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Invertible objects of D(O_X) for a ringed space

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aisejohan committed Nov 6, 2019
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@@ -11082,6 +11082,172 @@ \section{Duals}



\section{Invertible objects in the derived category}
\label{section-invertible-D-or-R}

\noindent
We characterize invertible objects in the derived category of
a ringed space (both in the case where the stalks of the
structure sheaf are local and where not).

\begin{lemma}
\label{lemma-category-summands-finite-free}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Set $R = \Gamma(X, \mathcal{O}_X)$. The category of
$\mathcal{O}_X$-modules which are summands of finite free
$\mathcal{O}_X$-modules is equivalent to the category of
finite projective $R$-modules.
\end{lemma}

\begin{proof}
Observe that a finite projective $R$-module is the same thing
as a summand of a finite free $R$-module.
The equivalence is given by the functor $\mathcal{E} \mapsto
\Gamma(X, \mathcal{E})$. The inverse functor is given by the construction of
Modules, Lemma \ref{modules-lemma-construct-quasi-coherent-sheaves}.
\end{proof}

\begin{lemma}
\label{lemma-invertible-derived}
Let $(X, \mathcal{O}_X)$ be a ringed space. Let $M$ be an object
of $D(\mathcal{O}_X)$. The following are equivalent
\begin{enumerate}
\item $M$ is invertible in $D(\mathcal{O}_X)$, see
Categories, Definition \ref{categories-definition-invertible}, and
\item there is a locally finite direct product decomposition
$$
\mathcal{O}_X = \prod\nolimits_{n \in \mathbf{Z}} \mathcal{O}_n
$$
and for each $n$ there is an invertible $\mathcal{O}_n$-module
$\mathcal{H}^n$ (Modules, Definition \ref{modules-definition-invertible})
and $M = \bigoplus \mathcal{H}^n[-n]$ in $D(\mathcal{O}_X)$.
\end{enumerate}
If (1) and (2) hold, then $M$ is a perfect object of $D(\mathcal{O}_X)$. If
$\mathcal{O}_{X, x}$ is a local ring for all $x \in X$ these condition
are also equivalent to
\begin{enumerate}
\item[(3)] there exists an open covering $X = \bigcup U_i$
and for each $i$ an integer $n_i$ such that $M|_{U_i}$
is represented by an invertible $\mathcal{O}_{U_i}$
module placed in degree $n_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
Assume (2). Consider the object $R\SheafHom(M, \mathcal{O}_X)$
and the composition map
$$
R\SheafHom(M, \mathcal{O}_X) \otimes_{\mathcal{O}_X}^\mathbf{L} M \to
\mathcal{O}_X
$$
To prove this is an isomorphism, we may work locally. Thus we may
assume $\mathcal{O}_X = \prod_{a \leq n \leq b} \mathcal{O}_n$
and $M = \bigoplus_{a \leq n \leq b} \mathcal{H}^n[-n]$.
Then it suffices to show that
$$
R\SheafHom(\mathcal{H}^m, \mathcal{O}_X)
\otimes_{\mathcal{O}_X}^\mathbf{L} \mathcal{H}^n
$$
is zero if $n \not = m$ and equal to $\mathcal{O}_n$ if $n = m$.
The case $n \not = m$ follows from the fact that $\mathcal{O}_n$ and
$\mathcal{O}_m$ are flat $\mathcal{O}_X$-algebras with
$\mathcal{O}_n \otimes_{\mathcal{O}_X} \mathcal{O}_m = 0$.
Using the local structure of invertible $\mathcal{O}_X$-modules
(Modules, Lemma \ref{modules-lemma-invertible}) and working locally
the isomorphism in case $n = m$ follows in a straightforward manner;
we omit the details. Because $D(\mathcal{O}_X)$ is symmetric monoidal,
we conclude that $M$ is invertible.

\medskip\noindent
Assume (1). The description in (2) shows that we have a candidate
for $\mathcal{O}_n$, namely,
$\SheafHom_{\mathcal{O}_X}(H^n(M), H^n(M))$.
If this is a locally finite family of sheaves of rings
and if $\mathcal{O}_X = \prod \mathcal{O}_n$, then we immediately
obtain the direct sum decomposition $M = \bigoplus H^n(M)[-n]$
using the idempotents in $\mathcal{O}_X$ coming from the product
decomposition.
This shows that in order to prove (2) we may work locally on $X$.

\medskip\noindent
Choose an object $N$ of $D(\mathcal{O}_X)$
and an isomorphism
$M \otimes_{\mathcal{O}_X}^\mathbf{L} N \cong \mathcal{O}_X$.
Let $x \in X$.
Then $N$ is a left dual for $M$ in the monoidal category
$D(\mathcal{O}_X)$ and we conclude that $M$ is perfect by
Lemma \ref{lemma-left-dual-derived}. By symmetry we see that
$N$ is perfect. After replacing $X$ by an open neighbourhood of $x$,
we may assume $M$ and $N$ are represented by a strictly perfect
complexes $\mathcal{E}^\bullet$ and $\mathcal{F}^\bullet$.
Then $M \otimes_{\mathcal{O}_X}^\mathbf{L} N$ is represented by
$\text{Tot}(\mathcal{E}^\bullet \otimes_{\mathcal{O}_X} \mathcal{F}^\bullet)$.
After another shinking of $X$ we may assume the mutually inverse
isomorphisms
$\mathcal{O}_X \to M \otimes_{\mathcal{O}_X}^\mathbf{L} N$ and
$M \otimes_{\mathcal{O}_X}^\mathbf{L} N \to \mathcal{O}_X$
are given by maps of complexes
$$
\alpha : \mathcal{O}_X \to
\text{Tot}(\mathcal{E}^\bullet \otimes_{\mathcal{O}_X} \mathcal{F}^\bullet)
\quad\text{and}\quad
\beta :
\text{Tot}(\mathcal{E}^\bullet \otimes_{\mathcal{O}_X} \mathcal{F}^\bullet)
\to \mathcal{O}_X
$$
See Lemma \ref{lemma-local-actual}. Then $\beta \circ \alpha = 1$
as maps of complexes and $\alpha \circ \beta = 1$ as a morphism
in $D(\mathcal{O}_X)$. After shrinking $X$
we may assume the composition $\alpha \circ \beta$ is homotopic to $1$
by some homotopy $\theta$ with components
$$
\theta^n :
\text{Tot}^n(\mathcal{E}^\bullet \otimes_{\mathcal{O}_X} \mathcal{F}^\bullet)
\to
\text{Tot}^{n - 1}(
\mathcal{E}^\bullet \otimes_{\mathcal{O}_X} \mathcal{F}^\bullet)
$$
by the same lemma as before. Set $R = \Gamma(X, \mathcal{O}_X)$. By
Lemma \ref{lemma-category-summands-finite-free}
we find that we obtain
\begin{enumerate}
\item $M^\bullet = \Gamma(X, \mathcal{E}^\bullet)$ is a bounded complex
of finite projective $R$-modules,
\item $N^\bullet = \Gamma(X, \mathcal{F}^\bullet)$ is a bounded complex
of finite projective $R$-modules,
\item $\alpha$ and $\beta$ correspond to maps of complexes
$a : R \to \text{Tot}(M^\bullet \otimes_R N^\bullet)$ and
$b : \text{Tot}(M^\bullet \otimes_R N^\bullet) \to R$,
\item $\theta^n$ corresponds to a map
$h^n : \text{Tot}^n(M^\bullet \otimes_R N^\bullet) \to
\text{Tot}^{n - 1}(M^\bullet \otimes_R N^\bullet)$, and
\item $b \circ a = 1$ and $b \circ a - 1 = dh + hd$,
\end{enumerate}
It follows that $M^\bullet$ and $N^\bullet$ define
mutually inverse objects of $D(R)$. By
More on Algebra, Lemma \ref{lemma-invertible-derived}
we find a product decomposition $R = \prod_{a \leq n \leq b} R_n$
and invertible $R_n$-modules $H^n$ such
that $M^\bullet \cong \bigoplus_{a \leq n \leq b} H^n[-n]$.
This isomorphism in $D(R)$ can be lifted to an morphism
$$
\bigoplus H^n[-n] \longrightarrow M^\bullet
$$
of complexes because each $H^n$ is projective as an $R$-module.
Correspondingly, using Lemma \ref{lemma-category-summands-finite-free} again,
we obtain an morphism
$$
\bigoplus H^n \otimes_R \mathcal{O}_X[-n] \to \mathcal{E}^\bullet
$$
which is an isomorphism in $D(\mathcal{O}_X)$. Setting
$\mathcal{O}_n = R_n \otimes_R \mathcal{O}_X$ we conclude (2) is true.

\medskip\noindent
If all stalks of $\mathcal{O}_X$ are local, then it is straightforward
to prove the equivalence of (2) and (3). We omit the details.
\end{proof}





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