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Integration of holomorphic differential forms along cycles embeds the first homology group as a full Euclidean lattice in $H^0(C,K_C{)}^{\vee }$. Fixing a base point, sum the path integrals to the points of an effective complex analytic divisor. Different path choices differ by that lattice. The Abel theorem for divisors identifies the fibres with linear equivalence of divisors.

A divisor class is an equivalence class of divisors under linear equivalence. Degree descends to divisor classes because every principal divisor has degree zero.

A divisor on an algebraic curve is a finite formal sumIts degree of a divisor is $\mathrm{deg}D={\sum }_p{n}_p$. It is an effective divisor, written $D\ge 0$, when every $n_p\ge 0$.

A meromorphic differential on a Riemann surface is a one-form represented in every local holomorphic coordinate $t$ by $a(t)dt$, where $a$ is a meromorphic function. Under a coordinate change $t=t(u)$, its coefficient becomes $a(t(u))t^{\prime }(u)$, so these expressions agree on overlapping charts. Since $t^{\prime }(u)$ is a nonvanishing holomorphic function for a valid coordinate change, the order of a zero or pole of the differential is independent of coordinate. Equivalently, for a fixed nonconstant meromorphic function $h$, any such differential can be written $fdh$ with $f$ meromorphic.

A canonical divisor is the divisor on an algebraic curveof a nonzero meromorphic differential on a Riemann surface $\eta$. If ${\eta }^{\prime }=f\eta$, then $({\eta }^{\prime })=(\eta )+(f)$, so all canonical divisors are related by linear equivalence of divisors. Their degree is

The Riemann-Roch theorem states that, for every divisor on an algebraic curve $D$ and any canonical divisor $K$ on the nonsingular projective curve $C$ of genus $g$,The term $\ell (K-D)$ measures the failure of the naive count of conditions on meromorphic functions.

For a divisor on an algebraic curve $D={\sum }_j{n}_j[p_j+\Lambda ]$, defineChanging the representatives $p_j$ changes the second coordinate only by an element of $\Lambda$. The two preceding contour arguments show that a principal divisor has degree zero and period lattice sum zero. Thus the formula is unchanged by linear equivalence of divisors and is a well-defined group homomorphism on the quotient.

It is nontrivial, since $[0]$ maps to $(1,0+\Lambda )$. In fact it is surjective: for any $d\in \mathbb{Z}$ and $z+\Lambda \in \mathbb{C}/\Lambda$, the divisor on an algebraic curve $(d-1)[0]+[z+\Lambda ]$ maps to $(d,z+\Lambda )$. Only well-definedness and a nontrivial homomorphism are required here; identifying its entire kernel would be an additional theorem.

The printed vanishing claim needs the restriction $i\ge 1$. On $C={\mathbb{P}}^1$ with $\mathcal{F}=\mathcal{O}$ and $D=n\{\infty \}$, Question 3 gives $h^0(\mathcal{F}(D))=n+1$ for every $n\ge 0$. Thus no bound on the degree of a divisor can force vanishing in degree zero. We prove finiteness in every degree and the corrected vanishing in every positive degree, uniformly over all divisors on an algebraic curve of large degree.

First take a finite morphism $\pi :C\to {\mathbb{P}}^1$ as in Question 3 and put $E={\pi }_{\ast }\mathcal{F}$. The direct image sheaf $E$ is coherent: on an affine chart its module is finitely generated over a finite algebra and hence over the chart's coordinate ring. The two affine-cover Čech cochain complexes for $\mathcal{F}$ on $C$ and for $E$ on ${\mathbb{P}}^1$ are identical. Therefore

We need an elementary presentation of any coherent sheaf $E$ on the projective line. Choose finitely many generators on each of its two standard affine charts. A generator on the first chart restricts to an element of the second chart's module localized at $t^{-1}$. Multiplication by a sufficiently large power of $t^{-1}$ brings that element into the second chart's module. Since the transition of $\mathcal{O}(n)$ is $t^n$, this makes the original generator extend to a global section of $E(n)$. A generator on the second chart similarly extends after clearing its denominator at $t=0$. Use one common exponent for all the finitely many generators. The extended sections generate $E(n)$ on both charts. Untwisting gives a surjectionwith coherent kernel $K$. The long exact sequence in sheaf cohomology and Question 3 show that $H^1(E)$ is a quotient of the finite-dimensional $H^1(P)$, since $H^2(K)=0$. Apply the same presentation argument to $K$ to obtain finiteness of $H^1(K)$. Then the segment $H^0(P)\to H^0(E)\to H^1(K)$ proves finiteness of $H^0(E)$. Higher groups vanish by Question 3. Hence every $H^i(C,\mathcal{F})$ is finite-dimensional.

Next establish a useful degree estimate without using the duality requested later. A global regular function on $C$ is constant: a nonconstant one would give a finite morphism $C\to {\mathbb{P}}^1$ by Question 1. Such a morphism is surjective, because its closed image contains the generic point, but a regular function misses $\infty$, a contradiction. Thus $H^0(C,{\mathcal{O}}_C)=k$. Write $g=h^1(C,{\mathcal{O}}_C)$, the genus of a smooth projective curve. For any divisor on an algebraic curve $A$ and point $P$, the local discrete valuation ring givesThe last term is a length-one skyscraper sheaf. Additivity of the Euler characteristic of a coherent sheaf and repeated addition or subtraction of points giveThe inequality follows because $h^1$ is nonnegative, and is sufficient for the following argument.

Fix $P\in C$. For $n\ge g+1$ this estimate gives at least two independent sections of ${\mathcal{O}}_C(nP)$, one of them the constant section. A nonconstant section is a rational function on an algebraic variety whose poles are confined to $P$. It must have a pole there, since a globally regular function is constant. Its extension $C\to {\mathbb{P}}^1$ is finite, and its inverse image of ${\mathbb{A}}^1$ is exactly $C\setminus \{P\}$. Consequently $U=C\setminus \{P\}$ is affine.

Let $T$ be the torsion subsheaf of $\mathcal{F}$ and let $V=\mathcal{F}/T$. The support of $T$ is finite; such a coherent sheaf is a finite direct sum of sheaves supported at points and has zero higher sheaf cohomology. The torsion-free sheaf $V$ is locally free: a finitely generated torsion-free module over a discrete valuation ring is free. Tensoring by any invertible sheaf is exact, so

The inclusions $V(nP)\subset V((n+1)P)$ have point-supported quotients. Their long exact sequences in sheaf cohomology therefore give surjections on $H^1$. Moreovera section regular away from $P$ has some finite pole order at $P$. Compute this direct limit on a finite affine cover of $C$. All its intersections with $U$ are affine by Question 2, and localization commutes with filtered direct limits. The Čech cochain complex consequently givesThe maps from $H^1(C,V)$ to these groups are surjective. Choose a finite basis of this starting vector space. Each basis vector becomes zero at some stage in the direct limit; one common stage kills the entire basis. Thus there exists $n_0\ge 0$ with $H^1(C,V(n_{0}P))=0$.

Now let $D$ be any divisor on an algebraic curve with $\mathrm{deg}D\ge n_0+g$. The estimate above gives $h^0({\mathcal{O}}_C(D-n_{0}P))\ge 1$. A nonzero section identifies $D-n_{0}P$ with an effective divisor $E$ up to linear equivalence of divisors. Hence $V(D)\cong V(n_{0}P+E)$. Adding $E$ gives a point-supported quotient of $V(n_{0}P+E)/V(n_{0}P)$, so its first sheaf cohomology is a quotient of $H^1(C,V(n_{0}P))=0$. We have proved uniform high-degree vanishing on a smooth projective curve:This bound concerns all high-degree divisors, not only multiples of a previously chosen divisor.

Let $V\subseteq U$ be open subsets, and let $s\in {\mathcal{F}}_3(V)$. The assumed flasque-kernel section-lifting lemma lifts $s$ to $\tilde{s}\in {\mathcal{F}}_2(V)$. Since ${\mathcal{F}}_2$ is a flabby sheaf, extend $\tilde{s}$ to a section on $U$, then project to ${\mathcal{F}}_3(U)$. Its restriction is $s$. Thus every restriction map of ${\mathcal{F}}_3$ is surjective, proving that the quotient of two flabby sheaves in a short exact sequence is flabby.

For the construction of sheaf cohomology, embed a sheaf of abelian groups $\mathcal{F}$ into the sheafby sending a section to all its germs. This is injective because a section with every germ zero is zero. Arbitrary families of stalk elements glue point by point, and restriction maps are projections; hence $C(\mathcal{F})$ is a flabby sheaf. Set ${\mathcal{Q}}^0=\mathcal{F}$, ${\mathcal{I}}^j=C({\mathcal{Q}}^j)$ and ${\mathcal{Q}}^{j+1}={\mathcal{I}}^j/{\mathcal{Q}}^j$. Compose the quotient map with the next embedding to obtain the flasque resolutionThe sheaf cohomology groups are the cohomology groups of the resulting cochain complex of global sections:The usual comparison of resolutions identifies this construction with the right derived functors of global sections; the construction above is the Godement resolution.

If $\mathcal{F}$ is itself a flabby sheaf, the quotient argument shows inductively that every ${\mathcal{Q}}^j$ is flabby. The assumed lifting result therefore makesan exact sequence for every $j$. The global-section complex is consequently exact in every positive degree. In particular,This establishes the vanishing from the section-lifting property rather than assuming it in advance.

For the smooth algebraic curve $X$, a divisor on an algebraic curve is a finite formal sum $D={\sum }_P{n}_{P}P$ of closed points with integer coefficients. At each closed point, the local ring of a smooth algebraic curve is a discrete valuation ring; write its normalized discrete valuation as $v_P$. For $f\in k(X{)}^{\ast }$ the principal divisor is $\mathrm{div}(f)={\sum }_P{v}_P(f)P$. Only finitely many coefficients are nonzero: zeros and poles are proper closed subsets in the Noetherian Zariski topology of a curve. The divisor class group is the abelian groupEquivalently, its elements are divisors on an algebraic curve modulo linear equivalence of divisors.

Fix $p_0\in X$ on a connected compact Riemann surface, and construct its Jacobian variety by periods as in the first solution. For a divisor on an algebraic curve $E={\sum }_p{n}_{p}p$ of degree zero, defineChanging the paths adds a period lattice element; changing $p_0$ adds a common integral multiplied by $\sum n_p=0$. The Abel theorem for divisors saysEquivalently, effective divisors on an algebraic curve of the same degree have the same Abelian sum exactly when they are equivalent under linear equivalence of divisors.

We need a meromorphic version of the Riemann bilinear relations. Let $\eta$ be a differential of the third kind, with residues $n_p$ at its simple poles, and let $\omega$ be a holomorphic differential form. Choose representatives of the cycles and paths avoiding the poles in the same cut polygon, and let $F(p)={\int }_{p_0}^p\omega$ there. ThenHere the equality is an equality of the chosen complex numbers, before taking any quotient. To prove it, remove small disks about the poles. Since $d(F\eta )=\omega \wedge \eta =0$ on the remaining polygon, its boundary integral is zero. Pairing opposite outer edges gives the left side, just as in the first solution. The clockwise boundary about $p$ contributes $-2\pi i{n}_{p}F(p)$ by the residue theorem. Moving these terms to the other side proves the formula and its sign.

Such an $\eta$ exists whenever $\sum n_p=0$. Here is the required existence argument. Let $S$ be the reduced support of $E$, and considerLocally the last map takes $cdz/z$ to $c$. By Serre duality, the obstruction in $H^1(K_X)\cong \mathbb{C}$ is dual to restriction of constant functions, so it is the sum of the residues. Thus the kernel of this sum is exactly the image of global meromorphic differentials. Alternatively, the Riemann-Roch theorem gives $h^0(K_X(S))-h^0(K_X)=|S|-1$, and the residue theorem identifies that image with the same codimension-one subspace. Subtracting a linear combination of the normalized holomorphic differential forms makes $A_i(\eta )=0$ for every $i$.

Suppose now that $\mathrm{AJ}(E)=0$. Let $u_i={\sum }_p{n}_p{\int }_{p_0}^p{\omega }_i$ with the paths used above. There are integer columns $m,n$ such that $u=m+\tau n$. The normalized meromorphic bilinear formula gives $B_i(\eta )=2\pi i{u}_i$. Thereforehas $a$-periods $-2\pi i{n}_i$ and $b$-periods $2\pi i{m}_i$. Its integrals around the poles are also in $2\pi i\mathbb{Z}$. These cycles generate the homology of the punctured surface, sois a single-valued, nowhere-zero holomorphic function away from the support of $E$. Near $p$, write ${\eta }^{\prime }=n_{p}dz/z+h(z)dz$. Then $f=z^{n_p}\exp (H(z))$ times a nonzero constant, with $H^{\prime }=h$. It extends meromorphically with order exactly $n_p$. Hence $\mathrm{div}(f)=E$.

Conversely, if $E=\mathrm{div}(f)$, take $\eta =df/f$. It has residues $n_p$ and every period lies in $2\pi i\mathbb{Z}$: continuation of a local logarithm of $f$ changes it by an integral multiple of $2\pi i$. Write $A_i(\eta )=2\pi i{r}_i$ and $B_i(\eta )=2\pi i{s}_i$. Applying the meromorphic bilinear formula to ${\omega }_i$ givesThus $u\in \Lambda$ and $\mathrm{AJ}(E)=0$, completing both directions without assuming the conclusion as a property of the Jacobian variety.

For completeness, the effective degree-$d$ version lives on the symmetric product of a curve $X^{(d)}=X^d/{\mathfrak{S}}_d$. The sum of integrals on $X^d$ is invariant under the finite symmetric group and descends to the Abelian sum map $u_d$. The quotient is locally described by elementary symmetric coordinates, so repeated points are included. The proof above identifies its fibre over $u_d(D)$ withExplicitly, a nonzero global section gives its effective zero divisor, and two sections give the same divisor precisely when their quotient is constant. These maps are holomorphic in local symmetric coordinates. More intrinsically, a family of effective divisors on an algebraic curve in this fixed class determines a line of sections of ${\mathcal{O}}_X(D)$, after locally trivializing any line bundle pulled back from the parameter space. Conversely such a line of sections gives its family of zero divisors on an algebraic curve. These operations are inverse and unaffected by that trivialization, proving the identification with the projective space in families as well as on points.

Finally this description identifies the analytic Jacobian variety with ${\mathrm{Pic}}^0(X)$. Every degree-zero line bundle has a meromorphic section: after twisting by a sufficiently large effective divisor, the Riemann-Roch theorem supplies a nonzero global section. Its divisor represents the original line bundle. The correspondence to the period quotient is well-defined and injective by the theorem. It is surjective because one may choose $g$ distinct points with independent evaluations of holomorphic differential forms; the derivative of the sum of their Abel integrals is then an isomorphism. Hence the subgroup generated by point differences contains an open subset of the connected complex torus, and must be the whole group. Local integration coordinates, or the exponential sequence for ${\mathcal{O}}_X$, give the same holomorphic identification. This justifies the subsequent use of degree-$d$ line bundles as points of a translate of the Jacobian variety.

Let $E$ be the elliptic curve, and choose a point $O\in E(k)$ to serve as identity. Its geometric genus is one and its canonical divisor $K$ has degree zero. For any divisor $D$ of degree one, the Riemann-Roch theorem givesSince $K-D$ has degree $-1$, its Riemann-Roch space is zero: a nonzero section would produce an effective divisor of negative degree. Hence $\ell (D)=1$. Choose a nonzero $f\in L(D)$. The divisor $D+(f)$ is effective of degree one, so over the algebraically closed field it is a single point $P$. Thus every degree-one divisor class has a point representative.

This representative is unique. If $P$ and $Q$ are linearly equivalent, choose $h$ with $(h)=P-Q$. Then $h\in L(Q)$. That space has dimension one and already contains the constants, so $h$ is constant and $P=Q$. Consequentlyis injective. It is also surjective: if $D$ represents a degree-zero divisor class, apply the preceding result to $D+O$ to obtain its unique point representative $P$, giving $[D]=[P-O]$.

The degree-zero Picard group of a curve is a group because divisors add commutatively, principal divisors form a subgroup, and degree passes to the quotient. DefineEquivalently, $R=P\oplus Q$ is the unique point satisfying the linear equivalence of divisors $P+Q\sim R+O$, where the plus signs in this equivalence denote divisor addition. The zero class corresponds to $O$, so $O$ is the identity. The inverse of $P$ is the unique point $P^{\prime }$ with $P+P^{\prime }\sim 2O$. Associativity follows because both $(P\oplus Q)\oplus R$ and $P\oplus (Q\oplus R)$ map to $\iota (P)+\iota (Q)+\iota (R)$, and $\iota$ is injective; commutativity follows in the same way. Thus this is an abelian group structure on $E(k)$, obtained directly by the elliptic curve group law from Riemann-Roch. The identity choice is part of the construction.

A divisor on a compact connected Riemann surface is a finite formal sum $D={\sum }_p{n}_{p}p$ with integer coefficients; its degree is ${\sum }_p{n}_p$. A canonical divisor is the divisor of zeros and poles of a nonzero meromorphic differential. Two divisors are linearly equivalent when their difference is the divisor of a nonzero meromorphic function. Such principal divisors have degree zero, but equal degree does not generally imply linear equivalence of divisors.

For example take distinct points $p,q$ on a genus-one surface. If $p-q$ were principal, its function would define a degree-one meromorphic map to the sphere, since it has just one simple pole. A degree-one map between compact Riemann surfaces is a biholomorphism, contradicting their different genera. Thus these degree-one divisors are not equivalent.

Define $L(D)=\{f\text{ meromorphic}:(f)+D\ge 0\}\cup \{0\}$ and $\ell (D)={\dim }_{\mathbb{C}}L(D)$. The Riemann-Roch theorem statesSince all holomorphic functions on the compact surface are constant, $\ell (0)=1$. Setting $D=0$ gives $\ell (K)=g$. Multiplication by a fixed meromorphic differential identifies $L(K)$ with holomorphic differentials, so their dimension is $g$. Setting $D=K$ then givesFor $g=2$, choose a nonzero holomorphic differential, whose effective canonical divisor has degree two. Since $\ell (K)=2$, $L(K)$ contains a nonconstant meromorphic function. Its pole divisor is bounded by that effective $K$, proving existence of a nonconstant function with total pole multiplicity at most two.

For a divisor on an algebraic curve $D={\sum }_p{a}_{p}p$ over an algebraically closed field, its degree isFor a nonzero rational function $f$, its principal divisor iswhere zeros have positive order and poles have negative order.

Now let $D={\sum }_i{a}_i{p}_i$ have degree zero on the projective line. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since ${\sum }_i{a}_i=0$, the expressionis homogeneous of degree zero and therefore defines a rational function on ${\mathbb{P}}^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, soThus every degree-zero divisor on ${\mathbb{P}}^1$ is principal. If $E$ and $E^{\prime }$ have the same degree, then $E-E^{\prime }$ is principal, whenceThis is the divisor class on the projective line.

Write $p_{\infty }=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; thenHence the rational differential has one double pole at infinity and no other zero or pole:which represents the canonical divisor of the projective line.

If $D\sim m{K}_{{\mathbb{P}}^1}$, then $D\sim -2m{p}_{\infty }$, and linear equivalence of divisors gives $L(D)\cong L(-2m{p}_{\infty })$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. ThereforeThis calculation uses only the rational functions on ${\mathbb{P}}^1$, rather than the Riemann-Roch theorem.

Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one finite morphism $f:C\to {\mathbb{P}}^1$. A degree-one finite morphism between smooth projective curves is an isomorphism. It would follow that $C\cong {\mathbb{P}}^1$ and hence that $C$ has genus zero, contrary to the hypothesis. Thus the principal divisor with one simple zero and one simple pole cannot occur on $C$:

The first assertion is false with the standard meaning of linear equivalence of divisors. For example, two distinct lines $D_1,D_2\subset {\mathbb{CP}}^2$ are smooth and linearly equivalent but intersect. Distinct fibers of a holomorphic map to ${\mathbb{CP}}^1$ are disjoint, so no map can have these lines as the fibers over $0$ and $\infty$. The ratio of defining sections gives only a meromorphic map, with an indeterminacy point at $D_1\cap D_2$. The assertion becomes true if the two sections have no common zero.

The requested cohomological conclusion is nevertheless valid. Linearly equivalent divisors define isomorphic holomorphic line bundles, and their Poincaré-dual Dolbeault classes both equal the First Chern class of that bundle. Hence $[D_1]=[D_2]$ in $H_{\bar{\partial }}^{1,1}(X)$.