Posted by: Steven Sam | October 26, 2009

## Exceptional sequences for the Grassmannian

Let K be a field of characteristic 0, and let V be a vector space over K of dimension n, and pick k < n. Let X be the Grassmannian Grass(k, V). We’ll briefly explore the (bounded) derived category of coherent sheaves of X, denoted ${\bf D}^b(X)$.

1. Derived category review

For those unfamiliar with derived categories, here’s a quick summary. If A is any Abelian category, set K(A) to be the category of (co)chain complexes of A with the morphisms being chain maps modulo homotopy equivalence. Chain maps which induce isomorphisms are formally inverted, and the result is the derived category ${\bf D}(A)$ of A. Usually we only want to consider bounded complexes, or at least complexes with finitely many nonzero (co)homology groups, and in this case we denote the category ${\bf D}^b(A)$. The category is equipped with a shift functor, which just shifts the degrees of a given complex.

One thing we can do is reformulate derived functors. Given a left exact functor $F \colon A \to B$, we define its right derived functor ${\bf R}f \colon {\bf D}^b(A) \to {\bf D}^b(B)$ as follows. Given an object X in A, an injective resolution $X \to I^\bullet$ of X becomes an isomorphism in ${\bf D}^b(A)$ (considering X as a complex with one nonzero term), so we define ${\bf R}F(X)$ to be the complex obtained by applying F to $I^\bullet$. Actually, we don’t need an injective resolution, we only need a resolution consisting of F-acyclic objects (i.e., the usual right derived functors of F vanish for them). To define ${\bf R}F$ on a general complex $C^\bullet$, we need to find a double complex $C^\bullet \to I^{\bullet, \bullet}$ which is term by term an injective resolution for each $C^i$ (these are called Cartan–Eilenberg resolutions). Then we apply F to the total complex of $I^{\bullet, \bullet}$. A similar story is true for right exact functors G, so we get left derived functors ${\bf L}G$. For notation, the left derived functor of the tensor product is denoted $\stackrel{\bf L}{\otimes}$.

The replacement for exact sequences are now exact triangles, which are written as $A \to B \to C \to A$. The relevant facts are that exact sequences of cochain complexes become exact triangles, and that if we try to calculate cohomology, exact triangles give long exact sequences. Most importantly, the total derived functors are exact in the sense that they preserve exact triangles.

We’ll need two facts. The first is the derived version of the projection formula. Let $f \colon X \to Y$ be a proper morphism of projective schemes, $E \in {\bf D}^b(X)$, and $F \in {\bf D}^b(Y)$. Then we have ${\bf R}f_* (E \stackrel{\bf L}{\otimes} {\bf L}f^* F) \cong {\bf R}f_*E \stackrel{\bf L}{\otimes} F$.

The second is flat base change. If we have a pullback diagram with u flat and f proper, then we have a natural isomorphism $u^*{\bf R}f_* E \cong {\bf R}g_* v^* E$

for all $E \in {\bf D}^b(Y)$.

2. Exceptional sequences and the Fourier–Mukai transform

For this post, we’ll look for a finite set of generators $\{E_1, \dots, E_N\}$ of ${\bf D}^b(X)$, i.e., using only the operations of mapping cones and shifting degrees (so we’re allowed to take kernels, cokernels, and direct sums also), we can obtain the isomorphism class of every object. More specifically, we will construct an exceptional sequence. This means the following things:

1. Every object in the sequence is exceptional: ${\rm Hom}(E_i, E_i) = K$ and ${\rm Ext}^k(E_i, E_i) = 0$ for k>0.
2. ${\rm Ext}^k(E_j, E_i) = 0$ for $k \ge 0$ and i < j.

Using the Schubert cell decomposition of the Grassmannian, it can be shown that the length of an exceptional sequence must equal the number of its Schubert varieties (and more generally, this is true for any homogeneous space for a semisimple group, see the Böhning reference below).

The main tool will be the Fourier–Mukai transform. Suppose that Y and Z are any varieties let $\pi_Y, \pi_Z$ be the projections from $Y \times Z$ to Y and Z, respectively. For $E \in {\bf D}^b(Y \times Z)$, we define a map $\Phi_{Y \to Z}^E \colon {\bf D}^b(Y) \to {\bf D}^b(Z)$ $F \mapsto {\bf R} \pi_{Z,*}({\bf L}\pi_Y^*F \stackrel{\bf L}{\otimes} E)$.

Since it is a composition of exact functors, the Fourier–Mukai transform is itself exact. Also, we can let the argument stay fixed and vary the superscript to get another exact functor. So if we have an exact triangle $E \to F \to G \to E$

in ${\bf D}^b(Y \times Z)$, then we get, for any element $A \in {\bf D}^b(Y)$, an exact triangle $\Phi^E_{Y \to Z}(A) \to \Phi^F_{Y \to Z}(A) \to \Phi^G_{Y \to Z}(A) \to \Phi^E_{Y \to Z}(A)$

in ${\bf D}^b(Z)$. The key point is that in the case that Y=Z, the Fourier–Mukai transform using the structure sheaf of the diagonal $\Phi_{Y \to Y}^{\mathcal{O}_\Delta}$ is the identity functor. To see this, let $i \colon X \to X \times X$ be the diagonal embedding. Letting $p_1,p_2$ be the two projections, then $\Phi_{Y \to Y}^{{\cal O}_\Delta}(A) = {\bf R}p_{2,*}({\bf L}p_1^*A \stackrel{\bf L}{\otimes}_{{\cal O}_{Y \times Y}} i_*{\cal O}_Y)$ $= {\bf R}p_{2,*}({\bf R}i_*({\bf L}i^*{\bf L}p_1^*A \stackrel{\bf L}{\otimes}_{{\cal O}_Y} {\cal O}_Y) = A$,

where in the last equality, we use that $p_1 \circ i = p_2 \circ i$ is the identity map. In light of this remark and the exactness remark, it makes sense to try to find a resolution for the diagonal if we’re trying to find a set of generators of ${\bf D}^b(X)$. So we’ll do just that.

3. Resolution of the diagonal

Let $\mathcal{R}$ be the rank k tautological subbundle of X, and let $\mathcal{Q}$ be the tautological quotient bundle, so that we have a short exact sequence $0 \to \mathcal{R} \to X \times V \to \mathcal{Q} \to 0$.

We’ll use $\boxtimes$ to denote the exterior tensor product of sheaves on X, i.e., if $p_i \colon X \times X \to X$ are the two projections, and F and G are two sheaves on X, then $F \boxtimes G = p_1^*F \otimes p_2^*G$ is a sheaf on $X \times X$.There is a natural map $p_1^* \mathcal{R} \to V \times X \times X \to p_2^* \mathcal{Q}$ where the first map identifies the trivial bundle V with $p_1^*V$ and is the corresponding inclusion, while the second map identifies V with $p_2^* V$ and is the corresponding projection. Set-theoretically, the zero set of this map is the diagonal $\Delta$, and we can check locally that it also defines the diagonal scheme-theoretically.

So let s be the section of ${\cal H}{\rm om}(p_1^* \mathcal{R}, p_2^*\mathcal{Q}) = \mathcal{R}^* \boxtimes \mathcal{Q}$ corresponding to this map. We get a Koszul complex $\displaystyle 0 \to \bigwedge^{k(n-k)} (\mathcal{R} \boxtimes \mathcal{Q}^*) \to \cdots \to \bigwedge^2 (\mathcal{R} \boxtimes \mathcal{Q}^*) \to \mathcal{R} \boxtimes \mathcal{Q}^* \to \mathcal{O}_\Delta$ (*)

which is exact: $\Delta$ and $X \times X$ are nonsingular, so $\Delta$ is locally a complete intersection in $X \times X$. This idea is due to Beilinson, who did it for the case of projective space. The argument was extended to Grassmannians by Kapranov.

4. Finishing things up

Now we can splice (*) into exact triangles. To use the Fourier–Mukai transform, we have to decide if we’re going from the first factor to the second, or vice versa (there is an asymmetry in the terms of (*), so this matters). Both choices will give a different set of generators. For now, let’s go from the second factor to the first. Since the characteristic of K is zero, we get the decomposition $\displaystyle \bigwedge^i(\mathcal{R}^* \boxtimes \mathcal{Q}) = \bigoplus_{|\lambda| = i} {\bf S}_\lambda(\mathcal{R}^*) \boxtimes {\bf S}_{\lambda'}(\mathcal{Q})$

where the sum is over partitions of size i, ${\bf S}_\lambda$ denotes a Schur functor, and $\lambda'$ is the transpose partition to $\lambda$. This decomposition is the dual Cauchy identity, and is a consequence, for example, of the dual Robinson–Schensted–Knuth correspondence. If the field has positive characteristic, then we only get a filtration of the LHS whose associated graded is the RHS. Of course, if k=1, this is irrelevant.

Now pick any object $A \in {\bf D}^b(X)$. If we want to generate A, we can apply the Fourier–Mukai transform to A using the exact triangles we got by splicing (*). This shows that the set $\{ \Phi^{E_\lambda}_{X \to X}(A) \}$ generates A, where $E_\lambda = {\bf S}_\lambda(\mathcal{R}^*) \boxtimes {\bf S}_{\lambda'}(\mathcal{Q})$ and the set is over all $\lambda$ which fit in the $k \times (n-k)$ rectangle. Ranging over all A, this is an infinite set, but we can simplify further: $\displaystyle \Phi^{E_\lambda}_{X \to X}(A) = {\bf R}p_{1,*}({\bf L}p_2^*(A) \stackrel{\bf L}{\otimes} {\bf L}p_1^*({\bf S}_\lambda(\mathcal{R}^*)) \stackrel{\bf L}{\otimes} {\bf L}p_2^*({\bf S}_{\lambda'}(\mathcal{Q})))$

by definition, and using the projection formula, this is isomorphic to ${\bf S}_{\lambda}(\mathcal{R}^*) \stackrel{\bf L}{\otimes} {\bf R}p_{1,*} {\bf L}p_2^*(A \stackrel{\bf L}{\otimes} {\bf S}_{\lambda'}(\mathcal{Q}))$.

Using flat base change (with the notation in the above diagram, we have $Z = {\rm Spec}(K)$, $Y = X$, $v = p_2$, and $g = p_1$), the above can be replaced by derived global sections: ${\bf S}_\lambda(\mathcal{R}^*) \otimes {\bf R}\Gamma(X; A \stackrel{\bf L}{\otimes} {\bf S}_{\lambda'}(\mathcal{Q}))$,

where the second factor is isomorphic to a cochain complex consisting of its cohomology groups with zero differentials. Hence we see that the Fourier–Mukai transform is a complex consisting of copies of ${\bf S}_\lambda(\mathcal{R}^*)$ in various degrees. So we see that $\{ {\bf S}_\lambda(\mathcal{R}^*) \}$ is a generating set, where $\lambda$ ranges over all partitions fitting inside of the $k \times (n-k)$ box.

On the other hand, if we did the Fourier–Mukai transform going from the first factor to the second factor, we would get $\{ {\bf S}_\lambda(\mathcal{Q}) \}$ as a set of generators, where $\lambda$ ranges over all partitions which fit inside of the $(n-k) \times k$ box.

The last thing to check is that we get an exceptional sequence. We have a partial ordering: ${\bf S}_\lambda(\mathcal{R}^*) \le {\bf S}_\mu(\mathcal{R}^*)$ whenever $\lambda_i \le \mu_i$ for all i. Extending this to a total order will do the trick. To check that all of the appropriate Ext groups vanish, we need to use the Borel–Weil–Bott theorem, but I will omit this task.

References

-Steven