Let \[ f_n:\widehat{\mathbb C}\to\widehat{\mathbb C} \] be an unbounded sequence in \(\operatorname{Aut}(\widehat{\mathbb C})\). Prove that, after passing to a subsequence, there are \(p,q\in\widehat{\mathbb C}\) such that \[ f_n(z)\longrightarrow p \] uniformly on compact subsets of \(\widehat{\mathbb C}\setminus\{q\}\). Give an example for which \(p=q=0\).
Solution
1. Choose matrix representatives \(A_n\in SL_2(\mathbb C)\) for \(f_n\).After passing to a subsequence, their norms satisfy\[ \|A_n\|\longrightarrow\infty. \]
The quotient map \(SL_2(\mathbb C)\to PSL_2(\mathbb C)\) has finite fibers \(\{\pm I\}\). If the norms of determinant-one representatives stayed bounded, a subsequence would converge in the finite-dimensional matrix space. Since determinant is continuous and equal to \(1\), the limit would still lie in \(SL_2(\mathbb C)\), so the corresponding Möbius transformations would have a convergent subsequence in \(\operatorname{Aut}(\widehat{\mathbb C})\). This contradicts unboundedness. Hence an unbounded subsequence of norms exists; pass to it.
2. After normalizing\[ B_n=\frac{A_n}{\|A_n\|}, \]and passing to a further subsequence, one has \(B_n\to B\) for a nonzerorank-one matrix \(B\).
The matrices \(B_n\) lie on the unit sphere of the finite-dimensional matrix space, which is compact. Thus a subsequence converges to some \(B\) with \(\|B\|=1\), so \(B\neq0\). Moreover \[ \det B_n =\frac{\det A_n}{\|A_n\|^2} =\frac1{\|A_n\|^2} \longrightarrow0. \] Hence \(\det B=0\). A nonzero singular \(2\times2\) matrix has rank one.
3. Let\[ q=\mathbb P(\ker B), \qquad p=\mathbb P(\operatorname{im}B). \]Then \(f_n(z)\to p\) for every \(z\neq q\).
Multiplying a matrix by a nonzero scalar does not change its projective transformation, so \[ f_n([v])=[A_nv]=[B_nv]. \] If \([v]\neq q\), then \(Bv\neq0\). Since \(B\) has rank one, every nonzero \(Bv\) spans the same line \(\operatorname{im}B\). Thus \[ [B_nv]\longrightarrow[Bv]=p. \]
4. The convergence is uniform on compact subsets of\(\widehat{\mathbb C}\setminus\{q\}\).
Let \(K\) be such a compact set. Choose unit-vector representatives for points of \(K\) and consider the compact set of all such representatives. Because \(K\) does not meet the projective kernel of \(B\), there is a constant \(c>0\) such that \[ \|Bv\|\ge c \] for every such unit representative \(v\). Operator-norm convergence \(B_n\to B\) is uniform on the unit sphere, so \(B_nv\to Bv\) uniformly for \([v]\in K\), and \(B_nv\) stays uniformly away from zero. Projectivization is uniformly continuous on that set, hence \[ [B_nv]\longrightarrow[Bv]=p \] uniformly on \(K\).
5. The sequence\[ f_n(z)=\frac1{nz} \]has \(p=q=0\).
Each \(f_n\) is a Möbius transformation of \(\widehat{\mathbb C}\). For every \(z\neq0\), \[ f_n(z)=\frac1{nz}\longrightarrow0, \] and also \(f_n(\infty)=0\). The convergence is uniform on compact subsets of \(\widehat{\mathbb C}\setminus\{0\}\) because such a compact set stays a positive spherical distance from \(0\). At the exceptional point, \(f_n(0)=\infty\). Thus both the limiting point and the exceptional point are \(0\).
For example, determinant-one representatives are \[ A_n= \begin{pmatrix} 0&i/\sqrt n\\ i\sqrt n&0 \end{pmatrix}, \] whose norms tend to infinity, so this is indeed an unbounded sequence in \(PSL_2(\mathbb C)\).