 Thanks a lot. So yesterday, we discussed the definition and some general properties of derived categories. And from the definition, one may think that derived category of coherent shapes is a very complicated category, and it is very hard to work with it. But today, we will start showing that in some cases, it is a quite simple story. And sometimes, you can simplify the category considerably. And these simplifications lead to very nice and down-to-earth descriptions of modular spaces of some vector bundles or subvarieties in your variety of interest. So let me give the main definition for today's talk. This is the definition of an exceptional object. So let E be an object of some triangulated category. It is called exceptional object if it comes from this object to itself, and its shifts are as simple as possible. So if home from E to E is the base field, which for simplicity consider to be the field of complex numbers. And if you consider, say, x i from E to E, then this is zero for all non-zero i. So there is one endomorphism. And of course, you cannot have less than one endomorphism because there is always the identity map from E to E. So this first condition means that any map from E to E is a multiple of the identity, and there are no extensions. Equivalently, there are no homes from E to shifts of E. So this is the definition. Let me give you some examples. First of all, let us assume that T is the bounded derived category of coherent shifts on some smooth projective variety. So this is d of x, as we denoted. And let us first consider the case of the structure shift of this variety. So this is a coherent shift. So it is an object of x. And as we discussed when we consider a derived category of some abelian category, then we can think of objects of the abelian category as objects of the derived category given by complexes with only one non-zero term. So let us discuss the structure shift when it is exceptional. So in this case, it is very easy to see when this holds, just because if you compute homes from OX to OX, this is the same as just H0 of OX. So this is C if and only if X is connected. And if you consider X i from OX to OX, this is H i. So the condition for the structure shift to be exceptional is when X is connected, and the structure shift doesn't have higher cohomology. Or equivalently, you can say that X does not have holomorphic differential forms. So OX is exceptional. It's my fault. So it is exceptional if and only if H0 greater or equal than 0 is 0. So a very simple condition. And for instance, it always holds when X is a funny variety. OK. Note also that if this condition holds, then also you can take any line bundle on this variety X. So let me maybe call these conditions star. So this is condition star. So if star holds, any line bundle is also exceptional. And the reason is that just if you compute X i from L to L, this is the same as H i on X of L dual tensor L. And this is just the structure shift. So you obtain just the same list of conditions. OK. So what about non-locally free objects? And also it is easy to find examples. So for instance, let X be a surface. And let C on X be a minus 1 curve. So which means that it is a rational curve with self intersection minus 1. Then you can take E to be the structure shift of this curve. Then it is also exceptional. Again, if you want to prove this, you need to compute X i from OC to OC. And now you can use, first of all, you can compute local X spaces, local X shifts. And this is easy to see that this is just OC when E is 0. And this is just the power of the normal bundle when I is greater than 0. So it means that since X is a surface and C is a curve, so the normal bundle is a line bundle. And it means that the only non-trivial X shift is X1. And it is isomorphic just to the normal bundle, which is by definition, in this case, is OC of minus 1. And C is C1. It is a smooth rational curve, as we assumed. And this shift doesn't have a cohomology. And this shift has only H0 equal to the base field. So this computation easily shows that the structure shift of a minus 1 curve is an exceptional object. And by the way, this second observation can be also generalized. In fact, if E is exceptional and L is a line bundle, then L tensor E is exceptional. And indeed, again, if you compute Xt from such a tensor product to itself, then you can use the same property to get rid of L. In fact, the other way to say the same is just to note that the tensor product with a line bundle is an auto-equivalence of category. And since exceptionality is an intrinsic property of an object in a given triangulated category, clearly any auto-equivalence of a triangulated category takes exceptional objects to exceptional objects. So in particular, tensor product with a line bundle does. OK. So maybe let me give some general observations. So maybe the following observation sometimes is useful. If E is exceptional and I assume that its rank is non-zero. Of course, I should explain what does it mean, what the rank of an object of some derived category of coherence shifts means. But this is, in fact, a simple thing. In fact, you can just extend the notion of a rank to arbitrary complex just by considering the alternative sum of ranks of its terms. So rank of F dot, by definition, it is just the sum of ranks of the terms of this complex. Or equivalently, you can write it as the sum of ranks of the cohomology shifts of this complex. So this is a well-defined linear function on objects of the category. And clearly, it is additive with respect to distinguished triangles. So it gives you also a linear function on the growth and degroup of the category. And that was essentially used in lectures of Arendt. OK. And I assume that we have such a situation. Then it is easy to see that the structure shift is also exceptional. In fact, it is, in some sense, better to use this property in the opposite direction. Basically, if you want to try to find some exceptional objects on your variety, then if you want this object to be of non-zero rank, then an necessary property is that the structure shift itself is exceptional. If the structure shift is not exceptional, then there is no chance to find other exceptional objects of non-zero rank. And the proof of this property is also very simple. Just know that, so what we are interested in is, so know that x i from e to e, as before, we can write it as the cohomology of e dual tensor, derived tensor product with e. No, no, no. I mean, probably I should say that the dual means our home from e to o x. So it is also derived dual of e. And by assumption, our variety is smooth, so everything is fine here. So we have this expression for the x spaces that we have to discuss. And know that we have a natural map from o x to e dual tensor e, given basically just by, I mean, you can also think about this as about local our home from e to e. And so it has a natural global section corresponding to the identity and the morphism of e. And it gives you a map like this. So this is like identity of e morphism. And also there is trace morphism from here to o x. And the composition of these two maps is easy to see that this is just the multiplication by the rank of e. So as soon as the rank is non-zero, the composition of these two maps is an isomorphism. And this means that the structure shift splits as a direct cement from this tensor product. So this means that if the rank is non-zero, then e dual tensor e can be written as o x plus some other direct cement. And therefore, the cohomology of this tensor product is also a direct sum of the cohomology of the structure shift and something else, which means that the cohomology of the structure shift is less than the cohomology of this tensor product. So if e is exceptional, then the cohomology of this tensor product is always the same as these x spaces. If e is exceptional, then these x spaces are as small as possible, which also means that the cohomology of o x is as small as possible, which means that it is also exceptional object by this property that we discussed. So once again, if your variety does not satisfy these properties, then there is no chance to find an exceptional object of non-zero rank. Of course, still it is possible that, for instance, you can take a surface which does not satisfy these properties, but still it may have a minus 1 curve. For instance, you can always blow up a point on such a surface and then you will get a minus 1 curve. And so you will get an exceptional object of zero rank. So of course, you can always have objects of zero rank, but they are also quite rare. You need some restriction of this sort. But for objects of non-zero rank, the restriction is also quite strict. So it's not always possible to find exceptional objects. OK, yeah, maybe a couple more things to say is that maybe let us consider the simplest possible situation. So let us take x to be the spectrum of the field c. So this is just a point. Then other exceptional objects in the corresponding category. And of course, there are. So the corresponding category, so this is d of point. And I will also write it as d of c. This is just the category of vector spaces. And of course, a one-dimensional vector space in this category is exceptional. And also any shift of it is also exceptional. Yeah, probably I should say here that also you can shift any exceptional object, and still it will be exceptional. Again, because the shift is an auto-equivalence of a category. And in fact, one can see that any exceptional object in this category is of this form. So any exceptional object is a shift of a one-dimensional vector space. And this is, in fact, very simple just because the category is very simple. The corresponding abelian category is semi-simple. And because of that, it is easy to see that any object of the derived category is just a direct sum of shifts of some vector spaces. So in particular, if the object doesn't have this simple form, then it can be represented as a direct sum of two non-zero objects. And this means that it has more endomorphisms that is required by these definitions. Because if it is a direct sum of two things, you can take identity of one cement or identity of the other cement. And this will be two linearly independent endomorphisms. So this is very simple in this case. And also, one can consider the situation when x is a curve, a projective curve. In this case, also, it is easier to classify all exceptional objects. So what one can check is that if e is exceptional in d of x, then there are two immediate implications. So first of all, x should be isomorphic to p1. So you cannot have exceptional objects on curves of positive genus. And of course, from this property, this property definitely doesn't hold for curves of positive genus, which means that there are no exceptional objects of non-zero rank in the derived categories. But on the other hand, if you have an object of, still, when you consider a smooth curve, the abelian category of coherent shifts is also quite simple. It has global dimension 1. In particular, it follows that any object here is a direct sum of shifts of some coherent shifts. And again, definitely any exceptional object should be decomposable. So the only chance for an object to be exceptional if it is a shift of a pure shift. And if it has zero rank, then it means that this is just a torsion shift. But if you have a torsion shift on a curve, then it always have non-trivial x1 to itself. So there are no exceptional objects of this type for curves of, for curves. So it means that if there is an exceptional object, then the curve is p1. And moreover, one can check that the object itself should be isomorphic just to some line bundle with some shift. So basically, besides the structure shift, there is nothing if you forget about these operations. So any exceptional object on p1 is just a line bundle with a shift. And this second part is also easy to prove. Again, just because by the same arguments as before, e should be a pure shift up to some shift. And it should be decomposable as a direct sum. But on the other hand, on p1, any coherent shift can be represented as a direct sum of a torsion shift and of a sum of line bundles. And as we have discussed, torsion shifts are never exceptional. So the only chance is when it is a line bundle. And in this case, definitely it is exceptional. So for curves, it is very easy to classify. In fact, for surfaces, the situation is already much more complicated. In fact, there is some classification of exceptional objects for the Lepetsis offices. This is a result by Kuleshov and Darlov. But for instance, if you consider a surface which is not a Lepetsis office, then it is quite hard to classify exceptional objects. But classification maybe is not the most interesting part of the story. It is more interesting how we can use exceptional objects when we have some of them. So in fact, if you have an exceptional object, you can in some sense split your category into parts, with one part being very simple and another part being a bit more simple than the whole category. And so let E be an exceptional object in some triangulated category T. Then we can define two subcategories. Maybe before doing that, let me prove the following claim. So let me come one step back for a moment. Let T be a triangulated category, and E be an arbitrary object for a moment. Then we can define a functor phi sub E from derived category of vector spaces to T, which will just take any object of this category, which can be thought of as a graded vector space, and just take it to this graded vector space with E. In fact, this is a very simple operation. Basically, if you have a graded vector space, it is just a direct sum of shifts of C. And this tensor product is the corresponding direct sum of shifts of E. So for instance, if V has dimension 3 in degree 0 and dimension 2 in degree 5, then this tensor product is just the direct sum of 3 copies of E plus the direct sum of, I forget how many copies of E shifted by 5. So it's a very simple operation. And this is a triangulated functor. And one can prove that exceptionality of E is equivalent to fully faithfulness of this functor. And so in other words, to give an exceptional object is more or less equivalent to specifying a subcategory of T equivalent to derived category of vector spaces. So let me prove this lemma. And for a proof, it is useful to note the following general fact. Let me leave it as an exercise. So let phi be a functor between two triangulated categories and assume that it has an adjoin functor. Say phi star or phi shriek be a left or right adjoint. Then by definition of an adjoint functor, we always have a morphism from the composition phi upper star phi to the identity functor of T1 and from identity of T1 to phi upper shriek phi. And one can check that phi. So this is the exercise to check that phi is fully faithful if and only if one of these composition is an isomorphism. In the situation of triangulated categories, it is usually quite easy to check whether a functor is fully faithful as soon as you have an adjoint functor. And for a functor of this type, it is very easy to write down the adjoint functor. So let me write down what is the right adjoint functor in this situation. So to understand this, we should compute homes from this stuff to some object f of our triangulated category of T. So by definition, this is just homes from v tensor e to f. And then we can use the adjunction between tensor product and home. And this will be just homes from v to homes r home from e to f. And if you look at this formula, then you'll just see that this is basically shows that this functor is the right adjoint. This means that the right adjoint functor can be defined by this formula. This is just r home from e. Let me just put dash here. So the right adjoint functor for this particular functor is just r home from the object e. And maybe let me leave it as an exercise to write down the formula for the left adjoint functor. OK, now let us use this formula for the adjoint functor. By this exercise, if we want to check that whether this functor is fully faithful or not, we should consider the composition. So we should compute the composition of this adjoint functor with our original functor pi e. And since the formulas are completely explicit, this is just r home from e to v tensor e. And since this tensor product is basically a direct sum with shifts, you can easily check that you can take this v tensor outside. So this is just r home from e to e tensor v. And it means that if you want this functor as a functor of v to be isomorphic to identity, the only chance is when this graded vector space is isomorphic to c. So this is equal to v if and only if r home from e to e is just c, which means that e is exceptional, which proves this lemma. OK, so as I said before, if you have an exceptional object, then it gives you a very simple subcategory in your triangulated category t. This is just the image of this functor pi sub e, which is equivalent to derived category of vector spaces, a very simple subcategory. But a good point is that you can also, in some sense, split this subcategory almost as a direct cement. So let me define orthogonal subcategories to this object as follows. So e perp, by definition, will be the full triangulated subcategory of t, consisting of all objects f and t such that r home from e to f is 0. And similarly, one can define another orthogonal subcategory by interchanging the positions of e and f in the previous definition. So I guess this subcategory is called left orthogonal of e, and this subcategory is called right orthogonal of e in the triangulated category t. And now what you can check is that for any object f in t, there are, so assuming that e is exceptional, then for any object of this triangulated category t, one can somehow split this f into some multiple of e. And an object in one of these art diagonals doesn't matter which. I mean, either in the first or in the second, which can be written as the following statement. So there is a distinguished triangle of the following form. So first of all, we have phi sub e of some vector space, which depends on this object f. So I will put it as an index here. There is a map to f. And then there is f prime. So probably I should put one more from here. But the condition is that this object is in the left orthogonal to e. And analogously, there is another distinguished triangle which is similar, but still you have some v prime sub f object here. And here you have some f double prime and f double prime shifted by 1. And this time with f double prime in the other orthogonal. So in some sense, you can represent any object as a gluing of an object in the orthogonal and a vector space. And you achieve the gluing by using some morphism between this object and the image of this vector space under this function phi sub e. And analogously here, you can glue it from some vector space and an object in the other orthogonal subcategory. And in fact, this is very easy to see why this is true. Just you can define the corresponding vector space by using one of the joint functions for this fully faithful embedding phi e. So maybe let me also try to the SLMA. And the proof is just straightforward. So this is the first part of the SLMA. And this is the second part of the SLMA. So for the first part, just define v sub f to be phi of f. Then we have this canonical morphism given by the co-unit of the adjunction. We have this canonical co-unit morphism. And then we can define f prime to be the colon of this morphism, which means that we can use one of the axioms of a triangulated category and extend this to a distinguished triangle like that. In fact, usually I will skip the last term, because you can easily reconstruct it from the first term. It is always a shift of the first term, so I will usually omit it. So we can extend this co-unit of the adjunction to a morphism like that. And so the only thing to check is that this object, f prime, belongs to this orthogonal subcategory. But if you want to check this, then this orthogonal subcategory, by the way, can be just rewritten as the kernel of the adjoint functor. So it is basically immediately follows from this definition. And another orthogonal is the kernel of the other adjoint functor. So if you want to check that f prime belongs to this orthogonal subcategory, then the only thing you need to check is that if you apply phi e upper shrink to this object, you will get 0. But to see this, let us just apply this functor to the whole triangle. Since the functor is triangulated, it takes a distinguished triangle to a distinguished triangle. And we will get a distinguished triangle of this form of f and then this term. And now note that since e is exceptional, the functor phi e is fully faithful. And as we discussed, in this case, the composition of this functor with its joint is isomorphic to identity. So this composition is isomorphic to identity, which means that this triple composition is isomorphic just to phi e upper shrink. And so it means that this morphism, which is induced by the co-unit of the adjunction, it means that this is an isomorphism. But as we discussed yesterday, if you have a distinguished triangle in which one morphism is an isomorphism, then it follows that the other term is 0. So the proof, as you see, is very straightforward. And you can prove the second statement in the same way by using left adjoint functor instead of the right adjoint functor. So you see that there is this splitting. And in fact, this is the simplest form of what we will discuss later. And I mean a general notion, which is called a semi-artogonal decomposition. This is just the simplest form of a semi-artogonal decomposition you can use. And of course, you can also iterate this simplification process. Now, in some sense, you started from a triangulated category t. If you manage to find an exceptional object in it, then the category somehow can be split into the subcategory into the image of this functor phi e, which I will denote just by using angle brackets. This is subcategory equivalent to d of c. And say one of the artogonal subcategories. Yeah, so now you decompose the category into a very simple piece and another piece. Now you can try to continue this process, this simplification procedure with this subcategory. To do this, you need to find an exceptional object here. So maybe let me call this object we found first, e1. Now you can try to find another exceptional object, say, in this subcategory. So maybe let me change it a bit. So let me put this simple subcategory on the left and consider another artogonal on the right here. Now I assume that we have found another exceptional object in this category. Then we can do the same construction. And this category will be decomposed into this subcategory generated by e2, which is, again, will be very simple. And to the artogonal in this subcategory to e2. And then you can go on. If you manage to find a sequence of exceptional objects, the first object in this category, the next object in this category, the next object in this category, then you can do several steps of such simplification. And in fact, in this way, you arrive to the following very natural notion, the notion of an exceptional collection. So an exceptional collection in t, in a triangulated category t, is a collection of objects. e1, em, such that, first of all, every object in this collection is exceptional. And second, every next object is contained in the artogonal to the previously constructed objects, which basically can be rewritten as saying that x dot from ei to ej equal to 0 for all i greater than j. So which means that if we write the collection in this order, then we don't have any x spaces from right to left in this exceptional collection. So these two properties ensure that you can apply this simplification procedure and split off m copies of the derived category of vector spaces. And of course, a priori, there may be something left. But let me say that an exceptional collection is full if the extra piece, which appears after splitting off this m copies of a very simple category, is 0. And this subcategory is nothing as just the intersection of these artogonals. So if the artogonal to e1 intersects with the artogonal to e2, artogonal to em, this is 0. And if you have a full exceptional collection, then this procedure shows you that you can represent any object in your triangulated category just by specifying a bunch of vector spaces and some gluing data of this type. Maybe let me give some examples of exceptional collections. Yeah, maybe the first example is the following one. So as we already discussed, in some sense, the simplest possible exceptional object in the derived category of coherent shifts is the structure shift of the variety in case the variety is connected and doesn't have holomorphic forms. So assume that we have such a variety. So assume that the property that was called star folds for x. And let us take t to be equal to d of x. Then we already know that the structure shift is an exceptional object. And moreover, any line bundle on this variety is an exceptional object. So you can try to ask whether some sequence of line bundles is an exceptional collection. And in some sense, it is even more natural to consider not just an arbitrary sequence of line bundles, but let us fix one line bundle and consider the sequence of its powers. So assume let us consider the following collection of objects in d of x. So a good question is when this collection is an exceptional collection. And of course, it is quite easy to say what are the necessary conditions. So since we already know that every object is exceptional, so the first part of the property holds. So we only have to check the second part. So basically, we have to compute x spaces from one power of this line bundle to another power. And of course, this reduces to the computation of the cohomology of some negative powers of the same line bundle. So the condition is that if you take some negative power of this line bundle, then this is 0 for i between 1 and m minus 1. So if all cohomologies of these negative powers of this line bundle vanish, then this is an exceptional collection. And of course, it is very easy to find a collection of this form when x is a project of space. If x is a project of space, we can take L just to be the line bundle of 1 and take m equal to n plus 1. And then if you take O x, let me write just O, O 1, O 2, and so on, up to O of n, then this is an exceptional collection in d of e. And a very good point about this exceptional collection is that it is full. This is a theorem which was proved 37 years ago, I think. So it's 79. And it was proved by Sacha Williamson, who is here. And this was like a very important observation after which this investigation of derived categories started. The theorem says that this is a full exceptional collection. Since, in fact, this is a very important result, because of that, let me give a proof. Maybe even not just one proof, but several proofs of this theorem. So there will be several proofs. So proof one. In fact, most part of, I mean, usually if you want to prove that some exceptional collection is full, this is usually a quite hard statement. And there are more or less two approaches to prove such a result. The first approach that was a original approach that Sacha used is to construct what is called the resolution of the diagonal. And this will be the last proof that I will discuss. But I mean, the main problem about this type of proof is that it is very rarely that you can do this. Basically, you can do this for projective spaces and for grass maniants only. I mean, for any other variety, if you want to construct a resolution of the diagonal, then probably it is easier to do it in the opposite direction. It is easier to first prove that your exceptional collection is full and then use some general machinery to construct a resolution of the diagonal. What is good about projective space is that you can construct such a resolution more or less directly. But all the other proofs basically use the same idea. So you start with your exceptional collection that you want to prove to be full. Then at the first part of the proof, you should construct more objects generated by this exceptional collection. So in fact, whenever you have some number of objects in a triangulated category, you can always consider the minimal triangulated subcategory of your category that contains all these objects. So to say, the subcategory generated by this set of objects. And the first step is usually to construct some interesting new objects in the category generated by this collection. And then when you manage to construct sufficiently many objects in this category, then after that, usually you use some kind of spectral sequence to check that there is nothing in the orthogonal to the subcategory, which is basically the definition of the fullness. Yeah, by the way, here we considered these right orthogonal subcategories. But you could also equivalently consider left orthogonal subcategories. Or you can even consider left orthogonal to some number of these objects and right orthogonal to the other objects. I mean, there are many ways to express this fullness condition. And again, one more equivalent formulation is that the subcategory of T generated by this collection of objects is itself T, which means that T is the minimal triangulated subcategory of T containing all these objects. OK, so let me give a couple of proofs following this approach. So first of all, in the first proof, as new objects contained in the subcategory generated by this collection, let us show that any line bundle is in the subcategory generated by this collection. And for this, we can just use what is called Cauchy complex for this projective space. So basically, this is the following complex. So we can take O. There is a canonical map from O to the direct sum of n plus 1 copies of O of 1. So basically, homes from O to O of 1 correspond to linear functions on the vector space corresponding to your projective space. So if the projective space is n dimensional, then the corresponding vector space is n plus 1 dimensional. So for instance, we can just choose the basis in the space of linear functions and consider the map given by these bases. And then we can consider which powers of this vector bundle. And this is a standard thing in algebraic geometry that we have a complex consistent of these which powers with the maps used by the same map as we had before. So in this case, if you take lambda 2 of this direct sum of O of 1, it will be a direct sum of O of 2. And the number of some n's will be the binomial coefficient n plus 1 choose 2. And then we will have O of 3 with the number of copies equal to n plus 1 choose 3 and so on. And in the end, we will have O of n. Again, here we should put n plus 1 choose n, which is n plus 1. And the last term will be O of n plus 1. So there is this very nice exact sequence, in fact. In this case, it is exact just because if you consider the simultaneous 0 locus of all these n plus 1 linear functions on the projective space, it will be empty. This means that the casual complex is exact. It is a regular sequence with empty 0 locus. And so if you look at this complex, you can note that the first part of this complex is contained in the subcategory generated by all these objects, just because every term is just some direct sum of objects of this type. And so it shows that the last object, which is quasi-isomorphic to the other part of the complex, belongs to this subcategory. So this means that O of n plus 1 is in the subcategory generated by these objects. So let me use this angle brackets notation to denote the subcategory generated by this collection. OK, so we did a bit of what we was going to check. So we proved that one more line bundle is already contained in the same subcategory. But now, for instance, we can start twisting this casual complex by other line bundles and apply the same argument. For instance, if we twist it by 1, then this part of the complex will consist of O of 1, O of 2, and so on up to O of n plus 1. And all these objects are already contained in this subcategory. And the last term will be O of n plus 2. And this complex will prove that O of n plus 2 is also in this subcategory. And you can go on and prove that all positive line bundles are in this subcategory. Now, if you want also to prove that the negative line bundles are in this subcategory, you should just consider negative twists of the same complex. If you twist it by O of minus 1, for instance, then you can see that in this part of the complex, if you twist by O of minus 1, you have O of 1 and so on up to O of n. And the first term will be O of minus 1. So it will follow that O of minus 1 is contained in this subcategory. And if you will twist further, if you consider further negative twists, then you will see that, in fact, all line bundles are contained in this subcategory. And thus, you have a very big collection of objects already in your category. And now, it's enough to consider the following spectral sequence, if you consider. Now, assume that some object F is contained in the orthogonal to this subcategory. Then you can compute Homes from O of k to F, maybe from O minus k. It will be a bit more convenient. Let me write x. Of course, this is the same as the cohomology of F shifted by k. And now, you can write a spectral sequence in which the terms will be the cohomology on Pn of the cohomology shifts of this twisted complex. And that will converge to next maybe to Hp plus q of F of k to what we want. And now, if you choose k sufficiently positive, then since this is the same as Hp of F tensor with O of k. So since the complex has only a finite number of cohomology shifts, because we are in the bounded category, for sufficiently positive twist, all these shifts will not have higher cohomologies. So the terms of the spectral sequence will vanish for q greater than 0. This will mean that the spectral sequence will be concentrated just in one line. And then it follows that it degenerates in this second term. And so if this stuff is 0, we assume that F in the orthogonal, it means that there are no x spaces from any object of this subcategory to F. And since we already checked that every line bundle there, this is also 0. So we know that these cohomologies are 0. So the spectral sequence should converge to 0. But on the other hand, as we checked, it converges in the second term. And it means that all these cohomology shifts have no cohomology after arbitrary positive twist, which of course means that all these cohomology shifts are 0. Because if you have a non-zero coherent shift, then if you twist it sufficiently by a sufficiently positive line bundle, it will have non-zero global sections. So altogether, this means that this object F is 0. So all its cohomology shifts are 0. Hence, it is 0 as well. So yeah, I see that I don't have time to give the other proof. So maybe let me just say what can you do in alternatively. Alternatively, you can prove that the structure shift of any point of your projective space is generated by these objects. And let me leave it as an exercise for any point P. And then you can check that again by computing, say, X spaces from an arbitrary object of the orthogonal category to the structure shift of this point, you can also finish the proof in a similar way. And the third proof that uses a resolution of the diagonal probably we will discuss tomorrow, or maybe during exercises if someone wants to. And yeah, let me stop here.