68.6 The alternating Čech complex

Let $S$ be a scheme. Let $f : U \to X$ be an étale morphism of algebraic spaces over $S$. The functor

$j : U_{spaces, {\acute{e}tale}} \longrightarrow X_{spaces, {\acute{e}tale}},\quad V/U \longmapsto V/X$

induces an equivalence of $U_{spaces, {\acute{e}tale}}$ with the localization $X_{spaces, {\acute{e}tale}}/U$, see Properties of Spaces, Section 65.27. Hence there exist functors

$f_! : \textit{Ab}(U_{\acute{e}tale}) \longrightarrow \textit{Ab}(X_{\acute{e}tale}),\quad f_! : \textit{Mod}(\mathcal{O}_ U) \longrightarrow \textit{Mod}(\mathcal{O}_ X),$

$f^{-1} : \textit{Ab}(X_{\acute{e}tale}) \longrightarrow \textit{Ab}(U_{\acute{e}tale}),\quad f^* : \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ U)$

see Modules on Sites, Section 18.19. Warning: This functor, a priori, has nothing to do with cohomology with compact supports! We dubbed this functor “extension by zero” in the reference above. Note that the two versions of $f_!$ agree as $f^* = f^{-1}$ for sheaves of $\mathcal{O}_ X$-modules.

As we are going to use this construction below let us recall some of its properties. Given an abelian sheaf $\mathcal{G}$ on $U_{\acute{e}tale}$ the sheaf $f_!$ is the sheafification of the presheaf

$V/X \longmapsto f_!\mathcal{G}(V) = \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Mor}}\nolimits _ X(V, U)} \mathcal{G}(V \xrightarrow {\varphi } U),$

see Modules on Sites, Lemma 18.19.2. Moreover, if $\mathcal{G}$ is an $\mathcal{O}_ U$-module, then $f_!\mathcal{G}$ is the sheafification of the exact same presheaf of abelian groups which is endowed with an $\mathcal{O}_ X$-module structure in an obvious way (see loc. cit.). Let $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ be a geometric point. Then there is a canonical identification

$(f_!\mathcal{G})_{\overline{x}} = \bigoplus \nolimits _{\overline{u}} \mathcal{G}_{\overline{u}}$

where the sum is over all $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ such that $f \circ \overline{u} = \overline{x}$, see Modules on Sites, Lemma 18.38.1 and Properties of Spaces, Lemma 65.19.13. In the following we are going to study the sheaf $f_!\underline{\mathbf{Z}}$. Here $\underline{\mathbf{Z}}$ denotes the constant sheaf on $X_{\acute{e}tale}$ or $U_{\acute{e}tale}$.

Lemma 68.6.1. Let $S$ be a scheme. Let $f_ i : U_ i \to X$ be étale morphisms of algebraic spaces over $S$. Then there are isomorphisms

$f_{1, !}\underline{\mathbf{Z}} \otimes _{\mathbf{Z}} f_{2, !}\underline{\mathbf{Z}} \longrightarrow f_{12, !}\underline{\mathbf{Z}}$

where $f_{12} : U_1 \times _ X U_2 \to X$ is the structure morphism and

$(f_1 \amalg f_2)_! \underline{\mathbf{Z}} \longrightarrow f_{1, !}\underline{\mathbf{Z}} \oplus f_{2, !}\underline{\mathbf{Z}}$

Proof. Once we have defined the map it will be an isomorphism by our description of stalks above. To define the map it suffices to work on the level of presheaves. Thus we have to define a map

$\left(\bigoplus \nolimits _{\varphi _1 \in \mathop{\mathrm{Mor}}\nolimits _ X(V, U_1)} \mathbf{Z}\right) \otimes _{\mathbf{Z}} \left(\bigoplus \nolimits _{\varphi _2 \in \mathop{\mathrm{Mor}}\nolimits _ X(V, U_2)} \mathbf{Z}\right) \longrightarrow \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Mor}}\nolimits _ X(V, U_1 \times _ X U_2)} \mathbf{Z}$

We map the element $1_{\varphi _1} \otimes 1_{\varphi _2}$ to the element $1_{\varphi _1 \times \varphi _2}$ with obvious notation. We omit the proof of the second equality. $\square$

Another important feature is the trace map

$\text{Tr}_ f : f_!\underline{\mathbf{Z}} \longrightarrow \underline{\mathbf{Z}}.$

The trace map is adjoint to the map $\mathbf{Z} \to f^{-1}\underline{\mathbf{Z}}$ (which is an isomorphism). If $\overline{x}$ is above, then $\text{Tr}_ f$ on stalks at $\overline{x}$ is the map

$(\text{Tr}_ f)_{\overline{x}} : (f_!\underline{\mathbf{Z}})_{\overline{x}} = \bigoplus \nolimits _{\overline{u}} \mathbf{Z} \longrightarrow \mathbf{Z} = \underline{\mathbf{Z}}_{\overline{x}}$

which sums the given integers. This is true because it is adjoint to the map $1 : \mathbf{Z} \to f^{-1}\underline{\mathbf{Z}}$. In particular, if $f$ is surjective as well as étale then $\text{Tr}_ f$ is surjective.

Assume that $f : U \to X$ is a surjective étale morphism of algebraic spaces. Consider the Koszul complex associated to the trace map we discussed above

$\ldots \to \wedge ^3f_!\underline{\mathbf{Z}} \to \wedge ^2f_!\underline{\mathbf{Z}} \to f_!\underline{\mathbf{Z}} \to \underline{\mathbf{Z}} \to 0$

Here the exterior powers are over the sheaf of rings $\underline{\mathbf{Z}}$. The maps are defined by the rule

$e_1 \wedge \ldots \wedge e_ n \longmapsto \sum \nolimits _{i = 1, \ldots , n} (-1)^{i + 1} \text{Tr}_ f(e_ i) e_1 \wedge \ldots \wedge \widehat{e_ i} \wedge \ldots \wedge e_ n$

where $e_1, \ldots , e_ n$ are local sections of $f_!\underline{\mathbf{Z}}$. Let $\overline{x}$ be a geometric point of $X$ and set $M_{\overline{x}} = (f_!\underline{\mathbf{Z}})_{\overline{x}} = \bigoplus _{\overline{u}} \mathbf{Z}$. Then the stalk of the complex above at $\overline{x}$ is the complex

$\ldots \to \wedge ^3 M_{\overline{x}} \to \wedge ^2 M_{\overline{x}} \to M_{\overline{x}} \to \mathbf{Z} \to 0$

which is exact because $M_{\overline{x}} \to \mathbf{Z}$ is surjective, see More on Algebra, Lemma 15.28.5. Hence if we let $K^\bullet = K^\bullet (f)$ be the complex with $K^ i = \wedge ^{i + 1}f_!\underline{\mathbf{Z}}$, then we obtain a quasi-isomorphism

68.6.1.1
$$\label{spaces-cohomology-equation-quasi-isomorphism} K^\bullet \longrightarrow \underline{\mathbf{Z}}[0]$$

We use the complex $K^\bullet$ to define what we call the alternating Čech complex associated to $f : U \to X$.

Definition 68.6.2. Let $S$ be a scheme. Let $f : U \to X$ be a surjective étale morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be an object of $\textit{Ab}(X_{\acute{e}tale})$. The alternating Čech complex1 $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F})$ associated to $\mathcal{F}$ and $f$ is the complex

$\mathop{\mathrm{Hom}}\nolimits (K^0, \mathcal{F}) \to \mathop{\mathrm{Hom}}\nolimits (K^1, \mathcal{F}) \to \mathop{\mathrm{Hom}}\nolimits (K^2, \mathcal{F}) \to \ldots$

with Hom groups computed in $\textit{Ab}(X_{\acute{e}tale})$.

The reader may verify that if $U = \coprod U_ i$ and $f|_{U_ i} : U_ i \to X$ is the open immersion of a subspace, then $\check{\mathcal{C}}_{alt}^\bullet (f, \mathcal{F})$ agrees with the complex introduced in Cohomology, Section 20.23 for the Zariski covering $X = \bigcup U_ i$ and the restriction of $\mathcal{F}$ to the Zariski site of $X$. What is more important however, is to relate the cohomology of the alternating Čech complex to the cohomology.

Lemma 68.6.3. Let $S$ be a scheme. Let $f : U \to X$ be a surjective étale morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be an object of $\textit{Ab}(X_{\acute{e}tale})$. There exists a canonical map

$\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \longrightarrow R\Gamma (X, \mathcal{F})$

in $D(\textit{Ab})$. Moreover, there is a spectral sequence with $E_1$-page

$E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits _{\textit{Ab}(X_{\acute{e}tale})}^ q(K^ p, \mathcal{F})$

converging to $H^{p + q}(X, \mathcal{F})$ where $K^ p = \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$.

Proof. Recall that we have the quasi-isomorphism $K^\bullet \to \underline{\mathbf{Z}}[0]$, see (68.6.1.1). Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$ in $\textit{Ab}(X_{\acute{e}tale})$. Consider the double complex $\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet )$ with terms $\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^ q)$. The differential $d_1^{p, q} : A^{p, q} \to A^{p + 1, q}$ is the one coming from the differential $K^{p + 1} \to K^ p$ and the differential $d_2^{p, q} : A^{p, q} \to A^{p, q + 1}$ is the one coming from the differential $\mathcal{I}^ q \to \mathcal{I}^{q + 1}$. Denote $\text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$ the associated total complex, see Homology, Section 12.18. We will use the two spectral sequences $({}'E_ r, {}'d_ r)$ and $({}''E_ r, {}''d_ r)$ associated to this double complex, see Homology, Section 12.25.

Because $K^\bullet$ is a resolution of $\underline{\mathbf{Z}}$ we see that the complexes

$\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^ q) : \mathop{\mathrm{Hom}}\nolimits (K^0, \mathcal{I}^ q) \to \mathop{\mathrm{Hom}}\nolimits (K^1, \mathcal{I}^ q) \to \mathop{\mathrm{Hom}}\nolimits (K^2, \mathcal{I}^ q) \to \ldots$

are acyclic in positive degrees and have $H^0$ equal to $\Gamma (X, \mathcal{I}^ q)$. Hence by Homology, Lemma 12.25.4 the natural map

$\mathcal{I}^\bullet (X) \longrightarrow \text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$

is a quasi-isomorphism of complexes of abelian groups. In particular we conclude that $H^ n(\text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))) = H^ n(X, \mathcal{F})$.

The map $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \to R\Gamma (X, \mathcal{F})$ of the lemma is the composition of $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \to \text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$ with the inverse of the displayed quasi-isomorphism.

Finally, consider the spectral sequence $({}'E_ r, {}'d_ r)$. We have

$E_1^{p, q} = q\text{th cohomology of } \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^0) \to \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^1) \to \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^2) \to \ldots$

This proves the lemma. $\square$

It follows from the lemma that it is important to understand the ext groups $\mathop{\mathrm{Ext}}\nolimits _{\textit{Ab}(X_{\acute{e}tale})}(K^ p, \mathcal{F})$, i.e., the right derived functors of $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F})$.

Lemma 68.6.4. Let $S$ be a scheme. Let $f : U \to X$ be a surjective, étale, and separated morphism of algebraic spaces over $S$. For $p \geq 0$ set

$W_ p = U \times _ X \ldots \times _ X U \setminus \text{all diagonals}$

where the fibre product has $p + 1$ factors. There is a free action of $S_{p + 1}$ on $W_ p$ over $X$ and

$\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F}) = S_{p + 1}\text{-anti-invariant elements of } \mathcal{F}(W_ p)$

functorially in $\mathcal{F}$ where $K^ p = \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$.

Proof. Because $U \to X$ is separated the diagonal $U \to U \times _ X U$ is a closed immersion. Since $U \to X$ is étale the diagonal $U \to U \times _ X U$ is an open immersion, see Morphisms of Spaces, Lemmas 66.39.10 and 66.38.9. Hence $W_ p$ is an open and closed subspace of $U^{p + 1} = U \times _ X \ldots \times _ X U$. The action of $S_{p + 1}$ on $W_ p$ is free as we've thrown out the fixed points of the action. By Lemma 68.6.1 we see that

$(f_!\underline{\mathbf{Z}})^{\otimes p + 1} = f^{p + 1}_!\underline{\mathbf{Z}} = (W_ p \to X)_!\underline{\mathbf{Z}} \oplus Rest$

where $f^{p + 1} : U^{p + 1} \to X$ is the structure morphism. Looking at stalks over a geometric point $\overline{x}$ of $X$ we see that

$\left( \bigoplus \nolimits _{\overline{u} \mapsto \overline{x}} \mathbf{Z} \right)^{\otimes p + 1} \longrightarrow (W_ p \to X)_!\underline{\mathbf{Z}}_{\overline{x}}$

is the quotient whose kernel is generated by all tensors $1_{\overline{u}_0} \otimes \ldots \otimes 1_{\overline{u}_ p}$ where $\overline{u}_ i = \overline{u}_ j$ for some $i \not= j$. Thus the quotient map

$(f_!\underline{\mathbf{Z}})^{\otimes p + 1} \longrightarrow \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$

factors through $(W_ p \to X)_!\underline{\mathbf{Z}}$, i.e., we get

$(f_!\underline{\mathbf{Z}})^{\otimes p + 1} \longrightarrow (W_ p \to X)_!\underline{\mathbf{Z}} \longrightarrow \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$

This already proves that $\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F})$ is (functorially) a subgroup of

$\mathop{\mathrm{Hom}}\nolimits ((W_ p \to X)_!\underline{\mathbf{Z}}, \mathcal{F}) = \mathcal{F}(W_ p)$

To identify it with the $S_{p + 1}$-anti-invariants we have to prove that the surjection $(W_ p \to X)_!\underline{\mathbf{Z}} \to \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$ is the maximal $S_{p + 1}$-anti-invariant quotient. In other words, we have to show that $\wedge ^{p + 1}f_!\underline{\mathbf{Z}}$ is the quotient of $(W_ p \to X)_!\underline{\mathbf{Z}}$ by the subsheaf generated by the local sections $s - \text{sign}(\sigma )\sigma (s)$ where $s$ is a local section of $(W_ p \to X)_!\underline{\mathbf{Z}}$. This can be checked on the stalks, where it is clear. $\square$

Lemma 68.6.5. Let $S$ be a scheme. Let $W$ be an algebraic space over $S$. Let $G$ be a finite group acting freely on $W$. Let $U = W/G$, see Properties of Spaces, Lemma 65.34.1. Let $\chi : G \to \{ +1, -1\}$ be a character. Then there exists a rank 1 locally free sheaf of $\mathbf{Z}$-modules $\underline{\mathbf{Z}}(\chi )$ on $U_{\acute{e}tale}$ such that for every abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$ we have

$H^0(W, \mathcal{F}|_ W)^\chi = H^0(U, \mathcal{F} \otimes _{\mathbf{Z}} \underline{\mathbf{Z}}(\chi ))$

Proof. The quotient morphism $q : W \to U$ is a $G$-torsor, i.e., there exists a surjective étale morphism $U' \to U$ such that $W \times _ U U' = \coprod _{g \in G} U'$ as spaces with $G$-action over $U'$. (Namely, $U' = W$ works.) Hence $q_*\underline{\mathbf{Z}}$ is a finite locally free $\mathbf{Z}$-module with an action of $G$. For any geometric point $\overline{u}$ of $U$, then we get $G$-equivariant isomorphisms

$(q_*\underline{\mathbf{Z}})_{\overline{u}} = \bigoplus \nolimits _{\overline{w} \mapsto \overline{u}} \mathbf{Z} = \bigoplus \nolimits _{g \in G} \mathbf{Z} = \mathbf{Z}[G]$

where the second $=$ uses a geometric point $\overline{w}_0$ lying over $\overline{u}$ and maps the summand corresponding to $g \in G$ to the summand corresponding to $g(\overline{w}_0)$. We have

$H^0(W, \mathcal{F}|_ W) = H^0(U, \mathcal{F} \otimes _\mathbf {Z} q_*\underline{\mathbf{Z}})$

because $q_*\mathcal{F}|_ W = \mathcal{F} \otimes _\mathbf {Z} q_*\underline{\mathbf{Z}}$ as one can check by restricting to $U'$. Let

$\underline{\mathbf{Z}}(\chi ) = (q_*\underline{\mathbf{Z}})^\chi \subset q_*\underline{\mathbf{Z}}$

be the subsheaf of sections that transform according to $\chi$. For any geometric point $\overline{u}$ of $U$ we have

$\underline{\mathbf{Z}}(\chi )_{\overline{u}} = \mathbf{Z} \cdot \sum \nolimits _ g \chi (g) g \subset \mathbf{Z}[G] = (q_*\underline{\mathbf{Z}})_{\overline{u}}$

It follows that $\underline{\mathbf{Z}}(\chi )$ is locally free of rank 1 (more precisely, this should be checked after restricting to $U'$). Note that for any $\mathbf{Z}$-module $M$ the $\chi$-semi-invariants of $M[G]$ are the elements of the form $m \cdot \sum \nolimits _ g \chi (g) g$. Thus we see that for any abelian sheaf $\mathcal{F}$ on $U$ we have

$\left(\mathcal{F} \otimes _\mathbf {Z} q_*\underline{\mathbf{Z}}\right)^\chi = \mathcal{F} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi )$

because we have equality at all stalks. The result of the lemma follows by taking global sections. $\square$

Now we can put everything together and obtain the following pleasing result.

Lemma 68.6.6. Let $S$ be a scheme. Let $f : U \to X$ be a surjective, étale, and separated morphism of algebraic spaces over $S$. For $p \geq 0$ set

$W_ p = U \times _ X \ldots \times _ X U \setminus \text{all diagonals}$

(with $p + 1$ factors) as in Lemma 68.6.4. Let $\chi _ p : S_{p + 1} \to \{ +1, -1\}$ be the sign character. Let $U_ p = W_ p/S_{p + 1}$ and $\underline{\mathbf{Z}}(\chi _ p)$ be as in Lemma 68.6.5. Then the spectral sequence of Lemma 68.6.3 has $E_1$-page

$E_1^{p, q} = H^ q(U_ p, \mathcal{F}|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p))$

and converges to $H^{p + q}(X, \mathcal{F})$.

Proof. Note that since the action of $S_{p + 1}$ on $W_ p$ is over $X$ we do obtain a morphism $U_ p \to X$. Since $W_ p \to X$ is étale and since $W_ p \to U_ p$ is surjective étale, it follows that also $U_ p \to X$ is étale, see Morphisms of Spaces, Lemma 66.39.2. Therefore an injective object of $\textit{Ab}(X_{\acute{e}tale})$ restricts to an injective object of $\textit{Ab}(U_{p, {\acute{e}tale}})$, see Cohomology on Sites, Lemma 21.7.1. Moreover, the functor $\mathcal{G} \mapsto \mathcal{G} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p))$ is an auto-equivalence of $\textit{Ab}(U_ p)$, whence transforms injective objects into injective objects and is exact (because $\underline{\mathbf{Z}}(\chi _ p)$ is an invertible $\underline{\mathbf{Z}}$-module). Thus given an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$ in $\textit{Ab}(X_{\acute{e}tale})$ the complex

$\Gamma (U_ p, \mathcal{I}^0|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p)) \to \Gamma (U_ p, \mathcal{I}^1|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p)) \to \Gamma (U_ p, \mathcal{I}^2|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p)) \to \ldots$

computes $H^*(U_ p, \mathcal{F}|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p))$. On the other hand, by Lemma 68.6.5 it is equal to the complex of $S_{p + 1}$-anti-invariants in

$\Gamma (W_ p, \mathcal{I}^0) \to \Gamma (W_ p, \mathcal{I}^1) \to \Gamma (W_ p, \mathcal{I}^2) \to \ldots$

which by Lemma 68.6.4 is equal to the complex

$\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^0) \to \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^1) \to \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^2) \to \ldots$

which computes $\mathop{\mathrm{Ext}}\nolimits ^*_{\textit{Ab}(X_{\acute{e}tale})}(K^ p, \mathcal{F})$. Putting everything together we win. $\square$

[1] This may be nonstandard notation

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