The modular lambda function: Y(2) biholomorphic to the twice-punctured plane, and the slit-plane quadrilateral

$\lambda$ is holomorphic on $\mathfrak{H}$ and satisfies $\lambda (-\bar{\tau })=\overline{\lambda (\tau )}$ by [F1]; by the $\mathrm{\Gamma }(2)$-invariance of [F2] and the local quotient charts of [F3], it descends to a holomorphic function $\bar{\lambda }$ on $Y(2)=\mathfrak{H}\mathrm{/}\mathrm{\Gamma }(2)$, which takes values in $\mathbb{C}\setminus \{0,1\}$ because the $e_j$ are always distinct.

Reduction to the quadrilateral. Fix $\tau \in \mathfrak{H}$. The set $S=\{\mathrm{\mid }c\tau +d\mathrm{\mid }:\gamma =(\begin{array}{cc}a& b\\ c& d\end{array})\in \mathrm{\Gamma }(2)\}$ contains $1$ (the identity), and the pairs with $\mathrm{\mid }c\tau +d\mathrm{\mid }\le 1$ are finite by Reduction of orbits to the standard domain; hence $S$ has a least positive element $m$, realized by some ${\gamma }_0\in \mathrm{\Gamma }(2)$, and ${\tau }_0={\gamma }_0\tau$ has maximal imaginary part in the $\mathrm{\Gamma }(2)$-orbit. Applying a power of $T^2\in \mathrm{\Gamma }(2)$, which adds an even integer and does not change the height, we may assume $\mathrm{\mid }\mathrm{\Re }{\tau }_0\mathrm{\mid }\le 1$. Maximality forces $\mathrm{\mid }2{\tau }_0+1\mathrm{\mid }\ge 1$ and $\mathrm{\mid }2{\tau }_0-1\mathrm{\mid }\ge 1$, since otherwise ${\gamma }_{+}{\tau }_0$ or ${\gamma }_{-}{\tau }_0$ would have strictly larger imaginary part; thus ${\tau }_0$ lies in the closure of $Q^{\circ }$, and every $\mathrm{\Gamma }(2)$-orbit meets that closure.

$\bar{\lambda }$ is injective by [F3]: if $\bar{\lambda }$ agrees at two classes, the underlying $\lambda$-values agree and the points lie in one $\mathrm{\Gamma }(2)$-orbit. It is surjective: let $z\in \mathbb{C}\setminus \{0,1\}$ and put $R(x)=\frac{256(x^2-x+1{)}^3}{x^2(x-1{)}^2}$. Since $j:X(1)\to \widehat{\mathbb{C}}$ is onto [F4], there is $\tau \in \mathfrak{H}$ with $j(\tau )=R(z)$ (the value is finite, so it is attained off the cusp); then $R(\lambda (\tau ))=j(\tau )=R(z)$ by [F4]. Put $H(x)=(x^2-x+1{)}^3$ and $D(x)=x^2(x-1{)}^2$. For $u\mathrm{\ne }0,1$, clearing denominators gives $D(u)H(v)-H(u)D(v)=0$, a degree-six polynomial in $v$ with nonzero leading coefficient $D(u)$. Let $f_1(u),\dots ,f_6(u)$ be the six substitutions of [F2]. The identity $D(u)H(v)-H(u)D(v)=D(u){\prod }_{r=1}^6(v-f_r(u))$ holds first for generic $u$, where the six roots are distinct by direct substitution and the leading coefficients agree. It then holds for every $u\mathrm{\ne }0,1$, since each coefficient is a rational function of $u$ and an identity outside finitely many exceptional values is a rational-function identity. Thus the same factorisation handles the repeated roots at special parameters, and its root set is exactly the displayed substitutions; so $z$ is one of $\lambda (\tau ),\frac{1}{\lambda (\tau )},1-\lambda (\tau ),\frac{1}{1-\lambda (\tau )},\frac{\lambda (\tau )}{\lambda (\tau )-1},\frac{\lambda (\tau )-1}{\lambda (\tau )}$. By [F2] each of these is $\lambda (\gamma \tau )$ for some $\gamma \in PS{L}_2(\mathbb{Z})$, so $z$ lies in the image of $\bar{\lambda }$.

Uniqueness and the interior. Let $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \mathrm{\Gamma }(2)$ with $c\mathrm{\ne }0$, so $c$ is even and $d$ is odd, and consider the open disc $\{\tau :\mathrm{\mid }c\tau +d\mathrm{\mid }<1\}=\{\tau :\mathrm{\mid }\tau +d\mathrm{/}c\mathrm{\mid }<1\mathrm{/}\mathrm{\mid }c\mathrm{\mid }\}$. Its centre $-d\mathrm{/}c$ is not $-1,0$ or $1$ (it is not an integer), and if it lies in $(-1,0)$ or in $(0,1)$ then its distances to the endpoints of the corresponding interval are at least $1\mathrm{/}\mathrm{\mid }c\mathrm{\mid }$, since each is a nonzero integer divided by $\mathrm{\mid }c\mathrm{\mid }$, so the disc lies inside the boundary disc with diameter $[-1,0]$ or $[0,1]$; if the centre lies outside $[-1,1]$, its centre is at distance at least $1\mathrm{/}\mathrm{\mid }c\mathrm{\mid }$ from the strip $\mathrm{\mid }\mathrm{\Re }\tau \mathrm{\mid }\le 1$. Hence $\mathrm{\mid }c\tau +d\mathrm{\mid }\ge 1$ on the closure of $Q^{\circ }$ and $\mathrm{\mid }c\tau +d\mathrm{\mid }>1$ on $Q^{\circ }$. If $c=0$ then $\gamma =\pm T^{2k}$ is translation by an even integer, and two points of the closure with $\mathrm{\mid }\mathrm{\Re }\mathrm{\mid }\le 1$ differ by $2k$ with $\mathrm{\mid }k\mathrm{\mid }\le 1$; for $k=\pm 1$ both have real part $\mp 1$, a boundary value. Therefore no two distinct points of $Q^{\circ }$ are $\mathrm{\Gamma }(2)$-equivalent, and no point of $Q^{\circ }$ is equivalent to a boundary point: two interior points related by $\gamma$ with $c\mathrm{\ne }0$ would give $\mathrm{Im}(\gamma \tau )<\mathrm{Im}\tau$ and, applying the same to ${\gamma }^{-1}$ and $\gamma \tau$ in the closure, the reverse weak inequality. This also excludes an interior-to-boundary identification.

By 1.1 and 2.1 the holomorphic map $\bar{\lambda }:Y(2)\to \mathbb{C}\setminus \{0,1\}$ is bijective, hence biholomorphic by [F5] (injectivity forces local degree one everywhere). The quotient map ${\pi }_2:\mathfrak{H}\to Y(2)$ is a covering by [F3], so $\lambda =\bar{\lambda }\circ {\pi }_2$ is a covering. The total space $\mathfrak{H}$ is connected: the straight segment between any two points stays in $\mathfrak{H}$. Every $\bar{\mathrm{\Gamma }}(2)$ element is a deck transformation by [F2]. Conversely, for a deck transformation $h$ and a fixed ${\tau }_0\in \mathfrak{H}$, [F3] gives $\gamma \in \bar{\mathrm{\Gamma }}(2)$ with $h({\tau }_0)=\gamma {\tau }_0$; connected-cover uniqueness [F6] then gives $h=\gamma$. Thus the deck group is exactly $\bar{\mathrm{\Gamma }}(2)$, acting transitively on each fibre by [F3] and freely by [F6], hence simply transitively; the covering is regular.

Boundary values and the slit plane. The vertical boundary edges are $T^{\pm 1}(iy)=iy\pm 1$. The rational substitution $s(x)=x\mathrm{/}(x-1)$ is its own inverse, so both edges have value $s(\lambda (iy))<0$. The semicircular edges are $L_{\pm }(iy)$, where $L_{+}=TST$ in $PS{L}_2(\mathbb{Z})$ and $L_{-}=L_{+}^{-1}$. The substitution for $L_{+}$ is $s\circ (1-x)\circ s=1\mathrm{/}x$, also its own inverse; hence both semicircular edges have value $1\mathrm{/}\lambda (iy)>1$. Hence boundary values avoid the slit plane. Now let $w$ lie in the slit plane. Since $\bar{\lambda }$ is onto, $w=\lambda ({\tau }^{\mathrm{\prime }})$ for some ${\tau }^{\mathrm{\prime }}$, whose orbit meets the closure of $Q^{\circ }$ by 1.2; choose $\tau$ in that closure with $\lambda (\tau )=w$ by $\mathrm{\Gamma }(2)$-invariance. By the boundary computation, $\tau$ is not on the vertical or semicircular boundary (those values are negative or greater than $1$, while $w\mathrm{\notin }(-\mathrm{\infty },0]\cup [1,\mathrm{\infty })$), so $\tau \in Q^{\circ }$. Thus $\lambda (Q^{\circ })$ contains the slit plane. Conversely, if $\tau \in Q^{\circ }$ and $\lambda (\tau )$ is real, then $-\bar{\tau }\in Q^{\circ }$ and $\lambda (-\bar{\tau })=\overline{\lambda (\tau )}=\lambda (\tau )$ by 1.1, so $\tau =-\bar{\tau }$ by 2.2, that is $\tau =iy$, and then $\lambda (\tau )\in (0,1)$. Hence $\lambda (Q^{\circ })$ is contained in the slit plane, the restriction is bijective onto it, and being injective holomorphic it is biholomorphic onto the slit plane by [F5].

Consistency check: $\lambda (1+i)=\lambda (T(i))=\frac{\lambda (i)}{\lambda (i)-1}=\frac{1\mathrm{/}2}{-1\mathrm{/}2}=-1$, a value on the vertical boundary, so $1+i$ is not an interior point of $Q^{\circ }$ and its image is not in the slit plane; this is exactly the boundary behaviour that distinguishes the image of the quadrilateral from the twice-punctured plane.