Transformation laws and S_3-action of the modular lambda function

Let $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \mathrm{\Gamma }(2)$, so $a\equiv d\equiv 1$ and $b\equiv c\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Put ${\omega }_1:=a\tau +b$, ${\omega }_2:=c\tau +d$, so $\gamma \tau ={\omega }_1\mathrm{/}{\omega }_2$ and ${\mathrm{\Lambda }}_{\gamma \tau }=(1\mathrm{/}{\omega }_2){\mathrm{\Lambda }}_{\tau }$ because $\mathbb{Z}+\mathbb{Z}({\omega }_1\mathrm{/}{\omega }_2)=(1\mathrm{/}{\omega }_2)(\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2)=(1\mathrm{/}{\omega }_2){\mathrm{\Lambda }}_{\tau }$. By [F2], ${\mathrm{\wp }}_{{\mathrm{\Lambda }}_{\gamma \tau }}(z)={\omega }_2^2{\mathrm{\wp }}_{{\mathrm{\Lambda }}_{\tau }}({\omega }_{2}z)$. Now ${\omega }_1-\tau =(a-1)\tau +b\in 2{\mathrm{\Lambda }}_{\tau }$ and ${\omega }_2-1=c\tau +(d-1)\in 2{\mathrm{\Lambda }}_{\tau }$; hence the half-periods $1\mathrm{/}2$, $\gamma \tau \mathrm{/}2={\omega }_1\mathrm{/}(2{\omega }_2)$ and $(1+\gamma \tau )\mathrm{/}2=({\omega }_1+{\omega }_2)\mathrm{/}(2{\omega }_2)$ of ${\mathrm{\Lambda }}_{\gamma \tau }$ correspond under the scaling to ${\omega }_2\mathrm{/}2$, ${\omega }_1\mathrm{/}2$ and $({\omega }_1+{\omega }_2)\mathrm{/}2$, which are congruent modulo ${\mathrm{\Lambda }}_{\tau }$ to $\frac{1}{2}$, $\frac{\tau }{2}$ and $\frac{1+\tau }{2}$ by [F1] and the evenness of $\mathrm{\wp }$. Therefore $\lambda (\gamma \tau )=\frac{{\omega }_2^2(e_3-e_2)}{{\omega }_2^2(e_1-e_2)}=\lambda (\tau )$.

For $T$ one has ${\mathrm{\Lambda }}_{\tau +1}={\mathrm{\Lambda }}_{\tau }$ and the half-period values of the basis $(1,\tau +1)$ are $e_1$, $\mathrm{\wp }((\tau +1)\mathrm{/}2)=e_3$, $\mathrm{\wp }((\tau +2)\mathrm{/}2)=e_2$, so $\lambda (\tau +1)=\frac{e_2-e_3}{e_1-e_3}=\frac{e_3-e_2}{(e_3-e_2)-(e_1-e_2)}=\frac{\lambda (\tau )}{\lambda (\tau )-1}$. For $S$ one has $S\tau =-1\mathrm{/}\tau$ and ${\mathrm{\Lambda }}_{S\tau }=(1\mathrm{/}\tau ){\mathrm{\Lambda }}_{\tau }$, so by [F2] with $c=1\mathrm{/}\tau$ the scaled $\mathrm{\wp }$ is ${\mathrm{\wp }}_{{\mathrm{\Lambda }}_{S\tau }}(z)={\tau }^2{\mathrm{\wp }}_{{\mathrm{\Lambda }}_{\tau }}(\tau z)$; the half-periods $1\mathrm{/}2=c\cdot (\tau \mathrm{/}2)$, $S\tau \mathrm{/}2=c\cdot (-1\mathrm{/}2)$, $(1+S\tau )\mathrm{/}2=c\cdot ((\tau -1)\mathrm{/}2)$ of ${\mathrm{\Lambda }}_{S\tau }$ therefore give the values $e_1^{\mathrm{\prime }}={\tau }^2\mathrm{\wp }(\tau \mathrm{/}2)={\tau }^2{e}_2$, $e_2^{\mathrm{\prime }}={\tau }^2\mathrm{\wp }(-1\mathrm{/}2)={\tau }^2{e}_1$ and $e_3^{\mathrm{\prime }}={\tau }^2\mathrm{\wp }((\tau -1)\mathrm{/}2)={\tau }^2\mathrm{\wp }((1+\tau )\mathrm{/}2)={\tau }^2{e}_3$; hence $\lambda (-1\mathrm{/}\tau )=\frac{e_3-e_1}{e_2-e_1}=1-\lambda (\tau )$.

The square lattice ${\mathrm{\Lambda }}_i=\mathbb{Z}+\mathbb{Z}i$ is invariant under multiplication by $i$, and the substitution $(m,n)\mapsto (-n,m)$ is a bijection of ${\mathbb{Z}}^2$ sending $mi+n$ to $i(mi+n)$, so the absolutely summable family $((mi+n{)}^{-6})$ [F3] equals its negative and $g_3=140G_6=0$ for $\tau =i$; also ${\mathrm{\wp }}_{{\mathrm{\Lambda }}_i}(iz)=-{\mathrm{\wp }}_{{\mathrm{\Lambda }}_i}(z)$ by [F2] with $c=i$, so $e_2=\mathrm{\wp }(i\mathrm{/}2)=-\mathrm{\wp }(1\mathrm{/}2)=-e_1$, and then [F3] gives $e_3=0$. Hence $\lambda (i)=\frac{e_3-e_2}{e_1-e_2}=\frac{e_1}{2e_1}=\frac{1}{2}$. For $\tau =iy$, $y>0$, the lattice ${\mathrm{\Lambda }}_{iy}$ is invariant under conjugation: the conjugate of $miy+n$ is $-miy+n\in {\mathrm{\Lambda }}_{iy}$, so $\mathrm{\wp }(\bar{z})=\overline{\mathrm{\wp }(z)}$ [F6]; the half-periods $1\mathrm{/}2$, $iy\mathrm{/}2$, $(1+iy)\mathrm{/}2$ are each congruent to their conjugates modulo ${\mathrm{\Lambda }}_{iy}$, so $e_1,e_2,e_3$ are real and $\lambda (iy)$ is real, while $\lambda (iy)\mathrm{\ne }0,1$ because the $e_j$ stay distinct [F3]. By [F6] each $e_j(iy)$ is continuous in $y$; $\lambda (iy)$ is therefore a continuous real function of $y\in (0,\mathrm{\infty })$ avoiding $0$ and $1$, so it lies in a single connected component of $\mathbb{R}\setminus \{0,1\}$; since $\lambda (i)=1\mathrm{/}2\in (0,1)$, it follows that $\lambda (iy)\in (0,1)$ for every $y>0$.

Let $X$ be the set of the six rational functions $x$, $1\mathrm{/}x$, $1-x$, $1\mathrm{/}(1-x)$, $x\mathrm{/}(x-1)$, $(x-1)\mathrm{/}x$ of an indeterminate $x$; these are pairwise distinct functions on $\mathbb{C}\setminus \{0,1\}$, and $\sigma (x):=x\mathrm{/}(x-1)$ and $\tau (x):=1-x$ satisfy ${\sigma }^2={\tau }^2=1$ and generate a group of order $6$ acting transitively on $X$ (the orbit of $x$ is exactly $X$), hence isomorphic to $S_3$. By 1.2, $\lambda (S\tau )=1-\lambda (\tau )=\tau (\lambda (\tau ))$ and $\lambda (T\tau )=\sigma (\lambda (\tau ))$, and for a word $\gamma$ in $S,T$ induction gives $\lambda (\gamma \tau )=f_{\gamma }(\lambda (\tau ))$ with $f_{\gamma }$ the corresponding composition in this group; since $\lambda (i)=1\mathrm{/}2$ and $\lambda (i+1)=-1$ by 1.2 and 1.3, it is not constant; its real restriction $\lambda (iy)$ cannot be constant by the identity theorem, so its continuous image is an interval with more than one point by the intermediate value theorem (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a,b]$ takes every value between $f(a)$ and $f(b)$). Thus distinct $f_{\gamma }$ give distinct functions $f_{\gamma }\circ \lambda$, so the assignment $\gamma \mapsto f_{\gamma }$ is a homomorphism $PS{L}_2(\mathbb{Z})\to S_3$ [F5]. Its kernel is $\{\gamma :\lambda (\gamma \tau )=\lambda (\tau )\}$, which contains $\bar{\mathrm{\Gamma }}(2)$ by 1.1 and therefore has index at most $6$; the image is generated by $\sigma ,\tau$ and has order $6$, so the index is exactly $6$ and the kernel is $\bar{\mathrm{\Gamma }}(2)$, giving $PS{L}_2(\mathbb{Z})\mathrm{/}\bar{\mathrm{\Gamma }}(2)\cong S_3$ [F4]. Hence $\lambda (\gamma \tau )$ runs over the displayed expressions, with coincidences allowed (for example $\lambda (i)=1\mathrm{/}2$ gives the three values $1\mathrm{/}2,2,-1$) as $\gamma$ runs over $PS{L}_2(\mathbb{Z})$.