Analogue of the Koszul complex for symmetric powers?

Question

Let $R$ be a commutative ring and let $M$ be a free $R$-module of rank $n$. Let $s: M \rightarrow R$ be a $R$-linear map.

Then, the Koszul complex is a complex

$0 \rightarrow \bigwedge^{n}M \rightarrow \bigwedge^{n-1}M \rightarrow \dots \bigwedge^{2}M \rightarrow M \rightarrow R \rightarrow 0 $

where the differentials are defined in terms of the map $s$.

I was wondering whether there is some analogue of the Koszul complex for symmetric powers?

I am sure that I am not the first person to ask this kind of question, and there must be some literature on this, but a quick google search doesn't come up with much. Intuitively, I do not expect the analogy to be strong, but surely you should be able to say $\textit{something}$ about the case for symmetric powers?

To put my question in context, I was considering the tautological exact sequence on the Grassmannian $Gr^{k,n}$

$1 \rightarrow \mathcal{R} \rightarrow \mathcal{O}_{Gr^{k,n}}^{\oplus n} \rightarrow \mathcal{Q} \rightarrow 0 $,

and I was looking at the Koszul complex

$0 \rightarrow \bigwedge^{n-k}\mathcal{R} \rightarrow \bigwedge^{n-k-1}\mathcal{R} \rightarrow \dots \bigwedge^{2}\mathcal{R} \rightarrow \mathcal{R} \rightarrow \mathcal{O}_{Gr^{k,n}} \rightarrow 0$.

For technical reasons, I want to compare this complex against a complex involving the tautological bundle $\mathcal{Q}$ (aiming for something like a quasi-isomorphism), but I am not even sure whether such a complex involving $\mathcal{Q}$ should exist? My guess is that, if it does exist, it involves symmetric powers of $\mathcal{Q}$, hence my question.

Any help would be much appreciated!

Answer 1

Not an answer, just an extended comment. The key defining feature of the differential on the Koszul complex is that it $\bigwedge M=\bigoplus_k \bigwedge^{k} M$ is an algebra under the wedge product, and the differential is the unique extension of $s$ which turns $\bigwedge M$ into a differential graded algebra, i.e. the differential satisfies the (graded) product rule, $d(ab)=d(a)b+(-1)^{\deg(a)}ad(b)$. Because $\bigwedge M$ is graded-commutative, this is enough to guarantee that $d^2=0$.

Unfortunately, in the actual commutative case, this is no longer the case. For example, if $R=\mathbb R$, and $M=R$, then $\operatorname{Sym}(M)\cong R[x]$, and if $s(x)=1$, then using the product rule to extend $s$ simply gives the ordinary derivative, which does not satisfy $d^2=0$.

This doesn't mean that some other construction cannot work, but the naive attempt at generalization fails.