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There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

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.

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.

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.

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 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.

On the standard affine charts $U_i=\{X_i\ne 0\}$, use the formal frame $e_i=X_i^m$ and glue rank-one free sheaves byThe transition functions are regular units on overlaps and satisfy the cocycle identity. This constructs the twisting sheaf on projective space $\mathcal{O}(m)$ as an invertible sheaf for every integer $m$, including negative $m$.