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How to complete $i^*i_*\mathcal{F}\to \mathcal{F}$ into an exact triangle for a smooth divisor $i: X\hookrightarrow Y$?

The cone of $i^*i_*\mathcal{F} \to \mathcal{F}$ is isomorphic to $\mathcal{F} \otimes \mathcal{O}_X(-X)[2]$. EDIT. Let me write an argument for a sheaf $F$. Consider the distinguished triangle $$ i^*...
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What is the cone of $\mathcal{F}\to i_*i^*\mathcal{F}$ for a divisor $i: D\hookrightarrow X$?

For any Cartier divisor $i \colon D \hookrightarrow X$ and any $F \in D^b(X)$ there is an exact triangle $$ F \otimes \mathcal{O}_X(-D) \to F \to i_*i^*F. $$ It can be obtained by tensoring the exact ...
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