
## 20.43 Perfect complexes

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

Definition 20.43.1. 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 20.43.2. 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 20.41.2. $\square$

Lemma 20.43.3. 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 20.43.2 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 20.43.4. 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

$\mathcal{E}^\bullet \to E \to C \to \mathcal{E}^\bullet [1]$

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 20.27.15). Thus $\mathcal{E}'$ is locally a direct summand of a finite free module by Modules, Lemma 17.16.11. 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 20.43.5. 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 20.41.2. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_ i}$ has finite Tor dimension by Lemma 20.42.3. 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 20.43.4. $\square$

Lemma 20.43.6. 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 20.43.7. 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 20.43.5, 20.41.4, and 20.42.6. 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 20.40.8. Then $M$ is isomorphic to the cone of $\alpha$ which is strictly perfect by Lemma 20.40.2. $\square$

Lemma 20.43.8. 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 20.43.9. 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 20.43.10. 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 20.43.11. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ be a perfect object of $D(\mathcal{O}_ X)$. Then $K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O}_ X)$ is a perfect object too and $(K^\vee )^\vee = K$. There are functorial isomorphisms

$K^\vee \otimes ^\mathbf {L}_{\mathcal{O}_ X} M = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$

and

$H^0(X, K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} M) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K, M)$

for $M$ in $D(\mathcal{O}_ X)$.

Proof. We will use without further mention that formation of internal hom commutes with restriction to opens (Lemma 20.36.3). In particular we may check the first two statements locally on $X$. By Lemma 20.36.9 to see the final statement it suffices to check that the map (20.36.9.1)

$K^\vee \otimes ^\mathbf {L}_{\mathcal{O}_ X} M \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$

is an isomorphism. This is local on $X$ as well. Hence it suffices to prove the lemma when $K$ is represented by a strictly perfect complex.

Assume $K$ is represented by the strictly perfect complex $\mathcal{E}^\bullet$. Then it follows from Lemma 20.40.9 that $K^\vee$ is represented by the complex whose terms are $(\mathcal{E}^{-n})^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-n}, \mathcal{O}_ X)$ in degree $n$. Since $\mathcal{E}^{-n}$ is a direct summand of a finite free $\mathcal{O}_ X$-module, so is $(\mathcal{E}^{-n})^\vee$. Hence $K^\vee$ is represented by a strictly perfect complex too. It is also clear that $(K^\vee )^\vee = K$ as we have $((\mathcal{E}^{-n})^\vee )^\vee = \mathcal{E}^{-n}$. To see that (20.36.9.1) is an isomorphism, represent $M$ by a K-flat complex $\mathcal{F}^\bullet$. By Lemma 20.40.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

$\bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-q}, \mathcal{F}^ p)$

On the other hand, then object $K^\vee \otimes ^\mathbf {L} M$ is represented by the complex with terms

$\bigoplus \nolimits _{n = p + q} \mathcal{F}^ p \otimes _{\mathcal{O}_ X} (\mathcal{E}^{-q})^\vee$

Thus the assertion that (20.36.9.1) is an isomorphism reduces to the assertion that the canonical map

$\mathcal{F} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{F})$

is an isomorphism when $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_ X$-module and $\mathcal{F}$ is any $\mathcal{O}_ X$-module. This follows immediately from the corresponding statement when $\mathcal{E}$ is finite free. $\square$

Lemma 20.43.12. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)_{n \in \mathbf{N}}$ be a system of perfect objects of $D(\mathcal{O}_ X)$. Let $K = \text{hocolim} K_ n$ be the derived colimit (Derived Categories, Definition 13.31.1). Then for any object $E$ of $D(\mathcal{O}_ X)$ we have

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, E) = R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee$

where $(K_ n^\vee )$ is the inverse system of dual perfect complexes.

Proof. By Lemma 20.43.11 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

$R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (20.36.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

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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