Lemma 36.3.4. Let X be a scheme. Let F : \textit{Mod}(\mathcal{O}_ X) \to \textit{Ab} be an additive functor and N \geq 0 an integer. Assume that
F commutes with countable direct products,
R^ pF(\mathcal{F}) = 0 for all p \geq N and \mathcal{F} quasi-coherent.
Then for E \in D_\mathit{QCoh}(\mathcal{O}_ X)
H^ i(RF(\tau _{\leq a}E)) \to H^ i(RF(E)) is an isomorphism for i \leq a,
H^ i(RF(E)) \to H^ i(RF(\tau _{\geq b - N + 1}E)) is an isomorphism for i \geq b,
if H^ i(E) = 0 for i \not\in [a, b] for some -\infty \leq a \leq b \leq \infty , then H^ i(RF(E)) = 0 for i \not\in [a, b + N - 1].
Proof.
Statement (1) is Derived Categories, Lemma 13.16.1.
Proof of statement (2). Write E_ n = \tau _{\geq -n}E. We have E = R\mathop{\mathrm{lim}}\nolimits E_ n, see Lemma 36.3.3. Thus RF(E) = R\mathop{\mathrm{lim}}\nolimits RF(E_ n) in D(\textit{Ab}) by Injectives, Lemma 19.13.6. Thus for every i \in \mathbf{Z} we have a short exact sequence
0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}(RF(E_ n)) \to H^ i(RF(E)) \to \mathop{\mathrm{lim}}\nolimits H^ i(RF(E_ n)) \to 0
see More on Algebra, Remark 15.86.10. To prove (2) we will show that the term on the left is zero and that the term on the right equals H^ i(RF(E_{-b + N - 1})) for any b with i \geq b.
For every n we have a distinguished triangle
H^{-n}(E)[n] \to E_ n \to E_{n - 1} \to H^{-n}(E)[n + 1]
(Derived Categories, Remark 13.12.4) in D(\mathcal{O}_ X). Since H^{-n}(E) is quasi-coherent we have
H^ i(RF(H^{-n}(E)[n])) = R^{i + n}F(H^{-n}(E)) = 0
for i + n \geq N and
H^ i(RF(H^{-n}(E)[n + 1])) = R^{i + n + 1}F(H^{-n}(E)) = 0
for i + n + 1 \geq N. We conclude that
H^ i(RF(E_ n)) \to H^ i(RF(E_{n - 1}))
is an isomorphism for n \geq N - i. Thus the systems H^ i(RF(E_ n)) all satisfy the ML condition and the R^1\mathop{\mathrm{lim}}\nolimits term in our short exact sequence is zero (see discussion in More on Algebra, Section 15.86). Moreover, the system H^ i(RF(E_ n)) is constant starting with n = N - i - 1 as desired.
Proof of (3). Under the assumption on E we have \tau _{\leq a - 1}E = 0 and we get the vanishing of H^ i(RF(E)) for i \leq a - 1 from (1). Similarly, we have \tau _{\geq b + 1}E = 0 and hence we get the vanishing of H^ i(RF(E)) for i \geq b + N from part (2).
\square
Comments (2)
Comment #8610 by nkym on
Comment #9424 by Stacks project on