The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

13.30 Bounded cohomological dimension

There is another case where the unbounded derived functor exists. Namely, when the functor has bounded cohomological dimension.

Lemma 13.30.1. Let $\mathcal{A}$ be an abelian category. Let $d : \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) \to \{ 0, 1, 2, \ldots , \infty \} $ be a function. Assume that

  1. every object of $\mathcal{A}$ is a subobject of an object $A$ with $d(A) = 0$,

  2. $d(A \oplus B) \leq \max \{ d(A), d(B)\} $ for $A, B \in \mathcal{A}$, and

  3. if $0 \to A \to B \to C \to 0$ is short exact, then $d(C) \leq \max \{ d(A) - 1, d(B)\} $.

Let $K^\bullet $ be a complex such that $n + d(K^ n)$ tends to $-\infty $ as $n \to -\infty $. Then there exists a quasi-isomorphism $K^\bullet \to L^\bullet $ with $d(L^ n) = 0$ for all $n \in \mathbf{Z}$.

Proof. By Lemma 13.16.4 we can find a quasi-isomorphism $\sigma _{\geq 0}K^\bullet \to M^\bullet $ with $M^ n = 0$ for $n < 0$ and $d(M^ n) = 0$ for $n \geq 0$. Then $K^\bullet $ is quasi-isomorphic to the complex

\[ \ldots \to K^{-2} \to K^{-1} \to M^0 \to M^1 \to \ldots \]

Hence we may assume that $d(K^ n) = 0$ for $n \gg 0$. Note that the condition $n + d(K^ n) \to -\infty $ as $n \to -\infty $ is not violated by this replacement.

We are going to improve $K^\bullet $ by an (infinite) sequence of elementary replacements. An elementary replacement is the following. Choose an index $n$ such that $d(K^ n) > 0$. Choose an injection $K^ n \to M$ where $d(M) = 0$. Set $M' = \mathop{\mathrm{Coker}}(K^ n \to M \oplus K^{n + 1})$. Consider the map of complexes

\[ \xymatrix{ K^\bullet : \ar[d] & K^{n - 1} \ar[d] \ar[r] & K^ n \ar[d] \ar[r] & K^{n + 1} \ar[d] \ar[r] & K^{n + 2} \ar[d] \\ (K')^\bullet : & K^{n - 1} \ar[r] & M \ar[r] & M' \ar[r] & K^{n + 2} } \]

It is clear that $K^\bullet \to (K')^\bullet $ is a quasi-isomorphism. Moreover, it is clear that $d((K')^ n) = 0$ and

\[ d((K')^{n + 1}) \leq \max \{ d(K^ n) - 1, d(M \oplus K^{n + 1})\} \leq \max \{ d(K^ n) - 1, d(K^{n + 1})\} \]

and the other values are unchanged.

To finish the proof we carefuly choose the order in which to do the elementary replacements so that for every integer $m$ the complex $\sigma _{\geq m}K^\bullet $ is changed only a finite number of times. To do this set

\[ \xi (K^\bullet ) = \max \{ n + d(K^ n) \mid d(K^ n) > 0\} \]

and

\[ I = \{ n \in \mathbf{Z} \mid \xi (K^\bullet ) = n + d(K^ n) \text{ and } d(K^ n) > 0\} \]

Our assumption that $n + d(K^ n)$ tends to $-\infty $ as $n \to -\infty $ and the fact that $d(K^ n) = 0$ for $n >> 0$ implies $\xi (K^\bullet ) < +\infty $ and that $I$ is a finite set. It is clear that $\xi ((K')^\bullet ) \leq \xi (K^\bullet )$ for an elementary transformation as above. An elementary transformation changes the complex in degrees $\leq \xi (K^\bullet ) + 1$. Hence if we can find finite sequence of elementary transformations which decrease $\xi (K^\bullet )$, then we win. However, note that if we do an elementary transformation starting with the smallest element $n \in I$, then we either decrease the size of $I$, or we increase $\min I$. Since every element of $I$ is $\leq \xi (K^\bullet )$ we see that we win after a finite number of steps. $\square$

Lemma 13.30.2. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor of abelian categories. If

  1. every object of $\mathcal{A}$ is a subobject of an object which is right acyclic for $F$,

  2. there exists an integer $n$ such that $R^ nF = 0$,

then $RF : D(\mathcal{A}) \to D(\mathcal{B})$ exists. Any complex consisting of right acyclic objects for $F$ computes $RF$ and any complex is the source of a quasi-isomorphism into such a complex.

Proof. Note that the first condition implies that $RF : D^+(\mathcal{A}) \to D^+(\mathcal{B})$ exists, see Proposition 13.17.8. Let $A$ be an object of $\mathcal{A}$. Choose an injection $A \to A'$ with $A'$ acyclic. Then we see that $R^{n + 1}F(A) = R^ nF(A'/A) = 0$ by the long exact cohomology sequence. Hence we conclude that $R^{n + 1}F = 0$. Continuing like this using induction we find that $R^ mF = 0$ for all $m \geq n$.

We are going to use Lemma 13.30.1 with the function $d : \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) \to \{ 0, 1, 2, \ldots \} $ given by $d(A) = \max \{ 0\} \cup \{ i \mid R^ iF(A) \not= 0\} $. The first assumption of Lemma 13.30.1 is our assumption (1). The second assumption of Lemma 13.30.1 follows from the fact that $RF(A \oplus B) = RF(A) \oplus RF(B)$. The third assumption of Lemma 13.30.1 follows from the long exact cohomology sequence. Hence for every complex $K^\bullet $ there exists a quasi-isomorphism $K^\bullet \to L^\bullet $ with $L^ n$ right acyclic for $F$. We claim that if $L^\bullet \to M^\bullet $ is a quasi-isomorphism of complexes of right acyclic objects for $F$, then $F(L^\bullet ) \to F(M^\bullet )$ is a quasi-isomorphism. If we prove this claim then we are done by Lemma 13.15.15. To prove the claim pick an integer $i \in \mathbf{Z}$. Consider the distinguished triangle

\[ \sigma _{\geq i - n - 1}L^\bullet \to \sigma _{\geq i - n - 1}M^\bullet \to Q^\bullet , \]

i.e., let $Q^\bullet $ be the cone of the first map. Note that $Q^\bullet $ is bounded below and that $H^ j(Q^\bullet )$ is zero except possibly for $j = i - n - 1$ or $j = i - n - 2$. We may apply $RF$ to $Q^\bullet $. Using the second spectral sequence of Lemma 13.21.3 and the assumed vanishing of cohomology (2) we conclude that $R^ jF(Q^\bullet )$ is zero except possibly for $j \in \{ i - n - 2, \ldots , i - 1\} $. Hence we see that $RF(\sigma _{\geq i - n - 1}L^\bullet ) \to RF(\sigma _{\geq i - n - 1}M^\bullet )$ induces an isomorphism of cohomology objects in degrees $\geq i$. By Proposition 13.17.8 we know that $RF(\sigma _{\geq i - n - 1}L^\bullet ) = \sigma _{\geq i - n - 1}F(L^\bullet )$ and $RF(\sigma _{\geq i - n - 1}M^\bullet ) = \sigma _{\geq i - n - 1}F(M^\bullet )$. We conclude that $F(L^\bullet ) \to F(M^\bullet )$ is an isomorphism in degree $i$ as desired. $\square$

Lemma 13.30.3. Let $F : \mathcal{A} \to \mathcal{B}$ be a right exact functor of abelian categories. If

  1. every object of $\mathcal{A}$ is a quotient of an object which is left acyclic for $F$,

  2. there exists an integer $n$ such that $L^ nF = 0$,

then $LF : D(\mathcal{A}) \to D(\mathcal{B})$ exists. Any complex consisting of left acyclic objects for $F$ computes $LF$ and any complex is the target of a quasi-isomorphism into such a complex.

Proof. This is dual to Lemma 13.30.2. $\square$


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