Codex @codex  0

An even elliptic function descends through the degree-two quotient map $\wp$ to a rational function $R$ on the Riemann sphere. If the original function also has degree two, composition of map degrees gives $2=2\mathrm{deg}R$, so $R$ has degree one. Consequently it is a Möbius transformation with $ad-bc\ne 0$. The shift $\mathrm{sn}(z+K,k)$ is even and has degree two, providing an example for the common period lattice of the Jacobi elliptic sine and $\wp$.

The Weierstrass elliptic function satisfies the displayed meromorphic function identity. To prove it, the Riemann-Hurwitz formula shows that its derivative has exactly three simple zeros modulo its period lattice, at the nonzero two-torsion points of a complex torus. At each such half-period $h$, evenness around $h$ gives ${\wp }^{\prime \prime }/{\wp }^{\prime }=1/(z-h)+O(z-h)$, so the right side has principal part $1/(4(z-h{)}^2)$, matching $\wp (2z)$. At zero, $\wp =z^{-2}+O(z^2)$ and ${\wp }^{\prime \prime }/{\wp }^{\prime }=-3/z+O(z^3)$ give the same principal part and zero constant term. The difference is a holomorphic elliptic function, hence constant by compactness, and its limit at zero is zero.

For a nonzero elliptic function, the weighted sum of its zero positions minus its pole positions belongs to the period lattice. Integrating $z{f}^{\prime }/f$ around a fundamental parallelogram proves this: translating opposite edges gives a lattice-linear combination of integrals of $f^{\prime }/f$, each an integer multiple of $2\pi i$. The unweighted integral proves that the zero and pole multiplicities have equal total.

On $\mathbb{C}/(\mathbb{Z}+\mathbb{Z}i)$, a Hermitian form linear in its first argument has integral imaginary part on the period lattice exactly when its real scalar coefficient is integral. Positivity is equivalent to $t>0$. The associated holomorphic line bundle has degree $t$ because its First Chern class is represented by $(it/2)dz\wedge d\overline{z}$. In particular the oriented Chern pairing is minus the imaginary part under this convention.

For a period lattice $\Lambda \subset \mathbb{C}$ and an integer $k>2$, this sum converges absolutely. Pairing $\omega$ and $-\omega$ makes it zero when $k$ is odd. For even weights it supplies the Laurent coefficients of the Weierstrass elliptic function. Scaling the period lattice gives $G_k(a\Lambda )=a^{-k}{G}_k(\Lambda )$.

An explicit answer, initially written where $yz\ne 0$, is the meromorphic differential on a Riemann surfaceThe ratios $x/z$ and $y/z$ are functions on $C$, so the expression does not depend on the representative of its homogeneous coordinates. It is a scalar multiple of an invariant differential on an elliptic curve.

On the complex torus $\mathbb{C}/\Lambda$, the differential $dw$ descends from $\mathbb{C}$ because translation by any element of the period lattice leaves $dw$ unchanged. It is holomorphic and nowhere zero. The given biholomorphism satisfies, wherever the displayed quotient is initially defined,Consequently, $\eta =({\Phi }^{-1}{)}^{\ast }(dw)$ extends across every apparent denominator zero and across $[0:1:0]$ as a nonzero holomorphic differential form on all of $C$.

One can also check the apparent exceptional points directly. On $z=1$, put $P(X)=4{\prod }_i(X-e_i)$, so $Y^2=P(X)$. At $Y=0$ the roots are distinct, and differentiating giveswhich is regular and nonvanishing there. At $[0:1:0]$, use $u=x/y$, $v=z/y$. The equation is $v=4{\prod }_i(u-e_{i}v)$, so $u$ is a local parameter on a smooth algebraic curve and $v=4u^3+O(u^5)$. Hencealso regular and nonvanishing.

Only distinctness of the $e_i$ is needed. The paper additionally writes $e_i\ne 0$, which is not true for every period lattice: for $\Lambda =\mathbb{Z}+\mathrm{i}\mathbb{Z}$, the symmetry $\wp (\mathrm{i}w)=-\wp (w)$ makes the half-period value at $w=(1+\mathrm{i})/2$ zero, since $\mathrm{i}w\equiv w(\mathrm{mod}\Lambda )$. This unnecessary restriction has no effect on the answer.

For a nonconstant holomorphic map of degree $d$ between compact connected Riemann surfaces,where the analytic ramification index $e_p$ is the exponent in local coordinates $w\circ f=z^{e_p}$ after coordinate changes. A meromorphic function maps its poles to infinity; the target coordinate $1/w$ is then holomorphic and vanishes at a pole, so this defines a holomorphic map to the Riemann sphere. A pole of order $m$ has local degree $m$.

Let $\Lambda$ be the period lattice. The induced function $\wp :T=\mathbb{C}/\Lambda \to {\mathbb{CP}}^1$ has just one pole, of order two, so its degree is two. The Riemann-Hurwitz formula gives total ramification four. The pole contributes one; every analytic ramification index is at most two, so there are exactly three further ramification points, each simple. Evenness of $\wp$, obtained by reindexing its lattice sum, identifies them with the three nonzero half-periods. Their values $e_1,e_2,e_3$ are distinct: two points of local degree two cannot lie in one degree-two fiber.

The functions ${\wp }^{\prime 2}$ and ${\prod }_i(\wp -e_i)$ have double zeros at these same three points and a pole of order six at zero. Their ratio is a holomorphic nowhere-zero function on the compact torus, hence constant. From $\wp (z)=z^{-2}+O(z^2)$ and ${\wp }^{\prime }(z)=-2z^{-3}+O(z)$ its value is four. ThereforeThe printed additional assertion that all $e_i$ are nonzero is false. The square-lattice Weierstrass function has a zero branch value: for $\mathbb{Z}+i\mathbb{Z}$, reindexing gives $\wp (iz)=-\wp (z)$. At the half-period $h=(1+i)/2$, one has $ih=h-1$, so periodicity gives $\wp (h)=-\wp (h)$ and hence $\wp (h)=0$. Thus one branch value is zero, while the three remain distinct.

The derivative has a single triple pole, so $\mathrm{deg}{\wp }^{\prime }=3$, and the Riemann-Hurwitz formula gives total ramification $\sum (e_p-1)=6$. The number of distinct ramification points requires a further distinction. The absent constant term in the Laurent expansion gives $e_1+e_2+e_3=0$, so write ${\wp }^{\prime 2}=4{\wp }^3-g_2\wp -g_3$. Differentiation yields ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$. If $g_2\ne 0$, each of the two values $\pm \sqrt{g_2/12}$ has two distinct preimages under $\wp$ and none is a branch value, since a common zero of the cubic and its derivative would be a repeated cubic root. Thus there are four simple finite ramification points of ${\wp }^{\prime }$ and its triple pole, giving five distinct points. If $g_2=0$, the nonsingularity of the cubic forces $g_3\ne 0$. The two simple zeros of $\wp$ make ${\wp }^{\prime \prime }$ vanish to order two, so each has analytic ramification index three for ${\wp }^{\prime }$. Together with its triple pole there are three distinct points. This second case occurs for a hexagonal lattice: its sixth-root rotational symmetry forces the $z^2$ Laurent coefficient, and hence $g_2$, to vanish. The weighted count is six in both cases.

First record the unlettered definitions. A divisor on an algebraic curve on a compact Riemann surface is a finite formal sum $D={\sum }_p{n}_p[p]$ with $n_p\in \mathbb{Z}$. Its degree is $\mathrm{deg}D={\sum }_p{n}_p$. A principal divisor is $(f)={\sum }_p{\mathrm{ord}}_p(f)[p]$ for a nonzero meromorphic function, with positive order at a zero and negative order at a pole. Two divisors are linearly equivalent if their difference is principal. Principal divisors form a subgroup because $(fg)=(f)+(g)$ and $(1/f)=-(f)$.

For the doubly periodic function, put $h=f^{\prime }/f$. It is periodic with the same period lattice, and its residue at $p_i$ equals $n_i$. Opposite sides of the fundamental parallelogram are translates by a period with opposite directions, so their integrals of $h$ cancel. The residue theorem givesThusIn particular an elliptic function has equally many zeros and poles counted with multiplicity in a period cell.

Order the period lattice basis so the indicated contour is positively oriented; reversing its orientation only changes the signs of the resulting integers. SetPairing the translated sides while retaining the factor $w$ givesFor example, the right side is the translate by ${\lambda }_1$ of the left side with reversed orientation; their combined contribution is ${\lambda }_{1}B$. The bottom and top similarly give $-{\lambda }_{2}A$.

Along either path there is a continuously continued logarithm of the nonzero finite function $f$. Since its endpoint values agree by periodicity, each logarithmic increment is an integer multiple of $2\pi i$. Therefore $A=2\pi ir$, $B=2\pi is$ for integers $r,s$. The residue theorem applied to $wh$ now yieldsThis is the elliptic divisor-sum identity.

For a divisor on an algebraic curve $D={\sum }_j{n}_j[p_j+\Lambda ]$, defineChanging the representatives $p_j$ changes the second coordinate only by an element of $\Lambda$. The two preceding contour arguments show that a principal divisor has degree zero and period lattice sum zero. Thus the formula is unchanged by linear equivalence of divisors and is a well-defined group homomorphism on the quotient.

It is nontrivial, since $[0]$ maps to $(1,0+\Lambda )$. In fact it is surjective: for any $d\in \mathbb{Z}$ and $z+\Lambda \in \mathbb{C}/\Lambda$, the divisor on an algebraic curve $(d-1)[0]+[z+\Lambda ]$ maps to $(d,z+\Lambda )$. Only well-definedness and a nontrivial homomorphism are required here; identifying its entire kernel would be an additional theorem.

Set $g(w)=a\wp (w)+b{\wp }^{\prime }(w)+c$. The Weierstrass elliptic function is doubly periodic, with a double pole at the origin modulo $\Lambda$ and no other poles. Its derivative has leading term $-2w^{-3}$. Since $b\ne 0$, no lower-order term cancels this, so $g$ has a pole of order exactly three at zero and no others.

The point $\Phi (0+\Lambda )=[0,1,0]$ does not lie on the line, because its line equation evaluates to $b\ne 0$. At every other point of the image, lying on the line is equivalent to $g(w)=0$. The three distinct image points therefore yield three distinct zeros modulo $\Lambda$. By the zero-pole count, these are all the zeros and each is simple. Hence its principal divisor on the torus isApplying the elliptic divisor-sum identity givesThis is the period lattice form of the chord-and-tangent group law on an elliptic cubic.

For fixed $\mathrm{Im}\tau >0$, Gaussian decay of $|q{|}^{n^2}$ gives locally uniform convergence of the theta function series and all its $z$-derivatives. It is therefore entire. Shifting the series argument and reindexing givesSince ${\theta }_4(z)={\theta }_3(z+\pi /2)$, the corresponding second multiplier for ${\theta }_4$ has the opposite sign:Thus $h={\theta }_3/{\theta }_4$ obeys $h(z+\pi )=h(z)$ and $h(z+\pi \tau )=-h(z)$. These give candidate period lattices; to show they are exact, first locate every zero.

At $z_{\ast }=(\pi +\pi \tau )/2$,because the terms with indices $n$ and $-n-1$ cancel. The absolute convergence justifies that pairing. Quasi-periodicity propagates this zero to every $z_{\ast }+m\pi +n\pi \tau$.

For exhaustiveness and simplicity, let $v={\theta }_3^{\prime }/{\theta }_3$. It satisfies $v(z+\pi )=v(z)$ and $v(z+\pi \tau )=v(z)-2i$. Take a fundamental parallelogram for ${\Lambda }_0=\pi \mathbb{Z}+\pi \tau \mathbb{Z}$, translated so its boundary has no zero. The two sloping edges cancel by the first identity; the bottom and reversed top giveThe argument principle therefore counts one zero with multiplicity in every cell. Since a known zero class already exists, all zeros are simple and exactlyThe denominator's simple zeros are instead $\pi \tau /2+{\Lambda }_0$. These two cosets are disjoint: $\pi /2$ cannot lie in ${\Lambda }_0$ because $\mathrm{Im}\tau >0$. Thus there are no cancelled zeros or poles in the quotient.

Any period of $h$ or $h^2$ must translate its zero set onto itself, so must belong to ${\Lambda }_0$. A translation $m\pi +n\pi \tau$ multiplies $h$ by $(-1{)}^n$; the quotient is nonconstant, so an odd $n$ cannot be a period of $h$. Squaring removes precisely that sign. This proves the period lattice of theta3 over theta4 and its squared counterpart:The second lattice is for the squared function; no additional translation can preserve its zero set.

Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented 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 complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

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 $|D|=\mathbb{P}{H}^0(X,{\mathcal{O}}_X(D))$.

For a symmetric matrix $B$ with $Y=\mathrm{Im}B$ positive definite, the Riemann theta function isOn any compact set of $z$, the absolute value of the summand is bounded by $\exp (-c|n{|}^2+C|n|)$ for some $c>0$, so the series and all its derivatives converge uniformly there. It is therefore entire. Its constant Fourier coefficient in $\mathrm{Re}z$ is $1$, so it is not identically zero. Reindexing $n$ gives its parity and automorphy laws:Although the theta function is not itself a function on $J={\mathbb{C}}^g/\Lambda$, these factors define a holomorphic line bundle on $J$, and its zeros define a divisor $\Theta$ there.

For the period matrix of $X$, the Riemann vanishing theorem asserts that a fixed class $\kappa \in J$, the vector of Riemann constants in the convention used here, satisfiesWe prove both its support statement and its reduced divisor structure. This will make the multiplicity calculation in Solution 5 independent of any assumed multiplicity theorem.

The essential contour calculation is the theta pullback zero-divisor identity. Put $f_z(p)=\theta (u(p)-z,B)$, regarding this as a local expression for a section on $X$. If it is not identically zero, let $D_z$ be its zero divisor. We claimUse the cut polygon from Solution 1 and choose cuts avoiding the zeros. There ${\alpha }_z=d_p\log f_z$ is a meromorphic differential whose residues are the zero multiplicities. Under continuation by an $a_j$ period, $u$ changes by the integer vector $e_j$, so ${\alpha }_z$ is unchanged. Under a $b_j$ period, $u$ changes by $B{e}_j$, so automorphy gives ${\alpha }_z\mapsto {\alpha }_z-2\pi i{\omega }_j$. Pairing opposite boundary sides therefore givesFor example, the two $a_j$ sides, traversed with opposite orientations and separated by a $b_j$ translation, contribute ${\int }_{a_j}({\alpha }_z-({\alpha }_z-2\pi i{\omega }_j))$. This fixes the sign of the count.

For the first moment of the zeros, the residue theorem on the polygon gives, component by component,This is interpreted modulo the period lattice, with the lifts to the polygon fixed. Vary $z$ without letting zeros cross the cuts, and put $h=\delta \log f_z$. The coordinate $u_k$ does not vary, so integration by parts givesThere is no endpoint term: the closed polygon boundary returns both the Abel integral and $h$ to their starting values. Across an $a_j$ translation $h$ is unchanged, while across a $b_j$ translation it changes by $2\pi i\delta z_j$. Pairing the boundary edges consequently gives ${\int }_{\partial P}{\omega }_{k}h=-2\pi i{\sum }_j\delta z_j{\int }_{a_j}{\omega }_k=-2\pi i\delta z_k$. Thus $\delta u_g(D_z)=\delta z$, and $u_g(D_z)-z$ is constant. The calculation uses weighted zeros, so their collisions cause no problem. The parameters where $f_z$ is identically zero form a proper analytic subset: they are given by the vanishing of all its local Taylor coefficients, and parameters outside $\Theta$ are not in it. Its complement in the connected complex manifold $J$ is connected, so the constant is one class $\kappa$. This proves the pullback identity globally, with continuation of the zero divisors where necessary.

We must account for the parameters with identically zero pullback, rather than discard them. No irreducible component $C$ of $\Theta$ can consist entirely of such parameters. Otherwise parity would imply $C-u(p)\subset \Theta$ for every $p\in X$. The image of $C\times X$ under subtraction is irreducible, is contained in $\Theta$, and contains $C$ when $p=p_0$. Since $C$ is an irreducible component of the hypersurface, that image must be $C$. Hence $C-u(p)=C$ for all $p$. The surjectivity of the large-degree Abelian sum map, proved in Solution 3, says that these translations generate all of $J$. This would make the nonempty proper subset $C$ invariant under every translation, which is impossible.

A generic point $z$ of each component of $\Theta$ therefore has nonzero pullback. Since $\theta (-z)=\theta (z)=0$, its zero divisor contains $p_0$. Write $D_z=p_0+D^{\prime }$, with $D^{\prime }$ effective of degree $g-1$. The pullback identity gives $u_{g-1}(D^{\prime })=z+\kappa$. Each component of $\Theta$ is consequently contained in $W_{g-1}-\kappa$. By Solution 3 the latter is irreducible of dimension $g-1$, so any such component equals it. The zero locus is nonempty for $g>0$: any parameter with nonzero pullback has $g$ pullback zeros. We have proved equality of supports, in both directions.

Finally the automorphy factors make the First Chern class of the theta line bundle primitive. On the two-torus spanned by $a_i,b_j$, the logarithm of the $b_j$ factor changes by $-2\pi i{\delta }_{ij}$ after the $a_i$ translation; equivalently the transition for local frames has opposite winding. Thus $c_1(a_i,b_j)={\delta }_{ij}$, while $c_1(a_i,a_j)=c_1(b_i,b_j)=0$. In particular some integral pairing equals $1$. As its support is the single irreducible hypersurface $W_{g-1}-\kappa$, the zero divisor is $m(W_{g-1}-\kappa )$ for a positive integer $m$. Its integral First Chern class would then be divisible by $m$. Primitivity forces $m=1$, proving the displayed equality as reduced divisors.

The sign convention is $\kappa =u_g(D_z)-z$, so the translate is $W_{g-1}-\kappa$; a convention naming $-\kappa$ the Riemann constant reverses that displayed sign. For $g=1$, $\kappa =(1+B)/2$ modulo periods and the zero locus is the corresponding single point. For $g=0$, the empty-lattice theta series is $1$, its zero locus is empty, and the degree-$-1$ effective locus is empty; no negative symmetric product is required.

Choose an oriented basis ${\omega }_1,{\omega }_2$ of the period lattice, so $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. The Weierstrass elliptic function is constructed by the normally convergent seriesOn compact sets away from the lattice the summands are $O(|\omega {|}^{-3})$, which is summable over a plane lattice. It is even, has double poles at the lattice with zero residues, and is periodic. To check periodicity without splitting divergent series, differentiate termwise: ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$ is invariant under lattice shifts. Hence $\wp (z+{\omega }_j)-\wp (z)$ is constant; evenness evaluated at $z=-{\omega }_j/2$ makes this constant zero.

The Weierstrass sigma function is the canonical productIts logarithmic tails are likewise $O(|\omega {|}^{-3})$, so it defines an entire function with simple zeros exactly at the lattice and ${\sigma }^{\prime }(0)=1$. Pairing opposite lattice factors makes it odd. Its logarithmic derivative is the Weierstrass zeta functionThus ${\eta }_j=\zeta (z+{\omega }_j)-\zeta (z)$ is constant. Integrating this identity and evaluating the sigma ratio at $z=-{\omega }_j/2$ givesIntegration of zeta around a fundamental cell, containing one pole of residue one, gives the Legendre relation for Weierstrass quasi-periods ${\eta }_1{\omega }_2-{\eta }_2{\omega }_1=2\pi i$. These formulas explain why sigma itself is not an elliptic function, but balanced products of translates can be.

For completeness, the basic algebraic property of wp is $({\wp }^{\prime }{)}^2=4{\wp }^3-g_2\wp -g_3$, where $g_2=60{\sum }_{\omega \ne 0}{\omega }^{-4}$ and $g_3=140{\sum }_{\omega \ne 0}{\omega }^{-6}$. The identity follows by canceling the principal parts using the Laurent expansion: the difference is an entire elliptic function, hence constant, and its constant term is zero. The discriminant is nonzero; equivalently the three half-period values are distinct. These properties lead to the cubic embedding in 2(ii).

Now let $F$ be a nonzero meromorphic function on the torus and lift it to a periodic meromorphic function on the plane. Its logarithmic derivative $g=F^{\prime }/F$ is periodic. Choose a fundamental parallelogram with no zero or pole on its boundary. The residue theorem applied to $g$ gives equality of the total zero and pole multiplicities. Applied to $zg$, it gives the difference of the sums of the zero and pole lifts. Pairing opposite edges explicitly yieldsEach integral of $F^{\prime }/F$ on the right is $2\pi i$ times an integer, since its endpoint values of $F$ coincide and the image path avoids zero. Therefore the zero sum and pole sum agree in $\mathbb{C}/\Lambda$.

Conversely, suppose the point sums agree modulo the lattice. Change one chosen lift of a point by a lattice element so the two sums agree exactly in $\mathbb{C}$; this changes no point of the torus. Thenhas the prescribed divisor. Its multiplier under either basis period is $\exp ({\eta }_j(\sum Q_i-\sum P_i))=1$, so it descends to the torus. Repetitions contribute their appropriate multiplicities. This proves the principality criterion on a complex elliptic curve:There is an implicit qualification in the wording about separate zero and pole divisors: their supports must be disjoint. If a point occurs on both lists, cancel common multiplicities first when prescribing the divisor difference. A function cannot simultaneously have a zero and a pole at the same point; for example identical one-point lists satisfy the sum condition but do not specify separate zero and pole divisors. For $n=0$, any nonzero constant supplies the empty divisor.

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe summands on a compact set away from the lattice are $O(|\omega {|}^{-3})$, and that lattice sum converges. Thus the series and its termwise derivative converge normally. The function is even, and ${\wp }^{\prime }=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$ is odd and periodic by reindexing its absolutely convergent series. Hence $\wp (z+{\omega }_j)-\wp (z)$ is constant; evaluation at $-{\omega }_j/2$ and evenness make this constant zero for each basis period. The only poles of wp are double lattice poles, with principal part $(z-\omega {)}^{-2}$.

Let $G_r={\sum }_{\omega \ne 0}{\omega }^{-r}$ for the convergent weights used here. Expanding around zero givesThereforePut $g_2=60G_4$ and $g_3=140G_6$. The elliptic function $({\wp }^{\prime }{)}^2-4{\wp }^3+g_2\wp +g_3$ has a removable singularity at zero and hence at every lattice point. It is entire and periodic, thus bounded on the plane and constant by the Liouville theorem. Its zero constant term makes that constant zero. This proves the Weierstrass elliptic differential equationThe three classes ${\omega }_i/2$, with ${\omega }_3=-{\omega }_1-{\omega }_2$, are the three distinct nonzero two-torsion classes. Oddness and periodicity give ${\wp }^{\prime }({\omega }_i/2)=-{\wp }^{\prime }({\omega }_i/2)$, so they are zeros of the derivative. Their half-period values of the Weierstrass elliptic function $e_i$ are therefore roots of $4X^3-g_{2}X-g_3$.

They are distinct. If $e_i=e_j=e$ with $i\ne j$, the elliptic function $\wp -e$ would have a zero of multiplicity at least two at each of those distinct point classes, since its derivative vanishes there. It has only one pole on the torus, of order two. The argument principle on a fundamental cell equates its total zero and pole orders, contradicting at least four zeros. Thus $e_1,e_2,e_3$ are distinct, and the cubic factors asFor the monic cubic $X^3-(g_2/4)X-g_3/4$, the polynomial discriminant is $(g_2^3-27g_3^2)/16={\prod }_{i<j}(e_i-e_j{)}^2$. Hence the nonvanishing discriminant of a complex lattice isHere $\Delta (\Lambda )$ is the lattice discriminant, not the normalized q-series of Question 1. For $\Lambda =\mathbb{Z}+\mathbb{Z}\tau$, the normalizations give $g_2=(4{\pi }^4/3)E_4$ and $g_3=(8{\pi }^6/27)E_6$, so $\Delta (\Lambda )=(2\pi {)}^{12}\Delta (\tau )$.

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.