Let \(T\subset\mathbb C\) be the triangle with vertices \[ 0,\qquad 1,\qquad 1+i. \] Give an explicit formula for a conformal map \[ f:T\to\mathbb H \] in terms of the Weierstrass \(\wp\)-function for a suitable lattice.
Solution
Take the square lattice \[ \Lambda=2\mathbb Z\oplus 2i\mathbb Z \] and write \(\wp=\wp_\Lambda\). Then an explicit conformal map is \[ \boxed{f(z)=-\wp(z)^2.} \]
1. The square-lattice Weierstrass function satisfies\[ \wp(iz)=-\wp(z), \qquad \wp(\overline z)=\overline{\wp(z)}. \]
Multiplication by \(i\) permutes \(\Lambda\). Reindexing the defining series of the Weierstrass function therefore gives \[ \wp(iz)=i^{-2}\wp(z)=-\wp(z). \] Likewise, complex conjugation permutes \(\Lambda\), and conjugating the defining series term by term gives \[ \wp(\overline z)=\overline{\wp(z)}. \] We will also use the usual identities \[ \wp(z+\lambda)=\wp(z) \quad(\lambda\in\Lambda), \qquad \wp(-z)=\wp(z). \]
2. The three sides of \(T\) are carried by \(f\) to the extended real axis.
On the horizontal side \(z=x\), \(0<x<1\), 1 gives \[ \wp(x)=\overline{\wp(x)}, \] so \(\wp(x)\) is real.
On the vertical side write \(z=1+iy\), \(0<y<1\). Then \[ \overline z=1-iy=2-z. \] Since \(2\in\Lambda\), periodicity and evenness give \[ \wp(\overline z)=\wp(2-z)=\wp(-z)=\wp(z). \] Together with 1, this again says that \(\wp(z)\) is real.
Finally, on the diagonal side write \(z=t(1+i)\), \(0<t<1\). Then \[ \overline z=-iz, \] and hence \[ \overline{\wp(z)} =\wp(\overline z) =\wp(-iz) =\wp(iz) =-\wp(z). \] Thus \(\wp(z)\) is purely imaginary there.
Consequently \(-\wp(z)^2\) is real on every open side. At \(z=0\) it tends to \(\infty\). At the opposite vertex, \[ i(1+i)=-1+i=(1+i)-2, \] so periodicity and 1 give \[ \wp(1+i)=\wp(i(1+i))=-\wp(1+i), \] and therefore \(\wp(1+i)=0\).
3. The map \(f\) is injective on the interior of \(T\).
Since \(\wp\) is even and has degree two on \(\mathbb C/\Lambda\) (one double pole), \[ \wp(a)=\wp(b) \quad\Longleftrightarrow\quad b\equiv \pm a\pmod\Lambda. \] Suppose \(z,w\in T\) and \(f(z)=f(w)\). Then \[ \wp(w)=\pm\wp(z). \] If the sign is positive, then \[ w\equiv\pm z\pmod\Lambda. \] If it is negative, 1 gives \(-\wp(z)=\wp(iz)\), so \[ w\equiv\pm iz\pmod\Lambda. \]
Write \[ z=x+iy, \qquad w=u+iv, \qquad 0<y<x<1, \quad 0<v<u<1. \] Because every lattice coordinate is an even integer, these inequalities rule out all four congruences except \(w\equiv z\): the coordinate differences in \(w-z\) lie in \((-1,1)\); both coordinates of \(w+z\) lie in \((0,2)\); the real coordinate of \(w-iz\) is \(u+y\in(0,2)\); and the imaginary coordinate of \(w+iz\) is \(v+x\in(0,2)\). Hence the only possible lattice congruence is \(w=z\).
4. The image of the interior of \(T\) lies in the upper half-plane.
By 1, \[ f(\overline z)=\overline{f(z)}. \] If \(f(z)\) were real at an interior point, then \[ f(z)=f(\overline z). \] The same fiber argument as in 3 would force \[ \overline z\equiv \pm z \quad\text{or}\quad \overline z\equiv\pm iz \pmod\Lambda, \] which is impossible for \(0<\operatorname{Im}z<\operatorname{Re}z<1\). Thus \(\operatorname{Im}f\) never vanishes in \(T\) and, since \(T\) is connected, has constant sign.
For \(z=re^{i\pi/8}\) with \(r>0\) small, the Laurent expansion \[ \wp(z)=z^{-2}+O(z^2) \] gives \[ f(z)=-z^{-4}+O(1)=ir^{-4}+O(1), \] whose imaginary part is positive for sufficiently small \(r\). Therefore \[ f(T)\subset\mathbb H. \]
5. The inclusion in 4 is onto.
The function \(f\) is holomorphic on \(T\) and injective by 3, so \(f(T)\) is a nonempty open subset of \(\mathbb H\).
By 2, \(f\) extends continuously to every point of \(\partial T\setminus\{0\}\) with real boundary value, while \[ f(z)\longrightarrow\infty \qquad(z\to0,\ z\in T). \] Hence a sequence in \(T\) that leaves every compact subset of \(T\) has image approaching the boundary \[ \mathbb R\cup\{\infty\} \] of \(\mathbb H\) in the Riemann sphere. Thus \(f:T\to\mathbb H\) is proper.
Its image is therefore closed in \(\mathbb H\) as well as open. Since \(\mathbb H\) is connected, \[ f(T)=\mathbb H. \] Together with injectivity, this proves that \[ \boxed{f(z)=-\wp_{\,2\mathbb Z\oplus2i\mathbb Z}(z)^2} \] is the required conformal map.