Divisor and residue laws for elliptic functions

Put $\phi (s,t):=a+s{\omega }_1+t{\omega }_2$. By [F19], $\mathrm{\mid }\phi (s,t)-\phi (s^{\mathrm{\prime }},t^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\mid }{\omega }_1\mathrm{\mid }\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }{\omega }_2\mathrm{\mid }\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$, so $\phi$ is continuous; the box $[0,1{]}^2$ is compact by [F16] and $P=\phi ([0,1{]}^2)$ is its continuous image, hence $P$ is compact and nonempty by [F16].

Put $\tau :={\omega }_2\mathrm{/}{\omega }_1$, so $\mathrm{Im}\tau >0$ by [F1], and let $\alpha (y):=(\mathrm{Re}\tau \mathrm{/}\mathrm{Im}\tau )y$, $\beta (y):=\alpha (y)+1$ on $[0,\mathrm{Im}\tau ]$, so that $T:=\{x+iy:0\le y\le \mathrm{Im}\tau ,\alpha (y)\le x\le \beta (y)\}=\{s+t\tau :0\le s,t\le 1\}$ with $T^{\circ }=\{s+t\tau :0<s,t<1\}$; the affine functions $\alpha \le \beta$ are real-analytic with Puiseux-analytic graphs, and the orientation-preserving similarity $\sigma (w):=a+{\omega }_{1}w$ has $\sigma (T)=P$, $\sigma (T^{\circ })=P^{\circ }$. Define ${\gamma }_1(t):=a+t{\omega }_1$, ${\gamma }_2(t):=a+{\omega }_1+t{\omega }_2$, ${\gamma }_3(t):=a+{\omega }_1+{\omega }_2-t{\omega }_1$, ${\gamma }_4(t):=a+{\omega }_2-t{\omega }_2$ for $t\in [0,1]$ and $\mathrm{\Gamma }:={\gamma }_1\ast {\gamma }_2\ast {\gamma }_3\ast {\gamma }_4$. Since $\sigma (0+t)=a+t{\omega }_1$, $\sigma (1+t\tau )=a+{\omega }_1+t{\omega }_2$, $\sigma (1+\tau -t)=a+{\omega }_1+{\omega }_2-t{\omega }_1$ and $\sigma (\tau -t\tau )=a+{\omega }_2-t{\omega }_2$, the four paths of $\mathrm{\Gamma }$ are the four pieces of $\sigma \circ {\gamma }_T$, where ${\gamma }_T$ is the boundary contour of [F3], up to the strictly increasing reparametrizations $y=t\mathrm{Im}\tau$ of the two graph pieces; by [F4] and the concatenation additivity of [F10] the integrals defining the indices agree, so the index assertions of [F3] give $n(\mathrm{\Gamma },q)=1$ for $q\in P^{\circ }$ and $n(\mathrm{\Gamma },q)=0$ for $q\in \mathbb{C}\setminus P$. In particular ${\mathrm{\Gamma }}^{\ast }=\mathrm{\partial }P$ is a closed contour avoiding $\mathrm{Zer}(f)\cup \mathrm{Pol}(f)$, so every zero and every pole of $f$ lies in $P^{\circ }$ or outside $P$.

Since ${\mathrm{\Gamma }}^{\ast }=\mathrm{\partial }P$ avoids $\mathrm{Zer}(f)\cup \mathrm{Pol}(f)$ by the given hypothesis, and $\mathrm{\Gamma }$ is null-homologous in $\mathbb{C}$ (the condition $n(\mathrm{\Gamma },p)=0$ for $p\in \mathbb{C}\setminus \mathbb{C}$ is vacuous), [F8] makes $\mathrm{\Gamma }$ admissible for the residue theorem in $\mathrm{\Omega }=\mathbb{C}$ both for $f$, whose pole set is $\mathrm{Pol}(f)$, and for $f^{\mathrm{\prime }}\mathrm{/}f$, whose pole set is $\mathrm{Zer}(f)\cup \mathrm{Pol}(f)$.

Write points of $P^{\circ }$ as $a+s{\omega }_1+t{\omega }_2$ with $s,t\in (0,1)$. If $u,v\in P^{\circ }$ and $u-v\in \mathrm{\Lambda }$, then $u-v=(m{\omega }_1+n{\omega }_2)$ with $m,n\in \mathbb{Z}$ and also $u-v=(s-s^{\mathrm{\prime }}){\omega }_1+(t-t^{\mathrm{\prime }}){\omega }_2$ with $s-s^{\mathrm{\prime }},t-t^{\mathrm{\prime }}\in (-1,1)$; uniqueness of the real coordinates from [F1] gives $m=s-s^{\mathrm{\prime }}$ and $n=t-t^{\mathrm{\prime }}$, so $m=n=0$ and $u=v$. Conversely, write $z-a=s{\omega }_1+t{\omega }_2$ with $s,t\in \mathbb{R}$, [F20] provides integers $m\le s<m+1$ and $n\le t<n+1$, so $z-(m{\omega }_1+n{\omega }_2)=a+(s-m){\omega }_1+(t-n){\omega }_2\in \{a+s^{\mathrm{\prime }}{\omega }_1+t^{\mathrm{\prime }}{\omega }_2:0\le s^{\mathrm{\prime }},t^{\mathrm{\prime }}<1\}\subseteq P$; hence every $\mathrm{\Lambda }$-orbit meets $P$, and it meets $P^{\circ }$ in at most one point. For a zero $c$ of $f$, the representative $w\in P$ of its orbit is again a zero by [F2], and $w\mathrm{\notin }\mathrm{\partial }P$ by the hypothesis, so $w\in P^{\circ }$; moreover the order is preserved since $f(c+\lambda +u)=f(c+u)$ for all $u$, so the germ of $f$ at a translate is the translated germ. Therefore the map sending a zero class to its representative in $P^{\circ }$ is a bijection from the classes of zeros of $f$ onto $\mathrm{Zer}(f)\cap P^{\circ }$ preserving multiplicity, and the same argument with poles in place of zeros gives a multiplicity preserving bijection from the classes of poles onto $\mathrm{Pol}(f)\cap P^{\circ }$.

For $t\in [0,1]$ the paths ${\gamma }_1(t)=a+t{\omega }_1$, ${\gamma }_2(t)=a+{\omega }_1+t{\omega }_2$, ${\gamma }_3(t)=a+{\omega }_1+{\omega }_2-t{\omega }_1$ and ${\gamma }_4(t)=a+{\omega }_2-t{\omega }_2$ are $C^1$ with ${\gamma }_1^{\mathrm{\prime }}={\omega }_1$, ${\gamma }_2^{\mathrm{\prime }}={\omega }_2$, ${\gamma }_3^{\mathrm{\prime }}=-{\omega }_1$, ${\gamma }_4^{\mathrm{\prime }}=-{\omega }_2$. For any function $g$ continuous on the trace of $\mathrm{\Gamma }$ the parametric formula [F9] and the concatenation identity of [F10] give ${\int }_{\mathrm{\Gamma }}gdz={\sum }_{j=1}^4{\int }_0^{1}g({\gamma }_j(t)){\gamma }_j^{\mathrm{\prime }}(t)dt$; moreover ${\gamma }_3$ is the reversal of the translated path ${\delta }_1(t):={\gamma }_1(t)+{\omega }_2$ and ${\gamma }_4$ is the reversal of ${\delta }_2(t):={\gamma }_2(t)-{\omega }_1$, so ${\int }_{{\gamma }_3}gdz=-{\int }_{{\delta }_1}gdz$ and ${\int }_{{\gamma }_4}gdz=-{\int }_{{\delta }_2}gdz$ by [F10].

On the open set $U:=\mathbb{C}\setminus \mathrm{Pol}(f)$ the function $f$ is holomorphic, and $U+\lambda =U$ for every $\lambda \in \mathrm{\Lambda }$ by [F2]. Fix $\lambda \in \mathrm{\Lambda }$ and let $T_{\lambda }(z):=z+\lambda$; the chain rule [F18] applied to $f=g$ and $T_{\lambda }$ at $z\in U$ gives $(f\circ T_{\lambda }{)}^{\mathrm{\prime }}(z)=f^{\mathrm{\prime }}(z+\lambda )\cdot 1$, while $f\circ T_{\lambda }=f$ by [F2], so $f^{\mathrm{\prime }}(z+\lambda )=f^{\mathrm{\prime }}(z)$ for all $z\in U$. Consequently $g:=f^{\mathrm{\prime }}\mathrm{/}f$, which is holomorphic on $U\setminus \mathrm{Zer}(f)$, satisfies $g(z+\lambda )=g(z)$ at every point of its domain.

By [F11] applied to $f$, $\mathrm{Pol}(f)$ is closed and discrete in $\mathbb{C}$ and every point of $\mathbb{C}$ has a neighbourhood meeting $\mathrm{Pol}(f)$ in at most one point. Since $f$ is not identically zero, [F12] makes $1\mathrm{/}f$ meromorphic, and its pole set is exactly $\mathrm{Zer}(f)$, with a zero of $f$ of order $m$ becoming a pole of order $m$ of $1\mathrm{/}f$; applying [F11] to $1\mathrm{/}f$ shows that $\mathrm{Zer}(f)$ is closed and discrete as well, and every point of $\mathbb{C}$ has a neighbourhood meeting each of $\mathrm{Zer}(f)$, $\mathrm{Pol}(f)$ in at most one point. Because $P$ is compact by step 1.1, finitely many such neighbourhoods cover $P$, so the sets $\mathrm{Zer}(f)\cap P$ and $\mathrm{Pol}(f)\cap P$ are finite, and so are their subsets in $P^{\circ }$.

The hypotheses of [F5] hold with $\mathrm{\Omega }=\mathbb{C}$: $f$ is meromorphic and not identically zero on the connected component $\mathbb{C}$ because it is nonconstant, $\mathrm{\Gamma }$ is admissible by step 1.3, and $f$ has no zero on ${\mathrm{\Gamma }}^{\ast }$ by the hypothesis; hence $\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{f^{\mathrm{\prime }}}{f}dz=Z(f,\mathrm{\Gamma })-P(f,\mathrm{\Gamma })$, with only finitely many nonzero terms in the weighted counts. By [F6] these counts are the sums over $\mathrm{Zer}(f)$ and $\mathrm{Pol}(f)$ weighted by $n(\mathrm{\Gamma },\cdot )$ and by the orders; by step 1.2 the index is $1$ on $P^{\circ }$ and $0$ off $P$, and by the hypothesis there is no zero or pole on $\mathrm{\partial }P$, so $Z(f,\mathrm{\Gamma })=Z_P:={\sum }_{c\in \mathrm{Zer}(f)\cap P^{\circ }}{\mathrm{ord}}_c(f),P(f,\mathrm{\Gamma })=P_P:={\sum }_{p\in \mathrm{Pol}(f)\cap P^{\circ }}{\mathrm{ord}}_p^{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}(f),$ both finite sums.

The hypotheses of [F7] hold with $\mathrm{\Omega }=\mathbb{C}$: $f$ is meromorphic on $\mathbb{C}$ with pole set $\mathrm{Pol}(f)$ and $\mathrm{\Gamma }$ is admissible by step 1.3. Hence ${\int }_{\mathrm{\Gamma }}f(z)dz=2\pi i{\sum }_{p\in \mathrm{Pol}(f)}n(\mathrm{\Gamma },p)\mathrm{Res}(f,p)$, and by step 1.2 all poles on $\mathrm{\partial }P$ are absent and the index is $1$ exactly at the poles in $P^{\circ }$, so ${\int }_{\mathrm{\Gamma }}fdz=2\pi i{\sum }_{p\in \mathrm{Pol}(f)\cap P^{\circ }}\mathrm{Res}(f,p)$.

Applying step 1.5 to $g:=f$, which is continuous on ${\mathrm{\Gamma }}^{\ast }$ because ${\mathrm{\Gamma }}^{\ast }=\mathrm{\partial }P$ has no pole of $f$, and using the $\mathrm{\Lambda }$-periodicity $f(a+{\omega }_2+t{\omega }_1)=f(a+t{\omega }_1)$ and $f(a+{\omega }_1+t{\omega }_2)=f(a+t{\omega }_2)$ from [F2] gives ${\int }_{{\delta }_1}fdz={\int }_0^{1}f(a+{\omega }_2+t{\omega }_1){\omega }_{1}dt={\int }_0^{1}f(a+t{\omega }_1){\omega }_{1}dt={\int }_{{\gamma }_1}fdz$, and, since ${\delta }_2(t)=a+t{\omega }_2$, ${\int }_{{\delta }_2}fdz={\int }_0^{1}f(a+t{\omega }_2){\omega }_{2}dt={\int }_0^{1}f(a+{\omega }_1+t{\omega }_2){\omega }_{2}dt={\int }_{{\gamma }_2}fdz$; with the reversal identities of step 1.5 the integrals over ${\gamma }_3,{\gamma }_4$ cancel those over ${\gamma }_1,{\gamma }_2$, so ${\int }_{\mathrm{\Gamma }}f(z)dz=0$.

Suppose that $\mathrm{Pol}(f)=\mathrm{\varnothing }$, so that $f$ is holomorphic on all of $\mathbb{C}$ and $\mathrm{\mid }f\mathrm{\mid }:\mathbb{C}\to \mathbb{R}$ is continuous. By step 1.4 every $z\in \mathbb{C}$ is $w+\lambda$ with $w\in P$ and $\lambda \in \mathrm{\Lambda }$, so $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }f(w)\mathrm{\mid }$ by [F2]; since $P$ is nonempty and compact by step 1.1, [F17] provides $M:={\max }_{w\in P}\mathrm{\mid }f(w)\mathrm{\mid }<\mathrm{\infty }$, hence $\mathrm{\mid }f(z)\mathrm{\mid }\le M$ for every $z\in \mathbb{C}$. Thus $f$ is a bounded entire function, and $f$ is constant by [F15].

Applying step 1.5 to the function $g=f^{\mathrm{\prime }}\mathrm{/}f$, which is holomorphic and hence continuous on the complement of $\mathrm{Zer}(f)\cup \mathrm{Pol}(f)$ and in particular on ${\mathrm{\Gamma }}^{\ast }$: by step 1.6, $g$ is $\mathrm{\Lambda }$-periodic on its domain, so $g(a+{\omega }_2+t{\omega }_1)=g(a+t{\omega }_1)$ and $g(a+{\omega }_1+t{\omega }_2)=g(a+t{\omega }_2)$ for $t\in [0,1]$; the same computation as in step 2.4 gives ${\int }_{\mathrm{\Gamma }}\frac{f^{\mathrm{\prime }}}{f}dz=0$.

Combining steps 2.3 and 2.4 gives $0={\int }_{\mathrm{\Gamma }}f(z)dz=2\pi i{\sum }_{p\in \mathrm{Pol}(f)\cap P^{\circ }}\mathrm{Res}(f,p)$, hence the residue sum vanishes.

Combining steps 2.2 and 3.1 gives $Z_P-P_P=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{f^{\mathrm{\prime }}}{f}dz=0$, so the two finite multiplicities of step 2.2 agree: $Z_P=P_P$.

Let $f$ be nonconstant and let $D:=P_P$ be the total pole multiplicity attached to $P$ by step 2.2. By step 2.5, $\mathrm{Pol}(f)\mathrm{\ne }\mathrm{\varnothing }$, so $D\ge 1$. If $D=1$, then $\mathrm{Pol}(f)\cap P^{\circ }$ consists of a single point $p$ with ${\mathrm{ord}}_p^{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}(f)=1$, so the pole is simple and [F13] gives a principal part $c_{-1}(z-p{)}^{-1}$ with $c_{-1}\mathrm{\ne }0$; by [F14], $\mathrm{Res}(f,p)=c_{-1}\mathrm{\ne }0$. The residue sum of step 3.2 then equals this single nonzero residue, contradicting step 3.2. Hence $D\ge 2$: counted with multiplicity, $f$ has at least two poles.

Let $b\in \mathbb{C}$ be a translation for which $\mathrm{\partial }{P}_b$ avoids $\mathrm{Zer}(f)\cup \mathrm{Pol}(f)$, where $P_b=\{b+s{\omega }_1+t{\omega }_2:0\le s,t\le 1\}$. The argument of steps 1.2, 1.3, 2.2, 2.3, 2.4 and 3.1 uses only this avoidance and the lattice periodicity, so it applies verbatim with $b$ in place of $a$ and yields $Z_{P_b}=P_{P_b}$ and vanishing residue sum for $P_b$. By step 1.4 applied to $P$, the number $Z_P$ is the total multiplicity of the zeros of $f$ on $T_{\mathrm{\Lambda }}$ (the sum of ${\mathrm{ord}}_c(f)$ over the finitely many zero classes), an invariant of $f$ and $\mathrm{\Lambda }$ alone, and the same holds for $P_P$; applied to $P_b$ the same identification gives $Z_{P_b}=Z_P$ and $P_{P_b}=P_P$. Hence both multiplicities and, by step 3.2 applied to each translate, the residue sum are independent of the translation.

Steps 4.1 and 3.2 prove clauses (1) and (2) for the given translate, step 5.1 proves clause (3), and steps 2.5 and 4.2 prove the two assertions of clause (4). ∎