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Dec 19, 2022 at 8:34 vote accept Saal Hardali
Dec 19, 2022 at 17:50
Dec 15, 2017 at 16:21 comment added David Ben-Zvi If we are given the tensor structure then we have the unit, and Hom from unit distinguishes ample and anti-ample, no?
Nov 13, 2017 at 12:30 comment added Sasha @LeoAlonso: I understand, but what I say just means that all positive powers of $L$ are contained in the subcategory generated by negative powers. And the other way round, all negative powers are contained in the subcategory generated by positive powers.
Nov 13, 2017 at 12:07 comment added Leo Alonso @Sasha What I mean is to generate the unbounded derived category $\mathbf{D}(Qco(X))$ as a triangulated category with coproducts, not $\mathbf{D}^b(Coh(X))$ as a triangulated category. The key point is that for every $\mathcal{F} \in Qco(X)$ a coproduct of negative tensors of $\mathcal{L}$ surjects onto $\mathcal{F}$, see SGA 6 p. 169. I guess this is the sign that distinguishes $\mathcal{L}$ from $\mathcal{L}^{-1}$.
Nov 13, 2017 at 11:32 comment added Sasha @Leo Alonso: Adding tensor powers does not help, since powers of an anti-ample line bundle $L$ generate the whole derived category. This follows easily by using the pullback of the (twisted) Koszul complex from a projective space; it allows to express $L^{i+1}$ by means of $L^i$, $L^{i-1}$, \dots, $L^{i-n}$, where $n = \dim X$.
Nov 13, 2017 at 9:19 history edited Leo Alonso CC BY-SA 3.0
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Nov 12, 2017 at 23:06 comment added Will Sawin I don't see how this can be true. In what way does $\mathcal O(1)$ on $\mathbb P^n$ generate the derived category, that $\mathcal O(-1)$ does not? Both of them do not generate on their own (even with shifts) but do if you allow tensor powers.
Nov 12, 2017 at 22:59 history answered Leo Alonso CC BY-SA 3.0