Convergence, zeros and quasi-periods of the Weierstrass zeta and sigma functions

(Uniform gap and finiteness in discs.) Put $A:=\mathrm{\mid }{\omega }_1{\mathrm{\mid }}^2$, $B:=\mathrm{Re}({\omega }_1\overline{{\omega }_2})$, $C:=\mathrm{\mid }{\omega }_2{\mathrm{\mid }}^2$ and $M:=\max (A,C)$. Expanding with [F2] gives $\mathrm{\mid }s{\omega }_1+t{\omega }_2{\mathrm{\mid }}^2=A{s}^2+2Bst+C{t}^2$ for real $s,t$, and $AC-B^2=(\mathrm{Im}({\omega }_1\overline{{\omega }_2}){)}^2>0$ because real-linear independence forbids $\mathrm{Im}({\omega }_1\overline{{\omega }_2})=0$. If $\mathrm{\mid }s\mathrm{\mid }\ge \mathrm{\mid }t\mathrm{\mid }$, write $x:=\mathrm{\mid }s\mathrm{\mid }$ and $\tau :=\mathrm{\mid }t\mathrm{\mid }\le x$: then $A{s}^2-2\mathrm{\mid }B\mathrm{\mid }\mathrm{\mid }s\mathrm{\mid }\mathrm{\mid }t\mathrm{\mid }+C{t}^2$ is a convex quadratic in $\tau$ whose values at $\tau =0$, at its vertex $\tau =\mathrm{\mid }B\mathrm{\mid }x\mathrm{/}C$ and at $\tau =x$ are $A{x}^2$, $x^2(AC-B^2)\mathrm{/}C$ and $(A-2\mathrm{\mid }B\mathrm{\mid }+C)x^2$, all at least $x^2(AC-B^2)\mathrm{/}C\ge x^2(AC-B^2)\mathrm{/}M$; the case $\mathrm{\mid }t\mathrm{\mid }\ge \mathrm{\mid }s\mathrm{\mid }$ is symmetric with $A$ and $C$ interchanged. Hence $\mathrm{\mid }s{\omega }_1+t{\omega }_2\mathrm{\mid }\ge \delta \max (\mathrm{\mid }s\mathrm{\mid },\mathrm{\mid }t\mathrm{\mid })$ with $\delta :=\sqrt{(AC-B^2)\mathrm{/}M}>0$: every nonzero lattice point has modulus at least $\delta$, while a lattice point $\lambda =m{\omega }_1+n{\omega }_2$ with $\mathrm{\mid }\lambda \mathrm{\mid }\le R$ has $\mathrm{\mid }m\mathrm{\mid },\mathrm{\mid }n\mathrm{\mid }\le R\mathrm{/}\delta$ by [F1] and the displayed inequality. Consequently $\mathrm{\Lambda }\cap \bar{D}(0,R)$ has at most $(2R\mathrm{/}\delta +1{)}^2$ elements, so every bounded set meets $\mathrm{\Lambda }$ in finitely many points; in particular $F_n:=\{\omega \in \mathrm{\Lambda }\setminus \{0\}:\mathrm{\mid }\omega \mathrm{\mid }\le n\}$ is finite for every $n$.

(The tail of $\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ and the pointwise bounds.) For $\mathrm{\mid }z\mathrm{\mid }\le R$ and $\mathrm{\mid }\omega \mathrm{\mid }\ge 2R$ one has $\mathrm{\mid }z-\omega \mathrm{\mid }\ge \mathrm{\mid }\omega \mathrm{\mid }-\mathrm{\mid }z\mathrm{\mid }\ge \frac{1}{2}\mathrm{\mid }\omega \mathrm{\mid }$, so $\mathrm{\mid }{g}_{\omega }(z)\mathrm{\mid }=\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}(\mathrm{\mid }z-\omega \mathrm{\mid }\mathrm{\mid }\omega {\mathrm{\mid }}^2)\le 2R^2\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ and, by [F5] applied with $p=2$ and $\mathrm{\mid }z\mathrm{/}\omega \mathrm{\mid }\le \frac{1}{2}\le 1$, $\mathrm{\mid }{F}_{\omega }(z)-1\mathrm{\mid }=\mathrm{\mid }1-E_2(z\mathrm{/}\omega )\mathrm{\mid }\le \mathrm{\mid }z\mathrm{/}\omega {\mathrm{\mid }}^3\le R^3\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$. By step 1.1, the nonzero lattice points with $\mathrm{\mid }\omega \mathrm{\mid }<1$ are finite, and for $k\ge 0$ the shell $2^k\le \mathrm{\mid }\omega \mathrm{\mid }<2^{k+1}$ has at most $(2^{k+2}\mathrm{/}\delta +1{)}^2$ points. Hence ${\sum }_{\omega \mathrm{\ne }0}\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ is bounded by the finite contribution from $\mathrm{\mid }\omega \mathrm{\mid }<1$ plus the convergent series ${\sum }_{k\ge 0}(2^{k+2}\mathrm{/}\delta +1{)}^2{2}^{-3k}$, so it converges and its finite-subset tails are arbitrarily small.

(The region $\mathbb{C}\setminus (\mathrm{\Lambda }\setminus \{0\})$ is a domain.) Every point of $\mathrm{\Lambda }$ is isolated in $\mathrm{\Lambda }$ by the gap $\delta$ of step 1.1, so $\mathbb{C}\setminus \mathrm{\Lambda }$ is open; and $\mathbb{C}\setminus (\mathrm{\Lambda }\setminus \{0\})$ contains the ball $B(0,\delta \mathrm{/}2)$ by the gap, hence is open as well. For path-connectedness, let $x,y\in \mathbb{C}\setminus (\mathrm{\Lambda }\setminus \{0\})$. By step 1.1 the segment $[x,y]$ meets $\mathrm{\Lambda }\setminus \{0\}$ in finitely many points $p_1,\dots ,p_N$. If $N=0$, the segment is already a path in the region. Otherwise choose $r>0$ smaller than $\delta \mathrm{/}3$ and than every distance from a $p_j$ to either endpoint. The discs $B(p_j,r)$ are disjoint, contain no other lattice point, and neither endpoint lies in them; they also exclude $0$ because each nonzero lattice point is at least $\delta$ from $0$. Replace the subsegment through each $p_j$ by one of the two arcs on $\mathrm{\partial }B(p_j,r)$ joining its endpoints. These arcs and the remaining straight pieces avoid $\mathrm{\Lambda }\setminus \{0\}$, so they form a path in the region from $x$ to $y$.

(Normal convergence away from the lattice.) Let $K\subseteq \mathbb{C}\setminus \mathrm{\Lambda }$ be compact and choose $R\ge {\sup }_{z\in K}\mathrm{\mid }z\mathrm{\mid }$. The finite set $F_0:=\{\omega \mathrm{\ne }0:\mathrm{\mid }\omega \mathrm{\mid }<2R\}$ contains every possible pole of a summand in $\bar{D}(0,R)$. For finite $F^{\mathrm{\prime }\mathrm{\prime }}\supseteq F^{\mathrm{\prime }}\supseteq F_0$, step 2.1 gives ${\sup }_K\mathrm{\mid }{\sum }_{\omega \in F^{\mathrm{\prime }\mathrm{\prime }}\setminus F^{\mathrm{\prime }}}{g}_{\omega }\mathrm{\mid }\le 2R^2{\sum }_{\omega \in F^{\mathrm{\prime }\mathrm{\prime }}\setminus F^{\mathrm{\prime }}}\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}\to 0$ as $F^{\mathrm{\prime }}$ grows. Thus the finite-subset net converges uniformly on $K$, independently of its exhaustion, to $E:={\sum }_{\omega \mathrm{\ne }0}{g}_{\omega }$; its cofinal partial sums $S_n$ are holomorphic on $\mathbb{C}\setminus \mathrm{\Lambda }$, so [F12] makes $E$ holomorphic there. For each fixed $\lambda \in \mathrm{\Lambda }$, omit the term $g_{\lambda }$ if $\lambda \mathrm{\ne }0$; every remaining summand is holomorphic on $B(\lambda ,\delta \mathrm{/}2)$ by the gap of step 1.1, and the same tail estimate gives locally uniform convergence on that ball. In particular $E$ is holomorphic near $0$, while near $\lambda \mathrm{\ne }0$ only $g_{\lambda }$ has a pole.

(The factors $F_{\omega }$: holomorphy, zeros, and the product lower bound.) By [F3] and [F5] each $F_{\omega }$ is holomorphic on $\mathbb{C}$ with $F_{\omega }(z)=0$ exactly when $z=\omega$, the zero at $\omega$ being simple because $F_{\omega }(z)=(1-z\mathrm{/}\omega )e^{z\mathrm{/}\omega +z^2\mathrm{/}2{\omega }^2}$ has a simple factor $1-z\mathrm{/}\omega$ and a nonvanishing exponential factor; in particular $F_{\omega }(0)=1$. Fix $z_0\in \mathbb{C}\setminus \mathrm{\Lambda }$ and put $F_0:=\{\omega \mathrm{\ne }0:\mathrm{\mid }\omega \mathrm{\mid }<2\mathrm{\mid }{z}_0\mathrm{\mid }\}$, a finite set by step 1.1; every factor $F_{\omega }(z_0)$ with $\omega \in F_0$ is nonzero because $z_0\mathrm{\ne }\omega$, so $c_0:={\prod }_{\omega \in F_0}\min (1,\mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid })>0$. For every finite $F$ one has $\mathrm{\mid }{\prod }_{\omega \in F}{F}_{\omega }(z_0)\mathrm{\mid }\ge c_0\exp (-2\mathrm{\mid }{z}_0{\mathrm{\mid }}^{3}T)>0$ where $T<\mathrm{\infty }$ is the total supremum of the finite sums ${\sum }_{\omega \in G}\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ from step 2.1: indeed ${\prod }_{\omega \in F\cap F_0}\mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid }\ge c_0$ because $\mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid }\ge \min (1,\mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid })$ for every $\omega$, while for $\omega \mathrm{\notin }{F}_0$ one has $\mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid }\ge 1-\mathrm{\mid }{F}_{\omega }(z_0)-1\mathrm{\mid }\ge 1-\mathrm{\mid }{z}_0{\mathrm{\mid }}^3\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}\ge \frac{1}{2}$ and hence $\log \mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid }\ge -2\mathrm{\mid }{z}_0{\mathrm{\mid }}^3\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ by $\log (1-t)\ge -2t$ on $[0,\frac{1}{2}]$, so ${\prod }_{\omega \in F\setminus F_0}\mathrm{\mid }{F}_{\omega }(z_0)\mathrm{\mid }\ge \exp (-2\mathrm{\mid }{z}_0{\mathrm{\mid }}^3{\sum }_{\omega \in F\setminus F_0}\mathrm{\mid }\omega {\mathrm{\mid }}^{-3})\ge \exp (-2\mathrm{\mid }{z}_0{\mathrm{\mid }}^{3}T)$.

(Principal parts and residues of $\zeta$.) Let $\lambda \in \mathrm{\Lambda }$. For $\lambda =0$, $\zeta (z)-z^{-1}=E(z)$ is holomorphic near $0$ by step 3.1. For $\lambda \mathrm{\ne }0$, split off $g_{\lambda }$: $\zeta (z)-\frac{1}{z-\lambda }=\frac{1}{z}+\frac{1}{\lambda }+\frac{z}{{\lambda }^2}+{\sum }_{\omega \mathrm{\ne }0,\lambda }{g}_{\omega }(z)\mathrm{.}$ The right side is holomorphic near $\lambda$ by step 3.1. Thus every lattice point is a simple pole of residue $1$; elsewhere $\zeta =1\mathrm{/}z+E$ is holomorphic. The lattice is discrete by step 1.1, so $\zeta$ is meromorphic on $\mathbb{C}$ with precisely these poles.

(${\zeta }^{\mathrm{\prime }}=-\mathrm{\wp }$.) The cofinal partial sums $S_n$ converge locally uniformly to $E$ on $\mathbb{C}\setminus \mathrm{\Lambda }$ by step 3.1 and are holomorphic there. Thus [F12] gives $E^{\mathrm{\prime }}={\lim }_n{S}_n^{\mathrm{\prime }}$ on that domain. Since $g_{\omega }^{\mathrm{\prime }}=-h_{\omega }$ by [F6] and the $h_{\omega }$-net converges normally there by [F4], $E^{\mathrm{\prime }}(z)=-{\sum }_{\omega \mathrm{\ne }0}{h}_{\omega }(z)=z^{-2}-\mathrm{\wp }(z)$. Differentiating $\zeta =1\mathrm{/}z+E$ yields ${\zeta }^{\mathrm{\prime }}=-\mathrm{\wp }$ on $\mathbb{C}\setminus \mathrm{\Lambda }$.

(Oddness of $\zeta$.) For $z\in \mathbb{C}\setminus \mathrm{\Lambda }$ one has $\zeta (-z)=-1\mathrm{/}z+{\sum }_{\omega \mathrm{\ne }0}{g}_{\omega }(-z)$ in the net sense of [F3], and $g_{\omega }(-z)=z^2\mathrm{/}({\omega }^2(-z-\omega ))=-z^2\mathrm{/}((-\omega {)}^2(z-(-\omega )))=-g_{-\omega }(z)$ by [F2]; since $\omega \mapsto -\omega$ is a bijection of $\mathrm{\Lambda }\setminus \{0\}$ preserving inclusion of finite sets, the reindexed net converges to $-E(z)$ by step 3.1. Therefore $\zeta (-z)=-1\mathrm{/}z-E(z)=-\zeta (z)$ for all $z\in \mathbb{C}\setminus \mathrm{\Lambda }$, which extends to $\mathbb{C}$ with poles matched.

(The product net converges: $\sigma$ is entire with $\sigma (0)=0$ and ${\sigma }^{\mathrm{\prime }}(0)=1$.) Fix $R>0$ and put $F_0:=\{\omega \mathrm{\ne }0:\mathrm{\mid }\omega \mathrm{\mid }<2R\}$; writing $u_{\omega }:=F_{\omega }-1$, step 2.1 gives ${\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }{u}_{\omega }\mathrm{\mid }\le R^3\mathrm{\mid }\omega {\mathrm{\mid }}^{-3}$ for $\omega \mathrm{\notin }{F}_0$, so the finite sums ${\sum }_{\omega \in G}{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }{u}_{\omega }\mathrm{\mid }$ are bounded over all finite $G$ and shrink to $0$ outside large finite sets by step 2.1; put $M_R$ for their supremum. For finite $F_0\subseteq F_1\subseteq F_2$ and $\mathrm{\mid }z\mathrm{\mid }\le R$ one has $\mid {\prod }_{\omega \in F_2}{F}_{\omega }(z)-{\prod }_{\omega \in F_1}{F}_{\omega }(z)\mid \le e^{M_R}(\exp ({\sum }_{\omega \in F_2\setminus F_1}{\sup }_R\mathrm{\mid }{u}_{\omega }\mathrm{\mid })-1)$ because $\mathrm{\mid }{\prod }_{\omega \in E}(1+u_{\omega }(z))-1\mathrm{\mid }\le \exp ({\sum }_{\omega \in E}\mathrm{\mid }{u}_{\omega }(z)\mathrm{\mid })-1$ for finite $E$ and $\mathrm{\mid }{\prod }_{\omega \in F_1}{F}_{\omega }(z)\mathrm{\mid }\le \exp ({\sum }_{\omega \in F_1}{\sup }_R\mathrm{\mid }{u}_{\omega }\mathrm{\mid })\le e^{M_R}$; the right-hand side is arbitrarily small for $F_1$ large. Thus the finite-subset net of the products, indexed by the directed set of finite subsets of $\mathrm{\Lambda }\setminus \{0\}$, is uniformly Cauchy on every closed disc; along the cofinal sequence $F_n$ of step 1.1 the partial products $Q_n={\prod }_{\omega \in F_n}{F}_{\omega }$ are holomorphic and converge uniformly on each closed disc to a limit $P$, which is holomorphic on $\mathbb{C}$ by [F12], and the net limit equals $P$ by cofinality and is independent of the exhaustion. Moreover $P(0)=1$, because every finite product equals $1$ at $0$ by $F_{\omega }(0)=1$ of step 3.2. Hence $\sigma (z)=zP(z)$ is entire with $\sigma (0)=0$, and the product rule of [F6] gives ${\sigma }^{\mathrm{\prime }}(0)=P(0)+0\cdot P^{\mathrm{\prime }}(0)=1$.

(Nonvanishing off $\mathrm{\Lambda }$ and the simple zeros at $\mathrm{\Lambda }\setminus \{0\}$.) For $z_0\in \mathbb{C}\setminus \mathrm{\Lambda }$, step 3.2 shows that all finite products ${\prod }_{\omega \in F}{F}_{\omega }(z_0)$ have modulus at least $c_0\exp (-2\mathrm{\mid }{z}_0{\mathrm{\mid }}^{3}T)>0$, and step 4.4 makes $P(z_0)$ the limit of that net of numbers, so $\mathrm{\mid }P(z_0)\mathrm{\mid }\ge c_0\exp (-2\mathrm{\mid }{z}_0{\mathrm{\mid }}^{3}T)>0$; in particular $P$ has no zero on $\mathbb{C}\setminus \mathrm{\Lambda }$. Fix ${\omega }_0\in \mathrm{\Lambda }\setminus \{0\}$ and let $P^{{\omega }_0}$ be the limit of the net of finite products over finite subsets of $\mathrm{\Lambda }\setminus \{0,{\omega }_0\}$: by the estimates of step 4.4 that sub-net converges uniformly on each closed disc as well, and ${\prod }_{\omega \in F}{F}_{\omega }=F_{{\omega }_0}(z){\prod }_{\omega \in F\setminus \{{\omega }_0\}}{F}_{\omega }(z)$ for every finite $F\ni {\omega }_0$ gives, passing to the limit, $P(z)=F_{{\omega }_0}(z)P^{{\omega }_0}(z)$ for all $z$. Applying step 3.2 to the family $\mathrm{\Lambda }\setminus \{0,{\omega }_0\}$ with $z_0={\omega }_0$ shows $P^{{\omega }_0}({\omega }_0)\mathrm{\ne }0$, and $F_{{\omega }_0}(z)=(1-z\mathrm{/}{\omega }_0)e^{z\mathrm{/}{\omega }_0+z^2\mathrm{/}2{\omega }_0^2}$ has the simple zero at ${\omega }_0$ by step 3.2 and [F5]; since the exponential and $P^{{\omega }_0}$ are nonzero at ${\omega }_0$, [F7] makes ${\omega }_0$ a simple zero of $P$, hence of $\sigma =zP$. Finally $0$ is a simple zero of $\sigma$ because $\sigma (z)=zP(z)$ with $P(0)=1\mathrm{\ne }0$, and there are no other zeros: outside $\mathrm{\Lambda }$ both $z$ and $P(z)$ are nonzero, and on $\mathrm{\Lambda }\setminus \{0\}$ the zeros just located are simple.

(Oddness of $\sigma$.) For every finite $F\subseteq \mathrm{\Lambda }\setminus \{0\}$ one has ${\prod }_{\omega \in F}{F}_{\omega }(-z)={\prod }_{\omega \in F}{E}_2(-z\mathrm{/}\omega )={\prod }_{\eta \in -F}{E}_2(z\mathrm{/}\eta )$ by [F2], and $F\mapsto -F$ is a bijection of the directed set of finite subsets preserving inclusion, so the two nets have the same limit by step 4.4: $P(-z)=P(z)$. Hence $\sigma (-z)=-zP(-z)=-zP(z)=-\sigma (z)$ for all $z\in \mathbb{C}$.

(Quasi-periodicity of $\zeta$.) Fix $j\in \{1,2\}$ and put $G_j(z):=\zeta (z+{\omega }_j)-\zeta (z)$ for $z\in \mathbb{C}\setminus \mathrm{\Lambda }$, a holomorphic function there. Near any $\lambda \in \mathrm{\Lambda }$, step 4.1 writes $\zeta (z)=1\mathrm{/}(z-\lambda )+{\psi }_{\lambda }(z)$ and $\zeta (z+{\omega }_j)=1\mathrm{/}(z-\lambda )+{\psi }_{\lambda +{\omega }_j}(z+{\omega }_j)$ with ${\psi }_{\lambda },{\psi }_{\lambda +{\omega }_j}$ holomorphic near $\lambda$ and $\lambda +{\omega }_j$ respectively, so $G_j$ extends holomorphically across $\lambda$; thus $G_j$ is entire. On $\mathbb{C}\setminus \mathrm{\Lambda }$ one has $G_j^{\mathrm{\prime }}(z)={\zeta }^{\mathrm{\prime }}(z+{\omega }_j)-{\zeta }^{\mathrm{\prime }}(z)=-\mathrm{\wp }(z+{\omega }_j)+\mathrm{\wp }(z)=0$ by step 4.2 and the periodicity of $\mathrm{\wp }$ in [F4]. The complement of $\mathrm{\Lambda }$ is dense in $\mathbb{C}$: for $k\ge 2$ and $\lambda \in \mathrm{\Lambda }$ the points $\lambda +{\omega }_1\mathrm{/}k$ lie outside $\mathrm{\Lambda }$, since $\lambda +{\omega }_1\mathrm{/}k\in \mathrm{\Lambda }$ would give ${\omega }_1\mathrm{/}k\in \mathrm{\Lambda }$, say ${\omega }_1=k(m{\omega }_1+n{\omega }_2)$ with integers $m,n$, whence $1=km$ and $0=kn$ by the uniqueness of the real coordinates in [F1], contradicting $k\ge 2$. Hence the entire function $G_j^{\mathrm{\prime }}$, which vanishes on the dense set $\mathbb{C}\setminus \mathrm{\Lambda }$, vanishes identically by continuity; so $G_j$ is constant on the domain $\mathbb{C}$ by [F6]. Evaluating at $-{\omega }_j\mathrm{/}2$, a point of $\mathbb{C}\setminus \mathrm{\Lambda }$ because ${\omega }_j$ is a primitive basis vector, and using the oddness of step 4.3 gives $G_j(-{\omega }_j\mathrm{/}2)=\zeta ({\omega }_j\mathrm{/}2)-\zeta (-{\omega }_j\mathrm{/}2)=2\zeta ({\omega }_j\mathrm{/}2)={\eta }_j$. Therefore $\zeta (z+{\omega }_j)=\zeta (z)+{\eta }_j$ for all $z\in \mathbb{C}\setminus \mathrm{\Lambda }$, with poles matched.

(A translated parallelogram with one pole.) Put $\tau :={\omega }_2\mathrm{/}{\omega }_1$, which has $\mathrm{Im}\tau >0$ by [F1], and define the closed parallelogram $T:=\{s+t\tau :0\le s,t\le 1\}$ of the boundary lemma with $\alpha (y):=(\mathrm{Re}\tau \mathrm{/}\mathrm{Im}\tau )y$ and $\beta (y):=\alpha (y)+1$ on $[0,\mathrm{Im}\tau ]$; define also $a:=-\frac{1}{2}({\omega }_1+{\omega }_2)$, the orientation-preserving similarity $\sigma (w):=a+{\omega }_{1}w$, and $P:=\sigma (T)=\{a+s{\omega }_1+t{\omega }_2:0\le s,t\le 1\}$ with interior $P^{\circ }$. The boundary contour ${\gamma }_T$ of $T$ is the concatenation of the bottom segment, the graph of $\beta$ with $y$ increasing, the top segment and the graph of $\alpha$ with $y$ decreasing; its image under $\sigma$ traverses the four sides of $P$ as the paths ${\gamma }_1(t):=a+t{\omega }_1$, ${\gamma }_2(t):=a+{\omega }_1+t{\omega }_2$, ${\gamma }_3(t):=a+{\omega }_1+{\omega }_2-t{\omega }_1$, ${\gamma }_4(t):=a+{\omega }_2-t{\omega }_2$ for $t\in [0,1]$, up to strictly increasing reparametrizations of the two graph pieces. By [F10] applied to $\sigma$, the index of the contour $\mathrm{\Gamma }:=\sigma \circ {\gamma }_T={\gamma }_1\ast {\gamma }_2\ast {\gamma }_3\ast {\gamma }_4$ (with the reparametrizations of [F11]) is $n(\mathrm{\Gamma },q)=1$ for $q\in P^{\circ }$ and $0$ for $q\mathrm{\notin }P$. A lattice point $\lambda =m{\omega }_1+n{\omega }_2$ lies in $P$ iff $s=m+\frac{1}{2}\in [0,1]$ and $t=n+\frac{1}{2}\in [0,1]$, that is iff $m=n=0$; hence $0\in P^{\circ }$ is the only lattice point in $P$ and every other lattice point lies outside $P$, with $n(\mathrm{\Gamma },\lambda )=0$ for $\lambda \mathrm{\ne }0$. Consequently ${\mathrm{\Gamma }}^{\ast }$ avoids $\mathrm{\Lambda }$; since $\zeta$ is holomorphic on $\mathrm{\Omega }\setminus \{0\}=\mathbb{C}\setminus \mathrm{\Lambda }$ and $0$ is a pole of $\zeta$ by step 4.1, $\zeta$ is meromorphic on the domain $\mathrm{\Omega }:=\mathbb{C}\setminus (\mathrm{\Lambda }\setminus \{0\})$ of step 2.2 with pole set $\{0\}$, and $\mathrm{\Gamma }$ is admissible for the residue theorem in $\mathrm{\Omega }$ because $n(\mathrm{\Gamma },p)=0$ for every $p\mathrm{\notin }\mathrm{\Omega }$ by the preceding index computation.

($P^{\mathrm{\prime }}=PE$ and ${\sigma }^{\mathrm{\prime }}\mathrm{/}\sigma =\zeta$.) For $z\in \mathbb{C}\setminus \mathrm{\Lambda }$ the finite product rule of [F6] and the identity $F_{\omega }^{\mathrm{\prime }}(z)\mathrm{/}{F}_{\omega }(z)=g_{\omega }(z)$, obtained by differentiating the displayed factor with [F5] and [F6], give $Q_n^{\mathrm{\prime }}(z)=Q_n(z)S_n(z)$ with $S_n={\sum }_{\omega \in F_n}{g}_{\omega }$, and $S_n\to E$ uniformly on compact subsets of $\mathbb{C}\setminus \mathrm{\Lambda }$ by step 3.1; on a compact $K\subseteq \mathbb{C}\setminus \mathrm{\Lambda }$ the factors $Q_n$ are uniformly bounded by $e^{M_K}$ as in step 4.4, so $Q_n^{\mathrm{\prime }}\to PE$ uniformly on $K$. The convergence $Q_n\to P$ is locally uniform on the open set $\mathbb{C}\setminus \mathrm{\Lambda }$, so [F12] gives $P^{\mathrm{\prime }}={\lim }_n{Q}_n^{\mathrm{\prime }}=PE$ there. Since $P$ is zero-free on $\mathbb{C}\setminus \mathrm{\Lambda }$ by step 5.1, the product and reciprocal rules of [F6] applied to $\sigma =zP$ give $\frac{{\sigma }^{\mathrm{\prime }}(z)}{\sigma (z)}=\frac{1}{z}+\frac{P^{\mathrm{\prime }}(z)}{P(z)}=\frac{1}{z}+E(z)=\zeta (z)(z\in \mathbb{C}\setminus \mathrm{\Lambda })\mathrm{.}$

(The four sides and the Legendre relation.) With $\mathrm{\Omega }$ as in step 5.4, the residue theorem [F9] applied to $\zeta$, whose only pole in $\mathrm{\Omega }$ is $0$ with $n(\mathrm{\Gamma },0)=1$ and residue $1$ by step 4.1, gives ${\int }_{\mathrm{\Gamma }}\zeta (z)dz=2\pi i$. On the other hand the parametric formula and additivity of [F11] give ${\int }_{\mathrm{\Gamma }}\zeta ={\sum }_{k=1}^4{\int }_{{\gamma }_k}\zeta$, and ${\gamma }_3$ is the reversal of the translated path ${\delta }_1(t):={\gamma }_1(t)+{\omega }_2$ while ${\gamma }_4$ is the reversal of ${\delta }_2(t):={\gamma }_2(t)-{\omega }_1$. By step 5.3, $\zeta ({\delta }_1(t))=\zeta ({\gamma }_1(t))+{\eta }_2$ and, using $\zeta (w-{\omega }_1)=\zeta (w)-{\eta }_1$ for $w={\gamma }_2(t)\in \mathbb{C}\setminus \mathrm{\Lambda }$, also $\zeta ({\delta }_2(t))=\zeta ({\gamma }_2(t))-{\eta }_1$; the parametric formula applied to each translated path (with derivative ${\omega }_1$, respectively ${\omega }_2$) yields ${\int }_{{\delta }_1}\zeta ={\int }_{{\gamma }_1}\zeta +{\eta }_2{\omega }_1$ and ${\int }_{{\delta }_2}\zeta ={\int }_{{\gamma }_2}\zeta -{\eta }_1{\omega }_2$. Since reversal negates integrals, ${\int }_{\mathrm{\Gamma }}\zeta ={\int }_{{\gamma }_1}\zeta +{\int }_{{\gamma }_2}\zeta -({\int }_{{\gamma }_1}\zeta +{\eta }_2{\omega }_1)-({\int }_{{\gamma }_2}\zeta -{\eta }_1{\omega }_2)={\eta }_1{\omega }_2-{\eta }_2{\omega }_1\mathrm{.}$ Comparing with ${\int }_{\mathrm{\Gamma }}\zeta =2\pi i$ gives ${\eta }_1{\omega }_2-{\eta }_2{\omega }_1=2\pi i$.

(Quasi-periodicity of $\sigma$.) Fix $j$ and put $H_j(z):=\sigma (z+{\omega }_j)\mathrm{/}\sigma (z)$, meromorphic on $\mathbb{C}$; by step 5.1 the only zeros of $\sigma$ are the lattice points, all simple, and $\sigma (z)=0$ exactly when $z\in \mathrm{\Lambda }=\mathrm{\Lambda }-{\omega }_j$, so $\sigma (z+{\omega }_j)$ has exactly the same simple zeros. At $\lambda \in \mathrm{\Lambda }$ write $\sigma (z)=(z-\lambda )u(z)$ and $\sigma (z+{\omega }_j)=(z-\lambda )v(z)$ with $u,v$ holomorphic and nonzero at $\lambda$ ([F7] applied to the simple zeros); then $H_j=v\mathrm{/}u$ is holomorphic near $\lambda$ with $H_j(\lambda )=v(\lambda )\mathrm{/}u(\lambda )\mathrm{\ne }0$. On $\mathbb{C}\setminus \mathrm{\Lambda }$ both $\sigma$ and $\sigma (\cdot +{\omega }_j)$ are zero-free, so $H_j$ is holomorphic and zero-free there as well; hence $H_j$ is entire and zero-free. The quotient rule of [F6] together with step 6.1 gives $\frac{H_j^{\mathrm{\prime }}(z)}{H_j(z)}=\frac{{\sigma }^{\mathrm{\prime }}(z+{\omega }_j)}{\sigma (z+{\omega }_j)}-\frac{{\sigma }^{\mathrm{\prime }}(z)}{\sigma (z)}=\zeta (z+{\omega }_j)-\zeta (z)={\eta }_j(z\in \mathbb{C}\setminus \mathrm{\Lambda }),$ an identity between entire functions, valid on the dense set $\mathbb{C}\setminus \mathrm{\Lambda }$ by step 5.3 and hence everywhere by continuity. It follows that $(H_j(z)e^{-{\eta }_{j}z}{)}^{\mathrm{\prime }}=e^{-{\eta }_{j}z}(H_j^{\mathrm{\prime }}(z)-{\eta }_j{H}_j(z))=0$ on $\mathbb{C}$ by [F5] and [F6], so $H_j(z)=C_j{e}^{{\eta }_{j}z}$ for a constant $C_j$ by [F6]. Evaluating at $-{\omega }_j\mathrm{/}2$ and using the oddness of $\sigma$ from step 5.2 and $\sigma ({\omega }_j\mathrm{/}2)\mathrm{\ne }0$: $-1=\frac{\sigma ({\omega }_j\mathrm{/}2)}{\sigma (-{\omega }_j\mathrm{/}2)}=H_j(-{\omega }_j\mathrm{/}2)=C_j{e}^{-{\eta }_j{\omega }_j\mathrm{/}2},\text{so}{C}_j=-e^{{\eta }_j{\omega }_j\mathrm{/}2},$ and therefore $\sigma (z+{\omega }_j)=-e^{{\eta }_j(z+{\omega }_j\mathrm{/}2)}\sigma (z)$ for all $z\in \mathbb{C}$; both sides vanish at lattice points.