Great so and thanks to these three guys for checking my slides for typos they I noticed they didn't find the one in their own name so that's a bad sign we'll see here we go okay so I'm going to start with the talk I'm not going to talk about Vuffle Whitting Theory for the first lecture and a half just about background material and I try to go simply as possible but of course it's not possible to describe everything in this field in intimate detail so there's going to be there's going to be stuff where I just maybe give a guide to how one would go about understanding something and so I'm going to set a bunch of exercises but maybe more suitable exercises are just flesh out the slides so there'll be slides that particularly interest you that you feel I didn't give enough detail a perfectly good exercise is to flesh that out or to start fleshing that out maybe go and look at some of the references and to talk to the the tutors about it or to talk to me about it so these are some references there's many more and in some kind of order that we use them exercise it'll be in green but as I say much better exercises this just to flesh out all the bits that I skate over so let's begin and so this course is going to be about coherent sheaves so I'm always going to be on a smooth projected variety over the complex numbers and is its dimension there'll be a an ample line bundle which I'll call 01 and its first showing class is h so it's just notation a coherent sheaf what I'll mean by a sheaf will always be a coherent sheaf so that means it's just locally finitely generated by the structure sheaf so over an open set u it has a presentation like this and then this is less an exercise than more sort of revision or go back and remember why that means that not just that it's finitely generated but that it's finitely presented so the kernel of that map the kernel of that map is also finitely generated and then why after finitely many steps the kernel becomes locally free so you get a resolution like this okay so the exercise is to go and make sure you're happy with this it uses the fact that x is smooth and if you want a reference griffus and harris of course it's analytic rather than algebraic but the proofs are the same griffus and harris have a I I really like their way of dealing with this it's very quick okay so you should think of a coherent sheaf as some kind of vector bundle with singularities where the singularities are the locus where this this rank drops and so this map this map here from one vector bundle to another this matrix where its rank drops f sort of picks up singularities and away from that it's a vector bundle its support is a subscheme of x defined by this ideal sheaf over an open set u it's just those functions which annihilate all the sections of f and define an ideal sheaf and therefore a subscheme called the support of f and f is really pushed forward from that support it's a it's a sheaf on the support whose push forward is the original f and then we talk about the dimension of a sheaf is the dimension of its support so i'll give some examples any questions so also f is a quotient of regular functions by this idea no no that so f is a module over the quotient so what you just said is the structure sheaf of the support and f is a module over the structure sheaf of the support so f is a sheaf on the support but it needn't be the structure sheaf so you could imagine f being a rank two vector bundle on a sub on a point or on a subscheme and then the support would be that subscheme this ideal sheaf would be the ideal sheaf of that subscheme but f itself would have f would be the push forward of something of rank two on that subscheme so it wouldn't be a quotient of the structure sheaf it would be a quotient in this case it would locally be two copies of the structure sheaf of that subscheme so the organisers if there's questions online yeah it's fantastic that people are able to attend this online but i can't believe they're really paying attention they're just looking at the internet yeah but they're they're reading the news or looking at the sport or they're like checking out england's chances of winning the euros they're not really paying attention okay and now sheaves are called pure so if it has dimension d it's called pure if it has no dimension strictly less subscheme sure okay you have to go back yeah so he says if r is equal to n in the first exercise no definitely not i mean f could actually be this guy right it could be r copies of the structure sheaf where r is the rank of f it could be r is not the rank of f in general but um f could well be just this sheaf and r could be as big as you like okay so um there's this notion of pure for sheaves i'll give an example in a minute and it implies that both the support and the structure sheaf of the support are also pure and exercise is to show that when when the support is an integral subscheme um then pure means that the the sheaf is torsion free on its support so it's the push forward from its support of a torsion free sheaf so a sheaf where with no torsion where there's no element to the structure sheaf that annihilates the section of the sheaf so some examples if d is a cartier divisor well x is smooth so if d is a divisor uh then the structure sheaf is pure and that's true even the divisor doesn't have to be reduced and you could take many copies of this example to answer some of the questions um so that then it won't be rank one on its support it can have higher rank on its support so for instance these are pure modules over the polynomials and two variables but this is not pure okay so if i take um the structure sheaf of this subscheme that i've drawn in red so where i've set x y to zero so i get the two axes and then i set y squared to zero so i i don't get the full y axis i only get its first order piece at the origin that's not pure because of this embedded point at the origin which gives you this zero dimensional sub module so this is a one dimensional sheaf but it has this zero dimensional sub module which is just the structure sheaf of the origin mapping into our sheaf or module uh by y okay and because um y times anything is zero x times y is zero and y squared is zero this really does uh define a well-defined module map uh from the structure sheaf so the this is this is sort of c x y divided by the maximal ideal x y okay any any questions about that okay now i should tell you about stability so um we have the Hilbert polynomial of a sheaf which is we twisted up many times here by t by that that uh polarization oh one t times and then we take the space of sections so or the holomorphic Euler characteristic for large t the two are the same and that's some polynomial in t and the leading coefficient is given by t to the d sorry the leading term is t to the d where d is the dimension of the support so usually when we think of something like a torsion-free sheaf of the whole variety that the leading term is t to the n and then the leading coefficient is roughly the volume of the manifold so like that h to the n divided by n factorial times the rank of the sheaf so you should think of up to some irrelevant constants this leading term is the rank of the sheaf um if if d is n so if f is torsion-free let's say or support f is if f has rank bigger than zero and then the reduced Hilbert polynomial is where we make that monic so we get rid of the leading term and then the slope there's different notions of slope but what i'm going to take is the leading term in the case where d is n so to start with let's assume f has rank bigger than zero so its dimension is n so it's supported over the whole variety then this this sort of second term here or i mean this first term is the same for all sheaves so the first interesting term here is called the slope and more generally if if d is less than n we just set that slope to be plus infinity because a n will be zero so if a n is zero we just set this to be plus infinity okay and then there's different notions of stability there's slope stability geziqa stability and others which which use this polynomial so slope stability only uses this leading term its first interesting term geziqa stability uses the whole polynomial and there's other things as well but let's start with slope stability so we have this this notion of slope so it's just the by reamon rock you can work out what it is and it's the degree of the sheaf so it's the first churn class the degree of the first churn class of the sheaf divided by the rank of the sheaf okay so up to some constants the the leading term here is the rank of the sheaf and then the sub leading term is the degree of the sheaf and then f is slope stable or semi-stable if and only if whenever you have a sub sheaf and a quotient like this which is non-trivial so neither a nor b should be zero then you want that the the slope of a should be the less than the slope of b so the brackets are what you might guess so if i want stability then i should have a strict inequality here and if i want semi-stability i'll allow the non-strict inequality uh where these these come from so it um so d here was the dimension of the sheaf so if if the sheaf is supported over the whole variety then d is n so this will be an and this will be an minus one this will roughly be the rank of the sheaf and this will roughly be the degree of the sheaf the first churn class dotted with the hyperplane class um and then the slope is the quotient of those so if the if the dimension of the sheaf is less than n then an will be zero the first term will be in lower degree so this d will be less than n and so an will be zero so then we just set the slope to be plus infinity all right did i answer your question i mean these are all given by riman rock formula they're all just these are all topological numbers yeah yeah they're they're the oiler characteristic of this um the sheaf twisted up many many times we only consider this for large tea for example when the sheaf is pure wouldn't you use a b minus one over a b as the you can that's a different notion of slope and that's relevant but we're not going to deal with it now yeah yeah that's absolutely right okay so we have this definition of stability which at the moment looks arbitrary but i will try and motivate it a little bit i'll show you it has good properties and i'll i'll say something about how you should think about where it comes from and why why it's there but for now just accept it for one more slide just accept it this is the definition that sub sheaves should have lower slope and quotient sheaves should have higher slope okay that's that's the definition of stability in particular if your slope semi stable you must be torsion free so maybe that should be an exercise but that's because if you had um a sub sheaf of lower dimension so if you had any torsion in your sheaf that would give you a sub sheaf of lower dimension but that would immediately have slope infinity because that would have a n is zero so you'd have slope plus infinity here and your f would have finite slope and so and your b would have finite slope so you would you would destabilize yeah maybe that's an exercise that that gray comment should be an exercise because you have to pick the right torque you need to pick the maximal torsion sub sheaf in order to violate this uh in particular slope f could be torsion i think i don't want to define slope stability using if f is torsion i don't want to define yeah yeah maybe i should have been more careful i would want to use your notion of slope at that point let's take f here to be um let's take f to have full dimension at the moment where's chaos yeah yeah i was trying to simplify things to make things to do one example which is simpler um but now everyone's finding the floor in that we'll go on to Gizika stability in a minute and everything will be hunky dory okay so exercise the seesaw inequality that the slope of a and the slope of b the relationship between the two is more or less the same as the relationship between the slope of a and the slope of f so roughly speaking this condition that the slope of a should be less than the slope of f is is very similar to the slope of a being sorry this condition the slope of a should be less than slope of b is very close to the condition that the slope of a should be less than the slope of f so it's just some rearrangement of this inequality using these numbers so good exercise to do if you've never done it it's called the seesaw inequality um but it's not quite the same so you don't you should not define stability as being that the slope of a is less than the slope of f um because they're not quite the same they're the same when the dimension of b is the same as the dimension of f so here and but here's an exercise which here's an example which will show you why they're not exactly the same so here this is the ideal sheaf of a let's say a closed point in x you should have a think about this example and uh this does not destabilize according to my definition but it does semi destabilize if you took the definition here so uh go i let you go away and think about it we don't want to get into a big discussion about okay so gizika stability so this is um going to be a bit more robust i think um what we say is very similarly instead of using slope which was really just the the leading coefficient of this the the first interest in coefficient of this reduced hillbett polynomial we're now going to use the whole hillbett polynomial for large t so you can see this is an extremely similar notion so your gizika semi stable if and only if the hillbett polynomial of the sub sheaf is less than the hillbett polynomial of the quotient sheaf where less or less equal are um described in the following way so you should concentrate on the first line so we say that a polynomial is less than another polynomial a monic polynomial is less than another monic polynomial if and only if well when they have the same degree you should it's just the obvious lexicographic ordering that p of t should be less than q of t for large t okay so you can imagine that's very closely related to the the first non-trivial coefficients satisfying the same inequality but we have to be careful and stick in this second condition that when them when q has lower degree then we should say it's bigger than p so that's confusing um because the inequality seems to go the wrong way round but it's because somewhere we divide it by zero it's to do with all the questions on the previous slide it's because you know some a n minus one is zero so when we divide by zero we should get plus infinity and so that that's the origin of this but basically what it means is um that you're destabilized either by things of higher so if you're if you have a sub sheaf with a bigger reduced polynomial then the quotient quotient sheaf that destabilizes you or if you have a sub sheaf of lower dimension than f then that destabilizes you so in other words if f is not pure then you're unstable so not so you can either take this as a definition or you can take the definition that you only insist on the first condition but you also ask that f must be pure and the two are equivalent okay so what you find is that gizika semi-stable sheaves are pure these are portion free sheaves sorry these are portion free sheaves not now no these are arbitrary sheaves now i i think they were probably yeah no no no you can have gizika stability for any sheaves yeah okay and again it's equivalent to the reduced hillbott polynomial of the sub sheaf being less than the reduced hillbott polynomial of the actual sheaf whenever the dimension of the quotient sheaf is is full is the same as the dimension of f but again it's not equivalent when that condition doesn't hold so you have to be careful okay so i want to say something about these oh maybe another exercise would be to show that um i just write it here i appreciate it's a bit small for the people um look checking the bbc news um but uh you could a good exercise would be yeah so check you're happy with these definitions that um slope semi-stable let's take the case where um maybe dim f is is the full dimension of x then slope semi-stable implies gizika semi-stable implies gizika stable implies uh slope stable did i get it all wrong bugger oh yeah oh yeah in france you in france you use the implications the other way don't you sorry this was the english way of writing it's amazing how you can't think at the board okay thank you please continue to do that okay so let me i'll leave that up just for a second uh let me um let me try and motivate these notions of stability which are a bit weird and show you some nice properties um firstly i'll just tell you a fact that they arise naturally when you try and form the moduli space by geometric invariant theory so you can try and form all these moduli spaces as quotients of quat schemes so you try and see your sheaves as quotients of some fixed big sheaf like many copies of the structure sheaf twisted by minus n for large n something like that so you try and see all your sheaves as quotients of a fixed sheaf so you manage to see them all in a quat scheme and then you have to divide by all the possible choices that got you that described it as a quotient in particular the automorphisms of the fixed sheaf and so you end up doing geometric invariant theory and that gives you a notion of semi-stability and you end up with this one another way of thinking about it is that um what it's saying is that quotient sheaves within reason should have more sections than sub sheaves all right so that that's not quite true that to leading order the amount of sections you have when you're twisted up by large t is given by the rank so you can't get away from that so that this this is not compatible with that right so the rank is what determines the number of sections to leading order but to sub leading order to the next order modulo the rank what determines how many sections you have is is this either the reduced Hilbert polynomial or the or the slope and what this is saying is that quotient sheaves should have roughly the right amount of sections according to their rank but then to next order the amount of sections they have they should well they should have more than the subs sheaves and that makes sense right if you have sections of f they give you just by law they give you sections of b so b has lots of sections whereas getting sections of of course many of them are zero because they lay in a but generally speaking is easier to get a section of b by just taking a section of f and projecting it what's harder is to get a section of a because your section of f have to satisfy lots of conditions to actually lie in a so you should think of f b as having more sections in a okay and that's roughly speaking what stability says and this is the generic situation so stability is a generic condition in fact it's a risky open condition and the generic sheaf is stable so if there's a single stable sheaf then there's a risky open you know in the space of all sheaves in some sense and those which are stable it's a risky open so they're dense so that that's that's why it's an important notion but now i want to show you why it's um that it has nice properties to motivate it i think to explain this a bit better but first there are any questions okay so um so i give you two nice properties of stability so one is this thing that's a bit like the shear lemma in representation theory and that if you have two stable sheaves of the same churn classes or same hillbett polynomial and then they satisfy this property that there's no homes between them unless of course they're the same and and when they're the same the only automorphisms they have is multiples of the identity okay so um i do this for slope stability but you could also do it for geysica stability the argument's very similar and very simple it's that if you do have a morphism from f to g you just factor it in this way um you just write the map has a kernel and an image and then the image so the image is a quotient of f and then it's a sub of g of g and that gives you this chain of inequalities that the slope of f should be less than the slope of the image because always as you go to the right slope should increase but the slope of the image should be the less than the slope of g because it's a sub but of course g and f have the same topological type so um their slope's the same so you get this contradiction here unless one of the exact sequences is trivial which means that phi is either either the kernel zero or the image is zero so either you have the zero map so we're in this case or you get that phi is an isomorphism so now you can think of f and g as being the same and in that case when f and g are the same then exercise run this argument again for uh phi minus multiples of the identity and pick find an appropriate essentially eigenvalue of phi to show that eventually this can't be an isomorphism for some lambda this can't be an isomorphism so it must be zero so phi is a multiple of the identity all right so that's one nice property of stability so you're choosing c to be your base field forever yeah yeah and then this is the really nice property of stability so this is really why it works and why we have this definition uh it's separateness essentially of the modulite space of stable sheaves so it's the one parameter criterion for separateness so it's the the moduli of sheaves of stable sheaves is housed off so here's the setup pick two families of sheaves parameterized by a curve so let's just say the affine line so we pick these two families of sheaves uh they they should vary nicely the right condition is flatness over the base and think of them as one parameter families of sheaves e t and f t and suppose they're all stable then what you find is if they're isomorphic away from the central fiber then they're isomorphic over the central fiber and more over the the families are identical the families are the same okay so hopefully it's clear to you this is this is saying that the the moduli space has a nice separateness property or house top property and then the sketch of the proof is that um when you take the homs down the fibers you get a line bundle away from the origin just by the previous slide and base change so on on the fibers the on the non-zero fibers um because e and t because the the sheaves e t and f t are isomorphic they have precisely multiples of the identity as their their homomorphisms from one to the other so you find that um the homs on the fibers are one-dimensional so what what that means is that the the relative homes the sheaf of relative homes is a line bundle away from the origin and it's torsion free and that's because of the way you define this you have to remember how this is defined this sheaf it's not defined fiber wise is defined over open sets in the base and so um if if the homes jump on the central fiber you might think that this sheaf would be a line bundle with maybe a little bit of torsion over the central fiber but that's not the case because you don't see if you only have extra homes on the central fiber you don't see that in this sheaf because this sheaf is defined by taking an open set of the origin downstairs and looking at the homes above that upstairs and there won't be any because there's just some homes on the central fiber but they don't extend to homes on the open set okay so it's an exercise to understand this statement here i've been a bit slack about it um but uh it's a good thing to go away and check you're happy with what happens is if the homes on the central fiber jump you won't see that in this sheaf you'll see it in the next one um so the the x one the relative x one sheaf will get some torsion it'll get um you know the structure sheaf of the origin or something in it but in this sheaf you won't see it okay so it's actually a line bundle on c and therefore it's the trivial line bundle on the affine line and so you can pick a nowhere vanishing section and the fact that it's nowhere vanishing means that on the central fiber it's a non-zero homomorphism from e0 to f0 and therefore again by the previous slide it must be an isomorphism so what you end up with is this section is an isomorphism on every fiber and so it's an isomorphism and so again obviously i'm i'm sketching this and i'm not writing in all the details so if if it interests you i hopefully this is enough to give you an idea of why it's true and if it interests you then go in and flesh this out and make it a theorem but i think more important is to give an example where this fails if you don't have the stability condition so go away and check you're happy that moduli of sheaves are just not is not a well behaved thing if you don't have stability you you get um hopelessly non-separated moduli okay and there's also another property which is when you take semi-stable sheaves and then you have to divide by something called s equivalence which i'm not going to go into um then uh the moduli spaces are proper and uh so that's a wonderful property and it's very important in many things but it's not so important in my lectures because in vaffawitton theory the moduli spaces which arise are not proper anyway so i'm not going to go into this but this is something very important that i don't have time to um go through okay so i want to do a little bit of defamation theory again i'm just going to give you the rough idea so let's start with vector bundles how do i get rid of this thing at the top right okay so let's start with locally free sheaves so vector bundles so these are made from gluing trivial vector bundles on affine open cover and you glue them over overlaps by transition functions which satisfy the co-cycle condition well you need to know that there is a moduli scheme so that comes from geometric invariant theory and then um the one there's this uh what's it called the one parameter criterion for separateness is this basically that uh if you if you look in heart shorn for the one parameter criterion for separateness then check that basically it's what i wrote down yeah so so it basically says that um brilliant thank you when you have a family of vector bundles um you want to know so separateness says that uh when you have a family you can fill in the central fiber uniquely and the one parameter criterion says you can you can test this just with smooth curves and that was what i did okay so now we can deform if we have a vector bundle we can deform it by infinitesimally altering these overlaps so i'm going to change my overlap my transition function sorry on the overlap i'm going to deform it by this i'm going to change it by this little guy here and then i'm i'm going to work to first order so i'm going to assume that t squared is zero so when you write down when you do this when you change the um transition functions over overlaps uh then you have to check that the co-cycle condition still holds when you do that mod t squared so let's set t squared to zero uh then what what you find is that these eijs here uh form a a check co-cycle so they're whatever it's called check co-closed and then moreover so so they they define an element in h1 of nd and moreover when you only consider them up to isomorphisms so when you consider two to be the same if there's an isomorphism bundle which takes one to the other or something then then you find you divide out by check co-boundaries so you end up with it that this check group here is the first order deformations of your bundle and then uh you can go further i mean i i've got to admit i haven't done this exercise recently but i promise i did it when i was your age so uh these things are hard but they're worth doing that i mean they're they're easy when you see them but they they take forever to think up anyway um so when you do the co-cycle condition to the next order then what you should find is you get this first order deformation here cupped with itself in h2 of nd and you know for that you've got to remember what the cut product is in check co-omology and nobody knows that so you know there's a these are easy exercises for me to state and they're kind of lengthy for you to do and you shouldn't worry about taking hours or days over them or discussing them with your colleagues okay and this is the obstruction to extending the deformation to second order okay and i'm going to do all this in more detail for sheaves by a different method in a second all right but there's a general principle that you what you find in these deformation problems is that you usually find a bunch of co-omology groups where let's say the zeroth co-omology groups are the infinitesimal automorphisms that's clearly the case here and people tend to call those t zero the deformations of the next co-omology group that's what we saw here these are tend to be called t1 and then the obstructions of the next co-omology group so let's do this for sheaves and here i really did the exercise so i'm honest and you'll see it's non-trivial to make it all fit together okay so i'm going to do deformations of sheaves here i'm going to work i see people are taking notes these i can give you these slides afterwards and i think they're going to be posted online without the pauses so um you will have something to look at okay so i'm going to work over spec of the dual numbers so i'm setting t squared to zero and then later when i do obstructions i'm going to go to next order by working over this a2 space and a0 is just the origin it's just c spec of c okay first order deformation of a sheaf is a sheaf over x times a1 spec of the dual numbers so it's x plus a little vector normal to x right it's x times by a little fat point i want things to be flat over a1 i'm not going to go over flatness just the lack of time and the sheaf over this thickened space should restrict to my original sheaf over x times the origin so that's what a first order deformation is and i want to show how to describe them okay so let's let's suppose we have a first order deformation and we'll go in the opposite direction in a minute okay so take a first order deformation restrict it to x then you get your original sheafy zero okay and so what you end up with is this exact sequence because the kernel of a restricting to x is multiplication by t okay and since t squared is zero what you find is the first map factors through e1 modulo t it kills anything in e1 that's been multiplied by t because t squared zero therefore it factors through here and this of course is e0 all right and then the exercise is to check that the result is an exact sequence so that you can make this exact on the left if you replace that e1 by e0 so flatness makes this an exact sequence all right so you should think of this as the x direction at this stage is not so important what's going on here what this really looks like is up to the x you know modulo all the stuff going on in the x direction what this looks like in the a1 direction is just this is that you know this this is the functions on a1 I restrict them to the origin and what's the kernel it's just a copy of c but it's it's those it's the functions multiplied by x oh sorry t so that's what's going on here those are your two e0s and this is the e1 okay up to what's going on in the x direction okay and now if we just take sections in the a1 direction so push down to x then this becomes this gives us an exact sequence on x originally this was an exact sequence on x times a1 but we can think of it by because the a1 directions are fine we can just take sections in that direction and we get an exact sequence on x okay so we get an extension so that's classified by this extension group all right so we get an element of this extension and the claim is that this completely classifies the first order deformation so I need to go backwards given one of these I claim I can produce a first order deformation e1 so conversely given a first order deformation I get an extension on x and I'm going to call these two maps iota and pi and that's fine but that's a sheaf on x and what I'm meant to produce is a sheaf on x times a1 so I need to make it not it's already an ox module I need it to be an ox brackets t over t squared module I need to tell you what the action of t is on this e1 all right but we know what the action of t should be on e1 from this exact sequence okay it should kill everything coming from the left because t squared is zero so it should kill e0 because this this multiplication by iota we're expecting to be multiplication by t so it should kill this guy in other words it should factor through here so you should project to here and then because of the action of t it should be multiplication by t so you should take this guy stick it there multiply by t and then end up in there all right so I probably confuse you completely but anyway my claim is that you can make this into an oxt over t squared module by making t act as this map and then the exercise is to show that's correct to show that the result is flat over a1 so there's two parts of this exercise first you should show that I have described an oxt over t squared module in other words that this map here has square zero and commutes with all the x you know it commutes with ox that bit sort of obvious it's an ox module map but you should show that t squared this this map here has square zero so it really defines the structure of an oxt over t squared module and then once you've done that once you have this module you should show it's flat over a1 okay and that's when you get everything in the right order it's completely trivial it's one line but getting everything in the right order is um is hard the first time you do it and you learn a lot okay so what you find is that first order deformations are given by this x1 so that's the tangent space of the modular space and another exercise is to relate that to the previous where the description I gave you for locally free shoes so um for vector bundles when you have a vector bundle this exact sequence on x splits locally all right so exact sequences of vector bundles are always locally split and so you should glue the splitings by um transition functions over overlaps um where the two splitings are not compatible over overlaps so you change them by instead of taking a direct sum here you change them by some map from here to here so you change them by this upper diagonal kind here okay and you should see this recovers the previous description okay so now to second order what I want to see is an obstruction if I have a first order deformation there should be an there might be an obstruction to extending it to second order and I want and that should lie in x2 now the next cormology group up and I want to see that so I really did the exercise perexine this once before do you remember this five years ago um so let's suppose we had managed to find a second order extension so we managed to find a deformation not just of my zero but really of my e1 my my sheaf over x times a1 I managed to find an extension flat over a2 to here okay call it e2 then we get an exact sequence here so I can take my sheaf over a2 I can restrict it to a1 the kernel are the things which are already multiplied by t squared because t squared is zero in a1 so what you find is that the kernel is by flatness is t squared times e0 but instead of restricting to a1 so this you know I think you can see this is um what I've written down there modulo what's going on in the x direction what I've written down in the a2 direction is just here's the a2 guy I can map it to the a1 guy and the kernel is just a single copy of c but instead instead of restricting to a1 I could restrict to a0 I could restrict to the origin so I could just map to the origin here and then the kernel will be t lots of you know c of t modulo t squared so this will be um essentially the functions on a2 but I multiply them by t under this map okay so let's put that on the diagram so they're the two different ways I can look at this sheaf it's on a2 I can either restrict it to a1 or to a0 so to a1 that's the horizontal one and to a0 is the vertical one okay and when I do that they fit together in the following way okay so um when I restricted vertically to a0 I had this very big kernel and the the smaller kernel sits inside it okay so the the t's you know in this diagram here uh the the the functions which are already multiplied by t certainly contain the functions which are already multiplied by t squared so they fit into this exact sequence um and similarly down here so once I restricted to a1 over here horizontally on a1 I could further restrict to a0 so it's obvious this restriction map to a0 to to the origin factors through restricting first to a1 and then restricting to the origin okay so that's all this diagram says so I end up with this all right so you recognize this vertical guy is the original description of the extension e1 in terms of an x class on the e0s this is the new one I've got this is t this is also the original extension of e0 by e0 to give e1 all multiplied by t okay so everything's familiar in this diagram but you know you need some time to absorb it and you won't do it right now and then you can look at what this says in terms of extension groups okay so um my e2 is defining me an extension group on x given by extensions from e1 to e0 so that's this guy all right when I uh go up to this row that's giving me the the extension I started with which defined e1 so the original extension which defined e1 was an extension from e0 to e0 and that's this guy here okay so my e2 is restricting to my e1 when I um look at what the extensions do on the kernel of the map to the origin all right and this sits inside an exact sequence so this is just the long exact sequence of extensions to e0 so um I take look at this vertical column here I take extensions of this sequence of this of these sheaves to e0 that gives me a long exact sequence of x groups okay and then the next x group along is x2 from this guy e0 to e0 okay and this is the co-boundary map and what you see is that um when I have an extension e2 that gives me this extension class which maps to e1 and therefore e1 must map to 0 here all right so when I when I have a second order extension in my sheave then the first order extension class must map to 0 in this x2 group uh because it comes from here and so what you find is that this e2 exists so I can find an extension here which maps to my extension e1 that I started with if and only if e1 maps to 0 in x2 here okay so we call this the obstruction class so the logic is probably a bit confusing here because I assumed e2 exists but I will go backwards in a minute okay but um given given my e1 class so given my first order deformation I consider under this co-boundary map I consider its image here okay so I'm only using e1 to define this part of the diagram because I'm only using this exact sequence always come back again does that mean someone's asked a question so I'm going to point anyway but for you guys um so given an e1 I can still form this right hand vertical sequence and therefore I can take this co-boundary map and so I can define the co-boundary of e1 as some class in x2 and I call that the obstruction space all right the obstruction element so this obstruction element in x2 is going to vanish whenever I have a flat e2 whenever I have a the sheaf to second order so that that's obvious from this this exact sequence of x groups okay and now I need to check the converse that when this obstruction class vanishes can I produce an e2 and probably I just set that as an exercise I can't remember um I see now we have the same problem again I shouldn't yeah so so what happens is whenever I go into the chat I can no longer move my slides like on the keyboard or with this so what did you do last time okay please just lean on the keyboard okay great thanks yeah yeah so a steeper question probably about it so in order the obstruction class for the first order extension was kind of quadratic right yeah it took the curve a little bit yeah that's on the next slide yeah but then in the second order extension it seemed like the obstruction class is linear because you take just more composition of the compounding map which is linear yeah but that co-boundary map is cut product with e yeah yeah you're going to see that I think I'm going to answer your question I'll check with you in a minute yeah well how do you do that ladies and gentlemen Andre right so I think this answers your question but I'll check with you in a minute so the right hand vertical exact sequence was given by the extension class e1 so we just go back right this guy the right hand vertical exact sequence is extension by e1 and and therefore this co-boundary map is cut product with e1 that's how that's how x'd works what's this called something x'd someone's description of x'd there's a name attached to this property sorry is it yonade or x'd yeah it is that's correct yeah so that's yonade's description of x'd so this the obstruction class is quadratic it's e1 cup e1 is that what your question was but e1 is the first order extension yeah yeah that's right and so you're saying that the second order extension has a fraction class equal to e1 cup e1 no so what I'm saying is the first order extension class defines the obstruction class and what this means is this vanishes if and only if my first order extension extends to a second order extension so this vanishes if and only if e1 lifts to an e2 okay so if this vanishes then there exists an e2 there's a whole choice of them but there exists an e2 and now you get a second order extension yeah that was for vector bundles so that was a special case of this and and there for vector bundles because everything was locally trivial there were no problems with doing these extensions locally the problem was did they glue globally so it was to do with the co-cycle condition but but these these two classes are the same if if your sheets are locally free then this is the same no let's check here so um of course I didn't do it but I said when you take the co-cycle condition to mod t cubed so you look at it whether it defines a second order extension then you find the obstruction is here yeah okay I'm kind of keenly aware that you can't if this is new to you you cannot absorb it in 10 minutes in a lecture this is this is a guide to what you should go away and you know work extremely hard on and suffer with and you know I have sympathy I've suffered with it in the past okay so we got to here so when the obstruction is zero I get an e2 I can lift my e1 to an e2 and that e2 gives me the central horizontal exact sequence okay and again again my claim is that defines me a second order extension so it defines an oxt modulo t cubed module where the t action is given again by pi composed iota iota composed pi okay so extension exercise show that is indeed a ct modulo t cubed module and show it's flat over over there and then this is a bit evil maybe don't do that okay so I don't think um I didn't do this for the sake of exposition or I didn't do this example because I expect you to understand it I did it to show you that um these exercises can be done and they're hard and they take time and uh you have to suffer um and I just wanted to be honest to show you that I have done the examples myself at some point in my life okay but yeah they're tough yeah absorption absorption class would characterize a second row or it's just a complete existence just existence because there's choices because there's another group down here right so in this long exact sequence of extension groups we should have it underneath we should have x1 e0 t0 so the given your e1 the choice of lifts to e2 I give it you know there's choices given by this x1 e0 t0 so they're not unique they're only unique up to the action of x1 so as you go to second order you can pick a first order deformation to deform it by so instead of picking a straight line deformation you can make it curve I don't know maybe that's not a very good way of saying it okay so you can do all these things I just wanted to do it because whenever you look at a book they always say oh yeah here's first order of deformations and you can show that second order deformations live in this group and then no one ever does it so there you go that's how you do it all right how are we doing for time uh we should stop yeah okay so I carry on next time there is a question does next go to third order imply that there is a need formula for the obstruction there yeah um yeah there is so I guess the question is if we expect h3 or s3 no I mean it also lies in x2 it lies in the same group so what was the obstruction to go into third order what's the formula yeah and it's very similar right um it's it's basically instead of e1 copy 1 I want to say it's e1 copy 2 but you have to interpret that correctly but it lies in the same group it lies in x2 e0 e0 so yeah that's important that the the obstructions and all the choices and deformations to each order they're all governed by the original e0 that's kind of interesting yeah okay so whoever asked that question that that's good you've set yourself an exercise I'll let you do it instead of me anything yeah anything else maybe this was a depressing note to end on yeah maybe the question is a bit off talking but I was wondering so these two so we can be fine are these stability conditions in the sense of bridges no not quite I mean yeah no they're not but they're not far off they're they're in the space of bridal and stability conditions there's something called the large volume limit and they're very very close to being stability conditions there but they're not quite for curves they are that's correct yeah yeah and then for surfaces the way you deal with the zero-dimensional sheaves is slightly different and it's to do with the fact you know that slope only sees the rank and the first churn class it doesn't even see the second churn class but for stable objects you have a Bogomolov inequality which means that you have some control on the second churn class and and that's dealt with slightly differently in bridal and stability from this you need to do a certain tilt where those zero-dimensional sheaves get shifted by a minus one or something like that and then work in that Abelian category I don't know if I'm saying words that mean anything to you but yeah that you you have to deal ever so slightly differently with the the zero-dimensional sheaves that is very close yeah other questions comments of a tool someone so let's thank each other


Hey what's it you guys welcome back to my channel if you're new here hi hello I'm Lillian and if you are new here make sure you hit subscribe button during the growing family we're almost at 6.2k and it would mean that I would be able to reach that goal by the end of this month. Today I'm going to be talking about my diagnosis story of being diagnosed with bipolar disorder. Hi I'm Lillian I'm 25 years old I'm diagnosed with bipolar type 1 and I hope the chair in my journey can help those of you who have symptoms of bipolar disorder go forward to get a diagnosis or seek professional help. So I was 13 when I was diagnosed with bipolar disorder but I wanted to put a lid up to the diagnosis I would manage and psychosis and the main year led me to do certain things that aren't normal I wasn't sleeping and if I did sleep it was for two three hours a night I wasn't eating I was taking on so many projects I couldn't keep up with that it was a really scary time in my life I was I felt on top of the world but I also had a sense of fear of crashing and the ultimately end up crashing but I was in hospital so the main year really took hold of my life and I was psychosis at the same time I mean I had delusions I had hallucinations and in experiencing that it just led me to a point where I made an attempt on my life because I was so unhappy but I was happy I was really elated but I was also struggling a lot with self-harm impulse control and stuff like that it wasn't just uh from main from my episode it was main year but it was also a sense of depression at the same time so I got I ended up getting sectioned under the mental health act after my mom took me to A&E because something wasn't right but she was right in thinking because obviously there was something wrong when I was in hospital I had one conversation with a psychiatrist because I was talking really fast I was taking on projects I was I was working all the time like busy I was making sure I had something to do and it just didn't start there was no end to it even being in hospital I was still taking on projects and doing things and just generally being impulsive it was so such a dangerous situation because I had delusions that people were watching what I was doing and that these people had who was buying on me and they was after me and that they was gonna hit me looking at it now I can tell it was a delusion but even in your experiencing it it's really hard to take step away pay step back and just really realize that it really realize that it's a delusion like I said I had one conversation with a psychiatrist and diagnosed me by her little type one and I started on lithium I was put on 2000 milligrams a day straight off it wasn't built at all so it wasn't a very good psychiatrist I believe he's actually lost his license now I might be wrong I don't think they know what I could be but when I was told to the psychiatrist I thought I was talking really fast I was cycling through everything like everything pops my brain I was saying it was just a roller coaster like this was everywhere so that's how I was diagnosed I was experiencing symptoms on a pretty intense level and I was diagnosed by her little type one but that's my diagnosis story if you have any questions about my pillow ask in the comments down below I'm always happy to answer thanks for watching this video and if you are new make sure you hit subscribe button like I said at the beginning if you have any video requests please leave them in the comments down below thanks for watching and I'll see you guys in my next video peace


Hi, I'm Rose with family recipes unfolded today We'll be joining my aunt Alex back home in Geneva, Illinois from where I am Berlin, Germany over zoom to make my grandma Muriel's famous Strawberry rhubarb pie recipe. Let's get started. But hi I'm Rose and I am on a zoom call for Family recipes unfolded with my lovely aunt Alex who is wearing a beautiful pinkly shirt and ready to make some pies I don't know what I'm doing and I'm excited to learn from you over zoom We never have too much pie You can never have too much pie Okay So quite a lot of flour here So about that much right? Yep, perfect. Okay Great. Okay. Now the easiest thing Rose is to do it in small pieces like either spoon it in or whatever But do it in smaller pieces to start with Here's my hand. Got the water and the egg in there. Now you need a tablespoon. I would say but you've got 15 milliliters One five of white vinegar white vinegar. Okay So what you want to do we're going to let it rest for a little bit while we make the filling Okay What you want to do is split it in two The dough Because we're only going to use half of it They're cute. So many strawberries are are we talking about cups of rhubarb and three cups of strawberries Which is 700 milliliters of each What you want to do is just you have you rinsed it already you rinse it off And so you have the icky part gone right and then we're going to chop it in about a half inch Side like the width of your finger Okay, so yeah, the leaves are poisonous. You don't want to leave them Yeah, you never want the leaf of a rhubarb plant. It's poisonous The easiest way to get your dough into the pie is you fold your dough in And shiny so now your last step is and you can't see it but All right, here's my pie again, right? It's so beautiful Now what you want to do is you just want to sprinkle the top with a little bit of sugar Oh, this is also satisfying We're going to be an hour An hour total so tell me again 200 for how long 15 to 20 minutes and then 175 for the other 35 to 40 Exactly and it might even take longer watch your crust Yay, but at the same time call me in an hour and a half because I want to see yours Okay, I will show you my pie. Thank you. This was so fun