 Solution

Let $S=k[X_0,\dots ,X_n]$ with its usual grading, and put $S(m{)}_d=S_{m+d}$. Define the twisting sheaf on projective space byMore explicitly, on $U_i=D_{+}(X_i)$ its sections are the homogeneous degree-$m$ part of $S_{X_i}$. They are a free rank-one module over $(S_{X_i}{)}_0=k[X_j/X_i:j\ne i]$, with generator $e_i=X_i^m$. These generators make sense for every integer $m$, since $X_i$ is invertible on this chart. Their transition rule isThe ratios are regular units on overlaps and satisfy the cocycle identity, so they glue to an invertible sheaf. In particular $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$.

For $n\ge 1$, a global section is a compatible homogeneous degree-$m$ rational expression lying in every $S_{X_i}$. Inside the fraction field of $S$,To justify this, write a fraction in lowest terms in the unique factorization domain $S$. Membership in $S_{X_i}$ forces every irreducible divisor of its denominator to be associated with $X_i$. Since there are at least two nonassociated variables, the denominator must be a unit. ThereforeThe degree-$m$ monomials are indexed by nonnegative integers $a_0,\dots ,a_n$ with sum $m$. The stars and bars count givesHere and below, an all-integer binomial shorthand must use the counting convention $\left(\genfrac{}{}{0ex}{}{a}{n}\right)=0$ when $a<n$, including negative $a$. The polynomial extension of binomial coefficients to negative upper arguments does not give the stated dimensions. In dimension zero, ${\mathbb{P}}^0$ is one point and every ${\mathcal{O}}_{{\mathbb{P}}^0}(m)$ is trivial, so its space of global sections is $k$ for every $m$; this case is treated separately.

For a closed inclusion $i:H\hookrightarrow {\mathbb{P}}^n$, direct image sheaf formation is exact: its stalk is the original stalk at a point of $H$ and zero outside $H$. It preserves flabby sheaves, since restrictions on the target are restrictions between the corresponding open subsets of $H$. Push forward a flasque resolution of $\mathcal{F}$; exactness gives a flasque resolution of $i_{\ast }\mathcal{F}$, and its complex of global sections is the original one becauseTaking cohomology of these identical complexes proves sheaf cohomology under a closed inclusion:This argument applies even without quasi-coherence.

We derive the remaining sheaf cohomology dimensions using the given eventual vanishing, the long exact sequence in sheaf cohomology, and induction on dimension. Multiplication by the linear equation of a hyperplane $H\cong {\mathbb{P}}^{n-1}$ gives the hyperplane exact sequence for twisting sheavesExactness can be checked in each chart by quotienting its polynomial ring by the equation of $H$; multiplication is injective because that equation is a non-zero-divisor. We use the closed-inclusion cohomology identification to replace the cohomology of the final sheaf by that on $H$.

The dimension-zero base is a point, whose global-section functor is exact. Thus all its higher sheaf cohomology is zero. For $n=1$, the relevant long exact sequence isWhen $m\ge 0$, restriction of homogeneous polynomials to the hyperplane point is surjective: after choosing that point as $[1:0]$, the section $X_0^m$ restricts to a generator. Therefore the consecutive $H^1$ groups are isomorphic for $m\ge 0$. The given vanishing for large $m$ propagates down to $m=-1$. For $m<0$, both displayed $H^0$ groups are zero, and the sequence becomesStarting with $H^1(\mathcal{O}(-1))=0$, induction downward gives $h^1(\mathcal{O}(m))=-m-1$ for $m\le -2$. For degrees $i>1$, the point contributes no cohomology to the exact sequence, so consecutive $H^i$ groups are isomorphic for every $m$; the given eventual vanishing makes them all zero. Hence the complete positive-degree formula holds for ${\mathbb{P}}^1$.

Now let $n\ge 2$ and assume the formulas for $H\cong {\mathbb{P}}^{n-1}$. For every $m$, the restrictionis surjective: for $m\ge 0$ extend a polynomial in the hyperplane coordinates to the same polynomial on ${\mathbb{P}}^n$, and for $m<0$ both groups vanish because $H$ has positive dimension. Consequently the exact sequence gives an injectionFor $2\le i\le n-1$, the induction hypothesis says $H^{i-1}(H,{\mathcal{O}}_H(m))=0$, and gives the same injection in degree $i$. For each fixed $m$, composing finitely many injections reaches a sufficiently large twist, whose higher cohomology is zero by the given hypothesis. ThereforeFor $i>n$, the two neighbouring hyperplane groups $H^{i-1}(H,{\mathcal{O}}_H(m))$ and $H^i(H,{\mathcal{O}}_H(m))$ are zero by induction. Consecutive $H^i$ groups on ${\mathbb{P}}^n$ are then isomorphic, and eventual vanishing again makes them zero. This derives vanishing above the dimension too, without needing a special projective-space cohomology theorem.

For top cohomology the exact sequence now reduces toWrite $b_n(m)=h^n({\mathcal{O}}_{{\mathbb{P}}^n}(m))$. The induction hypothesis and eventual vanishing give the recurrenceConsecutive top groups agree for $m\ge -n+1$, so descending from a large twist gives $b_n(m)=0$ for $m\ge -n$. Below this range, iterating the recurrence gives a finite sum:The last equality follows by summing Pascal's identity. These short exact sequences also prove that every group whose dimension is being counted is finite-dimensional. Thus, for $n\ge 1$,With the stated counting convention, the latter is the requested single binomial formula. The base ${\mathbb{P}}^0$ has $h^0=1$ for all twists and no higher cohomology; it is not governed by the negative-twist $H^0$ rule for positive-dimensional projective space.

Finally calculate the canonical line bundle of a smooth variety from its transition functions. On $U_0$ write $t_j=X_j/X_0$ for $1\le j\le n$. On $U_i$, $i>0$, take the coordinates $u_j=X_j/X_i$ in increasing order of $j\ne i$. ThusIn their wedge product, $d{u}_0=-t_i^{-2}d{t}_i$ and
$d{u}_j=t_i^{-1}d{t}_j-t_j{t}_i^{-2}d{t}_i$. Every term involving a second $d{t}_i$ vanishes, so reordering the remaining differentials givesThis computation is valid in every characteristic. Multiply the ordered local generator on $U_i$ by $(-1{)}^i$. The resulting generators ${\eta }_i$ satisfy ${\eta }_i=(X_i/X_0{)}^{-n-1}{\eta }_0$ on $U_i\cap U_0$, and hence the corresponding transition rule on every overlap. They are exactly the transitions of $\mathcal{O}(-n-1)$. Therefore the canonical bundle of projective space is

For a line bundle, Serre duality statesThe intermediate groups on both sides vanish. In the two outer degrees, the top group at twist $m$ has the same dimension as the global-section group at twist $-m-n-1$:The polynomial degree on the right is nonnegative precisely when $m\le -n-1$, and its dimension is $\left(\genfrac{}{}{0ex}{}{-m-1}{n}\right)$. Interchanging $m$ and $-m-n-1$ checks the other outer degree. On ${\mathbb{P}}^0$ both sides are one-dimensional. Thus all the formulas are consistent with Serre duality, which was used only for this final consistency check, not to prove the cohomology formulas.