## 20.47 Perfect complexes

In this section we discuss properties of perfect complexes on ringed spaces.

Definition 20.47.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{E}^\bullet$ be a complex of $\mathcal{O}_ X$-modules. We say $\mathcal{E}^\bullet$ is perfect if there exists an open covering $X = \bigcup U_ i$ such that for each $i$ there exists a morphism of complexes $\mathcal{E}_ i^\bullet \to \mathcal{E}^\bullet |_{U_ i}$ which is a quasi-isomorphism with $\mathcal{E}_ i^\bullet$ a strictly perfect complex of $\mathcal{O}_{U_ i}$-modules. An object $E$ of $D(\mathcal{O}_ X)$ is perfect if it can be represented by a perfect complex of $\mathcal{O}_ X$-modules.

Lemma 20.47.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E$ be an object of $D(\mathcal{O}_ X)$.

1. If there exists an open covering $X = \bigcup U_ i$ and strictly perfect complexes $\mathcal{E}_ i^\bullet$ on $U_ i$ such that $\mathcal{E}_ i^\bullet$ represents $E|_{U_ i}$ in $D(\mathcal{O}_{U_ i})$, then $E$ is perfect.

2. If $E$ is perfect, then any complex representing $E$ is perfect.

Proof. Identical to the proof of Lemma 20.45.2. $\square$

Lemma 20.47.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E$ be an object of $D(\mathcal{O}_ X)$. Assume that all stalks $\mathcal{O}_{X, x}$ are local rings. Then the following are equivalent

1. $E$ is perfect,

2. there exists an open covering $X = \bigcup U_ i$ such that $E|_{U_ i}$ can be represented by a finite complex of finite locally free $\mathcal{O}_{U_ i}$-modules, and

3. there exists an open covering $X = \bigcup U_ i$ such that $E|_{U_ i}$ can be represented by a finite complex of finite free $\mathcal{O}_{U_ i}$-modules.

Proof. This follows from Lemma 20.47.2 and the fact that on $X$ every direct summand of a finite free module is finite locally free. See Modules, Lemma 17.14.6. $\square$

Lemma 20.47.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E$ be an object of $D(\mathcal{O}_ X)$. Let $a \leq b$ be integers. If $E$ has tor amplitude in $[a, b]$ and is $(a - 1)$-pseudo-coherent, then $E$ is perfect.

Proof. After replacing $X$ by the members of an open covering we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet$ and a map $\alpha : \mathcal{E}^\bullet \to E$ such that $H^ i(\alpha )$ is an isomorphism for $i \geq a$. We may and do replace $\mathcal{E}^\bullet$ by $\sigma _{\geq a - 1}\mathcal{E}^\bullet$. Choose a distinguished triangle

$\mathcal{E}^\bullet \to E \to C \to \mathcal{E}^\bullet $

From the vanishing of cohomology sheaves of $E$ and $\mathcal{E}^\bullet$ and the assumption on $\alpha$ we obtain $C \cong \mathcal{K}[a - 2]$ with $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Applying $- \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}$ the assumption that $E$ has tor amplitude in $[a, b]$ implies $\mathcal{K} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{E}^{a - 1} \otimes _{\mathcal{O}_ X} \mathcal{F}$ has image $\mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{E}^ a \otimes _{\mathcal{O}_ X} \mathcal{F})$. It follows that $\text{Tor}_1^{\mathcal{O}_ X}(\mathcal{E}', \mathcal{F}) = 0$ where $\mathcal{E}' = \mathop{\mathrm{Coker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Hence $\mathcal{E}'$ is flat (Lemma 20.26.16). Thus $\mathcal{E}'$ is locally a direct summand of a finite free module by Modules, Lemma 17.18.3. Thus locally the complex

$\mathcal{E}' \to \mathcal{E}^{a - 1} \to \ldots \to \mathcal{E}^ b$

is quasi-isomorphic to $E$ and $E$ is perfect. $\square$

Lemma 20.47.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E$ be an object of $D(\mathcal{O}_ X)$. The following are equivalent

1. $E$ is perfect, and

2. $E$ is pseudo-coherent and locally has finite tor dimension.

Proof. Assume (1). By definition this means there exists an open covering $X = \bigcup U_ i$ such that $E|_{U_ i}$ is represented by a strictly perfect complex. Thus $E$ is pseudo-coherent (i.e., $m$-pseudo-coherent for all $m$) by Lemma 20.45.2. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_ i}$ has finite Tor dimension by Lemma 20.46.3. Thus (2) holds.

Assume (2). After replacing $X$ by the members of an open covering we may assume there exist integers $a \leq b$ such that $E$ has tor amplitude in $[a, b]$. Since $E$ is $m$-pseudo-coherent for all $m$ we conclude using Lemma 20.47.4. $\square$

Lemma 20.47.6. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $E$ be an object of $D(\mathcal{O}_ Y)$. If $E$ is perfect in $D(\mathcal{O}_ Y)$, then $Lf^*E$ is perfect in $D(\mathcal{O}_ X)$.

Lemma 20.47.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K, L, M, f, g, h)$ be a distinguished triangle in $D(\mathcal{O}_ X)$. If two out of three of $K, L, M$ are perfect then the third is also perfect.

Proof. First proof: Combine Lemmas 20.47.5, 20.45.4, and 20.46.6. Second proof (sketch): Say $K$ and $L$ are perfect. After replacing $X$ by the members of an open covering we may assume that $K$ and $L$ are represented by strictly perfect complexes $\mathcal{K}^\bullet$ and $\mathcal{L}^\bullet$. After replacing $X$ by the members of an open covering we may assume the map $K \to L$ is given by a map of complexes $\alpha : \mathcal{K}^\bullet \to \mathcal{L}^\bullet$, see Lemma 20.44.8. Then $M$ is isomorphic to the cone of $\alpha$ which is strictly perfect by Lemma 20.44.2. $\square$

Lemma 20.47.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. If $K, L$ are perfect objects of $D(\mathcal{O}_ X)$, then so is $K \otimes _{\mathcal{O}_ X}^\mathbf {L} L$.

Lemma 20.47.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. If $K \oplus L$ is a perfect object of $D(\mathcal{O}_ X)$, then so are $K$ and $L$.

Lemma 20.47.10. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $j : U \to X$ be an open subspace. Let $E$ be a perfect object of $D(\mathcal{O}_ U)$ whose cohomology sheaves are supported on a closed subset $T \subset U$ with $j(T)$ closed in $X$. Then $Rj_*E$ is a perfect object of $D(\mathcal{O}_ X)$.

Proof. Being a perfect complex is local on $X$. Thus it suffices to check that $Rj_*E$ is perfect when restricted to $U$ and $V = X \setminus j(T)$. We have $Rj_*E|_ U = E$ which is perfect. We have $Rj_*E|_ V = 0$ because $E|_{U \setminus T} = 0$. $\square$

Lemma 20.47.11. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E$ in $D(\mathcal{O}_ X)$ be perfect. Assume that all stalks $\mathcal{O}_{X, x}$ are local rings. Then the set

$U = \{ x \in X \mid H^ i(E)_ x\text{ is a finite free } \mathcal{O}_{X, x}\text{-module for all }i\in \mathbf{Z}\}$

is open in $X$ and is the maximal open set $U \subset X$ such that $H^ i(E)|_ U$ is finite locally free for all $i \in \mathbf{Z}$.

Proof. Note that if $V \subset X$ is some open such that $H^ i(E)|_ V$ is finite locally free for all $i \in \mathbf{Z}$ then $V \subset U$. Let $x \in U$. We will show that an open neighbourhood of $x$ is contained in $U$ and that $H^ i(E)$ is finite locally free on this neighbourhood for all $i$. This will finish the proof. During the proof we may (finitely many times) replace $X$ by an open neighbourhood of $x$. Hence we may assume $E$ is represented by a strictly perfect complex $\mathcal{E}^\bullet$. Say $\mathcal{E}^ i = 0$ for $i \not\in [a, b]$. We will prove the result by induction on $b - a$. The module $H^ b(E) = \mathop{\mathrm{Coker}}(d^{b - 1} : \mathcal{E}^{b - 1} \to \mathcal{E}^ b)$ is of finite presentation. Since $H^ b(E)_ x$ is finite free, we conclude $H^ b(E)$ is finite free in an open neighbourhood of $x$ by Modules, Lemma 17.11.6. Thus after replacing $X$ by a (possibly smaller) open neighbourhood we may assume we have a direct sum decomposition $\mathcal{E}^ b = \mathop{\mathrm{Im}}(d^{b - 1}) \oplus H^ b(E)$ and $H^ b(E)$ is finite free, see Lemma 20.44.5. Doing the same argument again, we see that we may assume $\mathcal{E}^{b - 1} = \mathop{\mathrm{Ker}}(d^{b - 1}) \oplus \mathop{\mathrm{Im}}(d^{b - 1})$. The complex $\mathcal{E}^ a \to \ldots \to \mathcal{E}^{b - 2} \to \mathop{\mathrm{Ker}}(d^{b - 1})$ is a strictly perfect complex representing a perfect object $E'$ with $H^ i(E) = H^ i(E')$ for $i \not= b$. Hence we conclude by our induction hypothesis. $\square$

Comment #637 by on

In the first line of the proof of Lemma 20.38.10, the word "us" should be "use."

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).