We're looking for a large set of exact sequences of vector bundles on Grassmannians. Here's the set up:

$V$ and $Q$ are complex vector spaces of dimensions $d$ and $r$ respectively $(d\geq r)$, and we're working on the Grassmannian $Gr(V,Q)$. For simplicity let's fix a trivialization of $det(V)$.

Now let $\alpha$ be a partition/Young diagram with at most $(r-1)$ rows and at most $(d-r)$ columns. Let $\beta$ be the Young diagram obtained from $\alpha$ by adding an extra row of length $(d-r)$ at the beginning. What we want is an exact sequence of vector bundles that goes $$\mathbb{S}_\alpha(Q)\otimes det(Q)^{-1} \rightarrow\;\; ... \;\;\rightarrow \mathbb{S}_\beta(Q) $$ ($\mathbb{S}$ denotes a Schur functor). For $r=1$ there's only one choice for $\alpha$, and the Koszul complex is the required sequence. For $d-r=1$ we have the short exact sequences $$\wedge^k Q \otimes det(Q)^{-1}\rightarrow \wedge^{k+1}V\ \rightarrow \wedge^{k+1} Q$$ We can also solve $r=2$ using Eagon-Northcott complexes. These known cases suggest that the exact sequence should have $(d-r+2)$ terms.

Does anyone know a general construction?

Update: we have a precise conjecture for the terms in the sequence. Let $\beta_0$ be the partition obtained from $\alpha$ by deleting the first column. Now define $\beta_i$ recursively as the partition obtained from $\beta_{i-1}$ by adding boxes to the $i$th column until it agrees with the $i$th column of $\beta$. In particular $\beta_{d-r}=\beta$. Then the terms in middle of the exact sequence should be $$...\rightarrow \wedge^{(|\beta| - |\beta_i|)} V \otimes \mathbb{S}_{\beta_i} Q \rightarrow... $$ If we fix a single point on the Grassmannian and split the tautological short exact sequence there then we can show that this works, which is pretty good evidence. Surely this isn't a new discovery?