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Translate the tangent hyperplane at a smooth point of the theta divisor to the identity of the Jacobian variety. Its annihilator defines the Gauss map. At $L={\mathcal{O}}_X(D)$ with $h^0(L)=1$, the derivative of the Abelian sum map identifies this annihilator with the unique line $H^0(K_X(-D))$. Thus $\Gamma (L)=[\omega ]$ implies $D\le \mathrm{div}(\omega )$. This connects the intrinsic geometry of the theta divisor to the canonical map of the curve.

Let $D={\sum }_p{r}_{p}p\in X^{(d)}$. The derivative of the Abelian sum map is most naturally stated using principal parts. The tangent space of the symmetric product of a curve at $D$ is $H^0({\mathcal{O}}_D(D))$, and the tangent space of the Jacobian variety is $H^0(K_X{)}^{\ast }$. With these identifications,Its dual is restriction of holomorphic one-forms to the length-$d$ divisor scheme. In particular,The last equality is the Riemann-Roch theorem. These formulas apply to repeated points as well as distinct ones.

Here is a direct proof of the repeated-point Abel differential formula. At a multiplicity-$r$ point, choose a local coordinate $t$ with $t(p)=0$. A nearby local divisor is described by the roots $t_1,\dots ,t_r$, or, without making an ordering choice, by their elementary symmetric functions $e_1,\dots ,e_r$:These $e_k$ are smooth local coordinates on the symmetric product of a curve, including at the repeated divisor. Under a first-order deformation, minus the logarithmic variation of this defining polynomial is the principal partThe sign can be checked at $r=1$: moving the point from $0$ to $ϵa$ gives $v=a/t$, and the Abel integral changes by $a\omega (p)$.

Write $\omega =({\sum }_{j\ge 0}{c}_j{t}^j)dt$ and its local primitive as $F(t)={\sum }_{j\ge 0}{c}_j{t}^{j+1}/(j+1)$. The local contribution to the Abelian sum map is ${\sum }_{i}F(t_i)$. Newton's identities, linearized at $e_1=\cdots =e_r=0$, giveSubstituting in the primitive therefore givesSumming over the support proves the theorem. In particular, a multiplicity-$r$ point tests the first $r$ coefficients of a holomorphic one-form, not just its value. The residue pairing between principal parts of order at most $r$ and differential jets through order $r-1$ is perfect. Thus the annihilator of the derivative is precisely the holomorphic one-forms vanishing to order at least $r_p$ at every $p$, namely $H^0(K_X(-D))$.

There is also a sheaf cohomology description that explains its kernel geometrically. The sequencegives a connecting homomorphism $\delta :H^0({\mathcal{O}}_D(D))\to H^1({\mathcal{O}}_X)$. Represent a principal part by meromorphic lifts on coordinate disks; their differences on overlaps give its Čech cohomology class. Under Serre duality, pairing this class with $\omega$ is the sum of the residues of $v_p\omega$, so $\delta$ is the displayed derivative. Exactness shows that its kernel is $H^0({\mathcal{O}}_X(D))/\mathbb{C}$, of dimension $h^0({\mathcal{O}}_X(D))-1$, in agreement with the tangent directions to the complete linear system of a divisor in Solution 2.

Two consequences will be useful below. If $d\ge 2g-1$, then $K_X(-D)$ has negative degree and no nonzero sections, so $u_d$ is a submersion everywhere. Its image is open, and it is also closed because $X^{(d)}$ is compact. Since the Jacobian variety is connected, $u_d$ is surjective. Thus the points $u(p)$ generate the Jacobian variety as a group. Together with Abel's theorem, this identifies the degree-zero Picard group with the Jacobian variety: every line bundle has a meromorphic section and thus a divisor representation, injectivity is Abel's theorem, and surjectivity follows from $u_d(D)-du(p_0)$.

For $g\ge 1$, one can choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$: at each step a surviving nonzero holomorphic one-form has only finitely many zeros, so choose a point where it does not vanish. Their sum $D$ has $h^0(K_X(-D))=1$ and hence $h^0({\mathcal{O}}_X(D))=1$. The derivative has rank $g-1$ there. Consequently $W_{g-1}=u_{g-1}(X^{(g-1)})$ is an irreducible subvariety of dimension $g-1$. This also covers $g=1$, when $X^{(0)}$ and $W_0$ are points.

Identify ${\mathrm{Pic}}^{g-1}(X)$ with the Jacobian variety using the base point and the translation of Solution 4. A line bundle $L$ of degree $g-1$ corresponds to the point $z=u(D)-\kappa$ for any divisor representation $L\simeq {\mathcal{O}}_X(D)$. The Riemann singularity theorem isIn particular, the theta divisor is singular exactly at the degree-$g-1$ line bundles with at least two independent sections. We prove the equality, including the upper bound on multiplicity.

Choose a fixed effective divisor $A$ of degree $N\ge g$, and a local holomorphic family $L_{\eta }$ representing nearby points of ${\mathrm{Pic}}^{g-1}(X)$. Such a family can be constructed locally from the holomorphic exponential sequence, by varying line-bundle transition functions through exponentials of Čech cohomology representatives. By Serre duality, $H^1(L_{\eta }(A))$ is dual to $H^0(K_X\otimes L_{\eta }^{-1}(-A))$, whose degree is $g-1-N<0$, so it vanishes. The Riemann-Roch theorem gives $h^0(L_{\eta }(A))=N$. By cohomology and base change for line bundles on a curve, these spaces form a holomorphic vector bundle $E$ of rank $N$. The spaces $H^0(L_{\eta }(A){|}_A)$ form another holomorphic vector bundle $F$ of rank $N$. Evaluation gives a holomorphic square matrix $T(\eta )$ locally, with exact sequenceConsequently $\det T(\eta )=0$ exactly when $L_{\eta }$ has a nonzero section, namely on the support of the theta divisor.

At the point $L$, put $r=h^0(L)=h^1(L)$, the equality being Riemann-Roch theorem. An invertible minor of size $N-r$ persists near this point. Holomorphic row and column operations, or the Schur complement, givewhere $M$ is an $r\times r$ holomorphic matrix. This is the Schur complement presentation of a theta singularity. Its linear term in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is the cup product mapTo see this concretely, vary transition functions $g_{ij}$ of $L$ to $g_{ij}(1+ϵ{\xi }_{ij})$. A section $s_i$ lifts to $s_i+ϵs_i^{\prime }$ precisely when its first-order compatibility equation can be solved; the obstruction is the cocycle ${\xi }_{ij}{s}_j$. This is $\xi ⌣s$. Passing to the kernel and cokernel of $T(0)$ gives exactly the derivative of the reduced matrix $M$, proving the assertion without assuming a multiplicity formula.

Before extracting the order of $\det M$, we verify that $\det T$ is a reduced equation of the theta divisor. Generic points of $W_{g-1}$ have $h^0(L)=1$ by Solution 3. At such a point, both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$ are one-dimensional. Their nonzero sections $s,t$ have nonzero product $st\in H^0(K_X)$. Choose $\xi \in H^1({\mathcal{O}}_X)$ with $⟨\xi ,st⟩\ne 0$ under Serre duality. The scalar $d{M}_0(\xi )$ is nonzero, so $\det T$ has a simple zero generically on the irreducible theta divisor. Since its support is exactly that irreducible hypersurface, it defines the reduced divisor. Solution 4 independently proved the same reduced structure for the Riemann theta function. On the smooth ambient Jacobian variety, local equations of the same reduced divisor differ by a nonvanishing holomorphic unit. Hence the orders of the local theta equation and $\det M$ agree everywhere.

Every entry of $M$ vanishes at the point, so each term in its determinant has order at least $r$. This proves ${\mathrm{mult}}_z\Theta \ge r$. To obtain equality, we exhibit an invertible cup-product direction for a line bundle on a curve. Set $V=H^0(L)$ and $W=H^0(K_X\otimes L^{-1})$, both of dimension $r$. There are $r$ distinct points $p_1,\dots ,p_r$ at which evaluation is an isomorphism for $V$: choose them inductively, since a surviving nonzero section has only finitely many zeros. The same holds for $W$. The nonvanishing of each evaluation determinant defines a nonempty open subset of the irreducible variety $X^r$. Their intersection, also avoiding the diagonals, is therefore nonempty. Choose points in it and local coordinates and compatible trivializations of $L$ and $K_X\otimes L^{-1}$ at each point.

Let $\xi$ be the connecting homomorphism class of the meromorphic principal parts ${\lambda }_i/t_i$ at these points, with all ${\lambda }_i\ne 0$. The residue form of Serre duality givesIf $S$ and $U$ are the two invertible evaluation matrices, the matrix of this pairing is $S^t\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_r)U$, and is invertible. Thus the cup product map $\xi ⌣-:H^0(L)\to H^1(L)$ is an isomorphism. Along the corresponding analytic parameter line,with nonzero leading coefficient. The order is at most $r$, completing the proof of the Riemann singularity theorem.

This argument also identifies the tangent cone to a theta divisor: its degree-$r$ equation is $\det (\xi ⌣-)$, which is not the zero polynomial. For $r=1$, the tangent hyperplane is $⟨\xi ,st⟩=0$, in agreement with the annihilator description of the derivative of the Abelian sum map. For $r\ge 2$ all first derivatives vanish, proving singularity. In genus $1$ the theta divisor is a single smooth point of multiplicity $1$. In genus $0$ the effective degree-$-1$ locus is empty, so there is no singularity statement at a point of it.

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.

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.