Verify that the spherical metric \[ \rho=\frac{2|dz|}{1+|z|^2} \] is invariant under \[ h(z)=\frac{z-1}{z+1}. \]
Show that the rotations \[ g_\theta(z)=e^{i\theta}z \] together with \(h\) generate the full group \[ SU(2)/(\pm I)\subset PSL_2(\mathbb C). \]
Solution
1. For\[ h(z)=\frac{z-1}{z+1} \]one has\[ h'(z)=\frac{2}{(z+1)^2}. \]
Differentiate the quotient: \[ h'(z) =\frac{(z+1)-(z-1)}{(z+1)^2} =\frac{2}{(z+1)^2}. \]
2. The spherical metric is invariant under \(h\).
First compute \[ 1+|h(z)|^2 =1+\frac{|z-1|^2}{|z+1|^2} =\frac{|z+1|^2+|z-1|^2}{|z+1|^2}. \] Since \[ |z+1|^2+|z-1|^2=2(1+|z|^2), \] we obtain \[ 1+|h(z)|^2 =\frac{2(1+|z|^2)}{|z+1|^2}. \] Therefore \[ \frac{2|h'(z)|\,|dz|}{1+|h(z)|^2} = \frac{2\cdot 2|dz|/|z+1|^2} {2(1+|z|^2)/|z+1|^2} =\frac{2|dz|}{1+|z|^2}. \] Thus \(h^*\rho=\rho\).
3. The rotation\[ g_\theta(z)=e^{i\theta}z \]is represented in \(SU(2)\) by\[ R_\theta= \begin{pmatrix} e^{i\theta/2}&0\\ 0&e^{-i\theta/2} \end{pmatrix}, \]while \(h\) is represented projectively by\[ H=\frac1{\sqrt2} \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix}. \]
Both matrices have determinant \(1\) and are unitary, hence lie in \(SU(2)\). Their associated Möbius transformations are \[ z\longmapsto \frac{e^{i\theta/2}z}{e^{-i\theta/2}} =e^{i\theta}z \] and \[ z\longmapsto\frac{z-1}{z+1}, \] respectively. Multiplying a matrix by \(-I\) does not change its Möbius transformation, so these represent the required elements of \(SU(2)/(\pm I)\).
4. Conjugating the rotations by \(H\) gives the one-parameter subgroup\[ X_\beta =H R_\beta H^{-1} = \begin{pmatrix} \cos(\beta/2)&i\sin(\beta/2)\\ i\sin(\beta/2)&\cos(\beta/2) \end{pmatrix}. \]
Since \[ H^{-1} =\frac1{\sqrt2} \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix}, \] direct multiplication gives \[ H R_\beta H^{-1} =\frac12 \begin{pmatrix} e^{i\beta/2}+e^{-i\beta/2}& e^{i\beta/2}-e^{-i\beta/2}\\ e^{i\beta/2}-e^{-i\beta/2}& e^{i\beta/2}+e^{-i\beta/2} \end{pmatrix}, \] which is the displayed matrix.
5. Every matrix in \(SU(2)\) has the form\[ U= \begin{pmatrix} a&b\\ -\overline b&\overline a \end{pmatrix}, \qquad |a|^2+|b|^2=1. \]
If the first row of a unitary determinant-one matrix is \((a,b)\), then the second row must be a unit vector orthogonal to it. The determinant condition fixes that row as \((-\overline b,\overline a)\), and unitarity gives \(|a|^2+|b|^2=1\).
6. Every such matrix admits an Euler factorization\[ U=R_\alpha X_\beta R_\gamma \]for suitable real \(\alpha,\beta,\gamma\).
Multiplication gives \[ R_\alpha X_\beta R_\gamma = \begin{pmatrix} e^{i(\alpha+\gamma)/2}\cos(\beta/2)& i e^{i(\alpha-\gamma)/2}\sin(\beta/2)\\ i e^{-i(\alpha-\gamma)/2}\sin(\beta/2)& e^{-i(\alpha+\gamma)/2}\cos(\beta/2) \end{pmatrix}. \]
Choose \(\beta\in[0,\pi]\) so that \[ \cos(\beta/2)=|a|, \qquad \sin(\beta/2)=|b|. \] When \(a,b\neq0\), write \[ a=|a|e^{iA}, \qquad b=|b|e^{iB}. \] The choices \[ \alpha=A+B-\frac\pi2, \qquad \gamma=A-B+\frac\pi2 \] make the upper row equal to \((a,b)\), and the lower row then has the required \(SU(2)\) form. If \(a=0\) or \(b=0\), the unused phase may be chosen arbitrarily and the same equations give a factorization.
7. Therefore the rotations \(g_\theta\) together with \(h\) generate\(SU(2)/(\pm I)\).
By 4, every \(X_\beta\) lies in the subgroup generated by the \(R_\theta\) and \(H\). By 6, every element of \(SU(2)\) is a product of three matrices from these generated one-parameter subgroups. Passing to projective classes shows that the Möbius transformations \(g_\theta\) and \(h\) generate the full subgroup \(SU(2)/(\pm I)\subset PSL_2(\mathbb C)\).