Consider Grassmanianns over fields of characteristic zero.

Let $i : Gr_{k-1,n} \rightarrow Gr_{k,n+1}$ be the `direct sum' map between Grassmannians. By universal property of Grassmannian, this map corresponds to the short exact sequence

$1 \rightarrow \mathcal{T}\oplus \mathcal{O}_{Gr_{k-1,n}} \rightarrow \mathcal{O}_{Gr_{k,n+1}}^{n+1} \rightarrow \mathcal{Q} \rightarrow 1$,

where $\mathcal{T}$ is the tautological subbundle of the Grassmaninan $Gr_{k-1,n}$.

This map induces a map of the bounded derived category of perfect complexes

$Ri_{*}: D^{b}\text{Perf}(Gr_{k-1,n}) \rightarrow D^{b}\text{Perf}(Gr_{k,n+1})$.

My question is: I would like to compute $Ri_{*}(S^{\lambda}\mathcal{T})$, where $S^{\lambda}$ is the Schur functor corresponding to a Young diagram $\lambda$ contained within rectangle of size $(k-1)\times (n-k+1)$.

This problem is motivated by the following:

A full exceptional collection of the bounded derived category of Grassmannian $Gr_{k,n+1}$ over fields of characteristic zero has been known since Kapranov (with generalisations to integers given by Efimov ). This is given by the Schur functors $S^{\lambda}\mathcal{T}$ of the tautological subbundle, indexed over the rectangle of size $k \times (n-k+1)$.

If we were to take the Verdier quotient of this with respect to the category generated by Schur functors living inside rectangle $\mathcal{U}$ of size $k \times (n-k)$, then the claim is that the resulting category is equivalent to the bounded derived category of Grassmannian $Gr_{k-1,n}$.

Indeed, the Young diagrams not contained in $\mathcal{U}$ are precisely those whose first component is of maximal length (i.e. n-k+1), but whose latter components are free to describe any Young diagram in rectangle of size $(k-1)\times (n-k+1)$. Intuitively therefore, this should correspond to $D^{b}\text{Perf}(Gr_{k-1,n})$.

The only map I can think of that could realise this equivalence is the composition $q \circ Ri_{*}$, where $q: D^{b}\text{Perf}(Gr_{k,n+1}) \rightarrow D^{b}\text{Perf}(Gr_{k,n+1})/\mathcal{U}$ is the projection given by the Verdier quotient. If this correct, then it should be the case that $Ri_{*}(S^{\lambda}\mathcal{T})$ is something like $U_{\cdot} \rightarrow S^{((n-k+1),\lambda)}\mathcal{T}$, where $U_{\cdot} \in \mathcal{U}$. So taking the quotient will kill off the $U_{\cdot}$ term.

Any help on this question will be much appreciated!



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.