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*}

15.90 Tricks with double complexes

This section continues the discussion in Homology, Section 12.23.

Lemma 15.90.1. Let

\[ (A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots ) \longrightarrow (B_0^\bullet \to B_1^\bullet \to B_2^\bullet \to \ldots ) \]

be a map between two complexes of complexes of abelian groups. Set $A^{p, q} = A_ p^ q$, $B^{p, q} = B_ p^ q$ to obtain double complexes. Let $\text{Tot}_\pi (A^{\bullet , \bullet })$ and $\text{Tot}_\pi (B^{\bullet , \bullet })$ be the product total complexes associated to the double complexes. If each $A_ p^\bullet \to B_ p^\bullet $ is a quasi-isomorphism, then $\text{Tot}_\pi (A^{\bullet , \bullet }) \to \text{Tot}_\pi (B^{\bullet , \bullet })$ is a quasi-isomorphism.

Proof. Recall that $\text{Tot}_\pi (A^{\bullet , \bullet })$ in degree $n$ is given by $\prod _{p + q = n} A^{p, q} = \prod _{p + 1 = n} A^ q_ p$. Let $C_ p^\bullet $ be the cone on the map $A_ p^\bullet \to B_ p^\bullet $, see Derived Categories, Section 13.9. By the functoriality of the cone construction we obtain a complex of complexes

\[ C_0^\bullet \to C_1^\bullet \to C_2^\bullet \to \ldots \]

Then we see $\text{Tot}_\pi (C^{\bullet , \bullet })$ in degree $n$ is given by

\[ \prod _{p + q = n} C^{p, q} = \prod _{p + q = n} C^ q_ p = \prod _{p + q = n} (B^ q_ p \oplus A^{q + 1}_ p) = \prod _{p + q = n} B^ q_ p \oplus \prod _{p + q = n} A^{q + 1}_ p \]

We conclude that $\text{Tot}_\pi (C^{\bullet , \bullet })$ is the cone of the map $\text{Tot}_\pi (A^{\bullet , \bullet }) \to \text{Tot}_\pi (B^{\bullet , \bullet })$ (We omit the verification that the differentials agree.) Thus it suffices to show $\text{Tot}_\pi (A^{\bullet , \bullet })$ is acyclic if each $A_ p^\bullet $ is acyclic.

Denote $f_ p : A_ p^\bullet \to A_{p + 1}^\bullet $ the given maps of complexes. Recall that the differential on $\text{Tot}_\pi (A^{\bullet , \bullet })$ is given by

\[ \prod \nolimits _{p + q = n} (f^ q_ p + (-1)^ p\text{d}^ q_{A_ p^\bullet }) \]

on elements in degree $n$. Let $\xi \in H^0(\text{Tot}_\pi (A^{\bullet , \bullet }))$ be a cohomology class. We will show $\xi $ is zero; the same argument works in other degrees. Represent $\xi $ as the class of an cocycle $x = (x_ p) \in \prod A^{p, -p}$. Since $\text{d}(x) = 0$ we find that $\text{d}_{A_0^\bullet }(x_0) = 0$. Thus there exists a $y_{-1} \in A^{0, -1}$ with $\text{d}_{A_0^\bullet }(y_{-1}) = x_0$. Then we see that $\text{d}_{A_1^\bullet }(x_1 + f_0(y_{-1})) = 0$. Thus we can find a $y_{-2} \in A^{1, -2}$ such that $-\text{d}_{A_1^\bullet }(y_{-2}) = x_1 + f_0(y_{-1})$. By induction we can find $y_{-p - 1} \in A^{p, -p - 1}$ such that

\[ (-1)^ p\text{d}_{A_ p^\bullet }(y_{-p - 1}) = x_ p + f_{p - 1}(y_{-p}) \]

This implies that $\text{d}(y) = x$ where $y = (y_{-p - 1})$. $\square$


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