The Stacks project

36.40 Detecting Boundedness

In this section, we show that compact generators of $D_\mathit{QCoh}$ of a quasi-compact, quasi-separated scheme, as constructed in Section 36.15, have a special property. We recommend reading that section first as it is very similar to this one.

Lemma 36.40.1. In Situation 36.9.1 denote $j : U \to X$ the open immersion and let $K$ be the perfect object of $D(\mathcal{O}_ X)$ corresponding to the Koszul complex on $f_1, \ldots , f_ r$ over $A$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $a \in \mathbf{Z}$. Consider the following conditions

  1. The canonical map $\tau _{\geq a}E \to \tau _{\geq a} Rj_*(E|_ U)$ is an isomorphism.

  2. We have $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], E) = 0$ for all $n \geq a$.

Then (2) implies (1) and (1) implies (2) with $a$ replaced by $a + 1$.

Proof. Choose a distinguished triangle $N \to E \to Rj_*(E|_ U) \to N[1]$. Then (1) implies $\tau _{\geq a + 1} N = 0$ and (1) is implied by $\tau _{\geq a}N = 0$. Observe that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], Rj_*(E|_ U)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(K|_ U[-n], E) = 0 \]

for all $n$ as $K|_ U = 0$. Thus (2) is equivalent to $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], N) = 0$ for all $n \geq a$. Observe that there are distinguished triangles

\[ K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i + e''_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e''_ i}, \ldots , f_ r^{e_ r}) \to \ldots \]

of Koszul complexes, see More on Algebra, Lemma 15.28.11. Hence $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], N) = 0$ for all $n \geq a$ is equivalent to $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K_ e[-n], N) = 0$ for all $n \geq a$ and all $e \geq 1$ with $K_ e$ as in Lemma 36.9.6. Since $N|_ U = 0$, that lemma implies that this in turn is equivalent to $H^ n(X, N) = 0$ for $n \geq a$. We conclude that (2) is equivalent to $\tau _{\geq a}N = 0$ since $N$ is determined by the complex of $A$-modules $R\Gamma (X, N)$, see Lemma 36.3.5. Thus we find that our lemma is true. $\square$

Lemma 36.40.2. In Situation 36.9.1 denote $j : U \to X$ the open immersion and let $K$ be the perfect object of $D(\mathcal{O}_ X)$ corresponding to the Koszul complex on $f_1, \ldots , f_ r$ over $A$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $a \in \mathbf{Z}$. Consider the following conditions

  1. The canonical map $\tau _{\leq a}E \to \tau _{\leq a} Rj_*(E|_ U)$ is an isomorphism, and

  2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], E) = 0$ for all $n \leq a$.

Then (2) implies (1) and (1) implies (2) with $a$ replaced by $a - 1$.

Proof. Choose a distinguished triangle $E \to Rj_*(E|_ U) \to N \to E[1]$. Then (1) implies $\tau _{\leq a - 1}N = 0$ and (1) is implied by $\tau _{\leq a}N = 0$. Observe that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], Rj_*(E|_ U)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(K|_ U[-n], E) = 0 \]

for all $n$ as $K|_ U = 0$. Thus (2) is equivalent to $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], N) = 0$ for all $n \leq a$. Observe that there are distinguished triangles

\[ K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i + e''_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e''_ i}, \ldots , f_ r^{e_ r}) \to \ldots \]

of Koszul complexes, see More on Algebra, Lemma 15.28.11. Hence $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], N) = 0$ for all $n \leq a$ is equivalent to $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K_ e[-n], N) = 0$ for all $n \leq a$ and all $e \geq 1$ with $K_ e$ as in Lemma 36.9.6. Since $N|_ U = 0$, that lemma implies that this in turn is equivalent to $H^ n(X, N) = 0$ for $n \leq a$. We conclude that (2) is equivalent to $\tau _{\leq a}N = 0$ since $N$ is determined by the complex of $A$-modules $R\Gamma (X, N)$, see Lemma 36.3.5. Thus we find that our lemma is true. $\square$

Lemma 36.40.3. Let $X$ be a quasi-compact and quasi-separated scheme. Let $P \in D_{perf}(\mathcal{O}_ X)$ and $E \in D_{\mathit{QCoh}}(\mathcal{O}_ X)$. Let $a \in \mathbf{Z}$. The following are equivalent

  1. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[-i], E) = 0$ for $i \gg 0$, and

  2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[-i], \tau _{\geq a} E) = 0$ for $i \gg 0$.

Proof. Using the triangle $ \tau _{< a} E \to E \to \tau _{\geq a} E \to $ we see that the equivalence follows if we can show

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[-i], \tau _{< a} E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P, (\tau _{< a} E)[i]) = 0 \]

for $i \gg 0$. As $P$ is perfect this is true by Lemma 36.18.2. $\square$

Lemma 36.40.4. Let $X$ be a quasi-compact and quasi-separated scheme. Let $P \in D_{perf}(\mathcal{O}_ X)$ and $E \in D_{\mathit{QCoh}}(\mathcal{O}_ X)$. Let $a \in \mathbf{Z}$. The following are equivalent

  1. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[-i], E) = 0$ for $i \ll 0$, and

  2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[-i], \tau _{\leq a} E) = 0$ for $i \ll 0$.

Proof. Using the triangle $ \tau _{\leq a} E \to E \to \tau _{> a} E \to $ we see that the equivalence follows if we can show

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[-i], \tau _{> a} E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P, (\tau _{> a} E)[i]) = 0 \]

for $i \ll 0$. As $P$ is perfect this is true by Lemma 36.18.2. $\square$

Proposition 36.40.5. Let $X$ be a quasi-compact and quasi-separated scheme. Let $G \in D_{perf}(\mathcal{O}_ X)$ be a perfect complex which generates $D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. The following are equivalent

  1. $E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$,

  2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = 0$ for $i \gg 0$,

  3. $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(G, E) = 0$ for $i \gg 0$,

  4. $R\mathop{\mathrm{Hom}}\nolimits _ X(G, E)$ is in $D^-(\mathbf{Z})$,

  5. $H^ i(X, G^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E) = 0$ for $i \gg 0$,

  6. $R\Gamma (X, G^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E)$ is in $D^-(\mathbf{Z})$,

  7. for every perfect object $P$ of $D(\mathcal{O}_ X)$

    1. the assertions (2), (3), (4) hold with $G$ replaced by $P$, and

    2. $H^ i(X, P \otimes _{\mathcal{O}_ X}^\mathbf {L} E) = 0$ for $i \gg 0$,

    3. $R\Gamma (X, P \otimes _{\mathcal{O}_ X}^\mathbf {L} E)$ is in $D^-(\mathbf{Z})$.

Proof. Assume (1). Since $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G, E[i])$ we see that this is zero for $i \gg 0$ by Lemma 36.18.2. This proves that (1) implies (2).

Parts (2), (3), (4) are equivalent by the discussion in Cohomology, Section 20.41. Part (5) and (6) are equivalent as $H^ i(X, -) = H^ i(R\Gamma (X, -))$ by definition. The equivalent conditions (2), (3), (4) are equivalent to the equivalent conditions (5), (6) by Cohomology, Lemma 20.47.5 and the fact that $(G[-i])^\vee = G^\vee [i]$.

It is clear that (7) implies (2). Conversely, let us prove that the equivalent conditions (2) – (6) imply (7). Recall that $G$ is a classical generator for $D_{perf}(\mathcal{O}_ X)$ by Remark 36.17.2. For $P \in D_{perf}(\mathcal{O}_ X)$ let $T(P)$ be the assertion that $R\mathop{\mathrm{Hom}}\nolimits _ X(P, E)$ is in $D^-(\mathbf{Z})$. Clearly, $T$ is inherited by direct sums, satisfies the 2-out-of-three property for distinguished triangles, is inherited by direct summands, and is perserved by shifts. Hence by Derived Categories, Remark 13.36.7 we see that (4) implies $T$ holds on all of $D_{perf}(\mathcal{O}_ X)$. The same argument works for all other properties, except that for property (7)(b) and (7)(c) we also use that $P \mapsto P^\vee $ is a self equivalence of $D_{perf}(\mathcal{O}_ X)$. Small detail omitted.

We will prove the equivalent conditions (2) – (7) imply (1) using the induction principle of Cohomology of Schemes, Lemma 30.4.1.

First, we prove (2) – (7) $\Rightarrow $ (1) if $X$ is affine. Set $P = \mathcal{O}_ X[0]$. From (7) we obtain $H^ i (X, E) = 0$ for $i \gg 0$. Hence (1) follows since $E$ is determined by $R\Gamma (X, E)$, see Lemma 36.3.5.

Now assume $X = U \cup V$ with $U$ a quasi-compact open of $X$ and $V$ an affine open, and assume the implication (2) – (7) $\Rightarrow $ (1) is known for the schemes $U$, $V$, and $U \cap V$. Suppose $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ satisfies (2) – (7). By Lemma 36.15.1 and Theorem 36.15.3 there exists a perfect complex $Q$ on $X$ such that $Q|_ U$ generates $D_\mathit{QCoh}(\mathcal{O}_ U)$. Let $f_1, \dots , f_ r \in \Gamma (V, \mathcal{O}_ V)$ be such that $V \setminus U = V(f_1, \dots , f_ r)$ as subsets of $V$. Let $K \in D_{perf}(\mathcal{O}_ V)$ be the object corresponding to the Koszul complex on $f_1, \dots , f_ r$. Let $K' \in D_{perf}(\mathcal{O}_ X)$ be

36.40.5.1
\begin{equation} \label{perfect-equation-detecting-bounded-above} K' = R (V \to X)_* K = R (V \to X)_! K, \end{equation}

see Cohomology, Lemmas 20.33.6 and 20.46.10. This is a perfect complex on $X$ supported on the closed set $X \setminus U \subset V$ and isomorphic to $K$ on $V$. By assumption, we know $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(Q, E)$ and $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K', E)$ are bounded above.

By the second description of $K'$ in (36.40.5.1) we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ V)}(K[-i], E|_ V) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K'[-i], E) = 0 \]

for $i \gg 0$. Therefore, we may apply Lemma 36.40.1 to $E|_ V$ to obtain an integer $a$ such that $\tau _{\geq a}(E|_ V) = \tau _{\geq a} R (U \cap V \to V)_* (E|_{U \cap V})$. Then $\tau _{\geq a} E = \tau _{\geq a} R (U \to X)_* (E |_ U)$ (check that the canonical map is an isomorphism after restricting to $U$ and to $V$). Hence using Lemma 36.40.3 twice we see that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(Q|_ U [-i], E|_ U) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Q[-i], R (U \to X)_* (E|_ U)) = 0 \]

for $i \gg 0$. Since the Proposition holds for $U$ and the generator $Q|_ U$, we have $E|_ U \in D^-_\mathit{QCoh}(\mathcal{O}_ U)$. But then since the functor $R (U \to X)_*$ preserves $D^-_\mathit{QCoh}$ (by Lemma 36.4.1), we get $\tau _{\geq a}E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$. Thus $E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$. $\square$

Proposition 36.40.6. Let $X$ be a quasi-compact and quasi-separated scheme. Let $G \in D_{perf}(\mathcal{O}_ X)$ be a perfect complex which generates $D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. The following are equivalent

  1. $E \in D^+_\mathit{QCoh}(\mathcal{O}_ X)$,

  2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = 0$ for $i \ll 0$,

  3. $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(G, E) = 0$ for $i \ll 0$,

  4. $R\mathop{\mathrm{Hom}}\nolimits _ X(G, E)$ is in $D^+(\mathbf{Z})$,

  5. $H^ i(X, G^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E) = 0$ for $i \ll 0$,

  6. $R\Gamma (X, G^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E)$ is in $D^+(\mathbf{Z})$,

  7. for every perfect object $P$ of $D(\mathcal{O}_ X)$

    1. the assertions (2), (3), (4) hold with $G$ replaced by $P$, and

    2. $H^ i(X, P \otimes _{\mathcal{O}_ X}^\mathbf {L} E) = 0$ for $i \ll 0$,

    3. $R\Gamma (X, P \otimes _{\mathcal{O}_ X}^\mathbf {L} E)$ is in $D^+(\mathbf{Z})$.

Proof. Assume (1). Since $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G, E[i])$ we see that this is zero for $i \ll 0$ by Lemma 36.18.2. This proves that (1) implies (2).

Parts (2), (3), (4) are equivalent by the discussion in Cohomology, Section 20.41. Part (5) and (6) are equivalent as $H^ i(X, -) = H^ i(R\Gamma (X, -))$ by definition. The equivalent conditions (2), (3), (4) are equivalent to the equivalent conditions (5), (6) by Cohomology, Lemma 20.47.5 and the fact that $(G[-i])^\vee = G^\vee [i]$.

It is clear that (7) implies (2). Conversely, let us prove that the equivalent conditions (2) – (6) imply (7). Recall that $G$ is a classical generator for $D_{perf}(\mathcal{O}_ X)$ by Remark 36.17.2. For $P \in D_{perf}(\mathcal{O}_ X)$ let $T(P)$ be the assertion that $R\mathop{\mathrm{Hom}}\nolimits _ X(P, E)$ is in $D^+(\mathbf{Z})$. Clearly, $T$ is inherited by direct sums, satisfies the 2-out-of-three property for distinguished triangles, is inherited by direct summands, and is perserved by shifts. Hence by Derived Categories, Remark 13.36.7 we see that (4) implies $T$ holds on all of $D_{perf}(\mathcal{O}_ X)$. The same argument works for all other properties, except that for property (7)(b) and (7)(c) we also use that $P \mapsto P^\vee $ is a self equivalence of $D_{perf}(\mathcal{O}_ X)$. Small detail omitted.

We will prove the equivalent conditions (2) – (7) imply (1) using the induction principle of Cohomology of Schemes, Lemma 30.4.1.

First, we prove (2) – (7) $\Rightarrow $ (1) if $X$ is affine. Let $P = \mathcal{O}_ X[0]$. From (7) we obtain $H^ i (X, E) = 0$ for $i \ll 0$. Hence (1) follows since $E$ is determined by $R\Gamma (X, E)$, see Lemma 36.3.5.

Now assume $X = U \cup V$ with $U$ a quasi-compact open of $X$ and $V$ an affine open, and assume the implication (2) – (7) $\Rightarrow $ (1) is known for the schemes $U$, $V$, and $U \cap V$. Suppose $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ satisfies (2) – (7). By Lemma 36.15.1 and Theorem 36.15.3 there exists a perfect complex $Q$ on $X$ such that $Q|_ U$ generates $D_\mathit{QCoh}(\mathcal{O}_ U)$. Let $f_1, \dots , f_ r \in \Gamma (V, \mathcal{O}_ V)$ be such that $V \setminus U = V(f_1, \dots , f_ r)$ as subsets of $V$. Let $K \in D_{perf}(\mathcal{O}_ V)$ be the object corresponding to the Koszul complex on $f_1, \dots , f_ r$. Let $K' \in D_{perf}(\mathcal{O}_ X)$ be

36.40.6.1
\begin{equation} \label{perfect-equation-detecting-bounded-below} K' = R (V \to X)_* K = R (V \to X)_! K, \end{equation}

see Cohomology, Lemmas 20.33.6 and 20.46.10. This is a perfect complex on $X$ supported on the closed set $X \setminus U \subset V$ and isomorphic to $K$ on $V$. By assumption, we know $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(Q, E)$ and $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K', E)$ are bounded below.

By the second description of $K'$ in (36.40.6.1) we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ V)}(K[-i], E|_ V) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K'[-i], E) = 0 \]

for $i \ll 0$. Therefore, we may apply Lemma 36.40.2 to $E|_ V$ to obtain an integer $a$ such that $\tau _{\leq a}(E|_ V) = \tau _{\leq a} R(U \cap V \to V)_*(E|_{U \cap V})$. Then $\tau _{\leq a} E = \tau _{\leq a} R(U \to X)_*(E|_ U)$ (check that the canonical map is an isomorphism after restricting to $U$ and to $V$). Hence using Lemma 36.40.4 twice we see that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(Q|_ U [-i], E|_ U) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Q[-i], R (U \to X)_* (E|_ U)) = 0 \]

for $i \ll 0$. Since the Proposition holds for $U$ and the generator $Q|_ U$, we have $E|_ U \in D^+_\mathit{QCoh}(\mathcal{O}_ U)$. But then since the functor $R(U \to X)_*$ preserves bounded below objects (see Cohomology, Section 20.3) we get $\tau _{\leq a} E \in D^+_\mathit{QCoh}(\mathcal{O}_ X)$. Thus $E \in D^+_\mathit{QCoh}(\mathcal{O}_ X)$. $\square$


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