 Thanks a lot. So yesterday, we discussed exceptional objects and exceptional collections. And we discussed an example of a full exceptional collection on the projective space. So if we take PN, maybe sometimes it is more convenient to denote it as a projectivization of some vector space. Then as we discussed, the collection of line bundles starting from O and going to O of N is a full exceptional collection. And we discussed two proofs of that statement. And maybe today I will give one more proof, which, in fact, is the original proof of this fact. But before doing that, let me mention some, maybe, general properties of full exceptional collections. So first of all, if in the same way as with exceptional objects, as soon as you have a full exceptional collection in some triangulated category. So if a full exceptional collection is given, I assume that we also have some auto-equivalence of this category. Then, of course, if you apply this auto-equivalence to every object of this full exceptional collection, this also will be a full exceptional collection. Of course, instead of considering an auto-equivalence, you could take an equivalence between two different categories. And then, of course, the same result holds. And this is, of course, more or less evident because it is easy to check that if you take the common orthogonal to all these objects, then this auto-equivalence takes it to the common orthogonal to their images. So it's more or less immediately. And in particular, if this triangulated category T is derived category of some algebraic variety, then, of course, one can take a twist by a line bundle to be such an auto-equivalence. So if you tensor every exceptional object by some fixed line bundle, then it will be a full exceptional collection. So in other words, if you have a full exceptional collection, you can twist it as you want. For instance, one of the possibilities to do this is to twist all these line bundles by o of minus n. Then you will get another exceptional collection, which is almost as good as the previous one. And of course, you can also consider other twists. But this is not all. So besides these trivial operations like these actions of auto-equivalences, there are many more exceptional collections than only those that can be obtained from the original by acting with auto-equivalences. Another possibility that you can use is the certain action of the brave group given by operations called mutations. So let me tell you what it is. So in fact, so if you have some exceptional object, then you can define the following operation in your triangulated category, so in T. Then you can define two operations. So the first operation is called the left mutation through E, and it is defined as follows. If you want to apply it to an object F of this triangulated category T, then what you do? So you consider the composition of the functor phi sub E with its right adjoint functor applied to F, then there will be a canonical morphism to F. We already used this morphism. And we can consider the cone of this morphism and define the left mutation of E through F as the cone of that morphism. So in other words, this is precisely the object F prime that we used yesterday. And a useful observation is that although, in general, taking a cone is not a factorial operation, but in this particular case it is. And in fact, this is just because this cone lives in the orthogonal subcategory to E, and you can check that fact it is well-defined up to a canonical isomorphism in this case. And of course, there is a similar version which is called the right mutation operation. In which case, instead of taking a morphism from this composition to F, you take a morphism. You use another adjoint functor. Again, there is a canonical unit of adjunction here, and you can take the cone. But it is more convenient instead of taking the cone, which is like extension of this morphism to a distinguished triangle in the right direction. It is more natural to extend it into the left direction, which is equivalent just to taking the cone and then shifting by minus 1. It's more convenient to use this definition here. So these two alterations are called left and right mutations. And they are well-defined. And the factors of this triangulated category T, and they have a number of nice properties. So first of all, maybe let me leave these properties as an exercise. So yeah, first of all, if you apply left mutation to any object of your triangulated category T, then you will get an object which necessarily lives in the orthogonal, in e-perp. And if you apply right mutation, then the image is in the other orthogonal subcategory. In fact, these two properties were more or less checked yesterday. In fact, we checked the first one, and the second one can be checked in a similar way. Second, if you apply the functor to the subcategory generated by e, it is 0, and the same for the right mutation. And now, if you restrict left mutation to the other orthogonal, then it will be an equivalence between this orthogonal and that orthogonal. And the right mutation functor is an equivalence in the opposite direction. These are mutually inverse equivalences. In particular, if you consider left or right orthogonal to an exceptional object as a subcategory of T, these two subcategories are equivalent. And equivalence can be realized by these mutation functors. OK. Now, you can use these mutation functors to construct new exceptional collections. So imagine that you have an exceptional collection consisting of some objects, EI, EI plus 1, so 1, EM. Then what you can do, you can apply the left mutation functor for this exceptional object and apply it to the object which is next in this sequence. So you can pass from this sequence to the sequence E1, E2, and so on, L, EI of EI plus 1, I, and so on, EM. And it is easier to see that this is still an exceptional collection. So maybe let me prove it. So what one should check to verify that this is an exceptional collection? So by definition, if you want to prove that something is an exceptional collection, you have to check that all the objects are exceptional and that they are semi-artogonal. Yeah, and in fact, if the original collection is full, then this new collection is also full. So how to check that the objects here, they should be semi-artogonal in one direction only. So the order is very important. So from right to left, there should be no homes or x from right to left. Like in this example, you see? Like in this example, there are plenty of maps from left to right. But there are no maps or extensions from right to left. The order is very important. OK, so for this collection, we have to check that all the objects are exceptional and to check this semi-artogonal condition. So to check that all of them exceptional, in fact, most of them are already present in the previous collection. So we already know that they are exceptional. And the only new one is this object, which is obtained by mutation. But to check that this object is exceptional, you can use this property that the left mutation factor gives you an equivalence between the right and the left-artogonal of your object. So for instance, this object was contained in the artogonal to EI, just because the first collection was exceptional. And because of that, our mutation factor is fully faithful when applied to this object. And this implies that this object is also exceptional. So it follows easily from this exercise. So exceptionality is very simple. Now you have to check semi-artogonality. And again, most of the objects in this new collection are just old objects from the previous collection. So they are automatically semi-artogonal. So you only have to check semi-artogonality between this new object and the old objects. So for instance, why there are no homes from this object to the objects that stand before it? In fact, this is very easy, just because by definition of this left mutation factor, this object lives in the subcategory generated by EI and EI plus 1. Just because when you write this defining triangle for the mutation, its first vertex is in the subcategory generated by E, which is EI in our case. And the second vertex is the object to which we apply our mutation factor. So it is in the subcategory generated by EI plus 1. So the cone is in the subcategory generated by these two. But from these two objects, there are no homes or x to previous objects, because the collection was exceptional. And by the same reason, there are no x from the objects starting from E plus 2. So from these objects, there are also no homes or x. So this is just because this object lives in this subcategory. And finally, you have to check that this pair is semi-arthogonal. But this also follows from this exercise. Just the image of the left, maybe even from this part. The image of the left mutation factor is contained in the left orthogonal to the corresponding object. I should say right orthogonal. So this is right orthogonal. And this is precisely the semi-arthogonality statement that you need. So this thing is more or less evident. And finally, you have to check that this exceptional collection is full if your original collection is full. So you have to check that it generates the whole category. But of course, since we know the fullness of this collection, it's enough to check that any object of this first collection is contained in the subcategory generated by this second collection. And almost all objects are already there by definition. And the only object you have to check that is contained is this object EI plus 1, the only object from the first collection that is not in the second. But again, if you look at the definition of the mutation, it is a distinguished triangle which looks like that. So some graded vector space tensor EI goes to EI plus 1 and then goes to this mutation. And since this object is contained in our category and object EI is also contained in our subcategory, this mutation triangle implies that the object EI plus 1 is also contained in this subcategory generated by this new exceptional collection. And so it is also full. So you see, when you apply mutations, you can construct new exceptional collections. And maybe a good exercise, yeah, yeah, of course, of course. So yeah. And yeah, of course, you can check that left mutation and right mutation are mutually inverse because of that operation. And also it is easy to see that they satisfy the braid group relation. So if you consider two pairs of objects in your exceptional collection that are far from each other, then mutating one pair and mutating the other pair commute just because these two operations are performed in different subcategories. And also if you take three objects, then you can like mutate first E3 to the left, then E2 to the left of E1. And you can do it in two different ways. And you can check that the result is the same. So maybe let me just, yeah, so let me write it also as an exercise. Mutations of exceptional collections satisfy braid group relations. A good exercise for an explicit computation is to take, say, P2 and the standard exceptional collection O1 and O2 and then make the following list of mutations. So let us first mutate O1 to the left of O. Then you will get this exceptional collection O and O2. Yeah, and then let us mutate O2 to the left of this. But then it will be exceptional collection which will look like this and O. And an exercise is just to check what are the objects that you will get in this way. So what is this exceptional collection explicitly? In fact, it is another very useful exceptional collection for a projective plane. And in fact, we will use it quite soon. But I think it is a good exercise to do all these mutations by hand. And it is not hard at all. Yeah, OK, so in general, you see that there is this huge group, braid group on N plus 1. I forgot the name, strings, yeah? Yeah, yeah. So the braid group on N plus 1 strings acts on the set of all exceptional collections in the n-dimensional projective space. And it is a good but open question is whether this action is transitive. So this is open question. Is the action of the braid group? No, it is not free. I will tell you. So it is not free, but people expect that it is transitive. Is the action of the braid group on the set of all exceptional collections, say D of PN, transitive? In fact, here, the only known thing is that the answer is positive for N at most 2. N less equal than 2, but not in general. In fact, a twist can easily be realized also by a braid group action. So if you start with this exceptional collection and just mutate the rightmost object to the leftmost position, then it is easier to check that up to a shift, you will get O of minus 1. So in some sense, the twist of the whole collection by O of minus 1 can be also realized in this way. But maybe I should say up to shifts here. OK, about action being free or not free, let me think more. I don't remember the answer. So probably for P2, it is free. In case of P2, it is free, but probably for bigger N is probably not known also. OK, so you see that whenever you have an exceptional collection, in fact, you have a huge number of these exceptional collections. And this, in some sense, is the reason why it is usually not very easy to construct an exceptional collection. Because if you want to construct something which is unique, then usually this is not so hard to do. But when you have plenty of possibilities, then you usually cannot do it at least easily. If there are too many mushrooms in a forest, then it is very hard to pick even one of them. This is true. If there is a unique mushroom in a forest, it is much easier to find it. Because you can use this uniqueness to trace it. Maybe it sounds a bit weird from a common sense point of view, but this is true mathematically. OK, so these are some general facts about exceptional collections. And now let me tell you another proof of the fullness of the standard exceptional collection of the Billionson's collection on the projective space by using what is called the resolution of the diagonal. So in fact, it is more convenient to start with this exceptional collection because of the following observation. So let us do the following construction. So let us consider, instead of just one projective space pn, let us consider pn cross pn, just the square of this projective space. And I will denote by p1 and p2 the two projections. And let us consider the full-end morphism of vector bundles on the product. So as usual, I will denote by v the vector space underlying this projective space. This is a vector space of dimension n plus 1. And then you can consider the pullback of o of minus 1 from the first vector. Then it embeds into the trivial vector bundle with fiber v. Basically, there is a canonical embedding of this sort, just the tautological embedding on this projective space pn. And you can pull back this embedding. And of course, the pullback of the trivial bundle is the trivial bundle. So you will get an arrow like that. On the other hand, on the second factor, there is a canonical quotient map from the trivial bundle with fiber v to the tautological quotient bundle, which by Euler exact sequence is amorphic to the twist of the tangent bundle. And if you pull back this quotient map from the second factor, you will get a surjective morphism to p2 upper star of t of minus 1. So you can consider this composition. So let us consider this composition. And this composition can be also thought of as a global section of the vector bundle, which is the tensor product of the dual of this bundle and that bundle. So we can take p1 upper star of o of 1. This is the dual of that bundle. And tensor it with p2 upper star of t of minus 1. And then this s can be thought of as a global section on pn cross pn of this vector bundle. Let us check where this global section vanishes. So clearly, the zero locus of this global section is precisely the zero locus of this composition of morphisms of vector bundles. But on the other hand, the kernel of the second arrow just by its definition is the pullback of o of minus 1 from the second factor. This is the kernel of that arrow. And so clearly, the composition vanishes at those points where the image of this sub-bundle coincides with the image of that sub-bundle. And which means that the zero locus is the diagonal in this product. So the zero locus of s is just the diagonal, the image of the diagonal embedding of pn. And in particular, you see that the rank of this vector bundle is equal to n. And it equals to the co-dimension of the zero locus. So rank of the vector bundle equals to the co-dimension of this zero locus. And because of that, the structure shape of the zero locus has a locally free resolution given by the casual complex corresponding to this section of this vector bundle. To be more precise, you should take exterior powers of the dual bundle and write down the corresponding, the maps between them induced by this section. So it will look like that. So on the right, but there will be the structure shape of the diagonal. Here you should put the dual bundle of that one, p1 upper star o of minus 1, tensor p2 upper star omega of 1. So the dual of t of minus 1 is omega of 1. Then you should put the second exterior power. But this tensor product is a tensor product of a line bundle and of some rank n vector bundle. If you take its exterior square, it is the same as to take the tensor square of the linear factor and tensor it with the exterior square of the second factor. So it will be p1 upper star o of minus 2, tensor p2 upper star omega 2 of 2, and so on. So there will be a resolution of this form. And now what you should do, you should think of this resolution as about a complex that is quite isomorphic to this shape. And now let us consider the 4MUKF factor corresponding to this shape and the 4MUKF factor corresponding to this complex. And let us apply it to an arbitrary object of the derived category of the projective space. And let us compare the results. So first of all, let us consider the 4MUKF factor corresponding to the structure shape of the diagonal. And let us apply to an arbitrary object of the derived category of the projective space. What's that? So by definition, you should pull back with respect to the first projection. Then you tensor product it with the structure shape of the diagonal. And then you push it forward to the second factor. Now if you use the projection formula and other properties of the factors, then you can rewrite it as the following. You should consider the composition of the second projection with the diagonal and the composition of the first projection with the diagonal. So take the derived pullback with respect to it and then push forward with respect to this guy. And this is just a way to rewrite this thing. And now since the composition of the diagonal embedded with the projections are identity morphisms. So this is identity. And this is also identity. Of course, the pullback with respect to the identity morphism is identity. So this is just f. So if you apply the 4MUKF factor corresponding to the structure shape of the diagonal, you always get the identity factor. On the other hand, you have this complex which is quasi-isomorphic to it. And it complex consists of terms which are quite simple. So let us see what the corresponding 4MUKF factors do. So if you take the kernel of 4MUKF factor to be something of this sort. So if you take O of minus t, tensor P2 pullback aminga t of t. And if you apply it to some object f, then again, if you substitute the definition of the 4MUKF factor, you should tensor the pullback of f with this object and then push forward. So now you can use the fact that pullback commutes with tensor products to combine this factor with that factor. And then you can use a base change formula to rewrite the whole composition. And you will see that this will be the following thing. So first of all, you should take the cohomology on Pn of f twisted by minus t. And then tensor it with omega t of t. So in fact, it will be just a tensor product of this graded factor space with this shift. And in fact, the fact that the structure shift of the diagonal is quasi-isomorphic to this resolution is equivalent to, can be rewritten as a certain spectral sequence. So it gives a spectral sequence in which whose second term is given by these objects. So you should take cohomology on Pn of f twisted by p. And then tensor with omega minus p of minus p. Here p will be between minus n and 0. And q will be between 0 and n. So it will be in a square of size n plus 1 times n plus 1. And this spectral sequence will converge to f sitting in degree 0. So this is a very useful spectral sequence. And we will apply it quite soon. Note also that, in fact, here we had one more possibility. In fact, we could also transpose the factors that we had here. So instead of pulling back o of minus 1 from the first factor and t of minus 1 from the second, we could do the other way around. And in that case, we would get another resolution of the diagonal, which would look very similarly. But all the factors will be transposed. And if you do the same computation in the second case, you will get another spectral sequence, which is very similar to that one, which will look like that. So you take the cohomology of f tensor omega minus p of minus p and then tensor o of p. And this also converges to f. So there are two spectral sequences. In some cases, one is more convenient. In other cases, another is more convenient. And both are called billions-on-spectral sequences. And for instance, it is easy to see that the second sequence very easily implies that the exceptional collection is full. So it gives another proof of fullness of this exceptional collection. Just because of this convergence statement, it is clear that every object of our derived category can be expressed from just objects o of p, where p sits in this interval. So we just take different o of p's. We tensor them with different factor spaces, which means that we just take the direct sums. And then if you look at what spectral sequence gives you, in fact, it proves that f is contained in the subcategory generated by these objects. So let me give you a couple of examples of how this spectral sequence can be used. So some applications. So let us consider, for instance, projective plane p2. And let us take, let f be a vector bundle, just some very explicit case, of rank 2 with the first churn class equal to 0 and the second churn class equal to 2. So let us consider such a vector bundle. Then let us apply the first spectral sequence to this vector bundle. Yeah, so maybe let us also assume that it is a stable vector bundle. And let us apply the first spectral sequence. So let us apply the first. Then what does it give? So let me first write very formally how should the second page log. So here we will have f twisted by minus 2, the second cachamology tensor. Amiga 2 for a projective plane is o of minus 3. And when you twist it by 2, it will be just o of minus 1. So maybe let me just write omega 2 of 2. This, in fact, omega n of n is isomorphic to o of minus 1 for any projective space. So here, the first column will look like that. Then the second column will consist of other twists or the cachamology of other twists, tensor with omega 1 of 1, the rightmost column like that. And sorry, this is not the second page. It is the first page of the spectral sequence. So the differentials go like that in this spectral sequence. So this is d1 differential. And at the next page, there will be also differentials which will go like that. So if you just look at this spectral sequence, maybe it doesn't look very nice. But let us now use the properties of f that we know. In fact, they ensure that at most positions of the spectral sequence, we have 0. So for instance, h0 of f, h0 of f twisted by minus 1, and h0 of f twisted by minus 2, all of them 0 by stability, just more or less by definition of a stable vector bundle. And also, we don't need it here. But let me also write that h0 of f of minus 3. Maybe not. Now, so it already shows you that in the first line, we have only 0s here. So everything is 0 in the first line. In the same way, one can consider the second line. So if you consider h2 of f of minus k, then you can rewrite it by ser duality as h0 of f dual twisted by k minus 3 by ser duality. And so for k equal 0, 1, and 2, here you have minus 3, minus 2, minus 1. And again, all these are also vanished by stability of f, by stability. So the top line of the spectral sequence is also 0. And this already means that the spectral sequence degenerates in the second term, because the other differentials will go from 0 to 0 only. And finally, one can check also by Riemann-Roch computation. One can easily find the dimensions of the vector spaces that stand here. And if you do this computation, you will see that here the dimension is 0, here the dimension is 2, and here the dimension is also 2. So in the end, this term also disappears. Only these two terms are left. And in the end, it follows that this looks like amorphism from O of minus 1 plus O of minus 1, or omega 1 of 1 plus omega 1 of 1. And in the end, it follows that this vector bundle f can be included into a short exact sequence of this form. It is just a co-kernel of a map from O of minus 1 plus O of minus 1 to omega of 1 plus omega of 1. And so in some sense, yeah, and you see that you can also do this construction in the opposite direction. You can start from any morphism of vector bundles like this and consider the co-kernel of this morphism. And for a general choice of a map, it will be a vector bundle with all these parameters. And so in the end, this gives you a description of the modular space of such vector bundles on P2. And this modular space can be just described as a certain JIT quotient of a certain parameter space for representations of a quiver. So let me write this. So let me just write what kind of data do we need to specify a map like this. So maybe let me call this map small f. So first of all, we have here a two-dimensional space. So in fact, the first term here, it is better to write as a certain two-dimensional vector space tensor O of minus 1. Just if you don't want to start by choosing a basis, it is more natural to write it in this way. And the second term can be written as B tensor omega of 1 of 1, where A and B are two-dimensional vector spaces. So the complex can be rewritten as this exact sequence can be rewritten in this more compact form. And so this morphism f is contained in the home space from A tensor O of minus 1 to B tensor omega 1 of 1. And this is the same as A dual tensor B tensor home from O of minus 1 to omega 1 of 1. So in other words, we can think about this morphism as about 2 by 2 matrix with elements being morphisms from O of minus 1 to omega 1 of 1. And finally, this space can be canonically identified with a space V corresponding such that P2 is P of V. So this is a three-dimensional vector space. So altogether, you get this vector space that parameterizes maps f. But of course, there is an action of the group GLA times GLB on this vector space, when we just change bases here and change bases there. So it acts on this tensor product. And clearly, if you act by an element of this group, it will not change the isomorphism class of the co-kernel. So altogether, this means that the modular space of our vector bundles on P2 of rank 2 with C1 equals 0 and C2 equal 2 can be written as a JT quotient of this form, A dual tensor B tensor G. And note also that to give such a data, to give an element in this vector space is equivalent to specifying a representation of a quiver with two vertices and three arrows. So in some sense, you can think of the space B as the space of arrows of such a quiver. And you can consider a representation of this quiver such that the vector space associated to the first vertex is A and the vector space associated with the second vertex is B. And then to give a representation of this quiver on these vector spaces is equivalent to giving an element F small like that. So a representation is equivalent to just this element. And so this shows that the modular space of these vector bundles is more or less the same as the modular space of representations of this quiver of dimension 2, 2. And this, in fact, in some sense, another explanation for this observation is that as you see, when we looked at this spectral sequence, we at some moment noticed that all the cohomology of F are equal to 0. From the categorical point of view, it means that our vector bundle F sits in the orthogonal sub category to 0. And this orthogonal subcategory is itself generated by an exceptional collection of length 2. You can take, for instance, O of minus 2 and O of minus 1. Or, in fact, another possible exceptional collection consists of O of minus 1 and Amigo of 1, those objects that appeared in this exceptional collection. And this triangulated subcategory, in fact, is equivalent to the derived category of representations of this quiver. So in fact, if you take this O perp in the derived category of P2, this is equivalent to the derived category of representations of this quiver with two vertices and three arrows. And so any object of this category, so for instance, our object F sits in this subcategory, as we discussed. So it can be also thought as an object of this representation, derived category of representations of the quiver. And in fact, this data of F small is precisely the representation corresponding to this object. And so you can use all the theory that you can develop for representations of quivers to describe geometrical objects, like vector bundles on a projective plane. For instance, you can consider different stability conditions on this derived category. You can consider variations of these stability conditions. And you can see what kind of birational transformations you will get for the corresponding model of spaces of objects from the geometrical point of view. And in many cases, this is a very useful thing to do. OK. So that was the first example. Maybe let me, I mean, of course, in this case, everything was very simple because we had a lot of vanishings in this spectral sequence. Of course, in general, you don't expect, say, that any column of the spectral sequence will vanish. So for instance, if you replace this condition that C2 is equal to 2 by an arbitrary second-chern class, then no longer you will have vanishing here. I mean, the vanishing appearing on this right blackboard still will hold true. So the first and the third line of the spectral sequence still will be 0. But it is no longer true that one of these three terms will vanish. So in general, you will get three terms. So you will not be able to interpret this spectral sequence as a representation of that very simple quiver. Instead, you will have to consider a more complicated quiver that will have now three vertices because you have three terms here. And it will have three arrows between the first and the second vertex and three arrows between the second and the third vertex. And it also will have three relations that will correspond just to commutations, commutativity of coordinates on the projective space. So let me denote these arrows by x1, x2, x3. These arrows by x prime, by x prime 2 and x prime 3. Then there will be three relations of the sort that x prime i xj equals x prime j xi. So there will be three relations. And you can consider representations of this more complicated quiver. And again, in fact, just the whole derived category of the projective plane is equivalent to derived category of representations of this quiver. So maybe let me write it like this. And so you can interpret objects of this derived category just as some complexes of representations of the quiver. And so you can also consider different stability conditions. And you can use all these machinery. So in some sense, whenever you have a full exceptional collection in the derived category of some variety, then from the categorical point of view, it does not differ from, in fact, a quiver with relations. And so you can use this algebraic point of view in a very effective way to describe some modular spaces and to describe some relations between different modular spaces of objects and so on. But of course, for instance, if you want to describe some particular modular space and you want to use a spectral sequence of this type, then of course, as you see from this simplest example, the description will be much easier if you have many vanishings in this spectral sequence. And in fact, so I mentioned here only two spectral sequences here. But in fact, you can write a spectral sequence for any exceptional collection in your triangulated category. So besides these two, there is a spectral sequence for any exceptional collection, for any full exceptional collection in your category. And so in fact, before applying this spectral sequence, you could first replace your original exceptional collection by, for instance, some mutation or some twist of this exceptional collection. And in some cases, for a mutated collection, the spectral sequence will look much simpler. So if you want to really apply this machinery, then you have to be careful with choosing your exceptional collection. So you first should choose exceptional collection in a smart way. And then the spectral sequence will be reasonably simple. And then you can use this to find an explicit and very nice description. Let me add one more remark here, and I will stop. So of course, to do it for a projective space is maybe not very impressive, because I mean, projective space is a reasonably simple variety. And of course, there are other technique that can be used to describe vector bundles on it. But there are other interesting examples when you can also construct a full exceptional collection, and in which cases other approaches just probably don't exist. So for instance, if you consider some FANA 3-folds, so for instance, you can take the most interesting FANA 3-fold with Picard number one that was originally in the FANA classification of FANA 3-folds, it was overlooked. So it is sufficiently complicated in some sense. But still, its derived category has a full exceptional collection, consistent of only four vector bundles. And for instance, you can use this exceptional collection to describe some Hilbert schemes of curves of small degree on this variety. And I don't know whether other techniques can produce similar results, for instance, in this particular case. So what I want to say just to finish is that this technique is very useful for such a description. But you have to be a bit careful when applying it because there are several choices to be done, and you have to do these choices in a smart way. OK, let me stop here.