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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 degree-zero complex analytic divisor $E$ on a compact Riemann surface is principal exactly when its Abel-Jacobi map of a compact Riemann surface is zero. Normalize a differential of the third kind with residues equal to the coefficients of $E$. The Riemann bilinear relations for a compact surface express its $b$-periods as $2\pi i$ times the Abel integrals. If those integrals are in the period lattice, subtract a suitable integral multiple of the normalized holomorphic differential forms; all periods then lie in $2\pi i\mathbb{Z}$, so exponentiation gives the required meromorphic function. Conversely, the logarithmic differential of a meromorphic function has integer residues and periods in $2\pi i\mathbb{Z}$, giving a zero Abel class by the same calculation.

If $K_X=(\omega )$ is represented by a nonzero rational differential, multiplication by $\omega$ identifies the Riemann-Roch space $L(K_X)$ with the vector space of holomorphic differential forms. The Riemann-Roch theorem gives $\ell (K_X)=g$ for a smooth projective curve of genus $g$.

A Riemann surface which is compact as a topological space. For a connected one of genus $g$, the Riemann-Roch theorem gives $g$ independent holomorphic differential forms. The Riemann bilinear relations for a compact surface make their normalized periods a full lattice in ${\mathbb{C}}^g$, and the Abel-Jacobi map of a compact Riemann surface turns this lattice into a complex torus encoding divisor classes.

For an effective divisor on an algebraic curve $D$ on a smooth projective curve, the Zariski tangent space to its symmetric product of a curve is $H^0({\mathcal{O}}_D(D))$. The derivative of the Abelian sum map is the connecting homomorphism of $0\to {\mathcal{O}}_X\to {\mathcal{O}}_X(D)\to {\mathcal{O}}_D(D)\to 0$. Its dual under Serre duality is restriction of holomorphic differential forms to $D$, including all jets at repeated points. Hence its rank is $\mathrm{deg}D+1-h^0({\mathcal{O}}_X(D))$, and its image annihilates $H^0(K_X(-D))$.

A meromorphic one-form on a compact Riemann surface with only simple poles is a differential of the third kind. Prescribed residues can be realized exactly when their sum is zero: in the residue exact sequence for $K_C(S)$, the obstruction in $H^1(C,K_C)\cong \mathbb{C}$ is the sum of residues. Subtracting a holomorphic differential form normalizes its $a$-periods. Integer residues and periods in $2\pi i\mathbb{Z}$ make the exponential of its integral a single-valued meromorphic function with those orders of zeros and poles.

The Dolbeault-Poincaré lemma makes this a locally exact complex of sheaves, with degree-zero kernel the holomorphic differential forms. The sheaves of smooth forms are fine sheaves by a partition of unity. The acyclic resolution theorem therefore computes the cohomology of ${\Omega }^p$ by the global Dolbeault operator complex.

This complex has the sheaves ${\Omega }_M^p$ of holomorphic differential forms and differential $\partial$. By the holomorphic Poincaré lemma, it resolves the complex constant sheaf.

On a star-shaped holomorphic coordinate neighbourhood, a closed holomorphic differential form of positive degree has a holomorphic primitive. Radial integration gives the homotopy operator. This proves exactness of the holomorphic de Rham complex on stalks.

For a smooth form $\theta =gd{z}_1\wedge \cdots \wedge d{z}_n$ of type $(n,0)$, the complex orientation gives $i^{n^2}\theta \wedge \overline{\theta }=2^n|g{|}^{2}d{x}_1\wedge d{y}_1\wedge \cdots \wedge d{x}_n\wedge d{y}_n$. Consequently its integral on a compact complex manifold is positive unless $\theta$ vanishes identically. Applying Stokes theorem to $d(\psi \wedge \overline{d\psi })$ proves that a holomorphic differential form of degree $n-1$ on a compact complex $n$-manifold is closed, without requiring a Kähler metric.

Let $X$ be a connected compact Riemann surface of geometric genus $g$, equivalently a smooth projective curve over $\mathbb{C}$, and put $V=H^0(X,K_X)$. Choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$ with $a_i\cdot b_j={\delta }_{ij}$. For a closed differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The underlying bilinear identity isIt holds for any two smooth closed one-forms. In particular, the Riemann bilinear relations for holomorphic differential forms arefor $\omega ,\eta \in V$ and $\omega \ne 0$ in the inequality. The inequality fixes the orientation and the sign convention.

To prove the identity, cut $X$ along the symplectic basis to a fundamental polygon $P$. On $P$, a closed one-form $\alpha$ has a primitive $F$. The Stokes theorem givesOpposite copies of each cut have opposite orientations, while their values of $F$ differ by the corresponding period of $\alpha$. Pairing those edges therefore leaves $A_i(\alpha ){\int }_{b_i}\beta -B_i(\alpha ){\int }_{a_i}\beta$ for the $i$th handle. Summing proves the identity. The same calculation can be made with a small disk about the polygon's vertices removed; its extra boundary terms tend to zero, so the vertices cause no additional term.

For holomorphic differential forms $\omega ,\eta$, their wedge product of differential forms vanishes: locally both are multiples of $dz$. This proves the first relation. If $\omega =f(z)dz$ and $z=x+iy$, thenA nonzero holomorphic differential form is nonzero on an open set, so the integral is strictly positive. Applying the identity to $\omega ,\overline{\omega }$ proves the second relation.

These relations first prove that the map $V\to {\mathbb{C}}^g$ taking $a$-periods is injective: if every $A_i(\omega )$ vanishes, the positive integral just computed would vanish. Since $\dim V=g$ by the Riemann-Roch theorem, it is an isomorphism. Consequently there is a unique basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$. Set ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. The first relation gives ${\tau }_{ij}={\tau }_{ji}$, and the second gives, for any nonzero column $c$,Thus the normalized period matrix of a complex torus is symmetric and has positive-definite imaginary part:

The integration map sends a cycle $\gamma$ to the functional $\omega \mapsto {\int }_{\gamma }\omega$ in $V^{\ast }$. In the normalized basis its image is the period latticeIndeed, the real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: its imaginary part is $Yy$, and $Y$ is invertible. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and has compact quotient. This proves thatis a well-defined $g$-dimensional complex torus, rather than merely a quotient by an arbitrary subgroup.

The relations also provide its canonical principal polarization of a complex torus. With the convention that a Hermitian form is conjugate-linear in its first argument, put $H(z,w)={\overline{z}}^{,t}{Y}^{-1}w$ and $E=\mathrm{Im}H$. Symmetry of $\tau$ givesIn particular $E$ is integral and unimodular on $\Lambda$, and it is compatible with the complex structure. These are exactly the positivity and integrality conditions for a Riemann form on a complex torus; they make the complex torus a Jacobian variety with a principal polarization of a complex torus. If instead one uses a first-argument-linear Hermitian form, the same alternating form is $-\mathrm{Im}H$. Changing the symplectic basis changes the coordinates but not this polarized quotient. For $g=0$, $V=0$ and the Jacobian variety is a point.

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 $X$ be a smooth projective curve over $\mathbb{C}$, let $D={\sum }_p{m}_{p}p$ be an effective divisor on an algebraic curve of degree $d$, and identify a translate of the Jacobian variety with ${\mathrm{Pic}}^d(X)$. The precise derivative theorem iswhere ${\delta }_D$ is the connecting homomorphism ofThe target is $T_{u_d(D)}\mathrm{Jac}(X)=H^1(X,{\mathcal{O}}_X)=H^0(X,K_X{)}^{\ast }$ under Serre duality. Its dual is restriction of holomorphic differential forms to the length-$d$ closed subscheme $D$:In particular, this includes derivatives of the local coefficients at repeated points, rather than simply evaluating once for each point of the support.

Here is a local proof of all these identifications. Near a point of multiplicity $m$, take a coordinate $z$ vanishing at that point. Nearby effective divisors are represented by monic polynomials of degree $m$. A first-order deformation has equationThese coefficients are the elementary symmetric coordinates on the symmetric product of a curve; they give $m$ independent tangent directions even at a repeated point. Dividing the variation of the equation by the original equation identifies the tangent direction with the principal part $v(z)/z^m$. Doing this at each point gives $H^0({\mathcal{O}}_D(D))$, a space of dimension $d$.

To see the derivative of the line bundle directly, use the local frame $1/(z^m-\epsilon v)$ for ${\mathcal{O}}_X(D_{\epsilon })$ and the frame $1$ outside the small disks. Relative to the frame $1/z^m$ at $\epsilon =0$, its transition multiplier changes byFirst-order changes of line bundle transition functions have the form $1+\epsilon c_{ij}$; changing frames adds a Čech coboundary. Thus the tangent space to the Picard group is $H^1({\mathcal{O}}_X)$, and the derivative is exactly the Čech cocycle of these principal parts, namely ${\delta }_D$.

This agrees with differentiating the Abel integrals, including the sign. If $\omega =f(z)dz$ and $F^{\prime }=f$, the sum of $F$ at the roots of $z^m-\epsilon v$ has derivativeFor example, expand $F={\sum }_{k\ge 1}{f}_{k-1}{z}^k/k$. The first-order Newton identities give ${\partial }_{\epsilon }\sum z_i^k=k{v}_{m-k}$ for $1\le k\le m$, and zero for $k>m$, yielding exactly the residue above. Equivalently the same formula follows by differentiating the contour integral for the sum over roots. Serre duality pairs the principal part class with $\omega$ by the sum of these residues. Locally this pairing is perfect between the principal parts $z^{-1},\dots ,z^{-m}$ and the jets $1,z,\dots ,z^{m-1}$ times $dz$. This proves the asserted dual restriction map.

The long exact sequence in sheaf cohomology now gives the kernel and rank without any reduced-support assumption:By duality,This is the geometric form of the derivative of the Abelian sum map. For distinct points, the image is spanned by the evaluation vectors of the canonical map; at repeated points, the corresponding osculating directions must also be included. The Riemann-Roch theorem confirms the rank as $g-h^0(K_X(-D))$. When $h^0(D)=1$, the Abelian sum map is an immersion at $D$. For $d=g-1$ and $h^0(D)=1$, its image is a hyperplane annihilated by the unique nonzero holomorphic differential form vanishing along $D$, which will be the tangent hyperplane to the theta divisor.

Assume $X$ is a connected compact Riemann surface of geometric genus $g\ge 1$. Work first in ${\mathrm{Pic}}^{g-1}(X)$, translating back to the Jacobian variety by tensoring with ${\mathcal{O}}_X(-(g-1)p_0)$ when desired. PutThis is the image of the Abelian sum map from $X^{(g-1)}$. The Riemann singularity theorem asserts that this image is a reduced Cartier divisor and thatHere multiplicity means the order of a local defining equation in the maximal ideal of the smooth ambient Jacobian variety. Consequently the singular points are precisely the classes with at least two independent sections. We will prove the multiplicity equality, including the nonvanishing of its proposed leading term.

Choose an effective divisor $E$ of degree $n\ge g$, and a local holomorphic family $\mathcal{L}$ of degree-$g-1$ line bundles about $L$ on $X\times U$, where $U$ is a small neighborhood in ${\mathrm{Pic}}^{g-1}(X)$. Such a family is obtained by varying the Čech cocycles of a line bundle. Since $\mathrm{deg}{L}^{\prime }(E)>2g-2$ for all $L^{\prime }\in U$, Serre duality gives $H^1(L^{\prime }(E))=0$; the Riemann-Roch theorem gives $h^0(L^{\prime }(E))=n$. The cohomology and base change for line bundles on a curve then identifiesas vector bundles of rank $n$. Evaluation defines a matrix $M:A\to B$. For each $L^{\prime }$, its exact sequence isThus $f=\det M$ vanishes exactly on $\Theta$.

The function $f$ is not identically zero. To exhibit a degree-$g-1$ line bundle with no section, choose $g$ distinct points $p_1,\dots ,p_g$ giving independent evaluation functionals on $H^0(K_X)$. Such points can be chosen successively because a nonzero holomorphic differential form cannot vanish everywhere. For $F=\sum p_i$, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_X(F))=1$. Choose $q$ outside $F$. The unique section of ${\mathcal{O}}_X(F)$ does not vanish at $q$, so ${\mathcal{O}}_X(F-q)$ has no section. Therefore the evaluation determinant genuinely cuts out a hypersurface. Its support is irreducible, since it is the image of the irreducible symmetric product of a curve $X^{(g-1)}$.

Fix now $L\in \Theta$, and put $r=h^0(L)=h^1(L)$, the equality following from the Riemann-Roch theorem. The matrix $M(L)$ has an invertible $(n-r)$ by $(n-r)$ block. After invertible holomorphic row and column operations, its local form iswhere $N$ is $r$ by $r$. More explicitly, eliminate the invertible block by its Schur complement; the discarded determinant is a holomorphic unit. Hence $f$ is a unit times $\det N$. Every entry of $N$ lies in the maximal ideal, so its order is at least $r$.

The linear term of $N$ is the cup-product mapTo verify this, deform the transition functions as $g_{ij}(1+\epsilon {\xi }_{ij})$ and try to lift a section $s$ to $s+\epsilon s^{\prime }$. Its first-order gluing equation has right side ${\xi }_{ij}{s}_j$, whose Čech cohomology class is exactly the obstruction to the lift. The kernel-to-cokernel derivative of the evaluation matrix gives that same obstruction. Under Serre duality, if $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$ are bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$, thenWe must show that this matrix is invertible for some $\xi$; otherwise the lower bound on multiplicity would not be an equality.

For each of these two $r$-dimensional section spaces, there are $r$ distinct points with independent evaluations. Choose them successively, using a section in the remaining kernel to find the next point. The set of such $r$-tuples is a nonempty open subset of $X^r$. The two open sets intersect because $X^r$ is irreducible, and we may avoid their diagonals. Choose $q_1,\dots ,q_r$ in that intersection and trivialize $L$ and $K_X$ there. Both matricesare invertible. Let $\xi \in H^0(K_X{)}^{\ast }=H^1({\mathcal{O}}_X)$ be the sum of the evaluation functionals at these points, with any nonzero weights $c_k$. Thenis invertible. It follows that the degree-$r$ polynomial $\det N_1$ is nonzero, so ${\mathrm{ord}}_{L}f=r$. This proves the needed invertible cup-product direction for a line bundle on a curve by an explicit construction.

We still need to identify this determinant hypersurface with the reduced image, rather than a multiple of it. Choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$. For their sum $D$, the Riemann-Roch theorem gives $h^0(D)=h^0(K_X-D)=1$. At its class, the calculation above gives order one. The determinant hypersurface has irreducible support, so its sole divisor on an algebraic curve coefficient is one everywhere. A local hypersurface in a smooth space with these divisor on an algebraic curve coefficients is reduced. Thus it is exactly the reduced image $W_{g-1}$, and the multiplicity formula holds for that image itself.

The argument gives the useful additional descriptionFor $r=1$, its nonzero linear equation is $⟨\xi ,st⟩=0$. This agrees with the derivative of the Abelian sum map: if $\mathrm{div}(s)=D$, then $st$ is the unique holomorphic differential form vanishing along $D$. For $r\ge 2$, the local equation has no linear term and the point is singular. When $g=1$, $W_0$ is the single trivial line bundle in ${\mathrm{Pic}}^0(X)$, with multiplicity one; the argument includes this case using the empty effective divisor. In genus zero there is no $W_{g-1}$ of effective divisors on an algebraic curve, so that hypothesis is necessary.