Rectangular lattices, real mapping, and inverse elliptic integrals

(Conjugation symmetry of $\mathrm{\wp }$ and ${\mathrm{\wp }}^{\mathrm{\prime }}$.) Since $\bar{\alpha }=\alpha \in \mathrm{\Lambda }$ and $\bar{\beta }=-\beta \in \mathrm{\Lambda }$, the lattice is conjugation-invariant. Conjugating the defining net of $\mathrm{\wp }$ termwise for $z\mathrm{\notin }\mathrm{\Lambda }$ and using continuity of conjugation and the local uniform convergence of the net ([F1, F4]) gives $\mathrm{\wp }(\bar{z})=\overline{\mathrm{\wp }(z)}$; differentiating this identity gives ${\mathrm{\wp }}^{\mathrm{\prime }}(\bar{z})=\overline{{\mathrm{\wp }}^{\mathrm{\prime }}(z)}$. In particular $\mathrm{\wp }$ is real on the real axis, and for $z=x+\beta \mathrm{/}2$ one has $\bar{z}=z-\beta \equiv z$ modulo $\mathrm{\Lambda }$, so $\mathrm{\wp }(\bar{z})=\mathrm{\wp }(z)$ and $\mathrm{\wp }(z)=\overline{\mathrm{\wp }(z)}$: $\mathrm{\wp }$ is real on the two lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=b\mathrm{/}2$.

(The half-period rectangle: no poles and no critical points inside.) Put $S=\{s\frac{\alpha }{2}+t\frac{\beta }{2}:0\le s,t\le 1\}$; its interior $S^{\circ }$ is $\{0<s,t<1\}$ in these coordinates. An interior point $z=s\alpha \mathrm{/}2+t\beta \mathrm{/}2$ is not in $\mathrm{\Lambda }$: if $z=m\alpha +n\beta$ with $m,n\in \mathbb{Z}$, then $s=2m$ and $t=2n$ by uniqueness of coordinates ([F5]), but $0<s,t<1$ excludes integers. It is not a zero of ${\mathrm{\wp }}^{\mathrm{\prime }}$: such a zero would satisfy $z\equiv c_j$ modulo $\mathrm{\Lambda }$ ([F2]) for one of $c_1=\alpha \mathrm{/}2$, $c_2=\beta \mathrm{/}2$, $c_3=(\alpha +\beta )\mathrm{/}2$, i.e. $(s,t)$ would differ from $(1,0)$, $(0,1)$, $(1,1)$ by even integers, forcing $s\in 2\mathbb{Z}+1$ or $t\in 2\mathbb{Z}+1$, again impossible for $0<s,t<1$. Consequently $\mathrm{\wp }$ is holomorphic on $S^{\circ }$, ${\mathrm{\wp }}^{\mathrm{\prime }}\mathrm{\ne }0$ there, and for $z,w\in S^{\circ }$ the equality $\mathrm{\wp }(z)=\mathrm{\wp }(w)$ forces $w\equiv \pm z$ by [F2]: a lattice shift $w=z+\lambda$ forces $s^{\mathrm{\prime }}-s,t^{\mathrm{\prime }}-t\in 2\mathbb{Z}$ with $\mathrm{\mid }{s}^{\mathrm{\prime }}-s\mathrm{\mid }<1$, hence $s=s^{\mathrm{\prime }}$, $t=t^{\mathrm{\prime }}$; and a sign choice $w=-z+\lambda$ forces $s+s^{\mathrm{\prime }}\in 2\mathbb{Z}$ with $0<s+s^{\mathrm{\prime }}<2$, hence $s+s^{\mathrm{\prime }}=2$ (impossible, as $s,s^{\mathrm{\prime }}<1$) or $s+s^{\mathrm{\prime }}=0$ (impossible). Thus $\mathrm{\wp }$ is injective on $S^{\circ }$.

(The Jacobi integrand: a normalized square root on $H$.) Let $H=\{z:\mathrm{Im}z>0\}$ and $P(\zeta )=(1-{\zeta }^2)(1-k^2{\zeta }^2)$. Its zeros are $\pm 1,\pm 1\mathrm{/}k$, all real, so $P$ is holomorphic and nowhere zero on the simply connected domain $H$; by [F8] there is a holomorphic square root $q$ of $P$ on $H$. For each real $x$ with $P(x)\mathrm{\ne }0$, [F11] supplies a local nonvanishing holomorphic square root $v$ of $P$ on a disc about $x$. The quotient $q\mathrm{/}v$ on the connected upper half-disc has square $1$, so is a constant sign; thus $q$ extends holomorphically across $x$. On $(-1,1)$ the extended function $q^2=P$ is positive, so $q$ is continuous and nowhere zero there with $q(x)=\pm \sqrt{P(x)}$: the sign is constant by [F9], and after replacing $q$ by $-q$ if necessary we may and do assume $q(x)>0$ for $-1<x<1$. Then $1\mathrm{/}q$ is holomorphic on $H$ and, by [F8], has a primitive $I_k(z)={\int }_0^{z}d\zeta \mathrm{/}q(\zeta )$ whose extension at $0$ is normalized by $I_k(0)=0$; it satisfies $I_k^{\mathrm{\prime }}(z)=1\mathrm{/}q(z)\mathrm{\ne }0$ on $H$.

(The exact real locus.) For $z\in \mathbb{C}\setminus \mathrm{\Lambda }$: $\mathrm{\wp }(z)\in \mathbb{R}$ iff $\mathrm{\wp }(z)=\overline{\mathrm{\wp }(z)}=\mathrm{\wp }(\bar{z})$ iff $\bar{z}\equiv \pm z$ modulo $\mathrm{\Lambda }$ by the fibre criterion ([F2]); and $\bar{z}\equiv z$ means $2i\mathrm{Im}z\in \mathrm{\Lambda }$, i.e. $\mathrm{Im}z\in \frac{b}{2}\mathbb{Z}$, while $\bar{z}\equiv -z$ means $2\mathrm{Re}z\in \mathrm{\Lambda }$, i.e. $\mathrm{Re}z\in \frac{\alpha }{2}\mathbb{Z}$ ([F4, F5]). Thus on $\mathbb{C}\setminus \mathrm{\Lambda }$ the real-value locus is exactly the union of the horizontal lines $\mathrm{Im}z\in \frac{b}{2}\mathbb{Z}$ and the vertical lines $\mathrm{Re}z\in \frac{\alpha }{2}\mathbb{Z}$; the excluded lattice points on those lines are poles by [F1].

(Monotonicity along the four edges and the order $e_2<e_3<e_1$.) The boundary $\mathrm{\partial }S$ consists of the images of the four segments $0\to c_1\to c_3\to c_2\to 0$. On the open segment from $0$ to $c_1$ (a real interval), $\mathrm{\wp }$ is real (step 1.1), has no pole and no critical point, so its derivative is continuous and nowhere zero, hence of constant sign ([F9]); since $\mathrm{\wp }(x)\to +\mathrm{\infty }$ as $x\to 0^{+}$ by the principal part in [F1] and $\mathrm{\wp }(c_1)=e_1$, this sign is negative and $\mathrm{\wp }$ decreases strictly from $+\mathrm{\infty }$ to $e_1$. Likewise along the open segment from $c_1$ to $c_3$ and from $c_3$ to $c_2$ and from $c_2$ to $0$, all lying on the real locus lines, $\mathrm{\wp }$ takes real values with nowhere-zero derivative and is thus strictly monotone; the segment from $c_2$ to $0$ is the imaginary interval because $c_2=ib\mathrm{/}2$, and $\mathrm{\wp }(iy)\to -\mathrm{\infty }$ as $y\to 0^{+}$ by [F1], so there $\mathrm{\wp }$ decreases to $-\mathrm{\infty }$. Because ${\mathrm{\wp }}^{\mathrm{\prime }}(c_j)=0$ ([F2]), Taylor expansion at $c_j$ using [F3] gives $\mathrm{\wp }(c_j+w)=e_j+\frac{1}{2}{\mathrm{\wp }}^{\mathrm{\prime }\mathrm{\prime }}(c_j)w^2+O(\mathrm{\mid }w{\mathrm{\mid }}^3)$ with ${\mathrm{\wp }}^{\mathrm{\prime }\mathrm{\prime }}(c_j)=6e_j^2-g_2\mathrm{/}2$ and, comparing the expansions of both sides of $({\mathrm{\wp }}^{\mathrm{\prime }}{)}^2=4(\mathrm{\wp }-e_1)(\mathrm{\wp }-e_2)(\mathrm{\wp }-e_3)$, ${\mathrm{\wp }}^{\mathrm{\prime }\mathrm{\prime }}(c_j)=2{\prod }_{i\mathrm{\ne }j}(e_j-e_i)$. At $c_1$ the incident edge towards $0$ has values $>e_1$ and the incident edge towards $c_3$ has values $<e_1$ (both signs by the second-order expansion, whose quadratic coefficient for the first edge is $+{\mathrm{\wp }}^{\mathrm{\prime }\mathrm{\prime }}(c_1)\mathrm{/}2$ and for the second $-{\mathrm{\wp }}^{\mathrm{\prime }\mathrm{\prime }}(c_1)\mathrm{/}2$), so $e_3<e_1$. At $c_2$ the incident edge towards $0$ has values $<e_2$ and the incident edge towards $c_3$ has values $>e_2$, so $e_3>e_2$. Hence $e_2<e_3<e_1$.

(Boundary values of $q$ and $I_k$.) Step 1.3 extends $q$ across every real point where $P\mathrm{\ne }0$. At a simple root $a$, write $P(z)=(z-a)g_a(z)$ with $g_a(a)=P^{\mathrm{\prime }}(a)\mathrm{\ne }0$ and use [F11] to choose a holomorphic unit $v_a$ with $v_a^2=g_a$. On the upper half-disc, $q(z)=\pm v_a(z)\sqrt{z-a}$, since the quotient has square $1$ and is constant on that connected set. As one passes from the interval to the left of $a$ to the interval to its right through the upper half-plane, $\sqrt{z-a}$ changes from $i\sqrt{\mathrm{\mid }x-a\mathrm{\mid }}$ to $\sqrt{\mathrm{\mid }x-a\mathrm{\mid }}$; the unit $v_a$ retains its sign continuously. Starting with $q>0$ on $(-1,1)$, this gives $q=-i\sqrt{\mathrm{\mid }P\mathrm{\mid }}$ on $(1,1\mathrm{/}k)$, $q=+i\sqrt{\mathrm{\mid }P\mathrm{\mid }}$ on $(-1\mathrm{/}k,-1)$, and $q=-\sqrt{P}$ on both tails. Here $P^{\mathrm{\prime }}(1)=-2(1-k^2)$, $P^{\mathrm{\prime }}(-1)=2(1-k^2)$, $P^{\mathrm{\prime }}(1\mathrm{/}k)=2(1-k^2)\mathrm{/}k$, and $P^{\mathrm{\prime }}(-1\mathrm{/}k)=-2(1-k^2)\mathrm{/}k$. Moreover $\mathrm{\mid }1\mathrm{/}q(z)\mathrm{\mid }\le C_a\mathrm{\mid }z-a{\mathrm{\mid }}^{-1\mathrm{/}2}$ near $a$ in $H$. Integrating on radial segments and circular arcs gives a finite boundary limit of $I_k$ with $\mathrm{\mid }{I}_k(z)-I_k(a)\mathrm{\mid }\le C_a^{\mathrm{\prime }}\mathrm{\mid }z-a{\mathrm{\mid }}^{1\mathrm{/}2}$; thus the primitive is continuous at each of the four branch points. Hence, using $I_k^{\mathrm{\prime }}(x+i0)=1\mathrm{/}q(x+i0)$ and $I_k(0)=0$: $I_k$ is strictly increasing on $(-1,1)$ with $I_k(\pm 1)=\pm K$; $I_k(x)=K+i{\int }_1^{x}ds\mathrm{/}\sqrt{(s^2-1)(1-k^2{s}^2)}$ for $1\le x\le 1\mathrm{/}k$, so $I_k(1\mathrm{/}k)=K+i{K}^{\mathrm{\prime }}(k)$ where $K^{\mathrm{\prime }}(k)={\int }_1^{1\mathrm{/}k}ds\mathrm{/}\sqrt{(s^2-1)(1-k^2{s}^2)}$; $I_k(x)=-K+i{\int }_x^{-1}ds\mathrm{/}\sqrt{(s^2-1)(1-k^2{s}^2)}$ for $-1\mathrm{/}k\le x\le -1$, so $I_k(-1\mathrm{/}k)=-K+i{K}^{\mathrm{\prime }}(k)$; and on the tails $I_k(x)=I_k(1\mathrm{/}k)-{\int }_{1\mathrm{/}k}^{x}ds\mathrm{/}\sqrt{(s^2-1)(k^2{s}^2-1)}$ and $I_k(x)=I_k(-1\mathrm{/}k)+{\int }_x^{-1\mathrm{/}k}ds\mathrm{/}\sqrt{(s^2-1)(k^2{s}^2-1)}$, with ${\int }_{1\mathrm{/}k}^{\mathrm{\infty }}ds\mathrm{/}\sqrt{(s^2-1)(k^2{s}^2-1)}=K$ and ${\int }_{-\mathrm{\infty }}^{-1\mathrm{/}k}ds\mathrm{/}\sqrt{(s^2-1)(k^2{s}^2-1)}=K$, so both tails tend to $i{K}^{\mathrm{\prime }}(k)$. Here ${\int }_1^{1\mathrm{/}k}ds\mathrm{/}\sqrt{(s^2-1)(1-k^2{s}^2)}=K(\sqrt{1-k^2})=K^{\mathrm{\prime }}$ under the substitution $s=(1-k^{\mathrm{\prime }2}{u}^2{)}^{-1\mathrm{/}2}$ with $k^{\mathrm{\prime }}=\sqrt{1-k^2}$, and ${\int }_{1\mathrm{/}k}^{\mathrm{\infty }}ds\mathrm{/}\sqrt{(s^2-1)(k^2{s}^2-1)}=K$ under $s=1\mathrm{/}(ku)$, both by the change-of-variables formula ([F9]). Finally, for $\mathrm{\mid }z\mathrm{\mid }=R>1\mathrm{/}k$ in $H$ one has $\mathrm{\mid }P(z)\mathrm{\mid }\ge (R^2-1)(k^2{R}^2-1)$, so the integral of $1\mathrm{/}q$ along the semicircle of radius $R$ is $O(1\mathrm{/}R)$; since $I_k(R)\to i{K}^{\mathrm{\prime }}(k)$ along the real axis, $I_k(z)\to i{K}^{\mathrm{\prime }}(k)$ uniformly as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ in $H$.

(The image lies in one half-plane.) $S^{\circ }$ is convex, hence connected, and $\mathrm{\wp }(S^{\circ })$ is connected; since ${\mathrm{\wp }}^{\mathrm{\prime }}\mathrm{\ne }0$ on $S^{\circ }$, $\mathrm{\wp }$ is an open map there and $\mathrm{\wp }(S^{\circ })$ is open ([F10]); and $\mathrm{\wp }(S^{\circ })\cap \mathbb{R}=\mathrm{\varnothing }$ by step 2.1, because no interior point lies on any of the lines $\mathrm{Re}z\in \frac{\alpha }{2}\mathbb{Z}$, $\mathrm{Im}z\in \frac{b}{2}\mathbb{Z}$. An open connected subset of $\mathbb{C}\setminus (\mathbb{R}\cup \{\mathrm{\infty }\})$ is contained in the upper or in the lower half-plane.

(The boundary maps onto $\mathbb{R}\cup \{\mathrm{\infty }\}$.) By step 2.2 the four open edges have images $(e_1,\mathrm{\infty })$, the interval between $e_1$ and $e_3$, the interval between $e_3$ and $e_2$, and $(-\mathrm{\infty },e_2)$, and with $e_2<e_3<e_1$ these intervals are $(e_1,\mathrm{\infty })$, $(e_3,e_1)$, $(e_2,e_3)$ and $(-\mathrm{\infty },e_2)$, which together with the endpoint values and $\mathrm{\wp }(0)=\mathrm{\infty }$ cover $\mathbb{R}\cup \{\mathrm{\infty }\}$ exactly once on the boundary circle.

($I_k$ maps $H$ biholomorphically onto the rectangle.) Let $R_0=(-K,K)\times (0,K^{\mathrm{\prime }})$ and $R>1\mathrm{/}k$. The real-segment path ${\gamma }_R:=I_k([-R,R])$ starts at $i{K}^{\mathrm{\prime }}-T_R$ and ends at $i{K}^{\mathrm{\prime }}+T_R$, where $T_R={\int }_R^{\mathrm{\infty }}ds\mathrm{/}\sqrt{(s^2-1)(k^2{s}^2-1)}>0$ tends to zero by step 2.3. It follows the boundary of $R_0$ counterclockwise except for the short top segment joining those endpoints. The image ${\delta }_R$ of the upper semicircle runs from $i{K}^{\mathrm{\prime }}+T_R$ back to $i{K}^{\mathrm{\prime }}-T_R$ and lies in a disc of radius $O(1\mathrm{/}R)$ about $i{K}^{\mathrm{\prime }}$ by step 2.3. For a fixed $w$ away from $\mathrm{\partial }{R}_0$, choose $R$ large enough that both ${\delta }_R$ and the missing top segment lie in a disc about $i{K}^{\mathrm{\prime }}$ disjoint from $w$. A straight-line homotopy in that disc deforms ${\delta }_R$ to the missing segment, so the closed full image contour ${\gamma }_R\ast {\delta }_R=I_k(\mathrm{\partial }{D}_R)$ has winding number $1$ about $w\in R_0$ and $0$ about $w\mathrm{\notin }\overline{R_0}$.

(The image of $S$ is exactly one half-plane and $\mathrm{\wp }{\mathrm{\mid }}_S$ is conformal.) Every boundary point of $\mathrm{\wp }(S^{\circ })$ is a limit $\mathrm{\wp }(z_n)$ with $z_n\in S^{\circ }$; by compactness of $\bar{S}$ pass to a subsequence with $z_n\to z_0\in \bar{S}$ and $\mathrm{\wp }(z_0)=w$ with the value $\mathrm{\wp }(0)=\mathrm{\infty }$ allowed. If $z_0\in S^{\circ }$ then $w\in \mathrm{\wp }(S^{\circ })$, which is impossible for a boundary point because $\mathrm{\wp }(S^{\circ })$ is open; hence $z_0\in \mathrm{\partial }S$ and, by step 3.2, $w\in \mathrm{\wp }(\mathrm{\partial }S)\subseteq \mathbb{R}\cup \{\mathrm{\infty }\}$. Thus $\mathrm{\partial }\mathrm{\wp }(S^{\circ })\subseteq \mathbb{R}\cup \{\mathrm{\infty }\}$, while by step 3.1 the set $\mathrm{\wp }(S^{\circ })$ is a nonempty open connected subset of one half-plane $U_{\pm }$, and $U_{\pm }\cap (\mathbb{R}\cup \{\mathrm{\infty }\})=\mathrm{\varnothing }$. If $w\in U_{\pm }\setminus \mathrm{\wp }(S^{\circ })$, join $w$ to a point $q\in \mathrm{\wp }(S^{\circ })$ by the segment $\gamma$ inside the convex set $U_{\pm }$ and let $t_0=\inf \{t:\gamma (t)\in \mathrm{\wp }(S^{\circ })\}$; then $\gamma (t_0)\in \mathrm{\partial }\mathrm{\wp }(S^{\circ })\cap U_{\pm }$, contradicting $\mathrm{\partial }\mathrm{\wp }(S^{\circ })\subseteq \mathbb{R}\cup \{\mathrm{\infty }\}$. Hence $\mathrm{\wp }(S^{\circ })=U_{\pm }$, and $\mathrm{\wp }:S^{\circ }\to U_{\pm }$ is injective with nowhere-zero derivative, hence biholomorphic, and in particular conformal.

(Derivative of the inverse branch.) Let $U=U_{\pm }$ and $f=\mathrm{\wp }{\mathrm{\mid }}_{S^{\circ }}^{-1}:U\to S^{\circ }$. Then $f$ is holomorphic, $\mathrm{\wp }(f(\zeta ))=\zeta$ for $\zeta \in U$, and the chain rule gives ${\mathrm{\wp }}^{\mathrm{\prime }}(f(\zeta ))f^{\mathrm{\prime }}(\zeta )=1$, so $f^{\mathrm{\prime }}(\zeta )=1\mathrm{/}{\mathrm{\wp }}^{\mathrm{\prime }}(f(\zeta ))$. Substituting $\mathrm{\wp }(f(\zeta ))=\zeta$ in the differential equation [F3] gives ${\mathrm{\wp }}^{\mathrm{\prime }}(f(\zeta ){)}^2=4{\zeta }^3-g_2\zeta -g_3$, and since $f^{\mathrm{\prime }}$ is nowhere zero its reciprocal $\zeta \mapsto 1\mathrm{/}{f}^{\mathrm{\prime }}(\zeta )={\mathrm{\wp }}^{\mathrm{\prime }}(f(\zeta ))$ is a holomorphic square root of $4{\zeta }^3-g_2\zeta -g_3$ on $U$; writing $(4{\zeta }^3-g_2\zeta -g_3{)}^{-1\mathrm{/}2}$ for that reciprocal root, every inverse branch satisfies $f^{\mathrm{\prime }}(\zeta )=(4{\zeta }^3-g_2\zeta -g_3{)}^{-1\mathrm{/}2}$.

(The four period integrals.) On each of the four real intervals between consecutive roots the polynomial $4x^3-g_{2}x-g_3=4(x-e_1)(x-e_2)(x-e_3)$ has constant sign; $4x^3-g_{2}x-g_3>0$ on $(e_1,\mathrm{\infty })$ and on $(e_2,e_3)$, while $g_3+g_{2}x-4x^3>0$ on $(-\mathrm{\infty },e_2)$ and on $(e_3,e_1)$. Each of these intervals is the image under $\mathrm{\wp }$ of one open edge of $\mathrm{\partial }S$. Since ${\mathrm{\wp }}^{\mathrm{\prime }}$ is nonzero on each open edge, the local inverse theorem [F11] extends the inverse branch across the corresponding real interval, mapping it bijectively onto that edge with $\mathrm{\mid }{f}^{\mathrm{\prime }}(x)\mathrm{\mid }=1\mathrm{/}\sqrt{\mathrm{\mid }4x^3-g_{2}x-g_3\mathrm{\mid }}$ by step 5.1, so the integral of the positive square root over the interval equals the length of the corresponding displacement: writing the inverse branch as $g$, ${\int }_a^{b}dx\mathrm{/}\sqrt{\mathrm{\mid }4x^3-g_{2}x-g_3\mathrm{\mid }}=\mathrm{\mid }{\lim }_{x\downarrow a}g(x)-{\lim }_{x\uparrow b}g(x)\mathrm{\mid }$ by the change-of-variables formula and strict monotonicity of $g$ ([F9]). The four displacements are: from $0$ to $c_1$, length $\alpha \mathrm{/}2$; from $c_2$ to $c_3$, length $\alpha \mathrm{/}2$; from $c_2$ to $0$, length $\mathrm{\mid }\beta \mathrm{\mid }\mathrm{/}2=\beta \mathrm{/}(2i)$; from $c_1$ to $c_3$, length $\beta \mathrm{/}(2i)$. The improper endpoints converge: at a simple root the integrand is $O(\mathrm{\mid }x-e_j{\mathrm{\mid }}^{-1\mathrm{/}2})$, and at infinity it is $O(\mathrm{\mid }x{\mathrm{\mid }}^{-3\mathrm{/}2})$. Hence the four displayed identities hold with the positive real square roots.

(The inverse sn and its double periodicity.) At a finite boundary branch point $a$ of $I_k$, use $z=a+t^2$. The expression $q(a+t^2)\mathrm{/}t$ is a holomorphic unit by the factorization in step 2.3; hence $J(t):=I_k(a+t^2)$ extends holomorphically to $t=0$ and $J^{\mathrm{\prime }}(0)=2\mathrm{/}(q(a+t^2)\mathrm{/}t){\mathrm{\mid }}_{t=0}\mathrm{\ne }0$. By [F11] its local inverse makes $z=a+t(w{)}^2$ holomorphic at the associated rectangle vertex. On the tails, the branch sign in step 2.3 gives $q(z)=-k{z}^2(1+O(z^{-2}))$: this follows by applying [F11] to the square root of $1-(1+k^2)\mathrm{/}(k^2{z}^2)+1\mathrm{/}(k^2{z}^4)$ near $1\mathrm{/}z=0$. Thus $I_k(z)=i{K}^{\mathrm{\prime }}+1\mathrm{/}(kz)+O(z^{-3})$, and $t\mapsto I_k(1\mathrm{/}t)$ extends holomorphically at $0$ with derivative $1\mathrm{/}k\mathrm{\ne }0$. Its local inverse shows that $1\mathrm{/}\mathrm{s}\mathrm{n}$ is holomorphic with a simple zero at $i{K}^{\mathrm{\prime }}$. Let $\mathrm{s}\mathrm{n}=I_k^{-1}:R_0\to H$; it is holomorphic ([F10]) and continuous on $\overline{R_0}\setminus \{i{K}^{\mathrm{\prime }}\}$ with real boundary values: the bottom edge $[-K,K]$ maps into $(-1,1)$ with $\mathrm{s}\mathrm{n}(0)=0$, $\mathrm{s}\mathrm{n}(\pm K)=\pm 1$; the left edge $-K+i(0,K^{\mathrm{\prime }})$ maps into $(-1\mathrm{/}k,-1)$ with $\mathrm{s}\mathrm{n}(-K+i{K}^{\mathrm{\prime }})=-1\mathrm{/}k$ (at the boundary branch value $-1\mathrm{/}k$); the right edge $K+i(0,K^{\mathrm{\prime }})$ maps into $(1,1\mathrm{/}k)$ with $\mathrm{s}\mathrm{n}(K+i{K}^{\mathrm{\prime }})=1\mathrm{/}k$; and the top edge $(-K,K)+i{K}^{\mathrm{\prime }}$ maps into $\mathbb{R}$ with $\mathrm{\mid }\mathrm{s}\mathrm{n}\mathrm{\mid }>1\mathrm{/}k$, with a simple pole at $i{K}^{\mathrm{\prime }}$ (from step 2.3, $I_k(z)=i{K}^{\mathrm{\prime }}+1\mathrm{/}(kz)+O(z^{-2})$ for large $z$, so near the omitted value $i{K}^{\mathrm{\prime }}$ the inverse behaves like $1\mathrm{/}(k(w-i{K}^{\mathrm{\prime }}))$). Since $\mathrm{s}\mathrm{n}$ is holomorphic on $R_0$ and real on each of the four open sides, the Schwarz reflection principle [F7] extends it across each side by reflection there: away from $i{K}^{\mathrm{\prime }}$ this is the stated holomorphic reflection, and at $i{K}^{\mathrm{\prime }}$ one applies the same principle to the reciprocal $u=1\mathrm{/}\mathrm{s}\mathrm{n}$, which is holomorphic near $i{K}^{\mathrm{\prime }}$ with $u(i{K}^{\mathrm{\prime }})=0$ and real boundary values, so $u$ reflects and $\mathrm{s}\mathrm{n}=1\mathrm{/}u$ reflects meromorphically; the reflected copies tile the plane, because the four reflections ${\sigma }_b(z)=\bar{z}$, ${\sigma }_t(z)=\bar{z}+2i{K}^{\mathrm{\prime }}$, ${\sigma }_l(z)=-2K-\bar{z}$, ${\sigma }_r(z)=2K-\bar{z}$ have compositions ${\sigma }_b{\sigma }_t:z\mapsto z-2i{K}^{\mathrm{\prime }}$ and ${\sigma }_r{\sigma }_l:z\mapsto z+4K$, and their reflected rectangle copies tile the plane. Across an open edge the extensions agree by reflection; at a vertex they agree with the holomorphic squared inverse just constructed, and at a reflected copy of $i{K}^{\mathrm{\prime }}$ they agree with its meromorphic pole extension. Thus the pieces glue on every edge and vertex, producing a single meromorphic function $\mathrm{s}\mathrm{n}$ on $\mathbb{C}$ satisfies $\mathrm{s}\mathrm{n}({\sigma }_{j}z)=\overline{\mathrm{s}\mathrm{n}(z)}$ for each reflection ${\sigma }_j$, so $\mathrm{s}\mathrm{n}(z-2i{K}^{\mathrm{\prime }})=\overline{\mathrm{s}\mathrm{n}({\sigma }_{t}z)}=\mathrm{s}\mathrm{n}(z)$ and $\mathrm{s}\mathrm{n}(z+4K)=\overline{\mathrm{s}\mathrm{n}({\sigma }_{l}z)}=\mathrm{s}\mathrm{n}(z)$: $\mathrm{s}\mathrm{n}$ is doubly periodic with periods $4K$ and $2i{K}^{\mathrm{\prime }}$ (and period $2i{K}^{\mathrm{\prime }}$ implies period $-2i{K}^{\mathrm{\prime }}$).