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We prove the Birkhoff–Grothendieck theorem by induction on the rank of the vector bundle $E$. Rank zero is immediate; rank one is the classification $\mathrm{Pic}({\mathbb{P}}_k^1)\cong \mathbb{Z}$ established above.

First there is a largest integer $a$ with $H^0(E(-a))\ne 0$. To see existence and boundedness directly, trivialize $E$ on the two standard affine charts: the associated finite projective modules over the principal ideal domains $k[t]$ and $k[t^{-1}]$ are free. Let $A(t)\in G{L}_r(k[t,t^{-1}])$ be its transition matrix, so a section of $E(-a)$ is a pair satisfyingIf $a$ exceeds the largest exponent in any entry of $A$, the right side has only strictly negative powers; the left side has nonnegative powers. Hence both vanish. For $a$ sufficiently negative, take a nonzero constant vector $p$ and put $q=t^{a}A(t{)}^{-1}p$; all its exponents are nonpositive, so it supplies a nonzero section. Thus a largest $a$ exists.

Such a section gives a nonzero map $\mathcal{O}(a)\to E$. It is injective, since it is nonzero generically and the source is torsion-free. Saturate its image. The resulting line subbundle is $\mathcal{O}(a+d)$ for the degree $d\ge 0$ of the effective divisor of zeros of the original map. If $d>0$, its inclusion in $E$ would give a nonzero section of $E(-a-d)$, contradicting maximality. Therefore $d=0$ andhas a locally free quotient. By induction, $Q\cong {⨁}_j\mathcal{O}(b_j)$.

Twist this exact sequence by $\mathcal{O}(-a-1)$. The outer groups $H^0(\mathcal{O}(-1))$ and $H^1(\mathcal{O}(-1))$ are zero by the Čech cohomology of twists on the projective line. The long exact sequence in sheaf cohomology therefore givesThe first is zero by the choice of $a$, so each $H^0(\mathcal{O}(b_j-a-1))$ vanishes. Thus $b_j\le a$ for every $j$. The extension class lies insince $a-b_j\ge 0$. The identification of this Ext functor with cohomology can be seen without additional duality: local splittings of a vector-bundle extension differ by a Čech cocycle in $\mathcal{H}om(Q,\mathcal{O}(a))$, and a coboundary changes them to compatible splittings. Vanishing therefore gives a global splitting. Consequentlycompleting the induction over any field $k$.

The degrees can be arranged in nonincreasing order and their multiset is unique. Indeed, for any such decomposition,so the dimensions of the twisted global sections recover the number of summands of every degree.

Use the finite map $f:C\to {\mathbb{P}}^1$ from the preceding solution. The inverse images of the two standard affine charts and of their intersection are affine, because $f$ is finite. The vanishing of quasi-coherent cohomology on an affine scheme therefore lets their two-term Čech cochain complex compute $H^i(C,E)$. It has terms only in degrees zero and one, provingThis actually applies to every quasi-coherent sheaf on $C$.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, define its genus of a smooth projective curve byThe preceding Serre duality also gives $g={\dim }_k{H}^0(C,{\Omega }_C^1)$. The condition on constants holds in particular for a geometrically connected smooth projective curve. It makes $g$ equal to the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. We state the theorem with this usual convention and then give the formula valid without that condition.

For a divisor on an algebraic curve $D$ and a canonical divisor $K$ with $\mathcal{O}(K)\cong {\Omega }_C^1$, the Riemann-Roch theorem isThe degree over an arbitrary field is weighted by residue degrees: ${\mathrm{deg}}_{k}D={\sum }_p{n}_p[k(p):k]$ when $D={\sum }_p{n}_{p}p$. Equivalently, for a line bundle $L$,We now prove it via Riemann-Roch via elementary modifications.

For any closed point $p$ and any line bundle $L$, the exact sequencehas a skyscraper sheaf as its quotient, with $k$-dimension $[k(p):k]$. This follows locally from a uniformizer of the discrete valuation ring at $p$: a local line-bundle generator modulo that uniformizer spans one copy of $k(p)$. The quotient has zero higher cohomology, as it is the direct image from a finite affine scheme. The long exact sequence in sheaf cohomology and the proved higher vanishing giveEvery line bundle has a nonzero rational section of a line bundle, whose local orders give a divisor on an algebraic curve $D$ with $L\cong \mathcal{O}(D)$. Iterating the displayed identity, adding or subtracting closed points with their multiplicities, givesFor the constants convention above, $\chi ({\mathcal{O}}_C)=1-g$. The dualizing sheaf on a smooth projective curve gives $h^1(\mathcal{O}(D))=h^0(\mathcal{O}(K-D))$, which converts this Euler-characteristic identity into the stated Riemann-Roch theorem. This proves the formula over arbitrary $k$, without assuming the closed points are $k$-rational.

For vector bundles, the corresponding formula isTo see it, saturate a rational rank-one subspace to a line subbundle and induct on rank. Both Euler characteristic of a coherent sheaf and the degree of the determinant are additive in the resulting short exact sequence. The line-bundle formula just proved starts the induction.

Taking $L={\Omega }_C^1$ and using duality gives $h^0({\Omega }_C^1)=g$, $h^1({\Omega }_C^1)=1$, and hence ${\mathrm{deg}}_k{\Omega }_C^1=2g-2$. In particular a line bundle of degree greater than $2g-2$ has zero first cohomology: its dual twisted by ${\Omega }_C^1$ has negative degree, and a negative-degree line bundle has no nonzero section, since any section would have an effective divisor of zeros.

If the term “curve over $k$” is used without requiring $H^0({\mathcal{O}}_C)=k$, the unconditional formula we proved isWriting $g=h^1({\mathcal{O}}_C)$ then replaces $1-g$ by $h^0({\mathcal{O}}_C)-g$. For example, ${\mathbb{P}}_{k^{\prime }}^1$ viewed as a $k$-curve for a finite separable extension $k^{\prime }/k$ has $h^0({\mathcal{O}}_C)=[k^{\prime }:k]$, so a literal constant term $1$ would be incorrect. Alternatively the arithmetic genus $p_a=1-\chi ({\mathcal{O}}_C)$ gives the formula with constant $1-p_a$. The constants hypothesis for the usual genus is recorded in Stacks Project, genus of a curve.

On the polydisc, put $b_i(z)=f_i(z)-f_i(a)$ and defineEach integral is along the straight coordinate segment, which remains in the polydisc. Holomorphic dependence on parameters and differentiation under the integral show that $F$ is a holomorphic function. For a fixed $j$, terms with $k<j$ do not depend on $z_j$, and the $k=j$ term contributes $b_j(z_1,\dots ,z_j,a_{j+1},\dots ,a_n)$. If $k>j$, use the assumed symmetry ${\partial }_j{b}_k={\partial }_k{b}_j$ to obtainThese differences telescope. ThereforeThis is the coordinate-integral primitive on a polydisc.

For a closed holomorphic one-form $\beta ={\sum }_i{c}_{i}d{z}_i$, closedness is exactly ${\partial }_i{c}_j={\partial }_j{c}_i$. The preceding construction plus ${\sum }_i{c}_i(a)(z_i-a_i)$ gives a local holomorphic function $G$ with $\partial G=\beta$. The kernel of $\partial$ on the structure sheaf of a complex manifold consists of locally constant functions: holomorphicity already gives $\bar{\partial }G=0$, and $\partial G=0$ then implies $dG=0$. Thus the holomorphic Poincaré lemma gives the short exact sequence of sheavesThe first object is the complex constant sheaf, and surjectivity is local, on stalks. It does not say that every global closed holomorphic one-form has a global primitive.

Now assume compactness. The constancy assertion is for a connected complex manifold; otherwise it holds on each connected component. A smooth complex-valued $u$ with $\partial \bar{\partial }u=0$ is a pluriharmonic function. Restricting to each complex coordinate line makes it a harmonic function of one complex variable. Choose $x_0$ where $|u|$ has its global maximum and a coordinate polydisc about $x_0$. The allowed harmonic maximum principle makes the first coordinate-line restriction constant. Every point on that line still has the same maximal modulus and the same value, so the second coordinate may then vary without changing $u$. Continue through all coordinates to conclude that $u$ equals $u(x_0)$ on the whole polydisc. The set where it has this value is closed and, by the same argument at each of its points, open. Connectedness therefore gives $u$ constant. In particular global holomorphic functions on a compact connected complex manifold are constant.

For the top-degree pairing, write locally $\theta =gd{z}_1\wedge \cdots \wedge d{z}_n$. With the complex orientation,It is everywhere nonnegative and is positive on an open set if $\theta$ is not identically zero. HenceThe phase factor is necessary: the unnormalized integral need not be a positive real number. This is the integral pairing of complex top forms.

If $\psi$ is a holomorphic differential form of degree $n-1$, then $d\psi =\partial \psi$ has type $(n,0)$ and $d\overline{d\psi }=0$. The graded Leibniz rule and Stokes theorem giveThe positive pairing just proved forces $d\psi =0$. Thus every holomorphic $(n-1)$-form on a compact complex $n$-manifold is closed, without a Kähler metric assumption.

On a compact complex surface, this proves closedness of all holomorphic one-forms. Sending such a form to its de Rham cohomology class defines the required inclusion. Indeed, if $\alpha =df$, its type $(1,0)$ gives $\bar{\partial }f=0$, so $f$ is a global holomorphic function and is constant. Therefore $\alpha =0$.

To prove the zero intersection, let $\alpha ,\beta$ be holomorphic one-forms and suppose $[\alpha ]=[\overline{\beta }]$. There is a smooth complex-valued $f$ with $\alpha -\overline{\beta }=df$. Its type components are $\partial f=\alpha$ and $\bar{\partial }f=-\overline{\beta }$. Closedness of $\beta$ gives $\partial \overline{\beta }=0$, hence $\partial \bar{\partial }f=0$. Compact pluriharmonic rigidity now makes $f$ constant and $\alpha =\beta =0$. The two spaces are therefore linearly independent in de Rham cohomology, and conjugation preserves their dimensions:For the other bound, take the long exact sequence in sheaf cohomology of the short exact sequence above. The map $H^0(M,\underline{\mathbb{C}})\to H^0(M,{\mathcal{O}}_M)$ is an isomorphism, componentwise. Since every global holomorphic one-form is closed, the relevant portion isIdentify the middle term with first de Rham cohomology, and the last with $H_{\bar{\partial }}^{0,1}(M)$ by the Dolbeault theorem. Taking dimensions givesThese are the Betti and Hodge bounds for compact complex surfaces.

A strict example is the scalar Hopf surface $X=({\mathbb{C}}^2\setminus \{0\})/⟨z\mapsto 2z⟩$. Polar coordinates identify it smoothly with $S^3\times S^1$: the dilation fixes the angular coordinate and shifts $\log |z|$ by $\log 2$. Its first Betti number is one, since its fundamental group is $\mathbb{Z}$. The proved lower bound then forces $h^{1,0}=0$, and the upper bound forces $h^{0,1}\ge 1$. Thus $h^{1,0}<h^{0,1}$ on this compact complex surface; no unsupported cohomology computation is needed.

The canonical bundle is $K_M={\bigwedge }^m(T^{1,0}M{)}^{\ast }$, the holomorphic line bundle of holomorphic top forms on a complex $m$-manifold. For an embedded complex submanifold $V$ of dimension $n$, the holomorphic normal bundle is the quotientTaking determinants and duals gives adjunction for a smooth submanifold:For a closed smooth hypersurface, $[V]={\mathcal{O}}_M(V)$ is its divisor line bundle, equipped with a holomorphic section whose zero divisor is $V$ with multiplicity one. If $f_{\alpha }$ locally defines $V$, local meromorphic generators $1/f_{\alpha }$ transform by the holomorphic units $f_{\alpha }/f_{\beta }$. The stated normal-bundle relation isEquivalently the conormal is ${\mathcal{O}}_M(-V){|}_V$. The closed-divisor interpretation is needed to define $[V]$; an arbitrary nonclosed local submanifold does not define that global divisor bundle.

For a holomorphic subbundle $F\subset E$, define $D_{F}s=\pi D_{E}s$ on sections of $F$, with $\pi$ acting on the bundle factor of a bundle-valued form. It obeys the Leibniz rule. Since $F$ is holomorphic, $D_E^{0,1}s={\bar{\partial }}_{E}s$ lies in $F$, so $D_F^{0,1}={\bar{\partial }}_F$. For $s,t\in F$, their inner products with the normal components of $D_{E}s,D_{E}t$ vanish; hence projecting preserves metric compatibility. The uniqueness proved for the Chern connection identifiesThis is the projected Chern connection.

Choose the adapted smooth unitary frame and retain Question 2's input-index convention. The connection matrix is skew-Hermitian. Its upper-left block is ${\theta }_1$ for $D_F$; writing the lower-left block as $-A$ forces the upper-right block to be ${\overline{A}}^t$:Holomorphicity of $F$ makes the normal part of $D_E^{0,1}s$ vanish, so $B={\overline{A}}^t$ has type $(1,0)$, and $A$ has type $(0,1)$. The matrix $B$ represents the second fundamental form of a holomorphic subbundle in this input-index convention. These types would appear in the opposite blocks with coefficient-column indices; they must not be swapped while retaining the printed minus-sign Cartan equation.

Take the upper-left block of the Cartan curvature equation with input indices. It isTherefore the curvature formula for a holomorphic subbundle isThe sign and the order of the factors both follow from the displayed block multiplication.

Now let $M$ be a complex torus and $V$ a closed smooth hypersurface. The torus has a translation-invariant flat Kähler metric, and $T^{1,0}M$ and $K_M$ are holomorphically trivial. Give $E=T^{1,0}M{|}_V$ its flat metric and apply the preceding formula to $F=T^{1,0}V$. Since ${\Theta }_E=0$,For a $(1,0)$ tangent vector $X$, the latter coefficient is $\mathrm{tr}(B(X)B(X{)}^{†})\ge 0$, so $i{\Theta }_{K_V}$ is a semipositive real $(1,1)$ form. Thus $K_V$ is a semipositive holomorphic line bundle. This proves the semipositive canonical bundle of a submanifold of a complex torus assertion directly from curvature. By adjunction for a smooth submanifold, $K_V\cong [V]{|}_V$.

Restrict a positive metric on $L$ to $V$. The tensor-product metric onhas curvature $i{\Theta }_{L{|}_V}+ai{\Theta }_{K_V}$, which is positive for every nonnegative integer $a$. HenceThe integer condition is implicit in a tensor power. No assertion of strict positivity of $[V]{|}_V$ is required.

For vanishing, write $E_a=L\otimes [V{]}^{\otimes a}$. The allowed Kodaira vanishing theorem says that on a compact Kähler manifold $X$, $H^i(X,K_X\otimes P)=0$ for $i>0$ when $P$ is a positive holomorphic line bundle. Since $K_M$ is trivial, its application on the torus gives $H^i(M,E_0)=H^i(M,L)=0$ for $i>0$.

For every $a\ge 1$, the divisor section gives the exact sequencewhere $i:V\hookrightarrow M$. Crucially,and $P_{a-1}$ is positive by the proved restriction statement. The induced Kähler metric makes $V$ compact Kähler, so Kodaira vanishing theorem on $V$ gives $H^i(V,E_a{|}_V)=0$ for all $i>0$. In the long exact sequence in sheaf cohomology, the segmentthen has zero outside terms by induction. ThereforeThis proves vanishing for divisor twists on a complex torus using restriction positivity and the divisor exact sequence, rather than presuming ambient positivity of every divisor twist.

Put $P={\mathbb{P}}_k^2$ and $S=k[t_0,t_1,t_2]$. A homogeneous polynomial $F$ of positive degree $d$ is nonzero, and $S$ is an integral domain. Multiplication by $F$ therefore gives an exact sequence of coherent sheaveswhere $i:X\hookrightarrow P$ is the closed immersion.

We compute the necessary cohomology of twisting sheaves on projective space. The standard three affine open subsets $D_{+}(t_i)$ and all their intersections are affine schemes. For a quasi-coherent sheaf, vanishing of quasi-coherent cohomology on an affine scheme and the acyclic cover theorem identify its sheaf cohomology with the Čech cohomology of this cover. In degree $m$, the Čech cochain complex isDecompose it into subcomplexes indexed by Laurent monomials $t_0^{a_0}{t}_1^{a_1}{t}_2^{a_2}$ with $a_0+a_1+a_2=m$. Let $N=\{i:a_i<0\}$. Such a monomial occurs in the summand for a nonempty index set $J$ precisely when $N\subseteq J$, since only the variables in $J$ are inverted.

If $N$ is empty, the monomial contributes the cochain complex of a full two-simplex: its degree-zero cohomology is $k$ and its higher cohomology vanishes. If $|N|=1$, the dimensions in degrees $0,1,2$ are $1,2,1$; the first map has rank one and the second map has rank one with the same kernel as the first map's image. If $|N|=2$, there is one copy of $k$ in each of degrees $1$ and $2$ and the map between them is an isomorphism. These two cases are acyclic over every field, including characteristic two. If $N=\{0,1,2\}$, only the degree-two term occurs, contributing one copy of $k$. It follows thatand that $H^0(P,{\mathcal{O}}_P)=k$, $H^0(P,{\mathcal{O}}_P(-d))=0$, and $H^2(P,{\mathcal{O}}_P)=0$. Moreover the Laurent-monomial description of top cohomology on projective space gives a basis of $H^2(P,{\mathcal{O}}_P(-d))$ consisting ofFor $d\ge 3$, subtracting one from each $b_i$ counts these as $\left(\genfrac{}{}{0ex}{}{d-1}{2}\right)$. For $d=1,2$ there are no such triples, and the formula $(d-1)(d-2)/2$ is also zero.

Use the long exact sequence in sheaf cohomology of the initial sequence, and cohomology under a closed immersion to identify the terms for $i_{\ast }{\mathcal{O}}_X$. The vanishings just proved giveThereforeThis proves the cohomology of a projective plane hypersurface without imposing smoothness, irreducibility or reducedness. In particular, the second number is an arithmetic genus, not a claim about the genus of a normalization.

For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

For the Sheaf of relative Kähler differentials on $X={\mathbb{P}}_{\mathbb{C}}^3$, the cotangent form of the Euler sequence isThe last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed Sheaf of relative Kähler differentials. The cohomology of twisting sheaves on projective space gives $H^q(X,{\mathcal{O}}_X(-1))=0$ for every $q$, while $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$ and its positive-degree groups vanish. The long exact sequence in sheaf cohomology therefore gives the cotangent-sheaf cohomology of projective space in this dimension:The isomorphism for $q=1$ is the connecting map from the constant global sections of ${\mathcal{O}}_X$. The Euler characteristic of a coherent sheaf is the alternating sum of dimensions, explaining the minus sign.

For the curve, put $S=\mathbb{C}[t_0,t_1,t_2,t_3]$. It is a unique factorization domain, so irreducibility of $F$ makes $(F)$ a prime ideal. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the integral domain $S/(F)$. It follows that $(F,G)$ is a regular sequence. Its Koszul complex is an exact sequence, and graded sheafification giveswhere $i:Z\hookrightarrow X$ is the closed immersion. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.

Let ${\mathcal{I}}_Z$ be the ideal sheaf of a closed subscheme. Split this Koszul resolution intoFor all integers $d$, $H^1(X,{\mathcal{O}}_X(d))=H^2(X,{\mathcal{O}}_X(d))=0$. Also $H^1(X,{\mathcal{O}}_X)=H^2(X,{\mathcal{O}}_X)=0$. The two long exact sequences in sheaf cohomology, together with sheaf cohomology under a closed inclusion, consequently identifyBy Serre duality, $H^3(X,{\mathcal{O}}_X(-m))\cong S_{m-4}^{\ast }$ for $m\ge 4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Thus the source has dimension $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$, and the two target spaces have dimensions $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The rank-nullity theorem proves the requested bound:

In fact equality holds. Under the Serre duality pairings, the dual of this map isIf $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. Thus the lower bound is attained for every pair allowed in the question. No smoothness assumption is needed. The regular sequence cuts out a projective complete intersection of dimension one. Also $H^0(X,{\mathcal{I}}_Z)=H^1(X,{\mathcal{I}}_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,{\mathcal{O}}_Z)=\mathbb{C}$. This is the genus of a complete-intersection space curve: the first cohomology has dimension $141$, equal to its arithmetic genus.

Order the index set of the open cover $\mathcal{U}=(U_a)$. The Čech cochain groups areand their differential is the alternating sum of restrictions:Terms obtained by deleting two indices cancel in pairs, so ${\delta }^2=0$. The Čech cohomology is ${\check{H}}^p=\ker \delta /\mathrm{im}\delta$, with zero incoming differential in degree zero. Its zeroth group is the group of global sections by sheaf gluing.

The acyclic cover theorem applies if every nonempty finite intersection is acyclic for $\mathcal{F}$. For a quasi-coherent sheaf on a variety it is sufficient that every such intersection is affine; on a separated variety, any affine open cover has this property. On ${\mathbb{P}}^n$, take the $n+1$ standard charts $U_i=D_{+}(X_i)$. Every intersection is a principal open in an affine chart, hence affine. The Čech complex has no terms above degree $n$, provingThis is the cohomological dimension bound from an affine cover and requires only quasi-coherence, not finite generation.

For the remaining projective calculations assume $n\ge 1$. There is a necessary zero-dimensional exception to the negative-twist assertion: ${\mathbb{P}}^0$ is a point, every twist is trivial there, and $H^0({\mathbb{P}}^0,\mathcal{O}(m))\cong k$ even for $m<0$.

Put $S=k[X_0,\dots ,X_n]$ with its usual grading. The twisting sheaf on projective space is the sheaf associated with the shifted graded module $S(m)$; on an intersection $U_I={\bigcap }_{i\in I}{U}_i$ its sections areOn $U_i$ a generator is $e_i=X_i^m$, with transition $e_j=(X_j/X_i{)}^m{e}_i$. These are regular units on overlaps and satisfy the cocycle relation, so they glue an invertible sheaf for every integer $m$, including negative $m$.

A global section is a compatible family of these homogeneous fractions, hence a single element of degree $m$ in ${\bigcap }_i{S}_{X_i}\subseteq \mathrm{Frac}S$. For $n\ge 1$, $X_0$ and $X_1$ are relatively prime in the unique factorization domain $S$, so $S_{X_0}\cap S_{X_1}=S$; therefore the full intersection is $S$. It follows thatFor $m\ge 0$ its dimension is $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. The proof uses all-chart compatibility; regularity on one chart alone would permit poles on its complement.

If some $d_i=0$, the ideal $(X_i^{d_i})$ contains $1$, so the requested containment is immediate. Otherwise every $d_i\ge 1$. A monomial $X_0^{a_0}\cdots X_n^{a_n}$ of degree ${\sum }_i{d}_i$ must have $a_i\ge d_i$ for some $i$: if not, its total degree would be at most ${\sum }_i(d_i-1)<{\sum }_i{d}_i$. Thus every degree-${\sum }_i{d}_i$ monomial, and hence every homogeneous polynomial of that degree, belongs to $(X_0^{d_0},\dots ,X_n^{d_n})$. This is monomial containment in an ideal of coordinate powers.

For $n>0$, a top-degree Čech cochain for $\mathcal{O}$ is a degree-zero Laurent polynomial on the full intersection. Write it with a common denominator asThe containment just proved gives $P={\sum }_i{Q}_i{X}_i^{d_i}$, with homogeneous $Q_i$ of degree ${\sum }_j{d}_j-d_i$. ConsequentlyThe $i$th term is regular on the intersection omitting $U_i$, and has degree zero. Give it the sign $(-1{)}^i$ in that component of the preceding Čech cochain. Its coboundary is the original fraction. Every top cochain is thus a coboundary, andThis is top Čech cohomology from missing-denominator monomials.

To obtain the negative-twist bound by induction on dimension, let $H\cong {\mathbb{P}}^{n-1}$ be a hyperplane, with inclusion $j$. Its equation gives the hyperplane exact sequence for twisting sheavesThe associated long exact sequence in sheaf cohomology and sheaf cohomology under a closed inclusion giveFor $n=1$, restriction $H^0({\mathbb{P}}^1,\mathcal{O})\to H^0(H,{\mathcal{O}}_H)$ is the surjection $k\to k$. Exactness and the already proved $H^1({\mathbb{P}}^1,\mathcal{O})=0$ give $H^1({\mathbb{P}}^1,\mathcal{O}(-1))=0$. The displayed surjections then give the same vanishing for every $m\ge 0$. This handles the point hyperplane without making a false negative-twist claim on ${\mathbb{P}}^0$.

For $n\ge 2$, the induction hypothesis in dimension $n-1$ gives $H^{n-1}(H,{\mathcal{O}}_H(m))=0$ whenever $m\ge -(n-1)$. Therefore consecutive top-degree groups are isomorphic for $m\ge 1-n$. Starting with $H^n({\mathbb{P}}^n,\mathcal{O})=0$ and stepping down through $m=0,-1,\dots ,1-n$ reaches $H^n({\mathbb{P}}^n,\mathcal{O}(-n))=0$. Stepping upward proves all positive twists too. We concludeThis is top-twist vanishing by hyperplane induction.

Finally, on $U_0$ write $x_a=X_a/X_0$ and ${\omega }_0=d{x}_1\wedge \cdots \wedge d{x}_n$. On $U_i$, let ${\omega }_i$ be the wedge of $d(X_a/X_i)$ for $a\ne i$, taken in increasing index order. For $i>0$, differentiating $X_0/X_i=x_i^{-1}$ and $X_a/X_i=x_a/x_i$ givesIn the wedge, all terms involving a second copy of $d{x}_i$ disappear; the surviving powers are $x_i^{-2}$ from $d(x_i^{-1})$ and $x_i^{-1}$ from the other $n-1$ differentials. Moving $d{x}_i$ into its original position produces the stated sign. Rescale each local generator by ${\eta }_i=(-1{)}^i{\omega }_i$. Then ${\eta }_j=(X_j/X_i{)}^{-n-1}{\eta }_i$ on every overlap, exactly the transition of the twist with $m=-n-1$. This canonical-form transition on projective space proves the canonical bundle of projective space:For line bundles the Serre duality pairing becomesIn particular $H^n(\mathcal{O}(-r))$ is dual to $H^0(\mathcal{O}(r-n-1))$, which is zero for $r\le n$, precisely the independently obtained bound. The case $r=0$ recovers $H^n(\mathcal{O})=0$, and $H^n(\mathcal{O}(-n-1))\cong k$ matches $H^0(\mathcal{O})\cong k$. Thus the calculated transition functions, global sections and top-degree vanishing agree with Serre duality.

Let ${\mathcal{A}}^{p,q}(X)$ be the space of smooth complex differential forms of type $(p,q)$. The Dolbeault operator satisfies ${\bar{\partial }}^2=0$, and the Dolbeault cohomology isFor $q=0$ the denominator is zero; forms outside the dimension range are zero.