Suppose we are given a complex of sheaves on $\mathbb P^N$ in which every term is direct sum of invertible sheaves:

$$ \mathcal F^\bullet = \dots \to \oplus_{j=1}^{n_{p-1}} \mathcal O (k_j^{p-1}) \overset{f^{p-1}}{\to} \oplus_{j=1}^{n_{p}} \mathcal O (k_j^p) \overset{f^{p}}{\to} \oplus_{j=1}^{n_{p+1}} \mathcal O (k_j^{p+1}) \to \dots $$

Say we want to compute its hypercohomology $H^r (\mathcal F^\bullet)$.

We can use the spectral sequence

$$ E_2^{p,q} = H^p_{f^p} \left(\oplus_j H^q (\mathcal O(k_j^\bullet)) \right) \implies H^{p+q}(\mathcal F^\bullet) $$

Since $H^q(O(k))$ is trivial in all gradings except $q= 0 $ or $q = N$, it's obvious that $$ E_2^{p,q} = \dots = E_{N+1}^{p,q} $$ In fact the only non-trivial step in this spectral sequence is $d_{N+1}^{p,N} : E_{N+1}^{p, N} \to E_{N+1}^{p+N+1, 0}$.

The question I'd like to ask is: Does there exist an easy, algorithmic way to determine $d_{N+1}^{p,N}$?

(I define "easy" to mean any method that is easier than, for example, writing down a Cech resolution on an open affine cover for each sheaf and doing a tedious calculation.)

Of course, there usually isn't a simple formula for the higher differentials in spectral sequences for bigraded complexes. But since the derived category of $\mathbb P^N$ is so simple, I'm hoping that I might be lucky.


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