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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.

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.

Throughout the algebraic questions, $C$ is a smooth projective connected algebraic curve over the algebraically closed field $k$. This is the curve convention needed for the Jacobian variety; for an arbitrary singular or nonproper curve the assertions need not hold. Choose $P_0\in C(k)$ and write $\pi :C\times T\to T$.

The degree-$d$ Picard functor of a curve iswith pullback as its action on morphisms. Equivalently, normalize $\mathcal{L}$ by tensoring with ${\pi }^{\ast }(\mathcal{L}{|}_{P_0\times T}{)}^{-1}$ and remember the resulting trivialization along $P_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C\times T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit; preservation of this trivialization forces that unit to be $1$. Thus normalized line bundles have effective descent and the displayed functor is already the appropriate sheaf, rather than just a list of classes over fields.

Here is a construction of its representing scheme. The symmetric product of a curve $C^{(r)}$ represents relative effective degree-$r$ Cartier divisors. It is projective, and its completed local ring at $\sum m_i{Q}_i$ is a power-series ring in the elementary symmetric coefficients of $m_i$ local parameters at each $Q_i$. Thus it is smooth of dimension $r$, in every characteristic. PutThis is open by the semicontinuity theorem for coherent cohomology. It is nonempty: choose $g$ points successively imposing independent vanishing conditions on $H^0(C,K_C)$, so that $H^0(C,K_C-D)=0$, and use Serre duality. The Riemann-Roch theorem gives $h^0({\mathcal{O}}_C(D))=1$ on $U$.

For any fixed line bundle $B$ of degree $g-d$, use a copy $U_B$ of $U$ with the universal normalized line bundle $\mathcal{O}(D)\otimes B^{-1}$. This represents exactly the open subfunctor on which $H^1({\mathcal{L}}_t\otimes B)=0$. Indeed, cohomology and base change for line bundles on a curve makes ${\pi }_{\ast }(\mathcal{L}\otimes B)$ a rank-one vector bundle. The evaluation morphismhas a unique relative zero Cartier divisor of degree $g$ and recovers the point of $U_B$. No choice of a global generator on $T$ is needed. Conversely that zero Cartier divisor recovers $\mathcal{L}$ modulo a bundle from $T$.

These charts cover: given any degree-$d$ line bundle $L$, choose a nonspecial effective degree-$g$ Cartier divisor $D$ and take $B=\mathcal{O}(D)\otimes L^{-1}$. On the overlap of $U_B$ and $U_{B^{\prime }}$, the unique effective representative of $\mathcal{O}(D)\otimes B^{-1}\otimes B^{\prime }$ is constructed by the same rank-one evaluation map. It therefore gives a regular transition morphism; uniqueness gives both the inverse and the cocycle identity. Gluing gives a scheme $J^d$ representing ${\mathcal{P}}^d$, with a universal normalized line bundle.

The construction is of finite type, not an infinite union of indispensable charts. Choose $m\ge 2g-1$ and a fixed line bundle $F$ of degree $m-d$. Every $L\otimes F$ has a nonzero section by the Riemann-Roch theorem, so the universal effective Cartier divisor gives a surjection $C^{(m)}\to J^d$. The inverse images of the charts cover the projective, hence quasi-compact, symmetric product of a curve $C^{(m)}$. A finite subcover gives finitely many charts for $J^d$. This surjection also makes $J^d$ irreducible. The charts show that it is smooth of dimension $g$.

For properness, apply the valuative criterion for properness. Let $R$ be a discrete valuation $k$-algebra with fraction field $K$, and let $L_K$ be a degree-$d$ line bundle on $C_K$. A rational section represents it by a Cartier divisor. Its closure in the regular surface $C_R$ is again a Cartier divisor, since regular local rings are factorial. Thus $L_K$ extends to a line bundle on $C_R$, whose fibre degree remains $d$ by constancy of line bundle degree in a family. Two extensions with the same generic class differ by a vertical Cartier divisor. The special fibre is irreducible, so that Cartier divisor is an integer multiple of the entire fibre, the principal Cartier divisor of a power of a uniformizer. The extensions therefore give the same normalized Picard class. This proves existence and uniqueness in the valuative criterion, hence separatedness and properness of the finite-type scheme $J^d$.

Consequently ${\mathrm{Pic}}_C^d={\mathrm{Jac}}_C^d$ is a smooth proper $k$-variety of dimension $g$. Tensoring by ${\mathcal{O}}_C(-d{P}_0)$ identifies it with $J^0$. On $J^0$ the tensor product and inverse of line bundles induce the algebraic group operations; it is the Jacobian variety, an abelian variety.

For a smooth projective algebraic curve of genus $g$, this smooth proper $g$-dimensional scheme represents degree-$d$ line bundles in families. Charts come from nonspecial degree-$g$ bundles: their first sheaf cohomology vanishes, so the Riemann-Roch theorem gives exactly one independent global section, whose zero Cartier divisor varies algebraically. Twisting by fixed line bundles gives charts covering every degree-$d$ class. For degree zero the tensor product gives the Jacobian variety; other components are its torsors.

A smooth projective curve over the complex numbers is determined up to isomorphism by its Jacobian variety together with its canonical principal polarization of a complex torus. For genus at least three, the branch locus of the theta Gauss map recovers the canonical curve by projective biduality theorem, or recovers the canonical rational normal curve and the branch points of a hyperelliptic curve. In genus two the theta divisor itself is the curve; in genus one the Jacobian variety is the curve after choosing an origin. The polarization is essential data in this statement.

For an abelian variety and an integer $N$ invertible in the base field, a torsion line bundle $M\in \hat{A}[N]$ has trivial pullback by $[N]$. Choose a trivialization. The descent action of the kernel $A[N]$ on that trivial line bundle acts through a character ${\chi }_M:A[N]\to {\mu }_N$. Define $e_N(P,M)={\chi }_M(P)$. This is independent of rescaling the trivialization and bilinear under translation and tensor product. A trivial character descends the trivial bundle, so $M$ is trivial. Since both groups have order $N^{2\dim A}$, the character map is an isomorphism and the Weil pairing for an abelian variety is perfect. The theta autoduality of a Jacobian turns it into a pairing on two copies of the Jacobian variety's torsion group.