex and right so my my stability condition Dependent on the choice of two complex numbers in the upper half plane. And so what happens if I if I Degenerate this so that Z2 is now some suddenly in the positive real line, which is not allowed Right and now I mean now this object as to satisfy said no longer satisfies this property Right as Z of S2 is suddenly not in age Right so what and so to complete this deformation property. What can we do? Well, the idea is now go to the derived category. Namely if you look at S2 shifted by one Right, which is just Means you take the complex that the zero everywhere except it has S2 in degree zero Here this is degree minus one then Z of this one Is minus C2 so this is an H right and some are you want and so the answers you try to form and Right so they try to form an abelian category containing As two shifted by one, but also lots of parts still coming from a Okay, so and so this is a little bit their motivation for a Lot of the construction in my in my talk today Why so I mean even in this quiver case where we already saw that we had strong deformation properties Of the stability condition to get an even stronger better Defamation behavior we need to go to the derived category Okay, and so really before I want to go Really go to the derived category I Want to define make one more definition purely within abelian categories and Because there are many of the definition in the triangulated category will be analog to that one and this is also something we've already In a way already seen so Given this Abelian category a pair of subcategories My subcategory always mean a full subcategory But this is called a torsion pair if it satisfies the following two conditions, so I'll write I'll start using distorted notation as I write form of from T comma F equal to zero Which means take any object in T any object in F, and there are no morphisms from T to F They have from from the former object to the second object right and any E in a fits into a Short exact sequence zero goes to T of E goes to E goes to F of E Goes to zero with T of E in T and F of E in F And I mean this should this should look quite similar to you because it's really a bit like a Notion of slopes ability with just two slopes Right where T are the same as tables of big slope and F the same as tables of small slopes you never have morphisms from stem of tables of big slope to small slope and this is a more than the Analog version of the hard on a simon filtration and So what are examples I mean So where the name comes from is when you take a the category of Korean chiefs modules over a ring and T are the torsion chiefs Every section is annihilated By some element in the structure ring and effort or some free chiefs And so I mean this is this property basically holds by definition Torsion free chiefs are those that don't have torsion subsets And this one fall follows because every chief has a maximum torsion subsets And the quotient is torsion free and of course there are I mean there are many variants of this one Right, I could say say X is a surface and I could say T are all the torsion chiefs supported in dimension zero And then F are all those chiefs that don't have a sub chief supported in dimension zero Or I could look at torsion chiefs supported in a given subscheme any any any games like that So there are many variants of that. Yeah, so it's it's autumn So it follows from this property that this decomposition is automatically unique and functorial And they are there there adjoins to the inclusion function and Have you give another example take? M Q reps and then T Call M representations supported at a fixed vertex I why it's so I mean I'll Typically when the quiver has no loop. This is just the extension closure of the simple vertex that Vertex I and then what is F these are all the river representation so that if I take The intersection of all the kernels of maps from I To J and I mean I should now I should really apologize to the notation because the end is notation I'm taking the intersection over all arrows starting from I so I is fixed and This part varies right and so I'm looking at representations where in a sense everything going did the representation is injective starting from This vector space behind right so here This is these are all in VI and if I take the intersection then I'm Presumed to get the zero six right and and similarly I could take a equal Q rep It could take F equal This part and then T the corresponding T is an exercise to that but then an additional example comes from slope stability right so assume we have assume B is a central charge on a with the hardware resume and property and now I take a cutoff phase in zero and one Then I let P to be Let me write this as a Greater than five Why these are all E in a so that all hardware cement filtration factors have face bigger than five I could write it as all the I Mean See my stables of face bigger than fine then taking the extension closure. I could also say that all the In a all quotients e on to be Have face greater than five right, so basically remember my picture was this I have some of the filtration of my category Into all these pieces of same are stables and now what I'm doing is I'm intentionally making this filtration much closer, right? I'm Just putting this part over here is a bigger than five It was corresponding to my choice of five and then correspondingly F is a Lesson a equal to five so the All hn factors have face less than or equal to five and then here I also have a cross line characterization like that By a sub-optics and then the claim is that this is a torsion pair, right? And I won't I won't prove this statement here, but It's not it's not difficult to prove when the key ingredient is of course the hard on them and filtration the hard on them and filtration and a home vanishing between I'm stable say my stable sheets between same I stable objects Right, so for example, how do you get the short exact sequence? Well T of e will but it will how do you get T of e well? You just look at the hard on them and filtration It will start with objects of big slope and at some point the face will stop being from Change from being bigger than five and less than equal to five and the corresponding filtration step there will be your T of e Okay, any Any questions of yeah, and then the strict inequality of the other yeah I have to take strict on one side and non strict on the other okay, so And our T structure are basically an analog of that for trying the later categories definition a T structure on a trying the later category D is a pair and the greater equal to zero and D less than or equal to zero of subcategories such that if I said the greater equal than one to be The greater equal to zero shifted by minus one then this is contained in the Greater equal to zero I also know that there is a There's always this sign shift here Just because if you shift the complex by a positive amount then it it means shifting the objects into negative degrees then there is harm from the less than or equal to zero to the Greater equal to one is equal to zero and any e fits into an Exact triangle less than they equal to zero goes to e goes to e Greater equal to one goes to e right where this of course is in here and this is in here right and the main example of course is when D is TV of a or The unbounded derived category in a or bonded on one side or the other and You said right the greater equal to zero are all those these So that h i of e is here for i less than zero On other words the homology is only known zero when I is greater equal than zero Right and so maybe Right and and and Sasha yesterday explained how to get these truncation functors and Maybe just one more comment right so if you have an Where does this how many things come from this is actually not quite as obvious as I may it may look like at first right, so if you have here something in D Less than or equal to zero then this is a context that you can basically to be assumed to be zero in positive degrees and The greater equal to one The context that you can assume to be zero in negative and zero degrees So why is there no morphism like that? Well, I mean how do you compute morphisms for example you can take an injective resolution of this one and This will still just be computed and this will still just be a complex in positive degrees And then the homomorphisms in the derived category are just morphism of complex Right or similarly you can take a project of resolution of that one And then this will still just be a complex concentrated in non non positive degrees right, I mean It is always possible if you were to do this the other way around Then in the derived category you can have morphisms like this that induce the zero morphism in homology right, this is exactly an M and Morphism from here to here would exactly be an x2 between these two two objects and why is that I mean compute an injective resolution Applying the harm. That's exactly the same as computing okay, and the The picture I like to draw is something like this. So you have the Less than or equal to zero looks like this The greater equal to one looks like this and then if you shift it by one If you undo the shift and Here I'm in my picture the shift by one is Just shifting everything to the left and so D The greater equal to zero would look something like this And so there is an intersection between the greater equal to zero and the less than or equal to zero so this is what's called the Heart of the t structure right and there It's for definition a The intersection of the greater equal to zero with the less than or equal to zero Is the heart of the t structure and the key thing is that this is automatically in a billing category I mean what is the exact structure and a short exact sequence and a is just a sequence of trying is it's just an exact Triangle in D where all the three objects happen to be in the heart and the extreme cases are Either that a is just zero But then that's called the semi orthogonal decomposition in this case the less than or equal to zero and the greater equal to zero are both Triangulated subcaterpies because they're closed on the shift and extensions and the other extreme case is that By the union right so that the this is the case We are basically the strip here in the middle as empty in the other extreme cases They're both the top part and the bottom part here is empty right so that the union of the Quater equal to minus and intersect the less than or equal to n that this gives you everything and That's called a bound the t structure and I mean Tom some bridge some favorite picture for this is always just Now the the category is just a strip and It's filtered by the shifts of your heart and some of the key lemma Is that a boundary t structures the equivalent is always determined by its heart right so it's equivalent to giving a full subcategory a of D with the following properties so one and if you have home from a Shifted by n to a shifted by m Then this is zero for n is bigger than m Right, so there are no negative x. This is again that this this picture that I drew here if you shift the left hand more down than the right one, then there are no more fisms and Right that corresponds to To this property here and so what corresponds to this property here is that every Right so for every object in my triangle at the category This has a filtration into homology sheets into homology objects of more precisely for each E and D there is a sequence of triangles the starting with zero E1 so there's a sequence of maps like this up to em equal e and So such that the following holds right so for each of these maps I can complete this to a triangle And so maybe let's call this a one Just a two so that a I that they are just in the Shifted out for some k i's And k one is bigger than a Two is bigger than and so on up to k m right and so in this in the standard example but these these a i's are just the and homology objects and Again, I have to do the sign shift h minus k i Of e right and and otherwise we just define these to be the homology objects with respect to our t structure, and I mean this Which I said I won't prove for you the equivalence, but again, it's it's not too hard using the using the octahedral axiom right, but I Mean maybe maybe two more comments on this filtration So on the one hand right even in the standard t structure thinking a lot of this filtration is often quite useful so for example, I mean basically almost Every time you see a spectral sequence argument If it's a simple argument and it cannot basically almost always be replaced by an argument using this using this filtration and The other comment I want to make is that this should now look really similar to the existence of other now salmon filtrations right, there are no Homes from object of bigger face to smaller face and This looks a little bit like the existence of our salmon filtration Right today is a sequence of maps like this so that when you compete when you complete each morphism To a triangle, right? This is the cone of this map here Then a this has to be in the shift in the shift of a and B The shifts have to be decreasing. Yeah, so here this is a triangle. This is a triangle Yeah, so this is a distorted error is a map from a 2 to e1 shifter by one not necessary. I can Yeah, yeah, so so if it has these two properties, then it's automatically abelian and the heart of about the t-struct I didn't so let me just Yes, so so so right if you if you take the bounded derived category of a and you take the standard t structure Then the heart is deep back you get back the original category and I mean How to go you from one way to the other given your your your t-structure you get a is the intersection Wait how to go from a to And your to your t-structure so here for d less than or equal to zero you have to take the extension closer of a a shifter by one a Shifter by two and so And similar here right so so just as this picture here suggests so maybe let me draw this over here. So this would be the less than or equal to zero and This here would be D greater equal to zero Okay, I mean so far So where did the right category of a's equivalent to which one so that's in general not true and there So so so there is a function somewhere if you work with DG categories or something like that Otherwise, it's a bit of a mess to construct a function, but the I mean But there's a more essential obstruction to write for the sort of question as whether DB of a Is isomorphic to D and some of the bigger obstruction is that? in here any xn right so if you have E and F in a Then any morphism to Right so harms like this. This is xn From e to F then in a billing category xn can always be written as a composition of x1 I say an x2 corresponds to a four-term short xx sequence and the corresponding short term Short xx sequences give you two x1's and this may not be true over here Right so this this xn must be x1 generated and then if that's true then basically under reasonable assumption This is an equivalent. Okay, so that Okay, so how to how to construct t structures? Well some of the main tool for constructing t structures on Divide categories of coherent chiefs as the following proposition, right? So let MT comma F Be a torsion pair in an abelian category then the following defines a Defines a t structure. So the I'm sharp greater equal to zero These are all the ease such that H i of e is equal to zero for I less than minus one and H minus one of e is in F D sharp. I guess I shouldn't write here because Some people can't see the sharp less than or equal to zero These are all the ease such that H i of e is equal to zero for I bigger than zero and H zero of e is in T It's a for completeness the corresponding hard the intersection of these two These are all the ease such that H zero of e is in T It's minus one of e is in F and H i of e is equal to zero otherwise, right? And so You can really think of this as The category of two-term complexes But any such ease isomorphic to a two-term complex Where the kernel of these in F and the co-kernel of these in T All right, so the description of objects is really concrete Just know that the the description of morphism is a little bit more is a little bit more subtle, right? Because you're really looking at morphisms in the derived category, not just morphism between ordinary morphisms between complexes right and what is the Let me also draw a picture for that one right so we have here the picture of our Original T structure where here we have a a shifted by one a shifted by minus one and Right, so this is some of filtration of the of the derived category You know, we also have the filtration of a into just two pieces. So we're T are those we think of those as objects with bigger face and F Similarly here we have T shifted by one and F shifted by one Right, and so my new heart is I'm out here This is my a show the extension closure of T and F shifted by one. Okay, any any any questions on this right and and I mean I should say that This works similarly, right, so I stated this Inside the bounded derived category Right, so here I stated this as a T structure in Db of a but this works similarly just Assuming that that a is the heart of a bounded T structure in D And I'm sorry here. I also wanted to insist that this this gives me a bounded T structure And let me let me leave the proof as an exercise, right? And so basically here. I'm taking this heart and I'm I'm I'm tilting it although Unfortunately would be a lie to claim that the name tilting comes from this picture over here Right and so in particular this this lets you give I mean each of the example of torsion pairs now Let's you construct a new case of a bounded T structure on and trying to let a category But also you can iterate this construction right by using this remark You start with a bounded derived category of a you tilt it you get a new heart If you find a new torsion pair in this new abealing category then you can keep going Okay, and so now I mean a Stability condition basically just puts these two notions together the notion of a Heart of a bounded T structure And the notion of Lobe stability in this abealing category a Free stability condition on a Triangulated category D is a pair sigma equal Z comma a ins D is the heart of a bounded T structure and The is a morphism from the Group homomorphism from the K group of a which is the same as the K group of D to see Group homomorphism that satisfies these two properties that I discussed last time So that yesterday so that if I have a non zero object then Z of E is in this See my closed upper half plane and Secondly that Z has the hard on a simon property So every object has a hard on a simon filtration with respect to Z and the way you should Think of this is that right? So I have my filtration of my derived category given my I'm a bounded T structure and now each of these pieces is filtered even more finally using slope stability Using slope stability for for this For this abealing category right of course I can shift is to get the similar filtration of a shifted by one And I somehow all these is hold together via the central charge that somehow organizes the These phases to lie in the complex plane and some of that's what the the second definition makes explicit a Pre-stability condition is a pair sigma equal Z comma P There now P is this fine slicing that I tried to sketch away there. Yes right so there It's a P of phi D is a full subcategory for all I in our with the following properties so that I have phi plus one is just a shift of P of phi. There are no more reasons from Same way going from P of phi one to B of phi two if phi one is bigger than phi two Right, and so these will be called the same. I stable They must table objects Objects of phase five, but this is the harm-managing that we saw before then for all Again, there's a filtration. There is a sequence of maps a sequence of Triangles right so there's a sequence of maps like this so that if I complete each map to a triangle by forming the cone then these cones are semi stable of decreasing phases and right so far these are These are all just conditions on P And then again Z is a map from the K group of D to see a Group homomorphism and the only thing that holds these together is that If I have a same as table object Then the central charge of that object is in the ray Corresponding to the phase five right so basically these What I what I sketch here as vertical strips these become these rays in the complex plane in the in the C stability picture why that the proposition is that Definition one and two are Equivalent and I won't prove this but let me just say how to go from one to the other Why it's so if I'm given this this construction that we've already Mean of this construction of course we implicitly have already seen examples of by my lecture yesterday well for this you said for Phi Phi in zero one You said P of Phi to be the C same as stable objects of a and phase five and P of Phi plus n is then forced to be P of Phi Shifted by n Or n in Z right so this defines P of Phi for all real numbers and Conversely how to go from two to one Now you have this fine filtration and somehow how do you get a value make this filtration Corsa? right, so you said a to be I'm So let me use this notation P of this interval from zero to one right so what I mean by this I take all the P of Phi's so that Phi is in Zero one and then I take the extension closer, so I think if Or equivalently I take all the ease so that all the Transiment filtration factor right by which I mean these objects AI all HN factors Have phase in in zero comma one in this semi-close interval Okay, and again if you if you actually want to prove that these two did Notions are equivalent you need to you need to play with the octahedral example right you take so for example I mean let's say you want to go from one to two you have this cause filtration into Chromology Objects and for each Chromology objects you have to how do I say one filtration and then you have to piece these together to a final filtration of all of E and For that you need the the octahedral axle Any any questions? And actually let me point out one more thing that right if you now let What is the corresponding D less than or equal to zero? this is actually P of Zero to plus infinity So there's and D Greater equal to zero this becomes P from minus infinity to one And where these notations are both interpreted in in the sense over here. It's again There's this vivid and unfortunate, but really unavoidable clash of notation that here We look at the lesson equal to zero and really this looks like P of greater than zero But that's just because when people invented T structures and when people invented slope stability they used Opposite science for anonymous behavior questions Okay, and I mean why do we why do we why do I insist on giving both definitions? There's some I mean this definition is the easier one to use if you actually want to construct them right in fact That's how I actually already have constructed stability condition if you if you go from the two constructions of slope stability yesterday But somehow I mean this is better at exhibiting the symmetry right so for example if you If you rotate the center charts a little bit then this corresponds to adding something to all the phases and so there you get an action by the rotation group of See Or rather the universal cover and that's harder to see over here So it's really worth having both both definition in mind all the time. Okay, and so I mean I've advertised that by going to the the derived category I Will get strong deformation properties Before we can do that one more ingredient is missing and let me select me Let me motivate this Right so what I'll explain now is what's called the Support property first Let's always fix a finite rank lattice lambda Together with something like a churn character from the K group to lambda right so and then Then we only consider Central charges that factor why it is we as we have K of X And this goes via V to lambda And I only want to consider group or more often from K of X to see that factor via this map and by app use of notation I'll know right see for I'll often write C for this map here over on the right K of D. Sorry right and The the examples you should have in mind is that I implicitly use this already so in this example a that acts as a smooth projective curve then I said lambda equal to C2 and we If I have an object in the derived category, I just send it to rank of E and Degree of E right so of course the K group of a Often algebraic curve of positive genius is much bigger than that But somehow I'm only interested in central charges that For example, don't distinguish between different skyscrapers chief of points. And that's I mean that will also be important for I mean on the one hand this means some of that my problem becomes more finer dimensional on the other hand It's also important for modular spaces later And then in the second example when you have Q reps Then I said lambda to be Just a free a building group Generated by one element for E for each vertex and we of We is just the dimension vector right again if you if if your quiver has If your quiver has cycles then the K group is can actually be much bigger than that But it always has this map just to the dimension vector, right and now we want The goal is we want a notion of stability of stability conditions such that Sigma Equals the P or the a deforms along with with any small deformation of Z inside but now a finer dimensional vector space Arm lambda to see right so I mean we we saw this already in this quiver example Whenever if I deform my center chart a little bit by deforming my z-eyes Then I can deform the notion of stability as long as least as long as this thing the upper half plane and This is the phenomenon. We want to generalize and So let's before we try to do this. Let's let's add a bit of thoughts on that so There's actually a natural topology on the set of pre stability condition. So maybe let me write this as pre step lambda Mainly it's the Causes Apologies such that the following functions are continuous, right? So first you have this map from Please stop of lambda to home from Namta to see that just a science that forgets P And just remembers this this linear data But then also there's a very natural notion of topology on the set of On the set of possible P's called slicing namely so for each e and There are two functions That's then Sigma that's then C of P and Phi plus of e and Phi minus of e Which are the maximum or? Minimal phase is appearing in the harassing infiltration Right, so I mean it of course some of the hardest infiltration of the itself may change because objects may become unstable and so on But at least the maximum minimal phase is appearing they should depend continuously on the stability condition Right, and then from this it follows that being Semi stable is a closed property because say my stable is equivalent to Phi plus being equal to Phi minus and So being stable should be an open property and Right, so if you have something that stable it should still be stable after small deformations And let's think about what this implies for our central charge I'd note that if if e is stable Then clearly I mean v of e Cannot be contained in the kernel of the central charge Because any stable object is in one of these rays Or if you go via the definition one if E is stable then one of its shift is in a and so it the central charge can never be zero right and so Somewhere here you have the kernel of these or this is now a picture inside Lambda R Right, and so if all these properties should hold right so this goal if the goal holds and This last property here should hold And not in nothing comes right This still has to hold Hold After small deformations of z right so in other words there must be a cone Around the kernel z that does not contain v of e for all the stable objects right so here This is we have stable objects must be outside This there must be a code like this so that we have stable objects outside of this cone and the kernel is inside this cone And so that's the that's exactly the definition of support property so definition The pre stability condition Satisfies the support property Satisfies support property There exists there is a quadratic form you from lambda tensor R to R That's that if e is say my stable then q of V of e Is greater equal to zero right so I think of this cone being this cone can be defined by a quadratic form and on the outside we have the same as tables and You restricted to the kernel is Negative definitely So that's that's the support property and what I'll show I mean what I hopefully try to convey to you today that the support property is Necessary if you want to have any hope of nice deformation property for stability condition and what I'll show tomorrow is that it's also sufficient so if you take the set of stability conditions Pre stability condition satisfying support property then you get a complex manifold and this this this goal that I formulated over the there holds Okay, thank you