Prove that \[ \lambda(i/2)=12\sqrt2-16. \]
Hint: for \(X_\tau=\mathbb C/(\mathbb Z\oplus\tau\mathbb Z)\), interpret \(\lambda(\tau)\) as the cross-ratio of the critical values of a degree-two map \(X_\tau\to\widehat{\mathbb C}\). For the square torus choose the critical values to be the roots of \(z^4+1\), and use a degree-two map for \(X_{i/2}\) whose critical values are \[ \{-1,-1/\sqrt2,1/\sqrt2,1\}. \]
Solution
Set \[ a=\frac1{\sqrt2}, \qquad \Lambda=\mathbb Z\oplus i\mathbb Z, \qquad \Lambda'=\mathbb Z\oplus \frac i2\mathbb Z, \] and let \[ q:X_i=\mathbb C/\Lambda\longrightarrow X_{i/2}=\mathbb C/\Lambda' \] be the natural degree-two quotient. For \[ c_1=\frac12, \qquad c_2=\frac i4, \qquad c_3=\frac12+\frac i4, \] write \(v_0=f_{i/2}(0)\) and \(v_j=f_{i/2}(c_j)\).
1. The square-torus normalization in the hint may be chosen so that\[ (v_1,v_2,v_3,v_0)=(-a,1,-1,a). \]
Let \(\wp_i\) be the Weierstrass function of the square lattice \(\Lambda\) and set \[ E=\wp_i(1/2)>0, \qquad u(z)=\frac{\wp_i(z)}E. \] For the square lattice, the marked half-period values are \[ u(0)=\infty, \qquad u(1/2)=1, \qquad u(i/2)=-1, \qquad u((1+i)/2)=0. \] Indeed, for \(\tau=i\) the square symmetry gives \(e_2=-e_1\) and \(e_3=0\).
Put \[ \zeta=e^{\pi i/4} \qquad\text{and}\qquad f_i(z)=\zeta\,\frac{u(z)+i}{u(z)-i}. \] At the four ramification points this gives, respectively, \[ \zeta, \quad i\zeta, \quad \zeta^{-1}, \quad -\zeta, \] which are exactly the four roots of \(w^4+1\).
The half-period addition formula for \(\wp_i\), using \((e_1,e_2,e_3)=(E,-E,0)\), gives \[ u(z+i/2)=\frac{1-u(z)}{1+u(z)}, \qquad u(z+1/2)=\frac{u(z)+1}{u(z)-1}. \] Consequently a direct calculation gives \[ f_i(z+i/2)=f_i(z)^{-1}. \]
At \(z=i/4\), evenness of \(u\) and the first translation formula imply \[ u(i/4)\in\{-1-\sqrt2,-1+\sqrt2\}. \] Along the open segment from \(0\) to \(i/2\), the value \(\wp_i(iy)\) is real and has no critical point; it runs from \(-\infty\) to \(-E\). Hence \[ u(i/4)=-1-\sqrt2. \] Substitution into the formula for \(f_i\) gives \[ f_i(i/4)=1. \] The second translation formula then gives \[ u(1/2+i/4)=-1+\sqrt2, \qquad f_i(1/2+i/4)=-1. \]
Finally set \[ g(w)=\frac12\left(w+w^{-1}\right). \] The construction in the hint is \(f_{i/2}\circ q=g\circ f_i\). Since \[ g(\zeta)=a, \qquad g(i\zeta)=-a, \qquad g(1)=1, \qquad g(-1)=-1, \] the marked critical values are \[ v_0=a, \qquad v_1=-a, \qquad v_2=1, \qquad v_3=-1. \]
2. After the Möbius change of coordinate\[ h(w)=\frac1{a-w}, \]the marked critical values are\[ h(v_0)=\infty, \qquad h(v_1)=a, \qquad h(v_2)=-(2+\sqrt2), \qquad h(v_3)=2-\sqrt2. \]
By 1, \[ h(v_0)=h(a)=\infty \] and \[ h(v_1) =\frac1{2a} =\frac1{\sqrt2} =a. \] Also \[ h(v_2) =\frac1{a-1} =-(2+\sqrt2), \] while \[ h(v_3) =\frac1{a+1} =2-\sqrt2. \] Postcomposition by \(h\) does not change the marked cross-ratio.
3. The modular lambda value is\[ \boxed{\lambda(i/2)=12\sqrt2-16}. \]
By the marked cross-ratio formula and 2, \[ \begin{aligned} \lambda(i/2) &=\frac{h(v_3)-h(v_2)}{h(v_1)-h(v_2)}\\ &=\frac{4}{2+\frac{3\sqrt2}{2}}\\ &=\frac{8}{4+3\sqrt2}\\ &=12\sqrt2-16. \end{aligned} \]
4. Q.E.D.
Steps 1 and 2 identify the suitably ordered critical values of the degree-two map, and step 3 evaluates their marked cross-ratio.