Laurent-monomial decomposition of the projective Cech complex

For a nonempty subset $\sigma =\{i_0<\cdots <i_p\}\subseteq \{0,\dots ,n\}$ put $U_{\sigma }=U_{i_0}\cap \cdots \cap U_{i_p}=D_{+}(x_{\sigma })$ with $x_{\sigma }={\prod }_{i\in \sigma }{x}_i$. By [F1] and [F2], $\mathrm{\Gamma }(U_{\sigma },{\mathcal{O}}_X(d))=S(d{)}_{(x_{\sigma })}$, the degree-$0$ part of $S(d)[x_{\sigma }^{-1}]$, which has as $A$-basis the Laurent monomials $x^e=x_0^{e_0}\cdots x_n^{e_n}$ with $e\in {\mathbb{Z}}^{n+1}$, ${\sum }_i{e}_i=d$ and $e_i\ge 0$ for every $i\mathrm{\notin }\sigma$; this is precisely the condition $N(e)\subseteq \sigma$ for $N(e)=\{i:e_i<0\}$. For $\sigma \subseteq \tau$ the restriction $S(d{)}_{(x_{\sigma })}\to S(d{)}_{(x_{\tau })}$ is the localisation inverting $x_{\tau \setminus \sigma }$ and sends the basis monomial $x^e$ to itself.

Taking the product over all $(p+1)$-element subsets $\sigma$, step 1.1 gives a decomposition of $A$-modules $C^p(\mathcal{U},{\mathcal{O}}_X(d))={\prod }_{\mathrm{\mid }\sigma \mathrm{\mid }=p+1}\mathrm{\Gamma }(U_{\sigma },{\mathcal{O}}_X(d))={⨁}_{\sum e_i=d}{K}^p(e),K^p(e)={⨁}_{\sigma \supseteq N(e),\mathrm{\mid }\sigma \mathrm{\mid }=p+1}A\cdot x_{\sigma }^e,$ where the factor $\mathrm{\Gamma }(U_{\sigma },{\mathcal{O}}_X(d))$ contributes to $K^p(e)$ exactly when $N(e)\subseteq \sigma$, and the Čech differential [F3], being a sum of restriction maps followed by the sign change of the ordered complex, sends the basis element $x_{\sigma }^e$ to $(-1{)}^{\mathrm{pos}(k,\sigma \cup \{k\})}{x}_{\sigma \cup \{k\}}^e$ for each $k\mathrm{\notin }\sigma$ and respects the grading by $e$; a term whose target satisfies $\sigma \cup \{k\}\supseteq ̸N(e)$ does not occur, and such a term never arises from a nonzero $x_{\sigma }^e$ with $\sigma \supseteq N(e)$. Hence $C^{\bullet }(\mathcal{U},{\mathcal{O}}_X(d))={⨁}_e{K}^{\bullet }(e)$ as complexes, and it suffices to compute the cohomology of each $K^{\bullet }(e)$.

Suppose $N(e)=\mathrm{\varnothing }$, so that $K^p(e)=A$ for every $(p+1)$-element subset $\sigma$, $p=0,\dots ,n$, and the differential is the alternating sum of the maps $x_{\sigma \setminus \{i\}}^e\mapsto x_{\sigma }^e$. Taking $M=A$ and $f_0=\cdots =f_n=1$ in the unit ideal of $A$ generated by $1$, the complex of [F5] has exactly these terms and this differential, hence is exact in positive degrees; therefore $H^q(K^{\bullet }(e))=0$ for $q\ge 1$, while $H^0(K^{\bullet }(e))=\ker {\delta }^0=A$, generated by the cochain whose component at every $i$ is $x_i^e$ (the diagonal class, corresponding to the global monomial section $x^e$ of ${\mathcal{O}}_X(d)$).

Suppose $N(e)=\{0,\dots ,n\}$, i.e. every $e_i<0$. Then the only subset $\sigma$ with $N(e)\subseteq \sigma$ is $\sigma =\{0,\dots ,n\}$ itself, so $K^p(e)=0$ for $p<n$ and $K^n(e)=A\cdot x_{\{0,\dots ,n\}}^e$; consequently $H^n(K^{\bullet }(e))=A$ and $H^q(K^{\bullet }(e))=0$ for $q\mathrm{\ne }n$.

Suppose $N(e)$ is nonempty and proper, and choose $v\mathrm{\notin }N(e)$, for instance the least such index. Since $v\mathrm{\notin }N(e)$ one has $N(e)\subseteq \sigma$ if and only if $N(e)\subseteq \sigma \cup \{v\}$, so the formula $(h\kappa {)}_{\sigma }=\{\begin{array}{ll}0,& v\in \sigma ,\\ (-1{)}^{\mathrm{pos}(v,\sigma \cup \{v\})}{\kappa }_{\sigma \cup \{v\}},& v\mathrm{\notin }\sigma ,\end{array}$ defines a homomorphism $h:K^{p+1}(e)\to K^p(e)$ for every $p\ge 0$. We verify $\delta h+h\delta =\mathrm{id}$ on $K^{\bullet }(e)$. For a $(p+1)$-element $\tau$ with $v\in \tau$, the sum $\delta (h\kappa {)}_{\tau }$ receives only the term with omitted index $v$, giving $(-1{)}^{\mathrm{pos}(v,\tau )}(h\kappa {)}_{\tau \setminus \{v\}}=(-1{)}^{\mathrm{pos}(v,\tau )}(-1{)}^{\mathrm{pos}(v,\tau )}{\kappa }_{\tau }={\kappa }_{\tau }$, while $(h\delta \kappa {)}_{\tau }=0$ by the definition of $h$. For $\tau$ with $v\mathrm{\notin }\tau$, put $t=\mathrm{pos}(v,\tau \cup \{v\})$. The terms in $\delta (h\kappa {)}_{\tau }$ and $(h\delta \kappa {)}_{\tau }$ that omit a fixed $i_j\in \tau$ cancel: the inserted-vertex position in $\tau \setminus \{i_j\}$ is $t$ if $i_j>v$ and $t-1$ if $i_j<v$, while the Cech position of $i_j$ in $\tau \cup \{v\}$ shifts by one exactly when $v<i_j$. The remaining term, which omits $v$ in $h\delta \kappa$, is $(-1{)}^t(-1{)}^t{\kappa }_{\tau }=(-1{)}^{2t}{\kappa }_{\tau }={\kappa }_{\tau }$. Hence the identity of $K^{\bullet }(e)$ is null-homotopic and $K^{\bullet }(e)$ is acyclic: $H^q(K^{\bullet }(e))=0$ for every $q\ge 0$.

By steps 3.1, 3.2 and 3.3 the cohomology of each summand $K^{\bullet }(e)$ is $A$ in degree $0$ when $N(e)=\mathrm{\varnothing }$, $A$ in degree $n$ when $N(e)$ is all of $\{0,\dots ,n\}$, and zero in all other degrees and cases. Since cohomology of complexes of $A$-modules commutes with direct sums by [F4], $H^q(C^{\bullet }(\mathcal{U},{\mathcal{O}}_X(d)))$ is the direct sum over the $e$ with $\sum e_i=d$ of these groups; the monomials $x^e$ with all $e_i\ge 0$ contribute $A$ to degree $0$ and the monomials $x^e$ with all $e_i<0$ contribute $A$ to degree $n$. This proves the asserted decomposition and the three cases.

Boundary and choice cases. For $n=0$ the cover has the single member $U_0=D_{+}(x_0)$ and $K^0(e)=A$ for every $e=(d)$: if $d\ge 0$ then $N(e)=\mathrm{\varnothing }$ and step 3.1 gives $H^0=A$, while if $d<0$ then $N(e)=\{0\}$ is all indices and step 3.2 with $n=0$ gives $H^0(K^{\bullet }(e))=H^n(K^{\bullet }(e))=A$; the two descriptions coincide in degree zero, as asserted. If $A=0$ then $S=0$, $\mathrm{Proj}S=\mathrm{\varnothing }$, all section modules vanish and every assertion reads $0=0$. The vertex $v$ in step 3.3 is chosen as the least index outside $N(e)$, and the cover order and the monomial bases are canonical, so no choice beyond those inherited from [F1], [F2] and [F5] is used; AC is declared in the statement.