Let \(\Gamma\subset\operatorname{Aut}(\widehat{\mathbb C})\) be finite. A rational map \[ f:\widehat{\mathbb C}\to\widehat{\mathbb C} \] is a quotient map for \(\Gamma\) if \(f(x)=f(y)\) if and only if \(\gamma(x)=y\) for some \(\gamma\in\Gamma\).
Find a quotient map for the dihedral group \(D_{2m}\) generated by \[ z\mapsto\zeta_m z, \qquad z\mapsto 1/z, \qquad \zeta_m=e^{2\pi i/m}. \]
Let \(\Gamma\subset\operatorname{Aut}(\widehat{\mathbb C})\) be the subgroup leaving \(\{0,1,\infty\}\) invariant. Give generators for \(\Gamma\) and find a quotient map normalized by \[ f(\infty)=\infty,\qquad f(\zeta_6)=0,\qquad f(-1)=1. \]
Solution
For part (a), a quotient map is \[ \boxed{F_m(z)=z^m+z^{-m}.} \]
For part (b), the group is generated by \[ s(z)=1-z, \qquad t(z)=\frac1z, \] and the required normalized quotient map is \[ \boxed{ J(z)=\frac{4(z^2-z+1)^3}{27z^2(1-z)^2}.} \]
1. We use the following finite-group criterion: if a finite group\(G\subset\operatorname{Aut}(\widehat{\mathbb C})\) leaves a nonconstantrational map \(F\) invariant and\[ \deg F=|G|, \]then \(F\) is a quotient map for \(G\).
Invariance gives one implication immediately: \[ y=\gamma(x),\ \gamma\in G \quad\Longrightarrow\quad F(y)=F(x). \]
For the converse, suppose \(F(x)=F(y)=c\). The set of critical values of \(F\) is finite. Also, the set of points with nontrivial \(G\)-stabilizer is finite: each nonidentity Möbius transformation has at most two fixed points, and \(G\) is finite. Hence we may choose regular values \[ c_j\longrightarrow c \] avoiding the images of all points with nontrivial stabilizer.
By the local form of a nonconstant holomorphic map, after passing to a subsequence if necessary there are points \[ x_j\to x, \qquad y_j\to y, \qquad F(x_j)=F(y_j)=c_j. \] For each \(j\), the point \(x_j\) has trivial stabilizer, so its orbit has exactly \(|G|\) distinct points. All of them lie in the fiber \(F^{-1}(c_j)\). Since \(c_j\) is a regular value and \(\deg F=|G|\), that fiber has exactly \(|G|\) points. Therefore \[ F^{-1}(c_j)=G\cdot x_j, \] so \(y_j=\gamma_j(x_j)\) for some \(\gamma_j\in G\).
Because \(G\) is finite, one element \(\gamma\) occurs along a subsequence. Taking limits gives \[ y=\gamma(x). \] Thus the fibers of \(F\) are exactly the \(G\)-orbits.
2. The map\[ F_m(z)=z^m+z^{-m} \]is invariant under the generators\[ r(z)=\zeta_m z, \qquad \iota(z)=1/z \]of \(D_{2m}\).
Since \(\zeta_m^m=1\), \[ F_m(\zeta_m z) =(\zeta_m z)^m+(\zeta_m z)^{-m} =z^m+z^{-m}. \] Also \[ F_m(1/z)=z^{-m}+z^m=F_m(z). \] Thus \(F_m\) is invariant under the whole group generated by \(r\) and \(\iota\).
3. The rational map \(F_m\) has degree \(2m\), equal to\(|D_{2m}|\); hence it is a quotient map.
Write \[ F_m(z)=\frac{z^{2m}+1}{z^m}. \] The numerator and denominator are coprime, so \[ \deg F_m=2m. \] The dihedral group generated by an order-\(m\) rotation and inversion has \(2m\) elements. The criterion in 1 now applies.
Equivalently, one can see the fibers directly: if \(F_m(z)=F_m(w)\) and \(u=z^m\), \(v=w^m\), then \[ u+u^{-1}=v+v^{-1}, \] so \[ (u-v)(uv-1)=0. \] Thus \(w=\zeta_m^kz\) or \(w=\zeta_m^k/z\), exactly the dihedral orbit.
4. The subgroup of Möbius transformations preserving\(\{0,1,\infty\}\) is generated by\[ s(z)=1-z, \qquad t(z)=1/z, \]and has order \(6\).
The map \(s\) swaps \(0\) and \(1\) and fixes \(\infty\), while \(t\) swaps \(0\) and \(\infty\) and fixes \(1\). These two transpositions generate all permutations of the three-point set.
A Möbius transformation is uniquely determined by its values on three distinct points, so the action on \(\{0,1,\infty\}\) is faithful. Hence this subgroup is isomorphic to \(S_3\) and has order \(6\).
5. The rational function\[ J(z)=\frac{4(z^2-z+1)^3}{27z^2(1-z)^2} \]is invariant under both generators \(s\) and \(t\).
Set \[ q(z)=z^2-z+1. \] For \(s(z)=1-z\), \[ q(1-z)=q(z), \] and the denominator \(z^2(1-z)^2\) is symmetric under \(z\leftrightarrow1-z\). Hence \(J(1-z)=J(z)\).
For \(t(z)=1/z\), \[ q(1/z)=\frac{q(z)}{z^2} \] and \[ \left(\frac1z\right)^2 \left(1-\frac1z\right)^2 =\frac{(1-z)^2}{z^4}. \] Therefore \[ J(1/z) =\frac{4q(z)^3/z^6}{27(1-z)^2/z^4} =\frac{4q(z)^3}{27z^2(1-z)^2} =J(z). \] Thus \(J\) is invariant under the whole group.
6. The map \(J\) has degree \(6\), so it is a quotient map for this\(S_3\)-group.
The numerator has degree \(6\) and the denominator degree \(4\). They have no common zero, since the roots of \(z^2-z+1\) are neither \(0\) nor \(1\). Hence \[ \deg J=6. \] By 4 the group also has order \(6\), and by 5 the map is invariant. The criterion in 1 therefore shows that its fibers are exactly the group orbits.
7. The quotient map \(J\) has the required normalization.
Since the numerator has degree two more than the denominator, \[ J(\infty)=\infty. \] For \(\zeta_6=e^{\pi i/3}\) one has \[ \zeta_6^2-\zeta_6+1=0, \] so \[ J(\zeta_6)=0. \] Finally, \[ J(-1) =\frac{4(1+1+1)^3}{27\cdot1\cdot(1+1)^2} =\frac{4\cdot27}{27\cdot4} =1. \]