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Here (at the bottom of page 13) it's stated that for a smooth map $f:X\to S$, the relative Spencer complex $$\Omega_{X/S}^\bullet \otimes_{f^{-1}\mathcal{O}_S} D_X$$ is a resolution of the transfer module $$D_{S\leftarrow X} \ = \ f^{-1}(D_S \otimes_{f^{-1}\mathcal{O}_S} \Omega_S^{-1}) \otimes_{f^{-1}\mathcal{O}_S} \Omega_X$$ by left $f^{-1}D_S$-modules.

  1. Why is this a resolution?
  2. Is there a similar resolution in the case where some fibres of $f$ are singular? If so, what can we say about it?

$$\text{}$$

It's easy to show that $\Omega_{X/S}^\bullet \otimes D_X$ is a complex, but I'm not sure exactly how to get a map $$\Omega_{X/S} \otimes_{f^{-1}\mathcal{O}_S} D_X\longrightarrow D_{S\leftarrow X}$$ of left $f^{-1}D_S$-modules. I know it should be a trivial consequence of $\Omega_X\otimes_{f^{-1}\mathcal{O}_S} f^{-1}\Omega_S\simeq \Omega_{X/S}$, but for some reason I can't get it to work. I would guess something like $$\omega \otimes P \longrightarrow \phi(P) \otimes \omega$$ might work for an appropriate map $\phi:D_X\to f^{-1}D_S$.

I'm also curious about what happens when the map $f$ is smooth except at a point where the fibre is singular.

Meow
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  • Aren't you looking for a map going the other way ? That is a map $D_{S\leftarrow X}\to\Omega_{X/S}\otimes_{f^{-1}\mathcal{O}_S}D_X$. – Roland Apr 23 '18 at 20:14
  • @Roland I'm trying to compute $Rf_*\mathcal{O}X$, for which one first needs to compute $D{S\leftarrow X}\otimes^L\mathcal{O}X$, so I definitely do need a locally free (or flat) resolution of $D{S\leftarrow X}$ (see for instance the bottom of p.13 in https://www.math.purdue.edu/~dvb/preprints/dmod.pdf). – Meow Apr 23 '18 at 20:38
  • I am not an expert in algebraic geometry nor I am an expert in the language of $D$-modules: however I remember an old paper of Francesco Succi (1966), "Alcune osservazioni sui teoremi di de Rham (Some observations on de Rham's theorems)", Rivista di Matematica della Universitá di Parma, II Serie, vol. 7, 35-46 (1966), MR0231305, Zbl 0178.57302, which deals with the relative de Rham cohomology. It is written in Italian ad perhaps the symbology is outdated: however I hope it could be of some help. – Daniele Tampieri Apr 29 '18 at 15:24

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