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Fix $p_0\in X$ and use the normalized holomorphic one-forms and period lattice of Solution 1. The Abel-Jacobi map of a compact Riemann surface isChanging paths changes the vector by a period, so this is well defined and holomorphic. Adding these vectors gives a symmetric map on $X^d$ and therefore the Abelian sum map $u_d:X^{(d)}\to \mathrm{Jac}(X)$. Locally, even where points coincide, symmetric holomorphic functions descend to the elementary symmetric coordinates of the symmetric product of a curve. For any divisor $D={\sum }_p{n}_{p}p$ of degree zero, define $u(D)={\sum }_p{n}_{p}u(p)$. Abel's theorem says that $D$ is a principal divisor if and only if $u(D)=0$. Equivalently, two effective divisors of the same degree have the same Abel sum exactly when they are linearly equivalent.

We first establish the residue version of the bilinear identity. Let $\eta$ be a meromorphic differential on a Riemann surface with only simple poles and total residue zero, and let $F={\int }_{p_0}^p\omega$ on the cut polygon, for a holomorphic one-form $\omega$. Excise small disks about the poles. Since $d(F\eta )=\omega \wedge \eta =0$ off the poles, the outer boundary integral is the sum of the positive small-circle integrals. The latter are $2\pi i{\mathrm{res}}_p(\eta )F(p)$. Pairing the outer edges as in Solution 1 therefore givesThe paths and the cut polygon are fixed together in this equality. Changing them changes the Abel vector by its period lattice, which is precisely the ambiguity we need.

For the sufficiency direction, we need a differential of the third kind with prescribed residues $n_p$. Let $S$ be the reduced support of $D$, consisting of $s$ points. When $s>0$, the Riemann-Roch theorem gives $h^0(K_X(S))=g-1+s$. The residue map from this space has kernel $H^0(K_X)$, of dimension $g$: a differential with at most a simple pole and residue zero has no pole. Its image therefore has dimension $s-1$. The residue theorem puts the image inside the $s-1$ dimensional space of tuples with sum zero, so it is all that space. In particular there is a meromorphic differential on a Riemann surface $\eta$ with residues $n_p$. Subtracting a combination of the normalized holomorphic one-forms makes all its $a$-periods zero. The bilinear identity then sayswhere on the right we temporarily choose the representative determined by the paths. If $D=0$, we simply take $\eta =0$.

Suppose now that $u(D)=0$ in the Jacobian variety, so the chosen vector is $m+Bn$, with $m,n\in {\mathbb{Z}}^g$. Replace $\eta$ byIts $a$-periods are $-2\pi i{n}_j$ and its $b$-periods are $2\pi i{m}_i$. The periods around the punctures are $2\pi i{n}_p$. These loops together generate the homology of the punctured Riemann surface, so every period of ${\eta }^{\prime }$ lies in $2\pi i\mathbb{Z}$. Consequently $f(p)=\exp ({\int }^p{\eta }^{\prime })$ is a single-valued nonzero holomorphic function off the support of $D$. At a support point, ${\eta }^{\prime }=n_{p}dt/t+$ a holomorphic differential, so $f=t^{n_p}$ times a nonvanishing holomorphic function. It extends meromorphically and has divisor exactly $D$. This proves sufficiency, including negative coefficients.

Conversely, if $D=\mathrm{div}(f)$, take $\eta =df/f$. Its residues are the coefficients of $D$, while its $a$- and $b$-periods are integer multiples of $2\pi i$, by the winding numbers of $f$ along these loops. Writing them as $2\pi ia$ and $2\pi ib$, the residue bilinear identity gives $u(D)=b-Ba\in \Lambda$. Thus $u(D)=0$ in the Jacobian variety, proving necessity and henceHere the displayed necessity of degree zero is also the residue theorem applied to $df/f$.

There is a useful geometric version of the conclusion. For an effective divisor $D$, the fiber of $u_d$ through $D$ consists precisely of the effective divisors linearly equivalent to $D$. A nonzero section of ${\mathcal{O}}_X(D)$ is a meromorphic function $f$ with $\mathrm{div}(f)+D\ge 0$, and its zero divisor as a section is $\mathrm{div}(f)+D$. Two such sections give the same effective divisor exactly when their ratio is a nonzero constant. Thus the fiber is the complete linear system of a divisor $|D|=\mathbb{P}{H}^0(X,{\mathcal{O}}_X(D))$.

A meromorphic differential on a Riemann surface has local expression $\omega =a(z)dz$ with a meromorphic coefficient. On overlapping coordinates it transforms by $a(z)dz=\tilde{a}(w)dw$. For a meromorphic function $f$, the coefficients $f^{\prime }(z)$ obey this transformation by the chain rule, so $df$ is such a differential. Two nonzero differentials have coefficient ratio $a/b$; the derivative factors cancel on overlaps, yielding a global meromorphic function $h$ with $\eta =h\omega$.

An invertible holomorphic coordinate change has a nonzero holomorphic derivative, so multiplying by its derivative changes no local zero or pole order. Thus the differential's divisor is well defined. For a compact connected surface, the canonical-degree theorem states ${\sum }_P{\mathrm{ord}}_P\omega =2g-2$, including negative orders for poles. Therefore $\boxed{g=1+\frac{1}{2}\sum _P{\mathrm{ord}}_P\omega }$.

The projective closure has equation $U^2-V^2-W^2=0$. Its gradient $(2U,-2V,-2W)$ cannot vanish at a projective point, so it is nonsingular. At infinity $W=0$ gives the two points $[1:1:0]$ and $[1:-1:0]$. In the supplied sphere parameter,This has double poles at $t=\pm 1$, the two infinity points, and simple zeros at $t=\pm i$. At the parameter infinity use $s=1/t$, giving $v=2s/(1-s^2)$ and $dv=2(1+s^2)(1-s^2{)}^{-2}ds$, which is regular and nonzero at $s=0$. Hence $dv$ extends meromorphically over the whole projective curve. Its order sum $2-4=-2$ gives genus zero, consistently with the stated biholomorphism from the sphere.