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For a well-order on the indices of an open cover, the ordered Čech cochain complex uses only strictly increasing tuples, with the alternating restriction Čech differential. It computes the same Čech cohomology as the complex using all tuples. If every finite intersection is nonempty and the coefficient constant sheaf has the same abelian group $A$ on each intersection, inserting the least index $\alpha$ gives a cochain homotopy: set $(hs{)}_I=s_{\alpha ,I}$ when $\alpha$ is not in $I$, and $(hs{)}_I=0$ otherwise. Expanding the alternating sums gives $\delta h+h\delta =\mathrm{id}$ in every positive degree, since the terms deleting an index of $I$ cancel and the term deleting $\alpha$ is $s_I$. Hence positive Čech cohomology vanishes in this situation; degree zero remains $A$.

Let $\mathcal{U}=(U_0,\dots ,U_m)$ be a finite affine open cover of the projective scheme $X$. For a coherent sheaf $\mathcal{F}$, form the Čech cochain complexThe two ways to omit any pair of indices have opposite signs, giving ${\delta }^2=0$. Its cohomology groups areA projective scheme over a field is a separated scheme. Finite intersections of its affine opens are affine, and a coherent sheaf is quasi-coherent. Higher sheaf cohomology of a quasi-coherent sheaf on an affine scheme vanishes. Thus this is an acyclic cover and the acyclic cover theorem identifies the displayed Čech cohomology with $H^p(X,\mathcal{F})$. In particular its degree-zero kernel glues compatible local sections to $\Gamma (X,\mathcal{F})$. A cover by $m+1$ opens also gives vanishing for $p>m$.

For the cohomology of twists on projective space, take $S=k[X_0,\dots ,X_r]$ with the usual grading and $U_i=D_{+}(X_i)$. On an intersection indexed by the nonempty set $J$,This follows directly from the construction of the twisting sheaf on projective space: sections of the twist are the degree-$n$ homogeneous fractions in the corresponding localization. The Čech differential preserves each Laurent monomial $X^a=X_0^{a_0}\cdots X_r^{a_r}$, where $a_i\in \mathbb{Z}$ and ${\sum }_i{a}_i=n$. Such a monomial appears precisely in summands withThe complex therefore decomposes as the direct sum of finite-dimensional cochain complexes indexed by these exponent vectors. Their differentials have only the alternating signs of the simplex incidence maps.

If $N(a)=\varnothing$, this is the ordinary unaugmented simplex cochain complex: its degree-zero kernel consists of the common value on every vertex and has dimension one, while all higher cohomology vanishes. If $\varnothing \ne N(a)\ne \{0,\dots ,r\}$, choose $v\notin N(a)$. A cochain homotopy inserting $v$ into the ordered index list, with the sign of that insertion, contracts this subcomplex. Insertion or deletion of $v$ preserves the condition $N(a)\subseteq J$; in $\delta h+h\delta$, terms inserting and deleting different vertices cancel in pairs, while the term inserting then deleting $v$ is the identity. Thus this entire monomial subcomplex is acyclic. Finally, if all $a_i<0$, the monomial occurs only in the full intersection, in degree $r$, and contributes one copy of $k$ there.

For $r\ge 1$, these three cases give the complete answer:In the first case counting nonnegative exponent vectors gives $h^0=\left(\genfrac{}{}{0ex}{}{n+r}{r}\right)$. In the top case write $a_i=-1-b_i$ with $b_i\ge 0$; then $\sum b_i=-n-r-1$, givingThis is the Laurent-monomial description of top cohomology on projective space. All intermediate degrees vanish for every twist, and the top degree vanishes when $n\ge -r$. If $r=0$, then ${\mathbf{P}}_k^0=\mathrm{Spec}k$ and every twist is trivial, so $H^0({\mathbf{P}}_k^0,\mathcal{O}(n))=k$ for every integer $n$, with all higher groups zero.