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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.

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.

The Torelli theorem concerns a connected smooth projective curve over $\mathbb{C}$ and its canonical principal polarization of a complex torus. Its conclusion isThe polarization of a complex torus is part of the hypothesis. We will reconstruct the curve intrinsically from the Jacobian variety and the theta divisor, using the Gauss map of a theta divisor, and prove the reconstruction in both the nonhyperelliptic and hyperelliptic curve cases. The needed biduality and ordinary-tangency facts about projective dual varieties are the general projective-duality results permitted here.

First explain why the polarized data determine the theta divisor up to translation. On $J=\mathrm{Jac}(X)$, take integral one-form classes ${\alpha }_i,{\beta }_i$ dual to the symplectic cycle basis. Pullback by the degree-one Abelian sum map identifies them with the corresponding classes on $X$; the Riemann bilinear relations giveOn $X^{g-1}$, the sum map pulls each one-form back to the sum of its pullbacks from the factors. It has generic degree $(g-1)!$ onto $W_{g-1}$: the Abel theorem for divisors and the existence of classes with $h^0=1$ give a unique effective unordered representative, and ordering gives the factorial. Integrating a wedge of $2g-2$ basis one-forms over the product, only terms assigning a matching pair ${\alpha }_i,{\beta }_i$ to each factor survive. Dividing by $(g-1)!$ gives integral one when precisely one matching pair is omitted, and zero for all other basis wedges. Hence the Poincaré dual class of $W_{g-1}$ isThis is exactly the integral form of the canonical principal polarization of a complex torus constructed by periods.

Its associated line bundle has one-dimensional section space. One can check this without any theta multiplicity assumption: in normalized period coordinates a representative has sections satisfyingExpand the first periodicity in a Fourier series $F(z)={\sum }_{k\in {\mathbb{Z}}^g}{c}_k{e}^{2\pi i{k}^{t}z}$. The second periodicity forces $c_k=c_0{e}^{\pi i{k}^t\tau k}$. Positive definiteness of $\mathrm{Im}\tau$ makes this series normally convergent, so the section space has dimension exactly one. Other line bundles giving the same polarization of a complex torus differ by a degree-zero line bundle. The map $x\mapsto T_x^{\ast }L\otimes L^{-1}$ is an isomorphism from $J$ to its dual abelian variety for a principal polarization of a complex torus: on lattices it is the unimodular alternating form. Thus all these line bundles are translates, and so are their unique zero divisors. In particular the given polarized data recover the reduced divisor on an algebraic curve $W_{g-1}$ up to translation. Translation does not change tangent hyperplanes after identifying tangent spaces with $T_{0}J$.

Put $V=T_{0}J=H^0(K_X{)}^{\ast }$ and $P=\mathbb{P}(V^{\ast })$. At a smooth class $L={\mathcal{O}}_X(D)\in \Theta$, the Riemann singularity theorem says $h^0(D)=1$. The derivative of the Abelian sum map identifies the translated tangent hyperplane with $H^0(K_X(-D){)}^{\perp }$. Thus the intrinsic Gauss map of a theta divisor isIn particularWe will see that this map is dominant and generically finite. Its function-field extension therefore determines an intrinsic finite cover: normalize $P$ in $\mathbb{C}(\Theta )$, obtaining $Z\to P$. Normalization in a finite extension is finite here. Let $B\subset P$ be its reduced codimension-one branch locus. Using this finite cover avoids counting omitted points of the rational Gauss map of a theta divisor as ramification. Everything in this construction is determined by the polarized Jacobian variety.

Suppose first that $g\ge 3$ and $X$ is nonhyperelliptic. Its canonical map embeds it as a nondegenerate curve $C\subset \mathbb{P}(V)$. For clarity, this follows directly from the Riemann-Roch theorem: $h^0(p)=1$ on a positive-genus curve, so $K_X$ has no base point. If a length-two effective divisor $E$ had $h^0(E)\ge 2$, a nonconstant meromorphic function with poles bounded by $E$ would give a map of degree at most two to ${\mathbb{P}}^1$. Degree one forces genus zero, and degree two is the hyperelliptic curve case. Hence $h^0(E)=1$, and the Riemann-Roch theorem gives $h^0(K_X(-E))=g-2$. The canonical sections therefore separate every pair of points and every tangent direction, proving the embedding.

Let $U\subset P$ be the open set of hyperplanes transverse to $C$. A hyperplane in $U$ cuts out $n=2g-2$ distinct points. These points span the hyperplane: if a second independent canonical section vanished at all of them, its zero divisor would have to equal the same degree-$2g-2$ divisor, and the ratio of the two sections would be a constant. Consequently some $g-1$ of the intersection points are independent. For the divisor $D$ formed by such a subset, $h^0(K_X(-D))=1$ and $h^0(D)=1$.

We need to know that every $(g-1)$-subset of a general hyperplane section has this property. Here is the monodromy argument, with its essential steps. The incidence space choosing one intersection point is a projective bundle over $C$ before restricting to $U$, so it is irreducible. The incidence space choosing two distinct ordered intersection points is likewise a projective bundle over $C\times C$ minus its diagonal, with fibre ${\mathbb{P}}^{g-3}$; it too is irreducible. Restricting to transverse sections preserves irreducibility and hence connectedness of these covers. Thus the monodromy action of a covering space on the $n$ intersection points is two-transitive. A general tangent hyperplane has just one contact of order two, by characteristic-zero projective biduality theorem; locally its intersection equation is $z^2=t$ together with the other simple points. A small loop in $t$ exchanges exactly those two points. Two-transitivity conjugates this transposition to every transposition, so the monodromy is the full symmetric group ${\mathfrak{S}}_n$.

It follows that the finite étale cover of $U$ choosing a $(g-1)$-subset is connected and irreducible. The condition of dependent evaluations is closed on that cover. It is proper because we have exhibited an independent subset; its image under the finite map is therefore a proper closed subset of $U$. Outside that image every subset is independent. By the Abel theorem for divisors, distinct such subsets have distinct Abelian sums: $h^0(D)=1$ means there is only one effective representative of its class. Thus the subset cover is birational to $\Theta$ with its Gauss map of a theta divisor, andMore precisely it is exactly the restriction of $Z\to P$ over $U$, since this étale cover is normal and has the same function field.

There can consequently be no branch component over $U$. Its complement is the irreducible hypersurface $C^{\ast }$ of tangent hyperplanes, the projective dual variety of $C$. Around a general point of $C^{\ast }$, the transposition of two roots acts nontrivially on $(g-1)$-subsets: choose a subset containing exactly one of those two roots. Hence the finite cover is ramified there. This proves the exact identityThe quoted biduality theorem now recovers $C=(C^{\ast }{)}^{\ast }=B^{\ast }$ as an embedded curve, and the canonical map identifies $X$ with $C$. This proves reconstruction in the nonhyperelliptic case, including special curves rather than only a general curve in moduli.

Now suppose that $X$ is a hyperelliptic curve of genus $g\ge 3$. Let $\pi :X\to {\mathbb{P}}^1$ be its degree-two map. The Riemann-Hurwitz formula gives $2g+2$ branch points $b_1,\dots ,b_{2g+2}$. Choose an affine coordinate with no branch at infinity and writeThe forms $dx/y,xdx/y,\dots ,x^{g-1}dx/y$ are holomorphic and independent; their orders at the two points at infinity show there are exactly $g$ of them. Thus the canonical map factors as $\pi$ followed by the degree-$g-1$ rational normal curve embeddingWrite $R=\nu ({\mathbb{P}}^1)$.

A general hyperplane section of $R$ consists of $g-1$ distinct points $x_1,\dots ,x_{g-1}$ avoiding the branch points. The corresponding canonical divisor on $X$ consists of the $g-1$ pairs ${\pi }^{-1}(x_j)$. Choosing one lift from each pair gives an effective divisor $D$ with $h^0(D)=1$: the evaluation matrix of the canonical forms has Vandermonde rank $g-1$. If instead a degree-$g-1$ sub-divisor contains an entire pair, it has at least two sections, because that pair is a fibre of $\pi$ and moves in a pencil. Every choice not taking exactly one point from each pair contains an entire pair. Therefore the generic smooth theta points over the hyperplane are exactly the independent choices of one lift in each pair, andThe cover of these choices is again the restriction of $Z$ to the open set of transverse sections of $R$ avoiding the $b_i$: the identification on a dense open follows from the Abel theorem for divisors, and the choice cover is finite étale and normal. Its components would correspond to components of the function field of the irreducible theta divisor, so it is connected.

There are precisely two types of codimension-one branching. If a simple root $x_j$ crosses a branch point $b_i$, its two lifts exchange, giving nontrivial monodromy on the choices. This yields the hyperplaneIf two roots collide away from the $b_i$, a local transverse parameter gives roots $x_0\pm \sqrt{t}$. Trivialize the two sheets of $\pi$ near $x_0$. Choices selecting different sheets for the two roots are exchanged when the roots exchange; hence this also causes branching. Its locus is $R^{\ast }$, the hypersurface of tangent hyperplanes to $R$. Away from these two loci all roots and all their lifts vary locally without branching, so there are no other branch components. ConsequentlyFor $g\ge 3$, $R^{\ast }$ is irreducible and nonlinear. Indeed, hyperplane sections of the degree-$g-1$ rational normal curve are binary forms of degree $g-1$, and the repeated-root hypersurface is their irreducible discriminant of degree $2g-4\ge 2$. Thus $B$ intrinsically separates into its single nonlinear component $R^{\ast }$ and its $2g+2$ linear components. Biduality recovers $R$, while dualizing $H_i$ recovers the point $\nu (b_i)$ on $R$.

These data recover the double cover uniquely. Identify $R$ with ${\mathbb{P}}^1$ and move infinity away from the recovered branch set. Its function field isAny other quadratic extension with the same simple branch points has a defining rational function whose ratio to this product has even valuation everywhere. On ${\mathbb{P}}^1$ such a rational function is a constant times a square, and every nonzero complex constant is a square. The quadratic extensions are therefore isomorphic; their smooth projective models are isomorphic. This proves the hyperelliptic reconstruction. The branch locus also distinguishes the two cases: the nonhyperelliptic locus has one irreducible branch hypersurface, whereas the hyperelliptic locus has the additional $2g+2$ linear components.

Finally handle the small genera. In genus two, the theta divisor is $W_1$, the image of $X$ itself. The Abelian sum map on $X$ is injective because a principal divisor $p-q$ with $p\ne q$ would give a degree-one map to ${\mathbb{P}}^1$. Its derivative is nonzero since the canonical bundle has no base point, so this compact injective immersion identifies $X$ with $W_1$. In genus one the same argument identifies $X$ with its Jacobian variety after choosing an origin. In genus zero the Riemann-Roch theorem supplies a function with one simple pole, giving $X\cong {\mathbb{P}}^1$. These cases complete the proof: the canonically polarized Jacobian determines the curve in every genus.

Let $V=H^0(C,{\Omega }_C^1)$, of complex dimension $g$, and choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(C,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. Integration gives a homomorphismWe first prove both injectivity and discreteness, rather than treating periods as a lattice by definition.

Cut the compact Riemann surface along this basis to a polygon. For closed one-forms $\omega ,\eta$, choose a primitive $F$ of $\omega$ on the cut polygon and apply Stokes theorem to $d(F\eta )=\omega \wedge \eta$. Pairing its boundary edges records the jumps of $F$ across the cuts and gives the Riemann bilinear relations for a compact surface:If a holomorphic differential form $\omega$ has all $a$-periods zero, take $\eta =\overline{\omega }$. The right side is zero, while in a holomorphic coordinate, for $\omega =f(z)dz$,Its integral is strictly positive unless $\omega =0$. Thus the $a$-period map $V\to {\mathbb{C}}^g$ is injective, hence an isomorphism. Choose a basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$, and put ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. Applying the same boundary identity to ${\omega }_i,{\omega }_j$, whose wedge product is zero, gives ${\tau }_{ij}={\tau }_{ji}$. Applying it to $\omega =\sum c_i{\omega }_i$ and its conjugate givesHence $\mathrm{Im}\tau$ is positive definite.

In the dual coordinates, $I(\sum m_i{a}_i+n_i{b}_i)=m+\tau n$. The real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: taking imaginary parts of a zero gives $(\mathrm{Im}\tau )y=0$, then $y=0$ and $x=0$. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and cocompact, and no nonzero integral cycle has zero image. Thus

The Abel-Jacobi map of a compact Riemann surface sends an effective complex analytic divisor $D=\sum Q_i$ to ${\sum }_i{\int }_{P_0}^{Q_i}({\omega }_1,\dots ,{\omega }_g)$ modulo $\Lambda$. Changing a path adds an integral period, so the map is well-defined. For two degree-$d$ complex analytic divisors, let $E=D-D^{\prime }={\sum }_P{n}_{P}P$, with $\sum n_P=0$, and putWe prove that $E$ is principal exactly when $v\in \Lambda$, including the construction of a meromorphic function.

First construct a differential of the third kind $\eta$ with residues $n_P$ at the support of $E$ and no other poles. For the reduced support $S$, the residue exact sequence isOn sheaf cohomology, the obstruction to prescribing residues belongs to $H^1(C,K_C)\cong \mathbb{C}$ by Serre duality; its pairing with the constant section of ${\mathcal{O}}_C$ is the sum of residues. This follows directly by integrating the local simple-pole terms around the boundary circles. Thus the obstruction is zero because $\sum n_P=0$, and the required $\eta$ exists. Subtract a linear combination of the normalized holomorphic differential forms so that all its $a$-periods are zero.

Take its poles and the integration paths inside the same cut polygon. Apply the boundary calculation above to a primitive $F_i(P)={\int }_{P_0}^P{\omega }_i$, removing small circles around the poles of $\eta$. Since $d(F_i\eta )=0$ away from the poles, the paired outer edges contribute $B_i(\eta )$, while the small circles contribute $2\pi i{\sum }_P{n}_P{F}_i(P)$ by the residue theorem. ThereforeThe same argument without normalizing $\eta$ gives the useful reciprocity identityThese identities use compatible cuts and paths; the final conclusions are unchanged by integral periods.

Suppose $v=m+\tau n$ with $m,n\in {\mathbb{Z}}^g$. DefineIts $a$-periods are $-2\pi i{n}_i$, and its $b$-periods are $2\pi i{m}_i$. Around a pole its integral is $2\pi i{n}_P$. These cycles generate the first homology group of the punctured surface, so every period of ${\eta }^{\prime }$ belongs to $2\pi i\mathbb{Z}$. Choosing an auxiliary point $Q$ away from the poles, setIt is single-valued away from the poles. Locally ${\eta }^{\prime }=n_{P}dt/t+\text{a holomorphic one-form}$, so $f=t^{n_P}$ times a nonvanishing holomorphic function. It extends meromorphically with $\mathrm{div}(f)=E$. This proves sufficiency and explicitly produces the required principal complex analytic divisor.

Conversely, if $E=\mathrm{div}(f)$, the meromorphic differential $df/f$ has residues $n_P$ and every period in $2\pi i\mathbb{Z}$, since each integral counts the winding number of $f$ about zero. Write its $a$-period vector as $2\pi ia$ and its $b$-period vector as $2\pi ib$, with $a,b\in {\mathbb{Z}}^g$. The reciprocity identity gives $v=b-\tau a\in \Lambda$. Since $E$ has degree zero, changing the base point cancels in its Abel sum; a base point at the support of $E$ causes no problem, because the Abel integrals use holomorphic forms. We have provedThis establishes the Abel theorem for divisors directly from the period and residue calculations.