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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!

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