Subgroup of Möbius transformations which are isometries with respect to the standard metric on the Riemann sphere I'm trying to find which subgroup of Mobius transformations are isometries with respect to the standard metric on the Riemann sphere (the one induced from the Euclidean metric on $\mathbb{R}^3$).
The question hints that the distance on the Riemann sphere corresponds to the distance function on $\mathbb{P}^1$ given by
$$d(L_1,L_2) = 2\sqrt{1 - \frac{|\langle v,w \rangle|^2}{||v||^2||w||^2}}$$
where $ v \in L_1\backslash\{0\}$ and $w \in L_2\backslash\{0\}$, and that a Mobius map corresponds to the action of a matrix $A \in GL_2(\mathbb{C})$ on lines in $\mathbb{C}^2$, and asks me to consider which $2x2$ matrices automatically preserve this expression for $d$.
I honestly have no idea where to start with this question, so I'd really appreciate whatever help you might be able to give.
 A: Note this could be all wrong.
It seems to me the isometries of the unit sphere in $\Bbb R^3$ are rotations. Hence it seems very clear that $R_t$ leaves that metric invariant, if $$R_t(z)=e^{it}z$$for some $t\in\Bbb R$.
I bet $\phi\circ R_t\circ\phi^{-1}$ is also an isometry; not willing to conjecture whether those are the only ones. (Of course there's also the question of giving a more intrinsic description. I bet that it's not hard to see that the $\phi\circ R_t\circ\phi^{-1}$ are precisely the Mobius transformatiions that fix exactly two points... (that would follow if the only transformations fixing $0$ and $\infty$ were the $R_t$.))
A: I wanted to answer this question despite it being 3 years old because I wondered the same thing and saw that the answers here were not clear. So it is in an easy to find spot, here is my answer. The subgroup of $PGL(\mathbb{C},2)$ is $PSU(2)$.
First, let us show that if $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is an isometry then it is unitary. We start by normalizing and so $\det A=ad-bc=1$. As noted by David C. Ullrich isometries must be elliptic, i.e. the characteristic constant $\lambda$ has $|\lambda|=1$ or equivalently $\operatorname{Tr} A=a+d\in(-2,2)$. However, it is not clear what $\phi$ is, in David's answer. If it is taken to be any Möbius transformation, then this is not correct because the fixed points of the isometry are where the axis of rotation intersects the sphere and therefore are antipodal.
If $c=0$, then $\infty$ is fixed and therefore $0$ is also fixed and $b=0$. So, $d=a^{-1}$ and $A$ is unitary.
If $c\neq0$, then the fixed points are $$z_0^+=\frac{a-d+\sqrt{(a+d)^2-4}}{2c}$$ and $$z_0^-=\frac{a-d-\sqrt{(a+d)^2-4}}{2c}.$$ We note that because $A$ defines an elliptic transformation, $(a+d)^2-4<0$. So we rewrite the fixed points as $$z_0^+=\frac{a-d+i\sqrt{4-(a+d)^2}}{2c}$$ and $$z_0^-=\frac{a-d-i\sqrt{4-(a+d)^2}}{2c}.$$ From this post, we have that $$z_0^+\overline{z_0^-}=\frac{|a-d|^2+2i\operatorname{Re}(a-d)\sqrt{4-(a+d)^2}+(a+d)^2-4}{4|c|^2}=-1.$$ Because $-1$ is real and $\sqrt{4-(a+d)^2}$ has to be non-zero we conclude that $\operatorname{Re}(a-d)=0$. So $a+d$ is real and $a-d$ is purely imaginary. Therefore, $d=\bar{a}$ and we get $$|a-\bar{a}|^2+(a+\bar{a})^2-4=4|\operatorname{Im}(a)|^2+4(\operatorname{Re}(a))^2-4=4|a|^2-4=-4|c|^2$$ or simply $|a|^2+|c|^2=1$. Using that $\det A=1$, we get $b=-\bar{c}$. Therefore, $A\in PSU(2)$.
The other direction follows from unitary matrices preserving the inner product used in defining the metric.
