• Question: Assume X is an algebraic manifold over $\mathbb{C}$ and $D$ is a simple normal crossings divisor. The logarithmic de Rham complex $\left(\Omega_{X}^{\bullet}\left(log\ D\right),d\right)$ (see for example Esnault-Viehweg, “Lectures on Vanishing Theorems”: 2.1, 2.2) is a complex whose terms are $\mathcal{O}_{X}$ -modules, but whose differentials seem to be only $\mathbb{C}$ -linear (for example $d\left(x_{1}dx_{2}\right)=dx_{1}\wedge dx_{2}$ but $x_{1}d\left(dx_{2}\right)=0$ ). Given another complex $\left(B^{\bullet},d^{\prime}\right)$ and morphism of complexes $f^{\bullet}:\left(\Omega_{X}^{\bullet}\left(log\ D\right),d\right)\to\left(B^{\bullet},d^{\prime}\right)$ I should have an induced map on hypercohomology. Given that the differentials d seem to be only $\mathbb{C}$ -linear, what are my restrictions on defining $f^{\bullet}$ ? id est, am I just thinking of each $f^{i}$ as a map of $\mathbb{C}$ -modules, or is there some additional structure I need to take into account?
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2$\begingroup$ I don't see the problem. Do everything in the category of sheaves of $\Bbb{C}$-vector spaces. Note that the usual de Rham complex (not logarithmic) is also not a complex of $\mathcal{O}_X$-modules. $\endgroup$– abxCommented Mar 20, 2014 at 14:11
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$\begingroup$ Taking @abx's comment a bit further, even the topological de Rham complex is not a complex of modules over the ring of functions. That's the whole idea of derivations: they are linear only w.r.t. constants. $\endgroup$– Alex DegtyarevCommented Mar 20, 2014 at 14:21
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$\begingroup$ @abx , AlexDegtyarev thanks for the answers. This is what I was thinking, but it felt fishy to me for some reason since a lot of places just refer to it as a "complex of $\mathcal{O}_X$-modules". $\endgroup$– aegbertCommented Mar 20, 2014 at 14:33
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