Lemma 36.9.2. In Situation 36.9.1. Let $M$ be an $A$-module and denote $\mathcal{F}$ the associated $\mathcal{O}_ X$-module. Then there is a canonical isomorphism of complexes

$\Psi : \mathop{\mathrm{colim}}\nolimits _ e \mathop{\mathrm{Hom}}\nolimits _ A(I^\bullet (f_1^ e, \ldots , f_ r^ e), M) \longrightarrow \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F})$

functorial in $M$ where the differentials on the $\mathop{\mathrm{Hom}}\nolimits$-complex are the contragredients of the differentials on $I^\bullet (f_1^ e, \ldots , f_ r^ e)$.

Proof. Recall that the alternating Čech complex is the subcomplex of the usual Čech complex given by alternating cochains, see Cohomology, Section 20.23. As usual we view a $p$-cochain in $\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F})$ as an alternating function $s$ on $\{ 1, \ldots , r\} ^{p + 1}$ whose value $s_{i_0\ldots i_ p}$ at $(i_0, \ldots , i_ p)$ lies in $M_{f_{i_0}\ldots f_{i_ p}} = \mathcal{F}(U_{i_0\ldots i_ p})$. On the other hand, a $p$-cochain $t$ in $\mathop{\mathrm{Hom}}\nolimits ^\bullet (I^\bullet (f_1^ e, \ldots , f_ r^ e), M)$ is a map $t : \wedge ^{p + 1}(A^{\oplus r}) \to M$. Write $[i] \in A^{\oplus r}$ for the $i$th basis element and write

$[i_0, \ldots , i_ p] = [i_0] \wedge \ldots \wedge [i_ p] \in \wedge ^{p + 1}(A^{\oplus r})$

For $t$ as above we set

$\Psi (t)_{i_0 \ldots i_ p} = (-1)^ p \frac{t([i_0, \ldots , i_ p])}{f_{i_0}^ e\ldots f_{i_ p}^ e}$

It is clear that $\Psi (t)$ is an alternating cochain. The rule above is compatible with the transition maps of the system as the transition map

$I^\bullet (f_1^ e, \ldots , f_ r^ e) \leftarrow I^\bullet (f_1^{e + 1}, \ldots , f_ r^{e + 1}),$

of (36.9.0.1) sends $[i_0, \ldots , i_ p]$ to $f_{i_0}\ldots f_{i_ p}[i_0, \ldots , i_ p]$. It is clear from the description of the localizations $M_{f_{i_0} \ldots f_{i_ p}}$ in Algebra, Lemma 10.9.9 that the rule $\Psi$ defines an isomorphism of cochain modules in degree $p$ in the colimit. To finish the proof we have to show that the map is compatible with differentials. To see this, for $t$ as above we compute

\begin{align*} d(\Psi (t))_{i_0 \ldots i_{p + 1}} & = \sum \nolimits _{j = 0}^{p + 1} (-1)^ j \Psi (t)_{i_0\ldots \hat i_ j \ldots i_{p + 1}} \\ & = (-1)^ p \sum \nolimits _{j = 0}^{p + 1} (-1)^ j t([i_0 \ldots \hat i_ j \ldots i_{p + 1}]) (f_{i_0} \ldots \hat f_{i_ j} \ldots f_{i_ p})^{-e} \end{align*}

Recall that the differentials on $I^\bullet (f_1^ e, \ldots , f_ r^ e)$ are the negative of the differentials on $K^\bullet (f_1, \ldots , f_ r)$. Thus

\begin{align*} \Psi (d(t))_{i_0 \ldots i_{p + 1}} & = (-1)^{p + 1} d(t)([i_0, \ldots , i_{p + 1}]) (f_{i_0} \ldots f_{i_{p + 1}})^{-e} \\ & = (-1)^{p + 1} t(d([i_0, \ldots , i_{p + 1}])) (f_{i_0} \ldots f_{i_{p + 1}})^{-e} \\ & = (-1)^{p + 1} t(-\sum \nolimits _{j = 0}^{p + 1} (-1)^ j f_{i_ j}^ e [i_0, \ldots , \hat i_ j, \ldots i_{p + 1}]) (f_{i_0} \ldots f_{i_{p + 1}})^{-e} \\ & = -(-1)^{p + 1} \sum \nolimits _{j = 0}^{p + 1} (-1)^ j t([i_0, \ldots , \hat i_ j, \ldots i_{p + 1}]) (f_{i_0} \ldots \hat f_{i_ j} \ldots f_{i_ p})^{-e} \end{align*}

The two formulas agree concluding the proof. $\square$

Comment #8621 by nkym on

I am not sure if the last equality in the proof reflects the fact that the diffential of $I$ is the negative of that of $K$. I am sorry if I am wrong.

There are also:

• 3 comment(s) on Section 36.9: Koszul complexes

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).