POTW Projection Formula for Ringed Spaces

Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
Show that if ##f : X \to Y## is a morphism of ringed spaces, ##\mathscr{F}## is an ##\mathcal{O}_X##-module and ##\mathscr{E}## is a locally free ##\mathcal{O}_Y##-module of finite rank, then for all ##p \ge 0##, there is an isomorphism $$R^pf_*(\mathscr{F}\otimes_{\mathcal{O}_X} f^*\mathscr{E}) \approx R^pf_*(\mathscr{F}) \otimes_{\mathcal{O}_Y} \mathscr{E}$$
 
Physics news on Phys.org
When ##p = 0##, the result follows from the projection formula ##f_*(\mathscr{F}\otimes f^*\mathscr{E}) \approx f_*(\mathscr{F}) \otimes \mathscr{E}##, which may be obtained by reducing to the case ##\mathscr{E} = \mathscr{O}_Y^n##. Both sides of the proposed isomorphism are delta functors (applied to ##\mathscr{F}##) that are effaceable. Indeed, if ##\mathscr{F}## is injective, then since ##f_*## has an exact left adjoint ##f^*##, ##f_*(\mathscr{F})## is injective. Also, by exactness of ##\otimes_{\mathscr{O}_X} f^*\mathscr{E}##, the tensor sheaf ##\mathscr{F} \otimes_{\mathscr{O}_X} f^*\mathscr{E}## is injective and hence ##f_*(\mathscr{F} \otimes_{\mathscr{O}_X} f^*\mathscr{E})## is injective. We then deduce ##R^pf_*(\mathscr{F}) = 0 = R^p f_*(\mathscr{F} \otimes_{\mathscr{O}_X} \mathscr{E})##, so that the delta functors are effaceable. It follows that these delta functors are universal. Since they agree at ##p = 0##, they are naturally isomorphic.
 
Back
Top