## 20.45 Perfect complexes

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

Definition 20.45.1. 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 20.45.2. 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 20.43.2. $\square$

Lemma 20.45.3. 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 20.45.2 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 20.45.4. 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 20.26.15). Thus $\mathcal{E}'$ is locally a direct summand of a finite free module by Modules, Lemma 17.17.3. 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 20.45.5. 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 20.43.2. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_ i}$ has finite Tor dimension by Lemma 20.44.3. 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 20.45.4. $\square$

Lemma 20.45.6. 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 20.45.7. 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 20.45.5, 20.43.4, and 20.44.6. 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 20.42.8. Then $M$ is isomorphic to the cone of $\alpha$ which is strictly perfect by Lemma 20.42.2. $\square$

Lemma 20.45.8. 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 20.45.9. 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 20.45.10. 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 $E|_{U \setminus T} = 0$. $\square$

Comment #637 by on

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

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