Moving the boundary of a fundamental parallelogram

(The shifted parallelogram is boundary-free for $\mathrm{\wp }$.) Every point of $\mathrm{\partial }{P}_a$ with $a=(1+i)\mathrm{/}4$ has the form $z=a+s+ti$ with $(s,t)\in \{0,1\}\times [0,1]\cup [0,1]\times \{0,1\}$, so $\mathrm{Re}z\in \{1\mathrm{/}4,5\mathrm{/}4\}$ or $\mathrm{Im}z\in \{1\mathrm{/}4,5\mathrm{/}4\}$. A lattice point of $\mathrm{\Lambda }$ has integer real and imaginary parts, hence cannot have real part or imaginary part equal to $1\mathrm{/}4$ or $5\mathrm{/}4$; so $\mathrm{\partial }{P}_a\cap \mathrm{\Lambda }=\mathrm{\varnothing }$. By [F2] the poles of $\mathrm{\wp }$ are exactly the lattice points, so $\mathrm{\wp }$ has no pole on $\mathrm{\partial }{P}_a$.

(The unshifted parallelogram meets the poles.) The four vertices of $P_0$ are $0,1,i,1+i$, all of which lie in $\mathrm{\Lambda }$ and in $\mathrm{\partial }{P}_0$; by [F2] each is a double pole of $\mathrm{\wp }$.

(Finitely many divisor classes.) Suppose first that $f$ is nonconstant. By [F5] applied to $f$, the pole set $\mathrm{Pol}(f)$ is closed and every point of $\mathbb{C}$ has a neighbourhood meeting it in at most one point; by [F5] applied to $1\mathrm{/}f$, whose pole set is $\mathrm{Zer}(f)$, the same holds for the zero set. Hence every point of $\mathbb{C}$ has a neighbourhood meeting $Z$ in at most two points, and since $P_0$ is compact by [F1] finitely many of these neighbourhoods cover $P_0$; therefore $Z\cap P_0=\{z_1,\dots ,z_N\}$ is finite. By [F4] the function $f$ is $\mathrm{\Lambda }$-periodic, so $Z$ is $\mathrm{\Lambda }$-invariant; conversely every $z\in Z$ can be written $z=s+ti$ with real $s,t$, and subtracting the integer parts $m\le s<m+1$, $n\le t<n+1$ given by [F6] puts $z-(m+ni)\in Z\cap P_0$. Hence $Z={\bigcup }_{j=1}^N(z_j+\mathrm{\Lambda })$: the zeros and poles of $f$ fall into the finitely many classes of $z_1,\dots ,z_N$ modulo $\mathrm{\Lambda }$.

(Generic translates avoid the divisor.) Let $U$ be a nonempty open ball of basepoints; we show that some $b\in U$ has $\mathrm{\partial }{P}_b\cap Z=\mathrm{\varnothing }$, treating the constant case at the end. Fix the finite list of step 1.3 and put $B_j:=z_j-\mathrm{\partial }{P}_0$ for $j=1,\dots ,N$; each $B_j$ is closed, being a translate of $\mathrm{\partial }{P}_0$, and contains no ball: if a ball $B(x,r)$ were contained in a segment with direction vector $q\mathrm{\ne }0$, then with $u:=iq\mathrm{\ne }0$ perpendicular to $q$ the two points $x$ and $x+\frac{r}{2\mathrm{\mid }u\mathrm{\mid }}u$ of the ball would both lie on the line of the segment, whose direction is $q$, forcing the nonzero vector $\frac{r}{2\mathrm{\mid }u\mathrm{\mid }}u$ to be a real multiple of $q$ and contradicting $u\perp q$. Consequently no ball is contained in $\mathrm{\partial }{P}_0$ or in any of its translates, which are finite unions of segments: if the closed sets $F_1,\dots ,F_m$ each contain no ball and a ball $B\subseteq F_1\cup \cdots \cup F_m$, then either $B\subseteq F_1$ or there is $y\in B\setminus F_1$, and since $F_1$ is closed some ball $B(y,\rho )\subseteq B\setminus F_1\subseteq F_2\cup \cdots \cup F_m$; iterating gives a ball inside some $F_i$, a contradiction. Hence each $B_j$ is closed with no ball inside it. For a basepoint $b\in U$ one has $\mathrm{\partial }{P}_b\cap Z\mathrm{\ne }\mathrm{\varnothing }$ iff $b\in B_j+\lambda$ for some $j$ and some $\lambda \in \mathrm{\Lambda }$: indeed $z\in \mathrm{\partial }{P}_b\cap Z$ means $z=z_j+\lambda =b+w$ with $w\in \mathrm{\partial }{P}_0$, that is $b=z_j+\lambda -w$. For $b\in U$ such a $\lambda$ must satisfy $\mathrm{\mid }\lambda \mathrm{\mid }\le \mathrm{\mid }b-z_j\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }\le C$ for a constant $C$ depending only on $U$ and the finitely many $z_j$ by [F1], so only finitely many $\lambda \in \mathrm{\Lambda }$ occur, and the set of bad basepoints in $U$ is the finite union $({\bigcup }_{j=1}^N{\bigcup }_{\lambda \in \mathrm{\Lambda },\mathrm{\mid }\lambda \mathrm{\mid }\le C}(B_j+\lambda ))\cap U$ of closed sets containing no ball. A finite union of sets each containing no ball does not contain $U$: successively avoiding each closed member leaves a nonempty open subball, as in the induction just described. Hence there is $b\in U$ with $\mathrm{\partial }{P}_b\cap Z=\mathrm{\varnothing }$, which is the claim. If $f$ is a nonzero constant then $Z=\mathrm{\varnothing }$ and every $b\in U$ works.

(The divisor law applies, and the explicit cases.) For a nonconstant $f$, step 2.1 provides a translate $P_b$ whose boundary avoids $\mathrm{Zer}(f)\cup \mathrm{Pol}(f)$, so the hypotheses of [F3] are met and the zero count, the pole count and the vanishing residue sum for $f$ hold for that parallelogram; for nonzero constant $f$ those counts are $0$ and the conclusions are trivial. Steps 1.1 and 1.2 are the explicit statements for $\mathrm{\wp }$ on $\mathrm{\Lambda }=\mathbb{Z}+i\mathbb{Z}$: the boundary of $P_{(1+i)\mathrm{/}4}$ carries no pole, while the boundary of the unshifted $P_0$ contains the four lattice vertices. This proves the three clauses of the Example.