For \(\lambda\in\widehat{\mathbb C}\setminus\{0,1,\infty\}\), let \[ S_\lambda=\widehat{\mathbb C}\setminus\{0,1,\infty,\lambda\}. \]
Prove that every holomorphic automorphism of \(S_\lambda\) is the restriction of a Möbius transformation.
Prove that \(\operatorname{Aut}(S_\lambda)\) always contains a subgroup isomorphic to \[ \mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z, \] and determine how this subgroup permutes \(\{0,1,\infty,\lambda\}\).
Find all \(\lambda\) for which \[ |\operatorname{Aut}(S_\lambda)|>4, \] and identify the automorphism group in each case.
Solution
Let \[ Q_\lambda=\{0,1,\infty,\lambda\}. \] Then \[ \boxed{ \operatorname{Aut}(S_\lambda) \cong \begin{cases} V_4,&\lambda\notin\{-1,\tfrac12,2,\tfrac{1+i\sqrt3}{2},\tfrac{1-i\sqrt3}{2}\},\\[1mm] D_8,&\lambda\in\{-1,\tfrac12,2\},\\[1mm] A_4,&\lambda\in\{\tfrac{1+i\sqrt3}{2},\tfrac{1-i\sqrt3}{2}\}, \end{cases} } \] where \(V_4\cong(\mathbb Z/2)^2\) and \(D_8\) denotes the dihedral group of order \(8\).
1. Every automorphism of \(S_\lambda\) extends to a Möbius transformation preserving \(Q_\lambda\).
Let \[ f:S_\lambda\to S_\lambda \] be biholomorphic. At each point of \(Q_\lambda\), the map \(f\) has an isolated singularity when regarded as a meromorphic map into \(\widehat{\mathbb C}\). It cannot be essential: by the Great Picard theorem, a meromorphic function with an essential singularity assumes every value of the sphere, with at most two exceptions, infinitely often near the singularity, whereas \(f\) omits all four values in \(Q_\lambda\). Thus every puncture is removable or a pole, and \(f\) extends to a meromorphic map \[ F:\widehat{\mathbb C}\to\widehat{\mathbb C}. \] The same argument extends \(f^{-1}\) to a meromorphic map \(G\) of the sphere. On the nonempty open set \(S_\lambda\), \[ G\circ F=\operatorname{id}, \qquad F\circ G=\operatorname{id}. \] By the identity theorem these equalities hold on the whole sphere. Hence \(F\) has degree \(1\) and is a Möbius transformation. Since \(F\) restricts to a bijection of the complements, it permutes \(Q_\lambda\).
Conversely, every Möbius transformation preserving \(Q_\lambda\) restricts to an automorphism of \(S_\lambda\). Thus \[ \operatorname{Aut}(S_\lambda) =\operatorname{Stab}_{\operatorname{PGL}_2(\mathbb C)}(Q_\lambda). \]
2. Three explicit involutions give a Klein four subgroup for every \(\lambda\).
Consider \[ A(z)=\frac{\lambda}{z}, \qquad B(z)=\frac{\lambda(1-z)}{\lambda-z}, \qquad C(z)=\frac{z-\lambda}{z-1}. \] Their actions on \(Q_\lambda\) are respectively \[ A:(0\ \infty)(1\ \lambda), \] \[ B:(0\ 1)(\infty\ \lambda), \] and \[ C:(0\ \lambda)(1\ \infty). \] Each is an involution. These are the three double transpositions, and the product of any two is the third. Hence \[ K=\{1,A,B,C\}\cong\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z. \] This also determines the permutation action requested in part (b).
3. The quotient by \(K\) is the stabilizer of the cross-ratio parameter under the sixfold \(S_3\) action.
A Möbius transformation is determined by the images of three points. Therefore the action of \[ \operatorname{Aut}(S_\lambda) \] on the four-point set \(Q_\lambda\) is faithful, so we may regard it as a subgroup of \(S_4\).
2. Thus the action factors through\[ S_4/K\cong S_3. \]
A permutation of \(Q_\lambda\) is induced by a Möbius self-map of that same four-point set exactly when the reordered quadruple has the same cross-ratio. Consequently there is an exact sequence \[ 1\longrightarrow K \longrightarrow\operatorname{Aut}(S_\lambda) \longrightarrow\operatorname{Stab}_{S_3}(\lambda) \longrightarrow1. \] In particular, \[ |\operatorname{Aut}(S_\lambda)| =4\,|\operatorname{Stab}_{S_3}(\lambda)|. \]
4. The only parameters with nontrivial \(S_3\) stabilizer are the harmonic and equianharmonic values.
Solving \(\lambda=1/\lambda\), \(\lambda=1-\lambda\), \(\lambda=\lambda/(\lambda-1)\), and \(\lambda=1/(1-\lambda)\) gives exactly five points of \[ \widehat{\mathbb C}\setminus\{0,1,\infty\} \] with nontrivial stabilizer under the six cross-ratio transformations.
The three harmonic values \[ \lambda\in\left\{-1,\frac12,2\right\} \] have stabilizer of order \(2\). Hence \[ |\operatorname{Aut}(S_\lambda)|=8. \] These three puncture configurations are Möbius-equivalent. For \(\lambda=-1\), the transformation \[ R(z)=\frac{z-1}{z+1} \] acts on \[ \{0,1,\infty,-1\} \] as the \(4\)-cycle \[ (0\ {-1}\ \infty\ 1). \] Thus the order-\(8\) subgroup of \(S_4\) contains a \(4\)-cycle and the normal Klein four subgroup; it is the dihedral group \[ D_8. \]
The two equianharmonic values are the roots of \[ \lambda^2-\lambda+1=0, \qquad \lambda=\frac{1\pm i\sqrt3}{2}. \] Their \(S_3\) stabilizer has order \(3\), so \[ |\operatorname{Aut}(S_\lambda)|=12. \] The exact sequence in 3 is then \[ 1\to V_4\to\operatorname{Aut}(S_\lambda)\to C_3\to1. \] As a faithful order-\(12\) subgroup of \(S_4\) containing the normal Klein four group and a \(3\)-cycle, this is \[ A_4. \]
For every other \(\lambda\), the \(S_3\) stabilizer is trivial, so \[ \operatorname{Aut}(S_\lambda)=K\cong V_4. \] This proves all three parts.