Let \[ f(z)=\frac{az+b}{cz+d} \] be a Möbius transformation. Show that the number of rational maps \(g:\widehat{\mathbb C}\to\widehat{\mathbb C}\) satisfying \[ g\circ g\circ g\circ g\circ g=f \] is \(1\), \(5\), or \(\infty\). Explain how to determine which alternative holds for a given \(f\).
Solution
1. Every rational map \(g\) satisfying \(g^5=f\) is itself a Möbius transformation.
The map \(g\) is nonconstant because \(f\) is nonconstant. For nonconstant rational maps on the Riemann sphere, degrees multiply under composition, so \[ 1=\deg f=\deg(g^5)=(\deg g)^5. \] Hence \(\deg g=1\), which means that \(g\) is Möbius.
2. Every fifth root \(g\) of \(f\) commutes with \(f\).
Since \(f=g^5\), \[ g\circ f=g^6=f\circ g. \] Consequently \(g\) permutes the fixed points of \(f\): if \(f(p)=p\), then \[ f(g(p))=g(f(p))=g(p). \]
3. If \(f\) is the identity, then it has infinitely many fifth roots.
Fix a primitive fifth root of unity \(\zeta\). For every \(t\in\mathbb C\), define \[ g_t(z)=t+\zeta(z-t)=\zeta z+(1-\zeta)t. \] This is the rotation of the affine plane about \(t\) through multiplier \(\zeta\), so \[ g_t^5(z)=t+\zeta^5(z-t)=z. \] Because \(\zeta\ne1\), distinct values of \(t\) give distinct maps \(g_t\). Thus the identity has infinitely many fifth roots.
4. If \(f\) is nonidentity and has exactly one fixed point, then it has exactly one fifth root.
Conjugating by a Möbius transformation preserves the equation \(g^5=f\) and gives a bijection between its solutions. A nonidentity Möbius transformation with one fixed point is conjugate to \[ F(z)=z+1. \] Let \(G^5=F\). By 2, \(G\) commutes with \(F\). Since \(\infty\) is the unique fixed point of \(F\), the map \(G\) fixes \(\infty\), and hence \[ G(z)=\alpha z+\beta \] for some \(\alpha\ne0\). The commutation relation gives \[ \alpha(z+1)+\beta=G(F(z))=F(G(z))=\alpha z+\beta+1, \] so \(\alpha=1\). Thus \(G(z)=z+\beta\), and therefore \[ G^5(z)=z+5\beta. \] The equality \(G^5=F\) forces \(\beta=1/5\). Hence there is exactly one fifth root.
5. If \(f\) has two distinct fixed points, then it has exactly five fifth roots.
After conjugating the two fixed points to \(0\) and \(\infty\), write \[ F(z)=\lambda z, \qquad \lambda\in\mathbb C^*,\quad \lambda\ne1. \] Let \(G^5=F\). By 2, \(G\) permutes the set \(\{0,\infty\}\).
If \(G\) fixes both points individually, then \(G(z)=\mu z\) for some \(\mu\in\mathbb C^*\), and \[ G^5=F\quad\Longleftrightarrow\quad \mu^5=\lambda. \] The polynomial \(X^5-\lambda\) has exactly five distinct roots in \(\mathbb C\), so this gives exactly five maps.
If instead \(G\) interchanges \(0\) and \(\infty\), then \(G(z)=\mu/z\) for some \(\mu\ne0\). Such a map satisfies \(G^2=\operatorname{id}\) and hence \(G^5=G\), which interchanges \(0\) and \(\infty\); it therefore cannot equal \(F\), which fixes both points. Thus there are no additional roots.
6. The alternative \(\infty\), \(1\), or \(5\) is determined directly from a matrix representing \(f\).
Let \[ A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}, \qquad \det A\ne0, \] represent \(f\) in \(\operatorname{PGL}_2(\mathbb C)\), and put \[ \Delta=(\operatorname{tr}A)^2-4\det A=(a-d)^2+4bc. \] Multiplying \(A\) by a nonzero scalar multiplies \(\Delta\) by its square, so the condition \(\Delta=0\) is independent of the chosen representative.
If \(A\) is scalar, then \(f=\operatorname{id}\), and 3 gives infinitely many roots. Otherwise, the fixed points of \(f\) are precisely the eigenlines of \(A\). Hence a non-scalar \(A\) has exactly one fixed point when its characteristic polynomial has a repeated root, equivalently when \(\Delta=0\); then 4 gives exactly one fifth root. If \(\Delta\ne0\), then \(A\) has two distinct eigenlines, so \(f\) has two distinct fixed points and 5 gives exactly five fifth roots.