Yeah, thanks a lot. Yeah, so yesterday we discussed exceptional collections, and I hope we have seen that having an exceptional collection is a very useful thing. It can be used for description of modular spaces of ships and for many other purposes. But on the other hand, there are many restrictive conditions for a variety to admit a full exceptional collection. So one of them is that the Hodge numbers of this variety should all be on the diagonal of the Hodge diamond. So of course, it is a natural question whether one can do something reasonable in case when there is no full exceptional collection. And in general, there is a certain generalization of this notion, which is quite useful. And we also will see some examples how it works. So let me start with the definition. So assume that T is a triangulated category, and assume we are given a subcategory in it. So let A be a full triangulated subcategory in a triangulated category T. And then it is called admissible if the embedding functor, which I denote by alpha, has an adjoint. It may have a left adjoint functor alpha star, which is a functor from T to A. Or a right adjoint functor alpha shriek from T to A. In fact, usually when a left adjoint functor exists, people say that the subcategory is left admissible. If a right adjoint functor exists, people say that the subcategory is right admissible. And if both adjoint functors exist, then the subcategory is called just admissible, maybe too site admissible, something like that. But in fact, if the ambient triangulated category is sufficiently nice, so I don't want to give a rigorous definition of that. But for instance, if it is the derived category of some smooth projective variety, then it is nice. Then existence of one adjoint functor implies existence of the other adjoint functor. So in some sense, in this situation, you don't need this distinction between left admissible, right admissible, and too site admissible subcategories. And in fact, in these lectures, we will never discuss triangulated categories that are not nice, so you can assume that these two properties are just equivalent in our cases. And in fact, maybe one more remark about this notion is that this existence of an adjoint functor, in fact, is an intrinsic property of the category A. So in fact, existence of adjoints, so to say, follows some intrinsic properties of this subcategory A. So in particular, if your category A is admissible in some nice triangulated category T, then if you can embed it into some other triangulated category T, then it also will be admissible there. This is also a useful observation. And you see this condition of existence of adjoint functors shows that an exceptional object gives you an example of such an admissible subcategory. So let me give this and maybe some other examples. When we say full subcategory, it means that the embedded functor full and faithful. In some sense, subcategory means faithful. And full subcategory means full and faithful. Yeah, so these two conditions can be phrased as alpha is full and faithful. OK, so examples. The first example is the one that we have seen earlier, that if E is an exceptional object, then the subcategory it generates, which is equivalent to derived category of vector spaces embedded by this functor phi sub E into T, is admissible. That's because we checked that there are adjoint functors in this case. And this is precisely the definition of admissibility. So another example is the following. That if A in T is admissible, then we can, analogously to the case of, again, of an exceptional object, we can define the orthogonal subcategories. So A perp is the set of all objects in T, such that homes from A to F is 0, which means that for any object of A, there are no homes to this object F. And you can also define the orthogonal on the other side in the same way by interchanging these two arguments. And then you can check that both these subcategories are also admissible. So again, in fact, in general, you should be a bit more accurate. So if the ambient category is not nice, then for these orthogonal, you can also prove only one side admissibility. So for this subcategory, you can prove left admissibility. And for this subcategory, you can prove right admissibility. But if T is nice, then they are just admissible. We will see a proof of this fact a bit later. Also, one more example is, again, very similar to what we had before. So if we have an outer equivalence of our triangulated category T, and if A is admissible, then phi of A is also admissible. And this is evident just because you can construct an adjoin factor just by conjugation of the adjoin factor for this embedding with this outer equivalence phi. And maybe this is not an example, but again, a remark that if T is nice and A in T is admissible, A is also nice. So for instance, if you have a subcategory which is left admissible in an admissible subcategory of a nice category, then it is admissible on both sides. So in particular, this shows that this class of triangulated categories, that admissible subcategories of derived categories of smooth and projective varieties, is a very good class of triangulated categories. So in some sense, this is what one should consider as non-commutative algebraic varieties. At least this is one of the approaches. OK, so this notion of an admissible subcategory is some analog of the notion of an exceptional object. And as well as, again, if you remember, for exceptional objects, we had an important operation which was called mutation. And of course, you can also define a mutation for an admissible subcategory in the very same way. So mutations. Basically, you can define the left mutation, functor through subcategory A, through an admissible subcategory A, as follows. If you want to apply it to an object F of your triangulated category T, then what you do, you apply alpha-alpha-Shrick to F. By a junction, there is a canonical morphism to F, and you take the cone of this morphism. This is called the left mutation. And analogously, you can define the right mutation as the cone of the natural morphism from F to alpha-alpha star of F. And it is convenient to shift by minus 1 here. So there are these two mutation operations, and they satisfy, more or less, the same properties that, as mutations, throw exceptional objects. So basically, each of these functor skills the corresponding subcategory. So if you apply left mutation through A to any object of A, you get 0 and the same for the right mutation. And on the other hand, if you restrict to these orthogonal subcategories, then you get an equivalence of categories. So L sub A, if you apply it to this orthogonal, then the image is in that orthogonal. And for R A, you get a functor in the opposite direction, and these two functors are mutually inverse equivalences between the corresponding orthogonal. In particular, these mutation functors give you an equivalence between the left and the right orthogonal of an admissible subcategory. Maybe one more remark is that, of course, if the admissible category A is generated by an exceptional object, then the corresponding mutation functors agree. And analogously, if your admissible subcategory is generated by several exceptional objects, the corresponding mutation functors can be written as compositions of mutation functors for these exceptional objects. L E 1 composed L E 2 L E k. And then analogously, the right mutation functor is the composition of the right mutations in the opposite order. If you have several objects, if you want to mutate something to the left of this exceptional collection, then you can first mutate through this object, then through that, and so on. So you can mutate through every object one by one. And analogously, for the right mutation, you should first mutate through E 1, then through E 2, and so on. So OK. And now we can define what is called SMR token of decomposition, which is an analog, I mean, an admissible subcategory as an analog of an exceptional object. And the SMR token of decomposition is an analog of a full exceptional collection. So it is a collection of subcategories, of full triangulated subcategories, A 1, A 2, and so on, AM in T, such that two properties are satisfied. So the first property is the semi-artigonality property. It says that there are no morphisms from A i to A j or i greater than j. And again, it means that if you pick up any object in this category and any object in this category, then there are no morphisms between these two objects. And since these categories are triangulated, with every object they contain all their shifts, it also follows there are no X spaces between objects of these categories. So this is just an analog of the semi-artigonality condition that we have seen before. And the second property can be written in two different ways. So maybe, first of all, let me give one version and then the other version. So the first version of the second property is that every subcategory A i is admissible. T is generated by all these subcategories, which means that T is the minimal triangulated subcategory in T that contains all these subcategories. And this version of this property is just an analog of what was the definition of a full exceptional collection. So the condition that every subcategory is admissible is an analog of the condition that every object was exceptional. And we have the same generation condition here. Or, ultimately, you can write a more precise version of this property. Sometimes it is more convenient to use it in this form that for every object of your ambient category T, there should be a chain of maps like this, f sub m goes to f sub m minus 1, f1, f0, and f0 equals f, such that if you take the cone of a map from fi to fi minus 1, then this cone is contained in the subcategory A i. So in fact, it is a good exercise to check that if one holds this chain of morphisms, let me call it star, for an object f is unique and functorial. Maybe I should say it's functorial, which means that if you have two different objects, say f and f prime, if you write two such chains with this property for f and for f prime, then any morphism from f to f prime induces, in a canonical way, a morphism between these chains. And in particular, from this, maybe I should end. In particular, if you are given an isomorphism between two objects f and f prime, then it induces isomorphisms between all these terms of the filtration, and in fact, also between the cones that appear here. So this diagram, including these cones, is functorial enough. This is very useful. I mean, in general, as we discussed probably in the first lecture, even the operation of taking a cone is not functorial in triangulated categories. But if you have the same orthogonality condition, then it is functorial. This is very useful. And the proof is very simple. So maybe I don't want to prove the, I mean, I want indeed to leave it as an exercise, but maybe let me show my conditions to and to prime a equivalent. I'm sorry, can you say it louder? In example one, do we have what? That the object is exceptional. Of course, if the category is equivalent to DFC, then of course it is. This is basically the definition of an exceptional object. In general, it maybe depends a bit in what sense it generates it. So for instance, there is a notion of strong generation when, I mean, yeah, so the question was whether it is true that if you have an object that generates an admissible subcategory, is it true that the object is exceptional? And the answer is that if also this condition holds, if the subcategory is equivalent to derived category of vector spaces, then of course the object is exceptional, just more or less by definition. But in general, I'm not sure that this is true. For instance, when we consider the notion of generation that is mentioned here, basically in which way we are allowed to generate objects, we have some initial set of objects, and then we can consider their shifts, their direct sums, and cones of morphisms between them. We can also allow one more operation. Namely, we can also allow to take direct summands. If we do it, then in fact, more or less any reasonable triangulated category can be generated by one object. This is a funny observation. But maybe one can also find. But I should actually think about this. Let us talk about this later. So let me explain why condition two prime and two are equivalent if condition one is satisfied. So first of all, let us deduce from two prime. Let us deduce two. So how do we check? No, maybe let me first do the other way. How do we check that if we have this condition, then why do we have such a chain of morphisms? In fact, we can construct it in a very explicit way. Assume that we have an object, F. And we want to construct a chain of morphisms like that. Then let us first set F0 to be F. And then let us define F1 to be the right mutation through A1 of F0 and so on. Let us define FI plus 1, FI to be the right mutation through AI of FI minus 1. So let us just apply a sequence of mutations through our subcategories to our object. Then note that whenever we have a right mutation, by the very definition, the right mutation is an object that fits into a distinguished triangle with this morphism, which stands on the left of this morphism. In particular, it comes with the canonical morphism to F. So in particular, it gives you this chain of morphisms. Fm goes to Fm minus 1, goes to F1, goes to F0, which is F. So by definition, we have this sequence of morphisms. Also, we have to check this condition, that the cones are in the corresponding categories. But again, by the very definition of a right mutation, if you take this, maybe let me denote this object by F prime, we have this distinguished triangle. Then the cone of this morphism is precisely this object. And so this object is contained in the subcategory A, which is AI in our case. So this second condition also holds. And the only thing left to check is that Fm equals 0. So we only have to check that the last object we have constructed is 0. But it is also quite easy because when we apply right mutation, the resulting object sits in the left orthogonal subcategory 2A. So the object F1 is contained in the orthogonal 2A1. The object F2 is contained in the intersection of the orthogonal 2A1 and the orthogonal 2A2, and so on. And the object Fm is contained in the intersection of orthogonals to all these subcategories. But the fact that these subcategories generate the whole category T means that this intersection of orthogonals is 0. Because otherwise, everything that we can generate from these subcategories will be itself orthogonal to this intersection of orthogonals. Yeah, so it will be orthogonal to all these objects. So by generation property, this is 0. And in particular, the object Fm is 0. So this is the proof in one direction. And the proof in the opposite direction is also quite simple. Yeah, basically here, what do we have to check? We have to check that the subcategories are admissible and that the category T is generated by these subcategories. But the generation property follows immediately from that triangle, yeah, just because we have these cones in these subcategories so we can start reconstructing first Fm is 0, then Fm minus 1 by this condition is contained in AM. Then Fm minus 2 is a cone of amorphous in between objects in AM minus 1 and AM and so on. So all together, it gives you an explicit reconstruction of an object F from objects in these categories as a sequence of cones. So the only thing that we have to check is that AI is admissible. But let me first show that A1 is admissible. Maybe it's more convenient to show. Let me show that AM is admissible. It is a bit more convenient. So to show that AM is admissible, we have to construct a functor which will be joined for its embedding. But let us just know that the functor that takes F to the object Fm minus 1 in this chain of morphisms is in fact is right joined to the embedding into T. So it is quite easy to check. And this shows that this subcategory is admissible. And therefore, its orthogonal subcategory is also admissible. And now it is clear that if you use this definition to prime, it follows that this category itself has a same orthogonal decomposition with components being all the other subcategories. And now if you apply the same argument to this category, you can deduce that AM minus 1 is admissible, and then you can go on. So in the end, you will prove that all components are admissible. So the two properties are equivalent. And in fact, you can also, of course, it is a good question how one can construct a semi orthogonal decomposition of a given category. And basically, the only, I mean, there is a very simple procedure which is, on the other hand, in some senses, the only one we know. So how one constructs SOD, semi orthogonal decomposition? So I will usually abbreviate a semi orthogonal decomposition to SOD. And to denote such a semi orthogonal decomposition, I will use this angle brackets notation. So when I will write it in this way, I will always assume that this is a semi orthogonal decomposition. This is just a notation. Yeah, and the easiest and, in some sense, the only known construction is the following. Just imagine we have a sequence of subcategories that enjoys almost the same properties with the exception of the last part. So let us forget this last part of the assumption. So let us assume that we have a collection of subcategories in t such that they are semi orthogonal and admissible for i greater than g. And assume that they are admissible. All these subcategories are admissible. Then what we can do, in fact, one can check that this collection of subcategories extends to a semi orthogonal decomposition. One can add one more component to this sequence of subcategories in such a way that it gives you a semi orthogonal decomposition. And there are, in fact, not only one way to do it, but, in fact, m plus 1 ways. Basically, you can add this new subcategory between any two adjacent subcategories in this sequence. Also, you can put it on the very left and on the very right. So basically, for instance, if you choose some integer i, then you can take the first subsequence here, the last subsequence here, and you can put in the middle the following subcategory. Basically, you should take the intersection of appropriate art diagonals. So this additional component should be artogonal. There should be no homes to these subcategories, which means that it should live in the intersection of these art diagonals. This is just the semi orthogonality condition for this sequence. And it should be also semi orthogonal to this sequence, but in the opposite direction. So you should take other art diagonals here. So you can take this intersection, and this is a semi orthogonal decomposition. And basically, the proof of this fact is analogous to the proof of this implication. Just if you want to decompose any object, then it is easiest to explain it in case when i equals m, which means that we put this additional component on the right-hand side of our collection, then the proof is just identical to this one. So we can construct a chain of morphisms in a very same way just by using these right mutation factors. And then the same argument shows that the last object lives in this intersection of art diagonals. And this is precisely what we are expected to put on the very right of this collection. And we just don't have this property. This intersection is not in general 0. But if we just add fm plus 1 equal to 0 here, then this will be precisely give you the sequence like in the property 2 prime for this collection of subcategories. So this is how you do in case when i equals m. And in the other case, basically, you can do more or less similar construction by using right mutations with respect to these subcategories and left mutations with respect to these subcategories. Basically, you can prove that this is a semi-artic node decomposition. But I'm afraid that I don't have time to explain this in more details. OK, so this is a notion of semi-artic node decomposition. And this is how one can construct it. And maybe we need some examples in geometrical situations. So first of all, of course, if we have a full exceptional collection, then if you define ai to be the subcategory generated by ai, then in a triangulated category t, then, of course, these subcategories form a semi-artic node decomposition. Just because the properties 1 and 2, the conditions 1 and 2, in this case, are satisfied by definition. And in this case, in fact, this exceptional collection is also usually denoted by Ivan by putting angle brackets around this sequence of exceptional objects. I mean, maybe it would be better to put more angle brackets here, but it doesn't make much sense. So we will skip amid this. OK, so this is one example. So whenever you have an exceptional collection, you have a full exceptional collection, you have a semi-artic node decomposition. But in fact, you can even do more if you have an exceptional collection, which is not full. Still, you can do the same thing, is a non-full exceptional collection. Then still, you can define ai to be the subcategory generated by ai. And then you can apply this construction. You can just add one more subcategory that is orthogonal to these exceptional objects in an appropriate way so you can put it in any place of this triangulated subcategory. So for instance, maybe let us put it on the left. Then t is just the intersection of the orthogonal. So this is just an application of this procedure. So you can just find some exceptional collection and then you can extend it to a semi-artic node decomposition. Maybe also I should explain that, of course, the notion of a semi-artic node decomposition is much more weak than the notion of an exceptional collection. There are many more varieties that have a semi-artic node decomposition. You don't need, for instance, to have hodge diamond supported on the diagonal. But still, there are some restrictions for an existence of a semi-artic node decomposition. So maybe this is like non-examples in some sense. So first of all, if you take x to be a smooth projective curve, then and assume that we have a semi-artic node decomposition for its derived category, say with two components, or maybe more components. And assume that the components are non-trivial, that both components are non-trivial. So maybe let me write both are non-zero. Then this is a result by Shinosuke Okawa, that x is p1, in fact. So if you have a curve of positive genus, then its derived category doesn't have any semi-artic node decomposition. And moreover, this is generated by one of the standard exceptional collections. So a1 is generated by some of k, and a2 is generated by some of k plus 1 by two line bundles. So the only possible semi-artic node decomposition for a curve is one of the standard exceptional collections for the projective line. There is nothing else. And there is another result by Bridgeland that says the following. So assume that x is a connected Calabio variety. Calabio means that its canonical class is trivial. Also, d of x has no non-trivial semi-artic node decomposition. And this, in fact, has extremely simple proof. Just if you have a semi-artic node decomposition, then let us assume that it has two components. It doesn't matter, in fact. Then we know that there are no homes from a2 to a1 just by definition. But now you can also use serduality. So by serduality, if you take, say, an object f1 in a1 and an object f2 in a2, in fact, it doesn't matter. Probably I should say it, that if you have any semi-artic node decomposition, then you can split the components into pieces, some piece in the beginning and some piece in the end. And if you consider the subcategories generated by these pieces, they will give you a semi-artic node decomposition with two pieces. Yes, so for instance, if you want to prove this property to prime for this decomposition, then just instead of considering the whole chain, you can consider only one intermediate object here.  So if you take a1 up to ai to be one category and ai plus 1 up to am to be the second category, then you can just write 0 goes to fi to f. Then this chain will give you a decomposition with respect to these subcategories. Basically, this will be contained in this subcategory, in this one, and the cone of that morphism of fi to f will be contained in this subcategory. So in some sense, if you want to check that there are no semi-artic node decomposition, it's enough to prove that there are no semi-artic node decomposition with two components. OK, just imagine that we have a semi-artic node decomposition with two components for a Calabiyao variety. Let us take an object in the first component and an object in the second component. Then if you compute Homes from f1 to f2, then you can use Serduality to compute this. By Serduality, this is the same as Homes from f2 to f1, tensored with the canonical class, and shifted by the dimension of x. And you should also dualize this vector space. This is duality. But the canonical class is the structure shift. This is a Calabiyao variety. So this is just f1, and then you also shift. So this shifted f1 object is definitely contained in a1, because a1 is a triangulated subcategory. It is closed under shifts. So this object is contained in a1, and this object is in a2. And now these two objects are in the order which ensures that there are no Homes by semi-artegonality. So this is 0. And this holds for any objects f1 and f2 in a1 and a2. And this simple computation means that, in this case, the composition is not only semi-artegonal, but it is completely orthogonal. So it implies that d of x, so probably we should write it like a product of this, or maybe a direct sum of these two categories, an orthogonal direct sum of these two subcategories. But it is easy to check that connected varieties don't have completely orthogonal decompositions. Just basically, you can imagine that you have a completely orthogonal decomposition for the derived category of a variety. Then, first of all, you can look at structure shifts of points. I mean, by definition, such direct sum decomposition should give you a direct sum decomposition of any object, such that one component in this subcategory and another component in this subcategory. Of course, if both components are non-zero, then such an object has a sufficiently big endomorphism algebra. So in particular, if your object has no non-trivial endomorphisms, then it cannot have a non-trivial direct sum decomposition. Then it should be contained either in one subcategory or in the other subcategory of this decomposition. So it follows that every structure shift of a point either sits in the first component or in the second component. So this is the first step. The second step is to see where the structure shift of the whole variety is contained. By this connectedness assumption, it also has no non-trivial endomorphisms. So it should also be contained in one of the subcategories. Assume, for instance, that it is in A1. But since there is a non-trivial morphism from the structure shift of the variety to the structure shift of any point, there is a non-trivial subjective morphism. It follows that the structure shift of any point should be contained in the same component of this direct sum decomposition. But then it easily follows that the other component is zero because objects in this other component are orthogonal to the structure shifts of all points and it is easily can be shown that this is impossible. So you see that there are quite many varieties that don't have a non-trivial semi-artogonal decomposition. Of course, in some sense it is a bit disappointing, but probably we should think that such varieties are like smallest blocks from which other triangulated categories can be constructed. So of course you should allow some non-trivial building blocks if you want to be able to construct something interesting. So, okay, maybe now let me give some positive results in the same direction. So let me talk about some fun varieties. So, of course, if you have a fun variety, then you always have a non-trivial semi-artogonal decomposition. Basically, as we have seen on Tuesday, if X is funner, if it's canonical, anti-canonical class example, then it follows that the structure shift of this funner variety is exceptional. And of course, whenever you have an exceptional object, you can use this procedure to construct a semi-artogonal decomposition. So you can always consider one component generated by this object, and the other component being the other part of the category. Of course, it doesn't give you much, but still it can be useful. And in fact, you can also try to find maybe some longer exceptional collection, not just consistent of one object, but maybe consistent of several objects, and then you can try to do the same. And let me explain what happens for funner three-folds. There is some interesting story related to this subject. And let me restrict to the case of Picard number one. Since, I mean, if you consider all funner three-folds, then there are too many of them. It's like 100 something. So it's hard to discuss 100 examples at the same time. But if you restrict to Picard number one case, which is in some sense, the most important part of the classification, then there are only 17 varieties like that. And as we discussed yesterday, you can divide them according to the value of the index, which is the maximal integer by which the canonical class is divisible. And the index can be equal to four, three, two, and one. So there are only four possible values for the index. And if you consider index four, then there is only one variety of this type. In fact, the general rule is that the smaller the index, the more complicated this part of the classification. So for the maximal possible index, there is only one funner variety, which is a projective space. For the next value of the index also, there is only one funner variety, which is a three-dimensional quadric. But in the last two cases, there are some interesting varieties. So basically, there are five varieties of index two. And there are 10 varieties of index one. And the natural parameter to classify varieties of index two is the degree of the generator of the canonical, of the PICAR group. So if you write PICAR as Z times H, so if you denote by H the generator of the PICAR group, then you can define the degree of this funner variety to be just H cubed, just the self-intersection of the generator of the PICAR group. Then in index two case, there are only five possibilities for this degree. So basically it can be equal to five, four, three, two, and one. And one can check that for each of these five possible values of the degree, there is one deformation family of funner varieties. So for instance, for D equal three, these are just cubic three-folds. For D equal four, it is just intersections of two quadrics. For D equal two, they are called quartic double solids. And one can also describe what happens for D equal five and D equal one. And I think that I will denote by a funner variety of index two and degree D, so let me call it YD. Okay, so there are five funner varieties of index two and there are 10 funner varieties of index one. And in this case, of course you can also consider the degree, but in some sense it is more natural to define the genus. I mean, in fact, the degree of such funner varieties is always even and it is more convenient to consider the genus which is defined by the formula that D is equal to two G minus two. In fact, this genus has a very nice geometrical meaning. If you take a hyperplane section of such a funner variety, it will be a K three surface by junction formula. And if you take another hyperplane section, you will get a curve which will be embedded by its canonical class. So it will be a canonical curve and G will be precisely the genus of this curve linear section of this funner variety. And for this genus, there are 10 possible values. So I will list them in a strange order. So first of all, first I will list possible even values of the genus. There is also G equal to two. And here let me list the other values, G equal nine, G equal seven, G equal five, and G equal three. So there are six possibilities for even genus and four possibilities for odd genus. It is funny that, of course it is clear that we cannot have G equal to one just because in this case D will be equal to zero which is impossible since we assume that H is ample. So it's self-intersection cannot be zero. And the funny thing is that the maximum possible index is 12, the maximum possible genus is 12, but genus 11 is a forbidden value. Okay, so there are several funner varieties like that and you can construct some exceptional, some semi-artogonal decompositions for them. So first of all, if you start with a funner three-fold of index two, then you can always, then besides OX, you can also consider OX of H. The line bundle corresponding to the generator of the PKAR group and they will always form an exceptional pair. So if you take YD, then OX, then O and O of H, they are exceptional objects and they are semi-artogonal. This is very easy to see, it follows from Kadaera vanishing in fact. And then you can extend this exceptional pair to a semi-artogonal decomposition. And let me call the additional subcategory by BYD. So we have a semi-artogonal decomposition that looks like this in this case. So we can split off two exceptional objects. And similarly in case of index one, so let me call the corresponding funner varieties by XG here. So we have YD here and XG here. In fact, in many cases, you can also construct an exceptional pair in the derived category of these varieties. So in fact, it was constructed by Mukai. So in fact, if G is greater or equal than six, then OG equals four, then there is an exceptional pair in this variety, in the derived category of this variety. But this time the second object is not a line bundle, but it is a certain vector bundle naturally associated to this picture. And so you can extend this to a semi-artogonal decomposition in the very same way. So you can define AXG to be the orthogonal subcategory and then you will have a subcategory and then you will have a semi-artogonal decomposition of this sort. And an interesting observation is that if you look at the two columns of this table, then the varieties that stand at the same line here have very similar non-trivial pieces of the semi-artogonal decompositions. In fact, what you can prove is the following. That if D is greater or equal than three, then there is a natural correspondence between the modular spaces of the corresponding finite varieties, YD and of X. So the corresponding genus is equal to two D plus two. X to D plus two. And this correspondence is in fact, so what does it mean? We have this modular space of varieties of index two and degree D and we have the modular space of varieties of genus two D plus two and index one and we have a certain correspondence here. So C and in fact, both projections are subjective. Both projections are objective, which means that to any variety of one type, there corresponds some variety of the other type. Objective. In fact, that the corresponding non-trivial pieces of derived categories are equivalent. YD equivalent to AX plus two. So I guess that my time is over and since we have strict time restrictions today, maybe let me stop here and then I will continue this story tomorrow.


Life is good all this month at Appalachian Wireless. Get the LG G5 for just one penny with signup or renewal of a two-year service agreement. That's almost $100 off the regular price. Better service, bigger savings. That's today's Appalachian Wireless. The Pike Fiscal Court is going on the offensive to help preserve its current form. At this time, I would like to make a motion that we challenge the legality on the vote of the Commissioner for America. At Tuesday's meeting, the court voted unanimously to challenge the validity of the vote to change the county's form of government to a commissioner-based body from its current magistrate-based government. The challenge will come by way of bringing in the question, the legality, of the petition to get the measure on the ballot. Magistrate Bobby Varney, who made the motion to challenge the results, said several of his constituents have expressed confusion with the ballot measure. He said his research indicated the court can pursue legal action in the matter. That was backed up by a statement by Assistant County Attorney John Doug Hayes, who said there are questions regarding the legality of the petition used to get the matter on the ballot. Hayes said the petition to get the measure on the ballot consisted of signatures from two separate petitions. So I think that's what Magistrate's party's motion is about, for me to challenge that legal issue or raise that legal issue with a circuit court judge who would rule. Yes, you can do that, but no, you couldn't do that. So, there it reminds me of the case. Varney said the intent is not to kill the commissioner effort, but to simply clarify whether the petition to get the effort on the ballot was legal. Also, he said to give the voters the chance to truly vote their conscience. Right at 30 percent of the people that voted in the election did not vote one way or the other on the ballot. A lot of people said, well, I voted for that because I don't want to change the commissioners. And I said, well, if you voted forward, you voted to change the commissioners. They didn't understand that. They were some older people told me that. So there was a lot of misconception there.  I'm okay with that, too. But I'm going to try everything I can to represent the people that come to me, you know. So that's the reason I've done that. The vote to change the county's fiscal court to commissioner based passed 57 percent to 43 percent with 20,153 votes being cast. Reporting in Pikeville for EKB News, I'm Chris Anderson.