A circular triangle is a region \(T\subset\widehat{\mathbb C}\) bounded by three circular arcs or lines. Let \(T_1,T_2\subset\widehat{\mathbb C}\) be two circular triangles with the same interior angles in the same cyclic order. Prove that there exists a Möbius transformation sending \(T_1\) to \(T_2\).
Solution
1. Label the vertices of \(T_i\) as \(a_i,b_i,c_i\) in the given cyclic order, and write the corresponding interior angles as\[ \pi\alpha,\qquad \pi\beta,\qquad \pi\gamma. \]For each \(i\) there is a conformal map\[ f_i:\mathbb H\longrightarrow T_i \]whose boundary extension sends \(0,1,\infty\) to \(a_i,b_i,c_i\), respectively.
The boundary of a circular triangle is a Jordan curve made from three circular arcs, so the Riemann mapping theorem and the Carathéodory boundary theorem give a conformal map from \(\mathbb H\) which extends homeomorphically to the boundary. The three preimages of the vertices occur on \(\widehat{\mathbb R}\) in the same cyclic order. Precomposing by an element of \(\operatorname{PSL}_2(\mathbb R)\) sends them to \(0,1,\infty\) in that order.
2. The Schwarzian derivatives of the two maps in 1 are equal:\[ S f_1=S f_2=Q(\alpha,\beta,\gamma), \]where\[ Q(\alpha,\beta,\gamma) =\frac12\left[ \frac{1-\alpha^2}{z^2} +\frac{1-\beta^2}{(z-1)^2} +\frac{1-\gamma^2-(2-\alpha^2-\beta^2)}{z(z-1)} \right]dz^2. \]
Schwarz reflection across each circular side continues \(f_i\) across the corresponding interval of \(\widehat{\mathbb R}\). Since reflection in a generalized circle is anti-Möbius and the Schwarzian is unchanged by postcomposition with a Möbius transformation, \(S f_i\) continues to a meromorphic quadratic differential on \(\widehat{\mathbb C}\) with possible poles only at \(0,1,\infty\).
At a prevertex whose image has interior angle \(\pi\delta\), the local reflection monodromy gives the same double-pole term as the model \(z\mapsto z^\delta\); hence the coefficient of \(dz^2/z^2\) is \[ \frac{1-\delta^2}{2}. \] The same statement includes the parabolic limiting case \(\delta=0\). Thus the double-pole coefficients at \(0,1,\infty\) are respectively \((1-\alpha^2)/2\), \((1-\beta^2)/2\), and \((1-\gamma^2)/2\).
There is a unique quadratic differential on \(\widehat{\mathbb C}\) having only those three double poles and those three leading coefficients. Indeed, after subtracting the displayed \(Q(\alpha,\beta,\gamma)\) the difference has at most simple poles at \(0,1,\infty\). Write it on \(\mathbb C\) as \(R(z)\,dz^2\). Then \[ R(z)=\frac{A}{z}+\frac{B}{z-1}+P(z) \] for a polynomial \(P\). Having at most a simple pole at \(\infty\) as a quadratic differential means \(R(z)=O(z^{-3})\). Hence \(P=0\), while the coefficients of \(z^{-1}\) and \(z^{-2}\) in the expansion at infinity force \(A+B=0\) and \(B=0\). Thus \(A=B=0\). The difference is therefore zero, and \(S f_i=Q(\alpha,\beta,\gamma)\) for both \(i\).
3. If two locally univalent holomorphic functions have the same Schwarzian derivative, then one is obtained from the other by postcomposition with a Möbius transformation.
Let \[ g=f_2\circ f_1^{-1}:T_1\longrightarrow T_2. \] The Schwarzian chain rule gives \[ S(f_2) =S(g\circ f_1) =(Sg\circ f_1)(f_1')^2+S(f_1). \] By 2 the two outer Schwarzians are equal, and \(f_1'\neq0\), so \(Sg=0\).
On any coordinate disk put \(u=g''/g'\). The equation \(Sg=0\) is \[ u'-\frac12u^2=0. \] If \(u=0\), then \(g\) is affine. Otherwise \[ u(z)=-\frac{2}{z-c} \] for a constant \(c\), so integration gives \[ g(z)=A+\frac{B}{z-c}, \] again a Möbius transformation. Thus \(g\) agrees locally with a Möbius transformation; by analytic continuation on the connected region \(T_1\), the same Möbius transformation works everywhere.
4. Consequently there is a Möbius transformation \(M\) with\[ M(T_1)=T_2. \]
By 3, \(f_2=M\circ f_1\) for some Möbius transformation \(M\). Since each \(f_i\) maps \(\mathbb H\) onto \(T_i\), \[ M(T_1)=M(f_1(\mathbb H))=f_2(\mathbb H)=T_2. \]