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This article is cited in 1 scientific paper (total in 1 paper) Dual exceptional collections on Lagrangian Grassmannians A. V. Fonarev Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 12, DOI: https://doi.org/10.4213/sm9880 
Bibliographic databases:
    § 1. IntroductionDerived categories of varieties are among the central objects in modern algebraic geometry. A skeptical reader might complain that the very notion of derived category is too abstract, and they might have a point. However, since the pioneering work of Beilinson [1], derived categories have become a great computational tool. Nowadays one would say that Beilinson constructed a full exceptional collection in the bounded derived categories of coherent sheaves on projective spaces. The following analogy is commonly used to explain the computational power of exceptional collections. Given a finite-dimensional real vector space $V$ with a positive definite symmetric bilinear form $\langle\,\cdot\,{,}\,\cdot\,\rangle$, one can find an orthonormal basis. That is, a basis consisting of vectors $(v_1,v_2,\dots,v_n)$ such that (i) $\langle v_i, v_i\rangle = 1$ for all $i=1,\dots, n$ and (ii) $\langle v_i, v_j\rangle = 0$ whenever $i\neq j$. The first condition says that the vectors are unit vectors, while the second is the orthogonality condition. Given such a basis, every vector $v\in V$ can easily be decomposed with respect to it:
$$
\begin{equation}
v=\langle v, v_1\rangle v_1+\langle v, v_2\rangle v_2+\dots +\langle v, v_n\rangle v_n.
\end{equation}
\tag{1.1}
$$
Let us now relax some of the conditions; for instance, symmetry. We still want a basis which consists of unit vectors, but we have to modify the orthogonality condition (ii). Let us say that a basis $(v_1,v_2,\dots,v_n)$ is semiorthonormal with respect to a nonsymmetric bilinear form $\langle\,\cdot\,{,}\,\cdot\,\rangle$ if (i) $\langle v_i, v_i\rangle = 1$ for all $i=1,\dots, n$ and (ii) $\langle v_i, v_j\rangle = 0$ whenever $i > j$. One obviously needs an adjustment to formula (1.1). Indeed, in (1.1) we explicitly used the fact that under the isomorphism $V^*\xrightarrow{\sim} V$ given by the form $\langle\,\cdot\,{,}\,\cdot\,\rangle$ the dual basis $(v^1,\dots, v^n)$ maps back to $(v_1,\dots, v_n)$. Since we actually have two isomorphisms this time, $V\to V^*$, $v\mapsto \langle v ,\cdot\,\rangle$, and $V\to V^*$, $v\mapsto \langle\,\cdot\,,v \rangle$, let us consider the second one and denote by $u_i\in V$ the images of the dual basis under the inverse map. That is, we have
$$
\begin{equation*}
\langle v_i, u_j\rangle=\delta_{ij} \quad \text{for all } 1\leqslant i,j\leqslant n.
\end{equation*}
\notag
$$
The vectors $u_1,\dots, u_n$ obviously form a basis, and for any $v \in V$ we have the desired formula
$$
\begin{equation}
v=\langle v, u_1\rangle v_1+\langle v, u_2\rangle v_2+\dots +\langle v, u_n\rangle v_n.
\end{equation}
\tag{1.2}
$$
What is less immediate, the vectors $(u_n, u_{n-1}, \dots, u_1)$ (notice the reverse order) form a semiorthonormal basis called the left dual to $(v_1,v_2,\dots,v_n)$. One rather indirect way to see that the latter holds is via mutations. Consider a semiorthonormal pair $(u, v)$, that is, $\langle u, u \rangle=\langle v, v\rangle = 1$ and $\langle v, u \rangle=0$. Let us define the new vector $\mathbb{L}_uv=v-\langle u, v\rangle u$, which is called the left mutation of $v$ through $u$. A simple calculation shows that $(\mathbb{L}_uv, u)$ is a semiorthonormal pair. As the name suggests, there is a sibling to the left mutation procedure: if one puts $\mathbb{R}_vu = u-\langle u, v\rangle v$, which is called the right mutation of $u$ through $v$, then the pair $(v, \mathbb{R}_vu)$ is semiorthonormal. It is a very nice exercise in linear algebra to check that left and right mutations of adjacent elements define an action of the braid group on $n$ strands on the set of all semiorthonormal bases. Moreover, the left dual basis to a semiorthonormal basis $(v_1,v_2,\dots,v_n)$ is given by
$$
\begin{equation}
\mathbb{L}_{v_1}\mathbb{L}_{v_2}\dotsb\mathbb{L}_{v_{n-1}}v_n, \quad \mathbb{L}_{v_1}\mathbb{L}_{v_2}\dotsb\mathbb{L}_{v_{n-2}}v_{n-1}, \quad \dots, \quad \mathbb{L}_{v_1}v_2, \quad v_1.
\end{equation}
\tag{1.3}
$$
We refer the reader to [2] for further insights and some interesting properties of this action. The linear-algebraic picture translates to triangulated categories in the following way. Instead of a vector space we consider a $\mathsf{k}$-linear triangulated category $\mathcal{T}$ such that $\operatorname{Ext}^\bullet(E, F)$ is finite-dimensional for all $E, F\in \mathcal{T}$ and treat $\operatorname{Ext}^\bullet_{\mathcal{T}}(\,\cdot\,{,}\,\cdot\,)$ as a kind of bilinear form. Then one says that an object $E\in \mathcal{T}$ is exceptional if ${\operatorname{Hom}(E, E)=\mathsf{k}}$ and $\operatorname{Ext}^i(E, E)=0$ for all $i\neq 0$. A collection of objects $(E_1,E_2,\dots, E_n)$ is called exceptional if every $E_i$ is an exceptional object and $\operatorname{Ext}^\bullet(E_i, E_j)=0$ for all $i>j$. Finally, a collection is called full if it generates the category in the sense that no proper strictly full triangulated subcategory of $\mathcal{T}$ contains all the $E_i$. If $(E_1,E_2,\dots, E_n)$ is a full exceptional collection in $\mathcal{T}$, then we will write $\mathcal{T}=\langle E_1,E_2,\dots, E_n\rangle$. The analogy with semiorthonormal bases should be clear by now. Assume that $\mathcal{T}$ has a full exceptional collection $(E_1,E_2,\dots, E_n)$ (which is rarely the case). What is most interesting is that not only every object in $\mathcal{T}$ can be obtained from the finite set of the $E_i$ by taking shifts and cones iteratively, but this procedure can be made rather explicit. Recall that the decomposition of every vector in terms of a given semiorthonormal basis could be done with the help of a left dual semiorthonormal basis by (1.2). Let us mimic its definition in the categorical case. Definition 1.1 (see [8]). A collection of objects $(E_n^\vee,E_{n-1}^\vee,\dots, E_1^\vee)$ is left dual to $(E_1,E_2,\dots, E_n)$ if It should not surprise the reader at this point that $(E_n^\vee,E_{n-1}^\vee,\dots, E_1^\vee)$ is again an exceptional collection, and it is full whenever the original one is. As in the linear algebraic case, there are operations of left and right mutation given, for an exceptional pair $(E, F)$, by the exact triangles
$$
\begin{equation*}
\mathbb{L}_EF \to \operatorname{Ext}^\bullet(E, F)\otimes E\xrightarrow{\mathrm{ev}} F\to \mathbb{L}_EF[1]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathbb{R}_FE[-1] \to E\xrightarrow{\mathrm{coev}} \operatorname{Ext}^\bullet(E, F)^*\otimes F\to \mathbb{R}_FE.
\end{equation*}
\notag
$$
These define an action of the braid group on $n$ strands on the set of all exceptional collections of length $n$ in $\mathcal{T}$, and a formula similar to (1.3) determines the left dual collection:
$$
\begin{equation*}
E_i^\vee=\mathbb{L}_{E_1}\mathbb{L}_{E_2}\dotsb\mathbb{L}_{E_{i-1}}E_i.
\end{equation*}
\notag
$$
Finally, instead of the decomposition (1.2) one has a spectral sequence which helps compute cohomological functors applied to objects in $\mathcal{T}$, which we will talk about in § 2.1. The bridge between exceptional collections and semiorthonormal bases is actually rather simple. Given $\mathcal{T}$ as above with a full exceptional collection $(E_1,E_2,\dots, E_n)$, one checks that the classes of $E_i$ form a basis in the Grothendieck group $K_0(\mathcal{T})$. In particular, the length of any full exceptional collection equals the rank of the latter, which must a posteriori be a free finitely generated abelian group. If one takes $\sum_i (-1)^i \dim_\mathsf{k} \operatorname{Ext}^i(\,\cdot\,{,}\,\cdot\,)$ as a bilinear form, then exceptional collections become ‘categorifications’ of the corresponding semiorthonormal bases. The first example of a full exceptional collection was given in the paper [1], in which Beilinson showed that
$$
\begin{equation*}
D^b(\mathbb{P}^n)=\langle \mathcal{O}, \mathcal{O}(1),\dots, \mathcal{O}(n)\rangle,
\end{equation*}
\notag
$$
where $D^b(X)$ stands for the bounded derived category of coherent sheaves on an algebraic variety $X$. Interestingly enough, he also showed that the left dual is (up to shifts) given by the collection $\langle \Omega^n(n), \Omega^{n-1}(n-1),\dots,\Omega^1(1),\mathcal{O} \rangle$. A long-standing conjecture states that the bounded derived category of coherent sheaves on a rational homogeneous variety admits a full exceptional collection. Though a lot of work has been done over the years, this conjecture has been established in a very limited number of cases. Say, for classical groups of Dynkin type $ABCD$ and homogeneous varieties of Picard rank $1$ the problem was fully resolved only for Grassmannians [3], quadrics [4], symplectic and orthogonal Grassmannians of planes [5], [6], Lagrangian Grassmannians [11] and in some sporadic cases. We refer the reader to [7] for further details. Since many explicit constructions realize varieties as subvarieties not in projective spaces, but in Grassmannians, exceptional collections on the latter become an important computational tool. One of our favourite recent examples can be found in [12], where the authors used exceptional collections to study moduli of Ulrich bundles. In order to use the tool’s maximum power, it is important to know the dual collection, and finding one is a task of its own. In the present paper we find the exceptional collections on Lagrangian Grassmannians that are dual to those constructed in [11]. We further use them to provide explicit resolutions of some very natural irreducible vector bundles. From now on we will be interested in the bounded derived category of coherent sheaves on $\operatorname{LGr}(n, V)$, that is, the Lagrangian Grassmannian of isotropic subspaces of dimension $n$ in a fixed $2n$-dimensional vector space $V$ over a field $\mathsf{k}$ of characteristic $0$ equipped with a nondegenerate symplectic form. Exceptional collections of maximum length (equal to the rank of the Grothendieck group) were constructed on all symplectic and orthogonal Grassmannians by Kuznetsov and Polishchuk in [7]. All these collections are conjecturally full; however, this was checked only for Lagrangian Grassmannians in [11]. The construction of Kuznetsov and Polishchuk is rather indirect: they start with certain collections of irreducible vector bundles, called blocks, which naturally form an exceptional collection in the equivariant derived category, and then pass to dual collections within each block (again, in the equivariant derived category). Each block must satisfy certain homological conditions which guarantee that the dual collections become exceptional in the nonequivariant derived category. Most of the hard work in [7] was related to checking that certain collections of irreducible equivariant vector bundles satisfy the block condition, which is a rather difficult problem in representation theory. While the block condition will be discussed in § 2.2, let us explain why isotropic Grassmannians are much harder than the classical ones. Full exceptional collections in the bounded derived categories of Grassmannians were constructed by Kapranov [4], who extended naturally Beilinson’s method. Consider $\operatorname{Gr}(k, V)$, the Grassmannian of $k$-dimensional subspaces in a fixed $N$-dimensional vector space $V$ over a field $\mathsf{k}$ of characteristic zero. Denote by $\mathcal{U}$ the tautological rank $k$ subbundle of the trivial bundle $V\otimes \mathcal{O}$. Kapranov showed that
$$
\begin{equation}
D^b(\operatorname{Gr}(k, V))=\bigl\langle\Sigma^\lambda \mathcal{U}^* \mid \lambda\in \mathrm{Y}_{k,N-k} \bigr\rangle,
\end{equation}
\tag{1.4}
$$
where $\lambda$ runs over the set of Young diagrams $\mathrm{Y}_{k, N-k}$ of height at most $k$ and width at most $N-k$, $\Sigma^\lambda$ is the corresponding Schur functor, and the order in this collection can be taken to be any linear order refining the partial inclusion order $\subseteq$ on the diagrams. Moreover, he constructed simultaneously the graded1 left dual to this collection. It is given by
$$
\begin{equation}
D^b(\operatorname{Gr}(k, V))=\bigl\langle\Sigma^{\lambda^\top} \mathcal{U}^\perp \mid \lambda\in \mathrm{Y}_{k,N-k} \bigr\rangle,
\end{equation}
\tag{1.5}
$$
where $\mathcal{U}^\perp = (V/\mathcal{U})^*$ and $\lambda^\top$ denotes the transposed diagram. Remark 1.1. The reader might wonder why we use this extra transposition and not index the collection by $\mathrm{Y}_{N-k, k}$, as Kapranov does. After all, $\{\lambda^\top \mid \lambda \in \mathrm{Y}_{k, N-k}\} = \mathrm{Y}_{N-k, k}$. The reason for this choice becomes clear in § 2.1, where we introduce our grading convention for dual exceptional collections. The duality relation can seem a little different from the one we described before:
$$
\begin{equation}
\operatorname{Ext}^\bullet(\Sigma^\lambda \mathcal{U}^*, \Sigma^{\mu^\top} \mathcal{U}^\perp)= \begin{cases} \mathsf{k}[-|\lambda|] & \text{if } \lambda=\mu, \\ 0 & \text{otherwise}, \end{cases}
\end{equation}
\tag{1.6}
$$
but this grading difference does not change much since the two definitions are equivalent up to shifts in the derived category. Actually, the grading choice in (1.6) is preferable since the dual collection then consists of vector bundles. Below, in Lemma 2.1, we give a formal definition of a graded dual exceptional collection, and this is an example of such a collection. We return to the case when $V$ is symplectic of dimension $2n$. Since $\operatorname{LGr}(n, V)$ is naturally embedded as a closed subvariety in $\operatorname{Gr}(n, V)$, and the tautological bundle on the latter restricts to the tautological bundle on the former, one can ask whether any of elements of Kapranov’s collection restrict to exceptional vector bundles on $\operatorname{LGr}(n, V)$. This is where surprising things happen. It turns out that up to twist the only Young diagrams for which $\Sigma^\lambda \mathcal{U}^*$ are exceptional on $\operatorname{LGr}(n, V)$ are ${\lambda \in \mathrm{Y}_{n,1}}$. The number of objects in any full exceptional collection on $\operatorname{LGr}(n, V)$ equals $\operatorname{rk} K_0(\operatorname{LGr}(n, V)) = 2^n$, and it is easy to see that one cannot form a sufficiently large exceptional collection using the latter bundles. Observe also that $\mathcal{U}^\perp$ restricts to $\mathcal{U}$ on $\operatorname{LGr}(n, V)$. The construction due to Kuznetsov and Polishchuk produces, for any $\lambda\in \mathrm{Y}_{h, n-h}$, where $0\leqslant h\leqslant n$, an equivariant non-irreducible (in general) vector bundle $\mathcal{E}^\lambda$ on $\operatorname{LGr}(n, V)$ with the following properties. First, $\mathcal{E}^\lambda$ belongs to the subcategory of the derived category $D^b(\operatorname{LGr}(n, V))$ generated by the $\Sigma^\mu \mathcal{U}^*$ for $\mu\subseteq \lambda$. Second, the $\mathbf{G}=\operatorname{Sp}_{2n}$-equivariant groups $\operatorname{Ext}^\bullet_\mathbf{G}(\mathcal{E}^\lambda, \Sigma^\mu \mathcal{U}^*)$ are equal to zero for all $\mu\subsetneq \lambda$, while $\operatorname{Ext}^\bullet_\mathbf{G}(\mathcal{E}^\lambda, \Sigma^\lambda \mathcal{U}^*) = \mathsf{k}$. A given diagram can belong to various sets $\mathrm{Y}_{h, n-h}$, but the resulting bundle does not depend on the choice of $h$ since the previous conditions characterize it fully and are independent of $h$. The first nontrivial example of such a bundle is the universal extension
$$
\begin{equation*}
0\to \mathcal{O} \to \mathcal{E}^2 \to S^2\mathcal{U}^* \to 0.
\end{equation*}
\notag
$$
Kuznetsov and Polishchuk showed that for any $0\leqslant h \leqslant n$ the bundles
$$
\begin{equation}
\langle\mathcal{E}^\lambda \mid \lambda\in \mathrm{Y}_{h,\,n-h}\rangle
\end{equation}
\tag{1.7}
$$
form an exceptional collection in $D^b(\operatorname{LGr}(n, V))$. As in the case of Grassmannians, one can linearly order them in any way compatible with the partial inclusion order on the corresponding Young diagrams. Let us denote by $\mathcal{F}^\lambda$ the bundle dual (in the usual sense) to $\mathcal{E}^\lambda$: Our first main result is Theorem 3.1. It is as follows. The bundles $\langle\mathcal{F}^{\lambda^\top} \mid \lambda\in \mathrm{Y}_{h,n-h} \rangle$ form a graded left dual exceptional collection to (1.7). We have formulated Theorem 3.1 in terms of transposed diagrams in order to show the parallel between our Lagrangian situation and the case of classical Grassmannians: compare the latter theorem with (1.4) and (1.5) keeping in mind that the bundle $\mathcal{U}^\perp$ is isomorphic to $\mathcal{U}=(\mathcal{U}^*)^*$ on $\operatorname{LGr}(n, V)$. It is easy to see that the bundle $\Sigma^\lambda\mathcal{U}^*$ belongs to the subcategory (1.7) whenever $\lambda\in \mathrm{Y}_{h,n-h}$. A nice geometric description (as a matter of fact, two of them) was given for the bundles $\mathcal{F}^\mu$ in [11]. Using this description, one can compute the graded dual spectral sequences and obtain the following result (Theorem 3.3) as an application of Theorem 3.1 (see § 3.2 for details). Let $\lambda\in \mathrm{Y}_{h,n-h}$ for some $0\leqslant h\leqslant n$. Then there is an exact $\operatorname{Sp}(V)$-equivariant sequence of vector bundles on $\operatorname{LGr}(n, V)$ of the form
$$
\begin{equation*}
\begin{aligned} \, 0 &\to \bigoplus_{\mu\in \mathrm{B}_{h(h+1)}}\mathcal{E}^{\lambda/\mu} \to \dots \to \bigoplus_{\mu\in \mathrm{B}_{2t}}\mathcal{E}^{\lambda/\mu} \to \dotsb \\ &\to \bigoplus_{\mu\in \mathrm{B}_4}\mathcal{E}^{\lambda/\mu} \to \bigoplus_{\mu\in \mathrm{B}_2}\mathcal{E}^{\lambda/\mu} \to \mathcal{E}^\lambda\to \Sigma^\lambda\mathcal{U}^*\to 0, \end{aligned}
\end{equation*}
\notag
$$
where $\mathrm{B}_{2t}$ denotes the set of balanced diagrams with $2t$ boxes and $\mathcal{E}^{\lambda/\mu}$ is a direct sum of certain bundles of the form $\mathcal{E}^{\nu}$; see § 3.2.4. The paper is organized as follows. In § 2 we collect all preliminaries. It contains no new material except, maybe, our convention for dual exceptional collections to exceptional collections indexed by a graded partially ordered set (see Lemma 2.1). Section 3 contains our main results. That is, we prove there Theorems 3.1 and 3.3. § 2. PreliminariesThroughout the paper we work over a field $\mathsf{k}$ of characteristic zero. 2.1. Dual exceptional collectionsIn this section we collect some preliminaries related to (dual) exceptional collections. The material is well known to specialists; however, since we work naturally with exceptional collections indexed by graded partially ordered sets, we introduce a certain convention in the definitions. This convention seems rather natural and useful, as we show in various examples. 2.1.1. Partially ordered setsRecall that a partially ordered set (poset) $\mathcal{P}$ is a set equipped with a binary relation $\preceq$, called a partial order, satisfying the following three properties: Reflexivity: $x\preceq x$ for all $x\in\mathcal{P}$. Antisymmetry: if $x\preceq y$ and $y\preceq x$, then $x=y$. Transitivity: is $x\preceq y$ and $y\preceq z$, then $x\preceq z$ for all $x,y,z\in\mathcal{P}$. Elements $x$ and $y$ are called comparable if either $x\preceq y$ or $y\preceq x$. If every two elements in $\mathcal{P}$ are comparable, then one calls $\mathcal{P}$ linearly ordered. If $x\preceq y$ and $x\neq y$, then one usually writes $x \prec y$. If $\mathcal{P}$ is partially ordered, its dual $\mathcal{P}^\circ$ is the set underlying $\mathcal{P}$ and equipped with the reverse relation: $x\preceq y$ in $\mathcal{P}^\circ$ if and only if $y\preceq x$ in $\mathcal{P}$. Let $x$ and $y$ be elements of a poset $\mathcal{P}$. One says that $y$ covers $x$ (written $x\lessdot y$) if $x\prec y$ and there is no element $z$ such that $x\prec z\prec y$. A grading function on $\mathcal{P}$ is a map $\rho: \mathcal{P}\to \mathbb{Z}$ with the following properties: A poset equipped with a grading function is called a graded poset. Of course, not all posets can be turned to graded ones. We will mainly be interested in finite posets. If $\mathcal{P}$ is finite and admits a grading function, then there is a rather natural choice for such a function: there exists a unique grading function $|\,{\cdot}\,|$ with the property that the smallest value it takes on each connected component is $0$.2 In what follows, by a graded poset we mean a finite poset equipped with this natural grading function. Moreover, all the posets of interest will contain a minimal element.Example 2.1. Let $\mathrm{Y}_{h,w}$ denote the set of Young diagrams of height at most $h$ and width at most $w$. This set can be identified with the set of integer sequences $(\lambda_1,\lambda_2,\dots,\lambda_h)$ such that $w\geqslant \lambda_1\geqslant \lambda_2\geqslant \dots\geqslant \lambda_h\geqslant 0$. There is a natural partial order on $\mathrm{Y}_{h,w}$ given by inclusion of diagrams: $\lambda\subseteq \mu$ if $\lambda_i\leqslant \mu_i$ for all $i=1,\dots, h$. With this partial order the poset $\mathrm{Y}_{h,w}$ is graded, and $|\lambda|=\lambda_1+\lambda_2+\dots+\lambda_h$ equals the number of boxes in the diagram $\lambda$. 2.1.2. Exceptional collectionsThroughout the paper we only work with triangulated categories which are $\mathsf{k}$-linear. Let $\mathcal{T}$ be such a category. Recall that an object $E\in\mathcal{T}$ is called exceptional if $\operatorname{Hom}(E, E)=\mathsf{k}$ and $\operatorname{Ext}^t(E, E)=\operatorname{Hom}(E, E[t])=0$ for all $t\neq 0$. Let $\mathcal{P}$ be a poset. An exceptional collection indexed by $\mathcal{P}$ is a collection of exceptional objects $\{E_x\}_{x\in\mathcal{P}}$ such that $\operatorname{Ext}^\bullet(E_x, E_y)=0$ unless $x\preceq y$. We denote by $\langle E_x \mid x\in \mathcal{P} \rangle$ the smallest strictly full triangulated subcategory in $\mathcal{T}$ containing all $E_x$. If $\mathcal{P}$ is finite, then one can always refine the order so that it becomes isomorphic to the poset $\{1, 2, \dots, l\}$. Under this isomorphism we obtain the usual definition of an exceptional collection: this is a collection of exceptional objects $(E_1,E_2,\dots, E_l)$ such that $\operatorname{Ext}^\bullet(E_j, E_i)=0$ for all $l\geqslant j>i\geqslant 1$. We allow ourselves to go over to some of the examples mentioned in the introduction in order to illustrate our grading conventions. Example 2.2. Let $V$ be an $n$-dimensional vector space over the field $\mathsf{k}$. Consider the Grassmannian $\operatorname{Gr}(k, V)$, and let $\mathcal{U}$ denote the tautological rank $k$ subbundle in the trivial bundle $V\otimes \mathcal{O}$. Kapranov showed in [3] that the bounded derived category of coherent sheaves $D^b(\operatorname{Gr}(k, V))$ admits a full exceptional collection indexed by the poset $\mathrm{Y}_{k, n-k}$, which was introduced in Example 2.1: 2.1.3. Mutations and dualityLet $(E, F)$ be an exceptional pair in $\mathcal{T}$. The left mutation $\mathbb{L}_EF$ of $F$ through $E$ is defined by the distinguished triangle
$$
\begin{equation*}
\mathbb{L}_EF\to \operatorname{Ext}^\bullet(E, F)\otimes E\xrightarrow{\mathrm{ev}} F\to \mathbb{L}_EF[1],
\end{equation*}
\notag
$$
where $\mathrm{ev}$ is the evaluation morphism. We define similarly the right mutation $\mathbb{R}_FE$ via the distinguished triangle
$$
\begin{equation*}
\mathbb{R}_FE[-1]\to E\xrightarrow{\mathrm{coev}} \operatorname{Ext}^\bullet(E, F)^*\otimes F\to \mathbb{R}_FE,
\end{equation*}
\notag
$$
where $\mathrm{coev}$ is the coevaluation morphism. One easily checks that both $(\mathbb{L}_EF, E)$ and $(F, \mathbb{R}_FE)$ are exceptional pairs. Moreover, $\langle E, F\rangle=\langle \mathbb{L}_EF, E\rangle=\langle F, \mathbb{R}_FE\rangle$.4 Example 2.3. Consider the Grassmannian $\operatorname{Gr}(k, V)$. Then the structure sheaf and the dual tautological bundle form an exceptional pair $(\mathcal{O}, \mathcal{U}^*)$. One quickly checks that $\mathbb{L}_{\mathcal{O}}\mathcal{U}^*\simeq \mathcal{U}^\perp$, where $\mathcal{U}^\perp=(V/\mathcal{U})^*$. Mutations of adjacent elements define an action of the braid group $\mathrm{Br}_l$ on $l$ strands on the set of exceptional collections in $\langle E_1,E_2,\dots,E_l\rangle$; see [9]. For a fixed collection there are two important elements in the orbit, namely, the dual collections. The left dual collection to $(E_1,E_2,\dots,E_l)$ is defined as
$$
\begin{equation*}
\bigl(\mathbb{L}_{E_1}\mathbb{L}_{E_2}\dotsb\mathbb{L}_{E_{l-1}}E_l,\, \mathbb{L}_{E_1}\mathbb{L}_{E_2}\dotsb\mathbb{L}_{E_{l-2}}E_{l-1},\, \dots,\, \mathbb{L}_{E_1}E_2,\, E_1\bigr).
\end{equation*}
\notag
$$
We denote it by $(E^\vee_l, E^\vee_{l-1}, \dots, E^\vee_1)$. The left dual exceptional collection can fully be characterized by the following three properties:
Example 2.4. The bounded derived category $D^b(\mathbb{P}(V))$ of the projective space $\mathbb{P}(V)$ has a full exceptional collection consisting of the line bundles $\langle \mathcal{O}, \mathcal{O}(1), \dots, \mathcal{O}(n)\rangle$, where $n$ is the dimension of $V$. Its left dual is given by $\langle \Omega^n(n), \Omega^{n-1}(n-1), \dots, \Omega^1(1), \mathcal{O}\rangle$, where $\Omega^i=\Lambda^i\Omega^1_{\mathbb{P}(V)}$. Similarly, the right dual collection is defined as
$$
\begin{equation*}
\bigl(E_l,\,\mathbb{R}_{E_l}E_{l-1},\, \mathbb{R}_{E_l}\mathbb{R}_{E_{l-1}}E_{l-2},\, \dots,\, \mathbb{R}_{E_l}\mathbb{R}_{E_{l-1}}\dotsb\mathbb{R}_{E_2}E_1\bigr)
\end{equation*}
\notag
$$
and is denoted by $({}^\vee\!{E}_l, {}^\vee\!{E}_{l-1}, \dots, {}^\vee\!{E}_1)$. It can fully be characterized by the following conditions:
Note that in a fixed exceptional collection $(E_1,E_2,\dots,E_l)$ one can replace any object $E_i$ by its shift $E_i[t]$ for any integer $t$. For instance, one can introduce an extra shift in the definitions of right and left mutations. The first two defining conditions for the left and right dual collections will not change, while the third condition will become slightly nicer:
$$
\begin{equation*}
\operatorname{Ext}^\bullet({}^\vee\!{E}_i, E_i)=\operatorname{Ext}^\bullet(E_i, E_i^\vee)= \mathsf{k}, \qquad 1\leqslant i\leqslant l.
\end{equation*}
\notag
$$
This convention is often reasonable, yet even in the case of a projective space, Example 2.4, the dual collection does not consist of vector bundles while the original collection does. Meanwhile, the definitions we have just given also have a downside. Imagine that $(E, F)$ is a fully orthogonal pair in $\mathcal{T}$. Then $\mathbb{L}_EF\simeq F[-1]$, while $\mathbb{R}_FE\simeq E[1]$. This is often inconvenient. We propose the following lemma-definition, which is tailored to the case of an exceptional collection indexed by a finite graded poset $\mathcal{P}$. The reader will immediately check that once the collection is linearly ordered, the left dual differs from the graded left dual by shifts of objects. Lemma 2.1. Let $\langle E_x\mid x\in\mathcal{P} \rangle$ be an exceptional collection indexed by a finite graded poset $\mathcal{P}$. For any $y\in \mathcal{P}$ there exists a unique (up to isomorphism) object $E^\circ_y\in \langle E_x\mid x\in\mathcal{P} \rangle$ such that The objects $E^\circ_y$ form an exceptional collection with respect to the opposite poset $\mathcal{P}^\circ$. This collection is called the graded left dual, and $\langle E^\circ_y\mid y\in\mathcal{P}^\circ \rangle=\langle E_x\mid x\in\mathcal{P} \rangle$. Remark 2.1. If the poset $\mathcal{P}$ is linearly ordered, then the definition of the graded left dual agrees with the definition of the left dual. Example 2.5. In Example 2.2 we saw that $D^b(\operatorname{Gr}(k, V))$ admits a full exceptional collection indexed by the poset $\mathrm{Y}_{k, n-k}$: where $\mathcal{U}$ is the tautological rank $k$ subbundle in $V$. The dual collection constructed by Kapranov, where $\mathcal{U}^\perp=(V/\mathcal{U})^*$, and $\lambda^\top$ denotes the transposed diagram, is a graded left dual. We leave the definition of the graded right dual to the reader, indicating that for the right dual the grading function should be taken for the opposite poset. 2.1.4. A categorical lemmaLet $\mathcal{T}=\langle E_x\mid x\in\mathcal{P}\rangle$ be a proper triangulated category generated by a graded exceptional collection. Denote by $\langle G_x\mid x\in\mathcal{P}^\circ\rangle$ its graded left dual. Assume that $F\colon\mathcal{T}\to \mathcal{T}'$ is an exact functor into another proper triangulated category $\mathcal{T}'$. Put $G'_x=F(G_x)$ for all $x\in\mathcal{P}^\circ$ and assume that $\langle G'_x\mid x\in\mathcal{P}^\circ\rangle$ form a graded exceptional collection in $\mathcal{T}'$. Denote by $\langle E'_x\mid x\in\mathcal{P}\rangle$ its graded right dual. Proof. Since the category $\mathcal{T}$ is saturated, the functor $F$ has a left adjoint $F^*$ (see [9]). In particular, We claim that $F^*(E'_x)\simeq E_x$. On the one hand $F^*(E'_x) \in \langle E_x\mid x\in\mathcal{P}\rangle = \langle {G_x\mid x\in\mathcal{P}^\circ}\rangle$. On the other hand, for any $y\in \mathcal{P}$ one has These are precisely the defining conditions of the graded right dual collection to $\langle G_x\mid x\in\mathcal{P}^\circ\rangle$, which is $\langle E_x\mid x\in\mathcal{P}\rangle$. The lemma is proved. 2.1.5. The spectral sequence associated with the graded dualAs it was stated in the introduction, dual collections provide a particularly nice computational tool. Let $(E_1,E_2,\dots, E_l)$ be an exceptional collection in $\mathcal{T}$. Recall that a cohomological functor from $\mathcal{T}$ to an abelian category $\mathcal{A}$ is an additive functor $F\colon\mathcal{T}\to \mathcal{A}$ which takes distinguished triangles to exact sequences. As usual, we denote by $F^i$ the composition $F\circ [i]$. Proposition 2.1 (see [10], § 2.7.3). Let $G\in \langle E_1,E_2,\dots,E_n\rangle$. Then there is a spectral sequence with the first page given by which converges to $F^{p+q}(G)$. We will be interested in the case when $\mathcal{T}=D^b(\mathcal{A})$ is the bounded derived category of an abelian category $\mathcal{A}$ (for instance, the bounded derived category of coherent sheaves on a smooth projective variety), and $F=\mathcal{H}^0$ is the usual $0$th cohomology functor. Assume that the exceptional collection $(E_1,E_2,\dots, E_l)$ consists of pure objects (for instance, of coherent sheaves). Then the spectral sequence (2.1) simplifies to
$$
\begin{equation}
E^{p,q}_1=\operatorname{Ext}^{-q}(G, E^\vee_{p+1})^*\otimes E_{p+1} \quad\Longrightarrow\quad \mathcal{H}^{p+q}(G).
\end{equation}
\tag{2.2}
$$
In the case of an exceptional collection indexed by a graded poset $\mathcal{P}$ the spectral sequence (2.2) becomes
$$
\begin{equation}
E^{p,q}_1=\bigoplus_{x\in\mathcal{P},\,|x|=p}\operatorname{Ext}^{-q}(G, E^\circ_{x})^*\otimes E_{x} \quad\Longrightarrow\quad \mathcal{H}^{p+q}(G).
\end{equation}
\tag{2.3}
$$
2.2. Exceptional collections on Lagrangian GrassmanniansLet $V$ be a $2n$-dimensional vector space over $\mathsf{k}$ equipped with a nondegenerate skew-symmetric bilinear form. We denote by $\operatorname{LGr}(n, V)$ the Lagrangian Grassmannian of maximal isotropic subspaces in $V$, and by $\mathcal{U}$ the tautological rank $n$ bundle on $\operatorname{LGr}(n, V)$. The generator of the Picard group of $\operatorname{LGr}(n, V)$ is $\mathcal{O}(1)\simeq \Lambda^n\mathcal{U}^*$. The Lagrangian Grassmannian comes with an action of the symplectic group $\mathbf{G}=\operatorname{Sp}_{2n}$. We denote by $\mathsf{P}$ the set5 of weakly decreasing integer sequences of length $n$:
$$
\begin{equation*}
\mathsf{P}=\{\lambda\in \mathbb{Z}^n \mid \lambda_1\geqslant \lambda_2\geqslant \dots\geqslant \lambda_n\}.
\end{equation*}
\notag
$$
Given $\lambda\in\mathsf{P}$, we denote by $\Sigma^\lambda$ the corresponding Schur functor. We follow the convention under which $\Sigma^{(p, 0, \dots, 0)}=S^p$. 2.2.1. Exceptional blocksIn what follows we consider subcategories in equivariant derived categories. Since a collection of equivariant objects generate a full triangualated subcategory both in the usual and the equivariant derived category, in the latter case we will use an additional subscript and write $\langle\,{\cdot}\,\rangle_\mathbf{G}$. It is well known that every irreducible equivariant vector bundle on $\operatorname{LGr}(n, V)$ is isomorphic to $\Sigma^\lambda\mathcal{U}^*$ for some $\lambda\in\mathsf{P}$. Moreover, these form an infinite full exceptional collection in the equivariant derived category:
$$
\begin{equation*}
D^b_\mathbf{G}(\operatorname{LGr}(n, V))=\langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\in \mathsf{P}^{\circ}\rangle_\mathbf{G},
\end{equation*}
\notag
$$
where $\mathsf{P}$ is treated as an infinite poset with the partial order6 given by
$$
\begin{equation*}
\lambda\preceq \mu \quad \text{if and only if}\quad \lambda_i\leqslant \mu_i \quad\text{for all } i=1,\dots, n,
\end{equation*}
\notag
$$
and $\mathsf{P}^\circ$ is its opposite. Any subset $S\subseteq \mathsf{P}$ with the induced partial order produces an exceptional collection $\langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\in S^{\circ}\rangle_\mathbf{G}$. If $S$ is finite and graded, then we denote by
$$
\begin{equation*}
\langle \mathcal{E}^\lambda \mid \lambda\in S\rangle_\mathbf{G} \quad \text{and} \quad \langle \mathcal{F}^\lambda \mid \lambda\in S\rangle_\mathbf{G}
\end{equation*}
\notag
$$
the graded right and left duals in $D^b_\mathbf{G}(\operatorname{LGr}(n, V))$ to $\langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\in S^{\circ}\rangle_\mathbf{G}$, respectively.7 Kuznetsov and Polishchuk came up with a very simple (yet hard to check in practice) condition under which the objects $\mathcal{E}^\lambda$ form an exceptional collection in the nonequivariant category (we do not distinguish between objects in the equivariant derived category and their images under the forgetful functor). Definition 2.1 (see [7], Definition 3.1). A subset $S\subset \mathsf{P}$ is called an exceptional block if for all $\lambda,\mu\in S$ the canonical map In plain words the block condition says that every extension between a pair of objects can uniquely be decomposed into a sum of equivariant extensions followed by homomorphisms. What is rather surprising is that even though the original objects $\Sigma^\lambda \mathcal{U}^*$ for $\lambda\in S$ almost never form an exceptional collection in the nonequivariant derived category, the objects of the right dual do form an exceptional collection in the nonequivariant derived category as long as $S$ is a block. Proposition 2.2 (see [7], Proposition 3.9). If $S\subset\mathsf{P}$ is an exceptional block, then the corresponding right dual objects form an exceptional collection in $D^b(\operatorname{LGr}(n, V))$, It turns out that it is natural to call exceptional blocks right exceptional blocks, and that it is useful to consider also left exceptional blocks. Definition 2.2 (see [11], Remark 2.12). A subset $S\subset \mathsf{P}$ is called a left exceptional block if for all $\lambda,\mu\in S$ the canonical map Recall that both the equivariant and nonequivariant categories have dualization functors compatible with the forgetful functor, sending $E$ to $\mathsf{RHom}(E, \mathcal{O})$, which is an anti-autoequivalence. Any anti-autoequivalence takes exceptional collections to exceptional collections (with respect to the opposite order) and left and right (graded) dual collections to right and left (graded) dual collections, respectively, so we immediately see that $S$ is a right exceptional block if and only if $-S$ is a left exceptional block, where $-S = \{-\lambda \mid \lambda \in S\}$ and $-\lambda = (-\lambda_n, -\lambda_{n-1},\dots, -\lambda_1)$. The latter follows from the isomorphism $(\Sigma^\lambda\mathcal{U}^*)^*\simeq\Sigma^{-\lambda}\mathcal{U}^*$. We conclude that the following statement, which is dual to Proposition 2.2, holds. Proposition 2.3. If $S\subset \mathsf{P}$ is a left exceptional block, then the corresponding left dual objects form an exceptional collection in $D^b(\operatorname{LGr}(n, V))$, Moreover, $\mathcal{F}^\lambda \simeq (\mathcal{E}^{-\lambda})^*$. The following theorem uses the term semiorthogonal decomposition. Recall that full triangulated subcategories $\mathcal{T}_1,\mathcal{T}_2\subseteq\mathcal{T}$ are called semiorthogonal (one writes $\mathcal{T}_1\subseteq\mathcal{T}_2^\perp$) if $\operatorname{Hom}_\mathcal{T}(X_2, X_1)=0$ for all $X_1\in\mathcal{T}_1$ and $X_2\in\mathcal{T}_2$. A semiorthogonal decomposition of a category $\mathcal{T}$ is a collection of full triangulated subcategories $\mathcal{T}_1,\mathcal{T}_2,\dots,\mathcal{T}_r$ such that $\mathcal{T}_i\subseteq \mathcal{T}_j^\perp$ for all $1\leqslant i < j\leqslant r$ and $\mathcal{T}$ is the smallest strictly full triangulated subcategory containing all the $\mathcal{T}_i$. The notation for a semiorthogonal decomposition is $\mathcal{T}=\langle\mathcal{T}_1,\mathcal{T}_2,\dots,\mathcal{T}_r\rangle$. Finally, for a full triangulated subcategory $\mathcal{B}\in D^b(\operatorname{LGr}(n, V))$ we denote by $\mathcal{B}(i)$ its image under the autoequivalence given by tensor product with the line bundle $\mathcal{O}(i)$. Theorem 2.1 (see [7], Theorem 9.2, and [11], Theorem 4). The set is a right exceptional block for all $h=0,\dots,n$. Moreover, there is a semiorthogonal decomposition where $\mathcal{B}_h = \langle \mathcal{E}^\lambda_{B_h} \mid \lambda \in B_h\rangle$. Let us make a few remarks. First, $B_h\cong\mathrm{Y}_{h, n-h}$ as (graded) posets. Second, the object $\mathcal{E}^\lambda_{B_h}$ for $\lambda\in\mathrm{Y}_{h, n-h}$ has the following homological description. It is an equivariant object such that
$$
\begin{equation}
\begin{gathered} \, \mathcal{E}^\lambda_{B_h} \in \langle \Sigma^\mu\mathcal{U}^*\mid \mu \subseteq \lambda\rangle_\mathbf{G} \subset D^b_\mathbf{G}(\operatorname{LGr}(n, V)), \\ \operatorname{Ext}^\bullet_\mathbf{G}(\mathcal{E}^\lambda_{B_h}, \Sigma^\mu\mathcal{U}^*) = \begin{cases} \mathsf{k}[-|\lambda|] & \text{if}\ \mu=\lambda, \\ 0 & \text{if}\ \mu\subsetneq\lambda. \end{cases} \end{gathered}
\end{equation}
\tag{2.5}
$$
In particular, it depends only on $\lambda$ (one only needs to know that $\lambda\in\mathrm{Y}_{h,n-h}$ for some $0\leqslant h \leqslant n$), so we will abbreviate $\mathcal{E}^\lambda=\mathcal{E}^\lambda_{B_h}$. Finally, by using the dualization anti-autoequivalence we obtain a semiorthogonal decomposition
$$
\begin{equation*}
D^b(\operatorname{LGr}(n, V))=\langle \mathcal{C}_n(-n), \mathcal{C}_{n-1}(-n+1),\dots, \mathcal{C}_1(-1), \mathcal{C}_0\rangle,
\end{equation*}
\notag
$$
where $\mathcal{C}_h = \langle \mathcal{F}^\lambda_{-B_h} \mid \lambda \in -B_h\rangle$, and $\mathcal{F}^\lambda$ for $\lambda\in\mathrm{Y}_{h, n-h}$ is an equivariant object which can be characterized by the following properties:
$$
\begin{equation}
\begin{gathered} \, \mathcal{F}^\lambda_{-B_h} \in \langle \Sigma^\mu\mathcal{U}\mid \mu \subseteq \lambda\rangle_\mathbf{G} \subset D^b_\mathbf{G}(\operatorname{LGr}(n, V)), \\ \operatorname{Ext}^\bullet_\mathbf{G}(\Sigma^\mu\mathcal{U}, \mathcal{F}^\lambda_{-B_h})= \begin{cases} \mathsf{k}[-|\lambda|] & \text{if}\ \mu=\lambda, \\ 0 & \text{if}\ \mu\subsetneq\lambda. \end{cases} \end{gathered}
\end{equation}
\tag{2.6}
$$
Again, since $\mathcal{F}^\lambda_{-B_h}$ depends only on $\lambda$, we will drop the subscript and write $\mathcal{F}^\lambda$. 2.2.2. Geometric constructionsWe need a geometric construction for the objects $\mathcal{F}^\lambda$ obtained in [11]. Since the cases $h=0$ and $h=n$ are trivial ($B_0$ and $B_n$ consist of a single weight $\lambda = (0, 0, \dots, 0)$, and $\mathcal{F}^\lambda=\mathcal{E}^\lambda=\mathcal{O}$), let us fix $h$, $0 < h < n$. Consider the diagram where $\operatorname{IGr}(h, V)$ denotes the Grassmannian of rank $h$ isotropic subspaces of $V$ and $\operatorname{IFl}(h, n; V)$ is the partial isotropic flag variety. Denote by $\mathcal{W}$ the universal rank $h$ bundle on $\operatorname{IGr}(h, V)$, as well as its pullback on $\operatorname{IFl}(h, n; V)$.Proposition 2.4 (see [11], Lemma 3.4 and Proposition 3.6). The bundles $\langle \Sigma^{\mu^\top}\!\mathcal{W}^* \mid \mu\in \mathrm{Y}_{n-h,h}\rangle$ form an exceptional collection in the nonequivariant derived category $D^b(\operatorname{IGr}(h, V))$. Denote its graded left dual by $\langle \mathcal{G}^\lambda \mid \lambda \in \mathrm{Y}_{n-h, h}^\circ\rangle$. Then From now on we will often identify the graded posets $\mathrm{Y}_{h, n-h}$ and $\mathrm{Y}_{n-h,h}$ via transposition of diagrams. The graded dual exceptional collection conditions for $\langle \mathcal{G}^\lambda \mid \lambda \in \mathrm{Y}_{n-h, h}^\circ\rangle$ become
$$
\begin{equation}
\begin{gathered} \, \mathcal{G}^\lambda \in \langle \Sigma^{\mu}\mathcal{W}^* \mid \mu\in \mathrm{Y}_{h,n-h}\rangle, \\ \operatorname{Ext}^\bullet(\Sigma^{\mu}\mathcal{W}^*, \mathcal{G}^\lambda) = \begin{cases} \mathsf{k}[-|\lambda|] & \text{if}\ \mu^\top=\lambda, \\ 0 & \text{if}\ \mu\in\mathrm{Y}_{h, n-h}\ \text{and}\ \mu^\top\neq \lambda. \end{cases} \end{gathered}
\end{equation}
\tag{2.8}
$$
§ 3. Main resultsWe continue to use the notation introduced in § 2.2. Our first goal is to prove Theorem 3.1. 3.1. Dual exceptional collections on Lagrangian GrassmanniansWe recall that characterizations of the objects $\mathcal{E}^\lambda$ and $\mathcal{F}^\lambda$ were given in (2.5) and (2.6), respectively. The first main result of the paper is the following theorem. Theorem 3.1. Let $0\leqslant h \leqslant n$. The exceptional collection $\langle \mathcal{F}^\mu \mid \mu\in \mathrm{Y}_{n-h, h}^\circ \rangle$ is the graded left dual to $\langle \mathcal{E}^\lambda \mid \lambda\in \mathrm{Y}_{h, n-h} \rangle$. That is, for $\mu\in \mathrm{Y}_{n-h, h}$ one has The proof of Theorem 3.1 takes the rest of this subsection. Our strategy is to compare the objects $\mathcal{E}^\lambda$ with the graded right dual exceptional collection to $\langle \mathcal{F}^\mu \mid \mu\in \mathrm{Y}_{n-h, h}^\circ \rangle$. Since the cases $h=0$ and $h=n$ are trivial (all the objects considered are isomorphic to $\mathcal{O}$), we assume that $0 < h < n$. Lemma 3.1. Let $\nu\in \mathrm{Y}_{h,n-h}$. Then the following subcategories coincide in $D^b(\operatorname{LGr}(n, V))$: where $\langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\subseteq \nu \rangle$ is the smallest strictly full triangulated subcategory in $D^b(\operatorname{LGr}(n, V))$ containing the corresponding objects. Proof. Since the objects $\mathcal{E}^\lambda$ form a graded right dual exceptional collection to $\langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\subseteq \nu \rangle_{\mathbf{G}}$ in $D^b_\mathbf{G}(\operatorname{LGr}(n, V))$, one has $\langle \mathcal{E}^\lambda \mid \lambda\subseteq \nu \rangle_\mathbf{G} = \langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\subseteq \nu \rangle_\mathbf{G}$. Once we apply the forgetful functor from $D^b_\mathbf{G}(\operatorname{LGr}(n, V))$ to $D^b(\operatorname{LGr}(n, V))$, we see that $\langle \mathcal{E}^\lambda \mid \lambda\subseteq \nu \rangle = \langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\subseteq \nu \rangle$ in the nonequivariant derived category.8 In a similar fashion one proves that
$$
\begin{equation*}
\langle \mathcal{F}^\mu \mid \mu\subseteq \nu^\top \rangle=\langle \Sigma^\mu\mathcal{U} \mid \mu\subseteq \nu^\top \rangle.
\end{equation*}
\notag
$$
It remains to show that $\langle \Sigma^\mu\mathcal{U} \mid \mu\subseteq \nu^\top \rangle = \langle \Sigma^\lambda\mathcal{U}^* \mid \lambda\subseteq \nu \rangle$, which follows from Lemma 3.7 in [11]. The lemma is proved. Proof of Theorem 3.1. Consider the isotropic Grassmannian $\operatorname{IGr}(h, V)$ with its tautological bundle $\mathcal{W}$. Fix $\nu\in \mathrm{Y}_{h,n-h}$ and consider the graded poset $\mathcal{P} = \{\lambda \mid \lambda \subseteq \nu\}$. By Proposition 2.4 the bundles $\langle \Sigma^\lambda\mathcal{W}^* \mid \lambda \in \mathcal{P}\rangle$ form an exceptional collection in $D^b(\operatorname{IGr}(h, V))$. Let us apply Lemma 2.2 in the following setup. Put $\mathcal{T} = \langle \Sigma^\lambda\mathcal{W}^* \mid \lambda \in \mathcal{P}\rangle$, $E_\lambda = \Sigma^\lambda\mathcal{W}^*$, $G_\lambda = \mathcal{G}_\lambda$, $\mathcal{T}'=D^b(\operatorname{LGr}(n, V))$ and $F = p_*q^*$, where $p$ and $q$ are as in (2.7). Denote by $\langle \widetilde{\mathcal{E}}^\lambda \mid \lambda\in \mathcal{P} \rangle$ the graded right dual collection to $\langle \mathcal{F}^\mu \mid \mu\subseteq \nu^\top \rangle$ in the nonequivariant derived category. By Proposition 2.4, $\mathcal{F}^\lambda = F(G_\lambda)$, so Lemma 2.2 is applicable. The conclusion is that $\operatorname{Ext}^\bullet(\widetilde{\mathcal{E}}^\lambda, p_*\Sigma^\mu\mathcal{W}^*)\simeq \operatorname{Ext}^\bullet(\Sigma^\lambda\mathcal{W}^*, \Sigma^\mu\mathcal{W}^*)$. A simple Borel-Bott-Weil computation shows that $p_*\Sigma^\mu\mathcal{W}^*\simeq \Sigma^\mu\mathcal{U}^*$ for $\mu\subseteq \nu$ (see [11], Lemma A.4). Finally, for $\mu\subseteq \nu$ we conclude that
Recall that our goal is to prove that $\widetilde{\mathcal{E}}^\nu\simeq \mathcal{E}^\nu$. It was shown in [7], Corollary 3.8, that for all $\lambda, \mu\in \mathcal{P}$ there is an isomorphism
$$
\begin{equation*}
\operatorname{Ext}^\bullet(\mathcal{E}^\lambda, \Sigma^\mu\mathcal{U}^*) \simeq \operatorname{Hom}(\Sigma^\lambda\mathcal{U}^*, \Sigma^\mu\mathcal{U}^*).
\end{equation*}
\notag
$$
It follows from the Lagrangian Borel-Bott-Weil theorem (see § 3.2.2) that
$$
\begin{equation*}
\operatorname{Hom}(\Sigma^\lambda\mathcal{U}^*, \Sigma^\mu\mathcal{U}^*)= \begin{cases} \mathsf{k} & \text{if}\ \lambda=\mu, \\ 0 & \text{if}\ \lambda\not\subseteq\mu. \end{cases}
\end{equation*}
\notag
$$
We conclude that
$$
\begin{equation}
\operatorname{Ext}^\bullet(\mathcal{E}^\nu, \Sigma^\mu\mathcal{U}^*)= \begin{cases} \mathsf{k} & \text{if}\ \mu=\nu, \\ 0 & \text{if}\ \mu\subsetneq \nu. \end{cases}
\end{equation}
\tag{3.2}
$$
Let $\nu\in\mathrm{Y}_{h, n-h}$. By Lemma 3.1 we know that $\widetilde{\mathcal{E}}^\nu, \mathcal{E}^\nu\in \langle \Sigma^\mu\mathcal{U}^* \mid \mu\subseteq \nu \rangle$. Choose a nontrivial element $\phi\in \operatorname{Ext}^\bullet(\mathcal{E}^\nu, \Sigma^\nu\mathcal{U}^*)\simeq \mathsf{k}$. It follows from the discussion preceding Corollary 3.8 in [7] that one has an exact triangle in $D^b(\operatorname{LGr}(n, V))$ of the form
$$
\begin{equation*}
\mathcal{E}^\nu \xrightarrow{\phi} \Sigma^\nu\mathcal{U}^*\to C_\phi\to \mathcal{E}^\nu[1],
\end{equation*}
\notag
$$
where the cone $C_\phi$ is in $\langle \Sigma^\mu\mathcal{U}^* \mid \mu\subsetneq \nu\rangle$. Applying the functor $\operatorname{Hom}(\widetilde{\mathcal{E}}^\nu,\cdot\,)$ to the previous triangle, from (3.1) we obtain a morphism $\psi\colon\widetilde{\mathcal{E}}^\nu\to \mathcal{E}^\nu$, which lifts a nontrivial element $\xi\in \operatorname{Hom}(\widetilde{\mathcal{E}}^\nu, \Sigma^\nu\mathcal{U}^*)\simeq \mathsf{k}$. Consider the cone of $\psi$:
$$
\begin{equation*}
\widetilde{\mathcal{E}}^\nu \xrightarrow{\psi} \mathcal{E}^\nu\to C_\psi\to \widetilde{\mathcal{E}}^\nu[1].
\end{equation*}
\notag
$$
On the one hand $C_\psi \in \langle \Sigma^\mu\mathcal{U}^* \mid \mu\subseteq \nu \rangle$ since both $\widetilde{\mathcal{E}}^\nu$ and $ \mathcal{E}^\nu$ belong to this subcategory. On the other hand it follows from the construction of $\psi$ and formulae (3.1) and (3.2) that $\operatorname{Ext}^\bullet(C_\psi, \Sigma^\mu\mathcal{U}^*) = 0$ for all $\mu\subseteq \nu$. Thus, $C_\psi=0$, and $\psi$ is an isomorphism. Theorem 3.1 is proved. Remark 3.1. Theorem 3.1 shows that the semiorthogonal decomposition (2.4) is self-dual up to a twist. Namely, if one applies the dualization anti-autoequivalence followed by the twist by $\mathcal{O}(n)$, then the $h$th block of the resulting decomposition is generated by the objects $\langle (\mathcal{E}^{\lambda})^* \mid \lambda \in \mathrm{Y}_{n-h, h}\rangle$. Since $(\mathcal{E}^{\lambda})^* \simeq \mathcal{F}^\lambda$, we see that the exceptional collection generating block $n-h$ is mapped to the graded left dual of the exceptional collection generating block $h$. Thus, the decomposition (2.4) is mapped to itself. Since $\mathcal{F}^\lambda$ form a graded left dual exceptional collection to $\mathcal{E}^\lambda$, there is a spectral sequence associated with it; namely, the spectral sequence (2.3) takes the following form. Corollary 3.1. For any object $G$ in the subcategory $\langle \mathcal{E}^\lambda \mid \lambda \in \mathrm{Y}_{h, n-h}\rangle$ there is a spectral sequence of the form In the following section we use this spectral sequence to prove Theorem 3.3. 3.2. Resolutions of irreducible equivariant bundlesIn the final subsection we produce nice resolutions for irreducible equivariant bundles on $\operatorname{LGr}(n, V)$ of the from $\Sigma^\lambda \mathcal{U}^*$ for $\lambda\in\mathrm{Y}_{h, n-h}$, for some $0\leqslant h\leqslant n$, in terms of bundles of the form $\mathcal{E}^\mu$. The proof is presented in § 3.2.5, and before that we introduce some material required for our computations. 3.2.1. Balanced diagramsLet $\lambda$ be a Young diagram. Most commonly, it is represented by a sequence of integers $\lambda=(\lambda_1,\dots,\lambda_k)$ such that $\lambda_1\geqslant\lambda_2\geqslant \dots \geqslant \lambda_k\geqslant 0$, where $\lambda_i$ is the length of the $i$th row of $\lambda$. There is an alternative description in terms of hook lengths. Let us say that $\lambda$ has rank9 $s$ if $\lambda_s\geqslant s$ and $\lambda_{s+1}\leqslant s$. Graphically, $s$ is the size of the largest square that fits into $\lambda$, or the length of the diagonal of $\lambda$. Let $a_i$ and $b_i$ denote the number of boxes to the right of the $i$th diagonal box (including itself) and below it (including itself), respectively. Classically, $a_i$ and $b_i$ are called the arm and leg length, respectively. One can alternatively say that $a_i=\lambda_i-(i-1)$ and $b_i=\max \{1\leqslant j\leqslant k\mid \lambda_j\geqslant i\} - (i-1)$. Equivalently, $b_i=(\lambda^\top)_i - (i-1)$. We write $\lambda=(a_1, a_2,\dots, a_s\mid b_1, b_2,\dots, b_s)$ if $\lambda$ has rank $s$ and its arm and leg lengths are $a_i$ and $b_j$, respectively. Definition 3.1. A diagram $\lambda=(a_1, a_2,\dots, a_s\mid b_1, b_2,\dots, b_s)$ is called balanced if for every $1\leqslant i\leqslant s$. The set of balanced diagrams with $2t$ boxes is denoted by $\mathrm{B}_{2t}$.10 Balanced diagrams are important to us for the following reason. Proposition 3.1 (see [13], Proposition 2.3.9). Let $E$ be a locally free module over a commutative ring of characteristic zero. Then Lemma 3.2. A diagram $\lambda=(\lambda_1,\dots,\lambda_k)$ of rank $s$ is balanced if and only if ${\lambda_s\geqslant s+1}$ and where, in particular, we assume that $\lambda_s\geqslant s+1$. Proof. Note that
$$
\begin{equation*}
(a_1, a_2,\dots, a_s\mid b_1, b_2,\dots, b_s)^\top=(b_1, b_2,\dots, b_s\mid a_1, a_2,\dots, a_s).
\end{equation*}
\notag
$$
In particular, $\lambda$ is symmetric, $\lambda=\lambda^\top$, if and only if $a_i=b_i$ for all $1\leqslant i\leqslant s$. It follows from Definition 3.1 that $\lambda$ is balanced if and only if $\lambda_s\geqslant s+1$ and $\mu=(\lambda_1-1, \lambda_2-1,\dots, \lambda_s-1, \lambda_{s+1},\dots, \lambda_{k-1},\lambda_k)$ is symmetric. Assume the latter.
The rank of $\mu$ equals that of $\lambda$, which is $s$. One can partition $\mu$ into three parts: a square of size $s$, everything to the right of it, and everything below it. The latter two parts are nothing but the diagrams $\mu^r = (\mu_1-s,\mu_2-s,\dots,\mu_s-s)$ and $\mu^b = (\mu_{s+1},\mu_{s+2},\dots,\mu_k)$. Under transposition the square maps to itself, while $\mu^r$ and $\mu^b$ get interchanged and transposed: $(\mu^\top)^r=(\mu^b)^\top$ and $(\mu^\top)^b=(\mu^r)^\top$. One immediately concludes that $\mu$ is symmetric if and only if $(\mu^r)^\top = \mu^b$. The lemma is proved. 3.2.2. The Lagrangian Borel-Bott-Weil theoremThe celebrated Borel-Bott-Weil theorem fully describes the cohomology of irreducible equivariant line bundles on rational homogeneous varieties. Since we only use it in one place, we formulate a much simplified version of it and refer the interested reader to Ch. 4 in [13]. Given a weakly decreasing sequence $\lambda\in\mathbb{Z}^n$, we denote by $-\lambda$ the sequence $(-\lambda_n, -\lambda_{n-1}, \dots, -\lambda_1)$, which is again weakly decreasing. The sum of weakly decreasing sequences is defined termwise. A weakly decreasing sequence is called nonsingular if the absolute values of all of its terms are positive and distinct. If $\lambda$ is nonsingular, then we denote by $\|\lambda\|$ the sequence obtained by taking all the absolute values of the terms of $\lambda$ and writing them in decreasing order. If $\lambda$ is nonsingular, then $\|\lambda\|-\rho$ is a weakly decreasing sequence with nonnegative terms, where $\rho=(n, n-1, \dots, 1)$. The set of weakly decreasing sequences $\mu\in\mathbb{Z}^n$ with nonnegative terms is identified with the set of dominant weights of $\operatorname{Sp}(V)$, where $V$ is a $2n$-dimensional symplectic vector space. We denote by $V^{\langle\mu\rangle}$ the corresponding irreducible representation of $\operatorname{Sp}(V)$. For instance, if $\mu=(0, 0,\dots, 0)$, then $V^{\langle\mu\rangle}=\mathsf{k}$. Theorem 3.2 (Lagrangian Borel-Bott-Weil theorem). Let $\lambda\in\mathbb{Z}^n$ be an integer weakly decreasing sequence. If $-\lambda+\rho$ is nonsingular, then where $\ell$ equals the number of negative terms in $-\lambda+\rho$ added with the number of pairs $1\leqslant i < j\leqslant n$ such that $(-\lambda+\rho)_i + (-\lambda+\rho)_j < 0$. Otherwise, the cohomology vanishes: $H^\bullet(\operatorname{LGr}(n, V), \Sigma^\lambda\mathcal{U})=0$. 3.2.3. A vanishing lemmaWe need the following result, which is definitely well known to experts. A proof is included for the sake of completeness. Proof. The proof is a direct application of Theorem 3.2. In order to show (1) we need to show that the weight $-\lambda+\rho$ is nonsingular if and only if $\lambda$ is balanced. Consider the strictly decreasing sequence
$$
\begin{equation}
-\lambda+\rho=\bigl(n-\lambda_n,\,(n-1)-\lambda_{n-1},\,\dots,\,2-\lambda_2,\,1-\lambda_1\bigr).
\end{equation}
\tag{3.4}
$$
Since $0\leqslant \lambda_i\leqslant n+1$ for all $i$, all the absolute values of the terms of $-\lambda+\rho$ are between $0$ and $n$. Recall that $-\lambda+\rho$ is nonsingular if and only if all the absolute values of its terms are distinct and positive. Assume the latter. Let $s$ denote the rank of $\lambda$. Then the first $n-s$ terms of (3.4) are positive, while the last $s$ terms are nonpositive. Since the absolute values of the latter cannot be zero (the weight being nonsingular), we conclude that $\lambda_i\geqslant s+1$ for $i=1,\dots,s$. Put $\mu_i=\lambda_i-(s+1)$. Then the sequence
$$
\begin{equation}
\bigl(n-\lambda_n,\,(n-1)-\lambda_{n-1},\, \dots,\,(s+1)-\lambda_{s+1},\,s+\mu_1,\,(s-1)+\mu_2,\,\dots,\, 1+\mu_s\bigr)
\end{equation}
\tag{3.5}
$$
differs from (3.4) only in the last $s$ terms: their signs have been changed, and their order has been reversed. Note that $(\lambda_{s+1}, \lambda_{s+2}, \dots, \lambda_n)=\lambda^b\in \mathrm{Y}_{n-s,s}$, while $\mu\in \mathrm{Y}_{s, n-s}$. The sequence (3.5) is very well known to anyone who has ever studied Kapranov’s work. It follows from § 2.7 in [3] that all terms in (3.5) are distinct if and only if $\lambda^b=\mu^\top$. The latter is equivalent, by Lemma 3.2, to saying that $\lambda$ is balanced.
Let us turn to (2). Assume that $\lambda$ is balanced. Since the absolute values of the terms of $-\lambda+\rho$ take all values between $1$ and $n$, by the Borel-Bott-Weil theorem the only nonzero cohomology is equal to $\mathsf{k}$. We only need to determine the degree in which it sits, and it equals the number of negative terms in $-\lambda+\rho$ plus the number of pairs $1\leqslant i < j\leqslant n$ such that $(-\lambda+\rho)_i + (-\lambda+\rho)_j < 0$. The first number equals $s$, and we need to compute the second. Note that if $1\leqslant j \leqslant n-s$, then both $(-\lambda+\rho)_i$ and $(-\lambda+\rho)_j$ are positive. Thus, we may assume that $n-s < j \leqslant n$. If $n-s < i\leqslant n$, then any $i<j\leqslant n$ contributes to the count since both terms are negative. There are $\frac{s(s-1)}{2}$ such pairs. Finally, assume that $1\leqslant i\leqslant n-s$ and $n-s < j\leqslant n$. Since $(-\lambda+\rho)_j < 0$, we need to determine when $(-\lambda+\rho)_i < -(-\lambda+\rho)_j$. Such pairs correspond precisely to inversions in the sequence (3.5). It is easy to show that the number of those equals $|\lambda^b|$ (see [3], § 2.7). We conclude that for a balanced $\lambda$ the only nontrivial cohomology sits in degree $s+s(s-1)/{2}+|\lambda^b|=s(s+1)/{2}+|\lambda^b|$. It remains to recall that $\lambda$ consists of $\lambda^b$, $\mu=(\lambda^b)^\top$, and a rectangle of size $s\times (s+1)$, so $|\lambda| = |\lambda^b| + |\mu| + s(s+1)=2|\lambda^b| + s(s+1)$. The lemma is proved. The following is a simple relative version of the previous lemma. As usual, all functors are derived. Corollary 3.2. Let $X$ be smooth projective variety, and let $\mathcal{V}$ be a rank $2k$ symplectic bundle on $X$. Let $\pi\colon\operatorname{LGr}_X(k, \mathcal{V})\to X$ be the relative Lagrangian Grassmannian. Denote by $\mathcal{U}\subset \pi^*\mathcal{V}$ the tautological bundle. Let $\nu \in \mathrm{Y}_{k, k+1}$. 3.2.4. Skew Schur functorsBefore we prove the second main result of the paper, we need to say a few words about skew Young diagrams. Let $\lambda$ and $\mu$ be two Young diagrams such that $\mu\subseteq \lambda$. Then one can define the skew Schur functor $\Sigma^{\lambda/\mu}$; see [13], § 2.1. The definition itself is not important for our paper. What is important is the decomposition (3.6). If $E$ is a free module over a commutative ring of characteristic zero and $\alpha$ and $\beta$ are two Young diagrams, then one has the direct sum decomposition
$$
\begin{equation*}
\Sigma^\alpha E\otimes \Sigma^\beta E\simeq \bigoplus_{|\kappa|=|\alpha|+|\beta|}(\Sigma^\kappa E)^{\oplus c(\alpha,\beta;\kappa)},
\end{equation*}
\notag
$$
where $c(\alpha,\beta;\kappa)$ are the celebrated Littlewood-Richardson coefficients. It turns out that one has a direct sum decomposition for skew Schur functors controlled by the same coefficients. Namely,
$$
\begin{equation}
\Sigma^{\lambda/\mu}E \simeq \bigoplus_{|\nu|+|\mu|=|\lambda|}(\Sigma^\nu E)^{\oplus c(\nu, \mu; \lambda)}
\end{equation}
\tag{3.6}
$$
(see [13], Theorem 2.3.6). If $c(\nu, \mu; \lambda)>0$, then we write $\nu\subseteq \lambda/\mu$. Observe that $\nu\subseteq \lambda/\mu$ implies that $\nu\subseteq \lambda$. In particular, if $\lambda\in \mathrm{Y}_{h,w}$, then $\nu\subseteq \lambda/\mu$ implies that $\nu\in\mathrm{Y}_{h, w}$. Inspired by the decomposition (3.6), for a pair of diagrams $\lambda, \mu\in \mathrm{Y}_{h,w}$, where $h+w=n$ and $\mu\subseteq\lambda$, we put
$$
\begin{equation}
\mathcal{E}^{\lambda/\mu}=\bigoplus_{\nu\subseteq \lambda/\mu}(\mathcal{E}^\nu)^{\oplus c(\nu, \mu; \lambda)}.
\end{equation}
\tag{3.7}
$$
We extend the definition, as usual, by setting $\mathcal{E}^{\lambda/\mu} = 0$ for $\mu\nsubseteq\lambda$. 3.2.5. ResolutionsWe are ready to present the main application of Theorem 3.1. Theorem 3.3. Let $\lambda\in\mathrm{Y}_{h,n-h}$ for some $0\leqslant h\leqslant n$. Then there is an exact $\operatorname{Sp}(V)$-equivariant sequence of vector bundles on $\operatorname{LGr}(n, V)$ of the form Proof. The vector bundle $\Sigma^\lambda\mathcal{U}^*$ lies in the subcategory $\langle \mathcal{E}^\lambda \mid \lambda \in \mathrm{Y}_{h, n-h}\rangle$. By Theorem 3.1 the collection $\langle \mathcal{F}^{\mu^\top} \mid \mu\in \mathrm{Y}_{h, n-h}\rangle$ is the graded left dual to the exceptional collection $\langle \mathcal{E}^\lambda \mid \lambda \in \mathrm{Y}_{h, n-h}\rangle$. The corresponding spectral sequence (3.3) is of the form
Let us compute $\operatorname{Ext}^\bullet(\Sigma^\lambda\mathcal{U}^*, \mathcal{F}^{\mu})$ for $\mu=\alpha^\top\in \mathrm{Y}_{n-h, h}$ using the isomorphism $\mathcal{F}^\mu\simeq p_*q^*\mathcal{G}^\mu$ given in Proposition 2.4, where $p$ and $q$ come from the diagram In the following we often simply write $\mathcal{U}$ instead of $p^*\mathcal{U}$ and $\Sigma^\lambda\mathcal{U}^*$ instead of $p^*\Sigma^\lambda\mathcal{U}^*$. One has a sequence of isomorphisms
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Ext}^\bullet(\Sigma^\lambda\mathcal{U}^*, \mathcal{F}^\mu) & \simeq \operatorname{Ext}^\bullet(\Sigma^\lambda\mathcal{U}^*, p_*q^*\mathcal{G}^\mu) \simeq \operatorname{Ext}^\bullet(\Sigma^\lambda\mathcal{U}^*, q^*\mathcal{G}^\mu) \\ & \simeq H^\bullet(\operatorname{IFl}(h, n; V), \Sigma^\lambda\mathcal{U}\otimes q^*\mathcal{G}^\mu) \simeq H^\bullet(\operatorname{IGr}(h; V), q_*(\Sigma^\lambda\mathcal{U})\otimes \mathcal{G}^\mu), \end{aligned}
\end{equation*}
\notag
$$
where the second isomorphism is given by adjunction, and the last one follows from the projection formula. As usual, all functors are derived.
Consider the spectral sequence of the composition of derived functors. Its second page is given by
$$
\begin{equation}
H^i(\operatorname{IGr}(h; V), R^jq_*(\Sigma^\lambda\mathcal{U})\otimes \mathcal{G}^\mu) \quad \Longrightarrow\quad H^{i+j}(\operatorname{IGr}(h; V), q_*\Sigma^\lambda\mathcal{U}\otimes \mathcal{G}^\mu).
\end{equation}
\tag{3.10}
$$
Let us compute $R^jq_*(\Sigma^\lambda\mathcal{U})$. Recall that we denote by $\mathcal{W}\subset \mathcal{U}$ the universal flag on $\operatorname{IFl}(h, n; V)$. There is a filtration of length $|\lambda|$ on $\Sigma^\lambda\mathcal{U}$ associated with the short exact sequence $0\to \mathcal{W}\to \mathcal{U}\to \mathcal{U}/\mathcal{W}\to 0$, whose $k$th associated quotient is isomorphic to
$$
\begin{equation}
\bigoplus_{\nu\subseteq \lambda,\,|\nu|=k}\Sigma^{\lambda/\nu}\mathcal{W}\otimes \Sigma^\nu(\mathcal{U}/\mathcal{W}),
\end{equation}
\tag{3.11}
$$
where if the height of $\nu$ is greater than $n-h$, then $\Sigma^\nu(\mathcal{U}/\mathcal{W})=0$ by convention.
Since for any $\nu\subseteq \lambda$ the bundle $\Sigma^{\lambda/\nu}\mathcal{W}$ is pulled back from $\operatorname{IGr}(h, V)$, we conclude by the projection formula that
$$
\begin{equation}
R^jq_*\bigl(\Sigma^{\lambda/\nu}\mathcal{W}\otimes \Sigma^\nu(\mathcal{U}/\mathcal{W})\bigr) \simeq \Sigma^{\lambda/\nu}\mathcal{W}\otimes R^jq_*\Sigma^\nu(\mathcal{U}/\mathcal{W}).
\end{equation}
\tag{3.12}
$$
Recall that the flag variety $\operatorname{IFl}(h, n; V)$ is isomorphic to $\operatorname{LGr}(n-h, \mathcal{W}^\perp/\mathcal{W})$, the relative Lagrangian Grassmannian over $\operatorname{IGr}(h, V)$. Since $\nu\subseteq \lambda$, one has $\nu\in\mathrm{Y}_{h,n-h}$. Thus, $\Sigma^{\lambda/\nu}\mathcal{W}\otimes R^jq_*\Sigma^\nu(\mathcal{U}/\mathcal{W})=0$ unless $\nu\in\mathrm{Y}_{n-h, n-h}$. Finally, if $\nu\in\mathrm{Y}_{n-h, n-h}$, then we are in the situation when we can apply Corollary 3.2. We conclude that if ${\nu\subseteq\lambda}$, then
$$
\begin{equation}
\Sigma^{\lambda/\nu}\mathcal{W}\otimes R^\bullet q_*\Sigma^\nu(\mathcal{U}/\mathcal{W}) \simeq \begin{cases} \Sigma^{\lambda/\nu}\mathcal{W}[-t] & \text{if}\ \nu\in B_{2t}, \\ 0 & \text{otherwise.} \end{cases}
\end{equation}
\tag{3.13}
$$
Using (3.12) and (3.13), we can compute $R^jq_*(\Sigma^\lambda\mathcal{U})$. Indeed, the spectral sequence associated with the filtration with the quotients (3.11) degenerates in the first page, and
$$
\begin{equation*}
R^jq_*(\Sigma^\lambda\mathcal{U}) \simeq \bigoplus_{\nu\subseteq \lambda,\, \nu\in \mathrm B_{2j}}\Sigma^{\lambda/\nu}\mathcal{W}.
\end{equation*}
\notag
$$
We are ready to get back to the spectral sequence (3.10). We have
$$
\begin{equation*}
\begin{aligned} \, & H^i(\operatorname{IGr}(h; V), R^jq_*(\Sigma^\lambda\mathcal{U})\otimes \mathcal{G}^\mu) \simeq \bigoplus_{\nu\subseteq \lambda,\, \nu\in \mathrm B_{2j}} H^{i}(\operatorname{IGr}(h; V), \Sigma^{\lambda/\nu}\mathcal{W}\otimes \mathcal{G}^\mu) \\ &\qquad\simeq \bigoplus_{\nu\subseteq \lambda,\, \nu\in \mathrm B_{2j}} \operatorname{Ext}^{i}(\Sigma^{\lambda/\nu}\mathcal{W}^*, \mathcal{G}^\mu) \simeq \bigoplus_{\nu\subseteq \lambda,\, \nu\in \mathrm B_{2j}} \bigoplus_{\kappa\subseteq\lambda/\nu} \operatorname{Ext}^{i}((\Sigma^{\kappa}\mathcal{W}^*)^{\oplus c(\kappa,\nu;\lambda)}, \mathcal{G}^\mu), \end{aligned}
\end{equation*}
\notag
$$
where the last isomorphism comes from (3.6). Since $\kappa\subseteq \lambda/\nu$ implies that $\kappa\subseteq \lambda$, using (2.8) we see that $\operatorname{Ext}^{i}(\Sigma^{\kappa}\mathcal{W}^*, \mathcal{G}^\mu)$ is equal to $\mathsf{k}$ if $\kappa=\mu^\top$ and $i=|\mu|$, and is $0$ otherwise. Finally,
$$
\begin{equation*}
H^{|\mu|}(\operatorname{IGr}(h; V), R^jq_*(\Sigma^\lambda\mathcal{U})\otimes \mathcal{G}^\mu) \simeq \bigoplus_{\nu\in B_{2j}}\mathsf{k}^{\oplus c(\nu, \mu; \lambda)},
\end{equation*}
\notag
$$
where $2j=|\nu|=|\lambda|-|\mu|$, are the only potentially nontrivial cohomology groups, and the spectral sequence (3.10) degenerates. We conclude that
$$
\begin{equation*}
H^{|\mu|+j}(\operatorname{IGr}(h; V), q_*\Sigma^\lambda\mathcal{U}\otimes \mathcal{G}^\mu)= \bigoplus_{\nu\in B_{2j}}\mathsf{k}^{\oplus c(\nu, \mu; \lambda)}.
\end{equation*}
\notag
$$
Let us return to the spectral sequence (3.9). From the computations above we know that the only nontrivial terms in it are
$$
\begin{equation}
E^{-|\lambda|+t,\,|\lambda|-2t}_1 \simeq \bigoplus_{\nu\in B_{2t}}\bigoplus_{\substack{\mu\subseteq\lambda/\nu \\ |\mu|=|\lambda|-2t}}(\mathcal{E}^{\mu})^{\oplus c(\nu, \mu; \lambda)} =\bigoplus_{\nu\in B_{2t}} \mathcal{E}^{\lambda/\nu},
\end{equation}
\tag{3.14}
$$
where the last equality comes from the definition in (3.7). (One might suspect that there is merely a filtration with the given associated graded quotient; however, the right-hand side is a direct sum of objects of the form $\mathcal{E}^\alpha$, where $\alpha\subseteq \lambda$ and $|\alpha| = |\lambda|-2t$, and all such objects are orthogonal in the derived category since the elements in the corresponding poset are incomparable.) We conclude that the spectral sequence contains at most one nontrivial term on each diagonal.
As the spectral sequence converges to $\mathcal{H}^\bullet(\Sigma^\lambda\mathcal{U}^*)$, one must have a long exact sequence of the form
$$
\begin{equation*}
\dots \to E^{-|\lambda|+t,\,|\lambda|-2t}_1\to \dots \to E^{-|\lambda|+1,\,|\lambda|-2}_1 \to E^{-|\lambda|, |\lambda|}_1 \to \Sigma^\lambda\mathcal{U}^* \to 0.
\end{equation*}
\notag
$$
Since any balanced diagram contained in $\lambda$ has height at most $h$, its width is at most $h+1$ and $E^{-|\lambda|+t, |\lambda|-2t}_1=0$ for $t>h(h+1)/2$. It remains to use (3.14) to obtain the required long exact sequence (3.8). Theorem 3.3 is proved. Remark 3.2. One can also obtain the resolutions (3.8) by working in the equivariant derived category. Example 3.2. We present examples of resolutions established in Theorem 3.3. Case $h=1$. In this case $\lambda$ consists of just a single row of length at most $n-1$. If $\lambda = (0)$, then $\mathcal{E}^\lambda\simeq\mathcal{O}$. If $\lambda = (1)$, then $\mathcal{E}^\lambda\simeq\Sigma^\lambda\mathcal{U}^*=\mathcal{U}^*$. The interesting case is when $\lambda=(p)$ for some $2\leqslant p \leqslant n-1$. Since $(2)$ is the only nontrivial balanced diagram of height one and $(p)/(2)$ = $(p-2)$, the resolution (3.8) takes the form
$$
\begin{equation*}
0\to \mathcal{E}^{(p-2)}\to \mathcal{E}^{(p)}\to S^p\mathcal{U}^* \to 0.
\end{equation*}
\notag
$$
As a consequence, we see that $\mathcal{E}^{(p)}$ has a filtration with the associated graded quotients of the form $S^{p-2t}\mathcal{U}^*$ for all $0\leqslant t\leqslant p/2$. Case $w=1$. In this case $\lambda$ consists of a column of length at most $n-1$. Since there are no nontrivial balanced diagrams of width $1$, we conclude that for $\lambda = (t)^\top$ there is an isomorphism Case $\lambda=(3,1)$. This is the first case when resolution (3.8) has a length greater than $1$. Note that $(2)\in \mathrm{B}_2$ and $(3,1)\in \mathrm{B}_4$ are the only balanced diagrams contained in $\lambda$. Moreover, the only nontrivial Littlewood-Richardson coefficients for $(3,1)/(2)$ are $c((2), (2); \lambda)=c((1,1), (2); \lambda) = 1$. Thus, one has a resolution of the form Since $\mathcal{E}^{(1,1)}\simeq \Lambda^2\mathcal{U}^*$ and $\mathcal{E}^{(2)}$ fits into a short exact sequence $0\to \mathcal{O}\to \mathcal{E}^{(2)}\to S^2\mathcal{U}^*\to 0$, with a little bit of work one can show that $\mathcal{E}^{(3,1)}$ is an extension of the form
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