Let $\mathbb{C} \cup \{\infty\}$ be the Riemann sphere and let$$S^2 = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$$be the unit sphere in $\mathbb{R}^3$. The stereographic projection with center $(0, 0, 1) \in S^2$ is an explicit bijection between the unit sphere and the Riemann sphere provided by the following map:$$p: S^2 \overset{\sim}{\to} \mathbb{C} \cup \{\infty\}, \quad (x, y, z) \mapsto {{x + iy}\over{1 – z}}.$$For any invertible matrix$$g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{GL}_2(\mathbb{C})$$there is an associated fractional linear transformation$$\phi_g: \mathbb{C} \cup \{\infty\} \to \mathbb{C} \cup \{\infty\},$$of the Riemann sphere, given by$$\phi_g(z) := {{az + b}\over{cz + d}} \quad (\text{in particular, we have }\phi_g(\infty) = {a\over c} \text{ and }\phi_g\left(-{d\over c}\right) = \infty, \text{ if }c \neq 0).$$One may use the bijection $p$ to transport the map $\phi_g$ to the unit sphere. That is, we consider the following composite map$$F_g: S^2 \overset{p}{\to} \mathbb{C} \cup \{\infty\} \overset{\phi_g}{\to} \mathbb{C} \cup \{\infty\} \overset{p^{-1}}{\to} S^2.$$Question. What is the precise relationship between the following two things?
- A continuous and surjective group homomorphism $\pi: \text{SU}(2) \twoheadrightarrow \text{SO}(3)$ with kernel $\{\pm \text{Id}\}$.
- The map $g \mapsto F_g$.
Best Answer
The fact that using matrices for fractional linear transformations creates a group action (turns matrix multiplication into function composition) may seem like divine inspiration if projective spaces are not mentioned to motivate the situation.
Given a vector space $V$, there is an associated projective space $\mathbb{P}^1(V)$ which collects together all of the "lines," i.e. one-dimensional subspaces. When $V=\mathbb{R}^n$ or $\mathbb{C}^n$ we write $\mathbb{RP}^{n-1}$ or $\mathbb{CP}^{n-1}$. In the case that $n=2$, we have a complex projective line $\mathbb{CP}^1$.
We may put an equivalence relation on $\mathbb{C}^2\setminus0$, saying $v\sim w$ whenever $v$ and $w$ generate the same complex one-dimensional subspace (i.e. $\mathbb{C}v=\mathbb{C}w$). Since $(x,y)\sim\lambda(x,y)$ for any $\lambda\ne0$, we may scale any vector to $(1,z)$ (if its $x$ coordinate was nonzero) or $(0,1)$ (if $x=0$). Thus, one way of interpreting $\mathbb{CP}^1$ is the Riemann sphere $\widehat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$. Topologically, this is indeed a sphere, and is the one-point compactification of the complex plane $\mathbb{C}$. To speak in a more uniform manner, we can interpret $(1,\infty)$ to be the special point $(0,1)$.
Since $\mathrm{GL}_2(\mathbb{C})$'s action on $\mathbb{C}^2\setminus 0$ commutes with scalar multiplication, and the one-dimensional subspaces (equivalence relations of $\sim$) are precisely the orbits of $\mathbb{C}^\times$'s action, there is an induced action of $\mathrm{GL}_2(\mathbb{C})$ on $\mathbb{CP}^1$. Indeed, if $(x,y)\sim(1,z)$ we have
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot (x,y)=(ax+by,cx+dy)\sim\left(\frac{az+b}{cz+d},1\right) $$
Thus, using $\mathbb{C}\cup\{\infty\}$ as a set of representatives for the elements of $\mathbb{CP}^1$, the induced action of $\mathrm{GL}_2(\mathbb{C})$ is by linear fractional transformations.
Stereographic projection creates a diffeomorphism $\widehat{\mathbb{C}}\simeq S^2\subset\mathbb{R}^3$. Explicitly, think of $\mathbb{C}\times\mathbb{R}$ as three-space $\mathbb{R}^3$. The line through the north pole $(0,1)$ and a complex number $(z,0)$ is parametrized by $t(z,0)+(1-t)(0,1)=(tz,1-t)$. In order for a point on the line to be on the unit sphere centered at the origin, we require $t^2|z|^2+(1-t)^2=1$ which after some algebraic manipulation implies that $t=2/(|z|^2+1)$, and therefore the point is
$$ z\mapsto \left(\frac{2z}{|z|^2+1},\frac{|z|^2-1}{|z|^2+1}\right).$$
This creates a map $\mathbb{C}\to S^2$. We can extend this to $\widehat{\mathbb{C}}$ if we can define where to send $\infty$, which can be done by taking the limit $z\to\infty$, yielding the north pole $(0,1)$.
Thus, we can transport the action of $\mathrm{GL}_2(\mathbb{C})$ over to $S^2$, but not every such matrix will act in a way that comes from rotating $S^2$ within $\mathbb{R}^3$. In fact, it will do so precisely when it acts by an isometry on the sphere $S^2$. (Rotations restrict to isometries of course, and it is a standard exercise to prove isometries come from orthogonal transformations, but $\mathrm{GL}_2(\mathbb{C})\to\mathrm{O}(3)$ must land in the connected component $\mathrm{SO}(3)$ since $\mathrm{GL}_2(\mathbb{C})$ is connected.)
Over at Qiaochu's question here I provide an elementary explanation of why matrices from $\mathrm{SU}(2)$ do indeed induce rotations on the space $\mathbb{R}^3$ via stereographic projection. I also mention how the three stabilizer subgroups of $\mathrm{SU}(2)$ associated to the special points $0,1,i$ correspond to one-parameter subgroups of coordinate axis rotations in $\mathbb{R}^3$ (and also to Pauli matrices), which implies the map $\mathrm{SU}(2)\to \mathrm{SO}(3)$ is surjective, and also mention the kernel is $\{\pm I\}$.
To sum up, have a map $F:\mathrm{GL}_2(\mathbb{C})\to\mathrm{Diff}(S^2)$ which restricts to a map $\mathrm{SU}(2)\to\mathrm{SO}(3)$ (note that $\mathrm{SO}(3)$ can be identified with a subgroup of $\mathrm{Diff}(S^2)$ due to the earlier remark about isometries of $S^2$ corresponding to transformations of $\mathbb{R}^3$).
Say we have two solutions $\alpha,\beta$ so to the short exact sequence
$$ 1\to K\to G\to H\to 1. $$
That is, $\alpha,\beta:G\to H$ are onto Lie group homomorphisms with kernel $K$ (a Lie subgroup). Then they are related by an automorphism $\gamma$ of $H$. In particular, consider $\overline{\alpha},\overline{\beta}:G/K\to H$ (afforded by the first isomorphism theorem) and define $\gamma=\overline{\beta}\circ\overline{\alpha}^{-1}$, so that $\beta=\gamma\circ\alpha$.
In the case of $H=\mathrm{SO}(3)$, every automorphism $\gamma$ is inner, i.e. is conjugation $\gamma_g(h)=ghg^{-1}$ for some rotation $g\in\mathrm{SO}(3)$. To see this, note that any automorphism $\gamma$ of $\mathrm{SO}(3)$ induces one of its lie algebra $\mathfrak{so}(3)\cong(\mathbb{R}^3,\times)$ which must be by a rotation $g$ since $\mathrm{SO}(3)$ is the symmetry group of the three-dimensional cross product. Note that $\mathfrak{so}(3)\cong(\mathbb{R}^3,\times)$ is not just an isomorphism of lie algebras but also $\mathrm{SO}(3)$-representations, so $\gamma$ and $g$ agree acting on $\mathfrak{so}(3)$ (where the adjoint action is used). Then since the matrix exponential $\exp:\mathfrak{so}(3)\to\mathrm{SO}(3)$ is surjective (as $\mathrm{SO}(3)$ is compact, but it can be proved as an elementary exercise too), their actions can be lifted to $\gamma$ and conjugateion-by-$g$ on $\mathrm{SO}(3)$, which must agree too.
(I'm sure there's a better way than this, but this is what comes to mind.)
By the way, other matrices in $G=\mathrm{GL}_2(\mathbb{C})$ can be interpreted as acting on $S^2\subset\mathbb{R}^3$ too, but we require some more freedom. In particular, we have an Iwasawa decomposition $G=KAN$ with subgroups $K=\mathrm{SU}(2)$, $A$ the positive real diagonal matrices, and $N$ the unitriangular complex matrices. We've already seen the interpretation of $\mathrm{SU}(2)$ acting, $A$ acts on $\mathbb{C}$ via scalings / dilations / homotheties, which can be interpreted as the restriction of corresponding scalings of $\mathbb{R}^3$, and finally $N$ acts on $\mathbb{C}$ by translations, which can be interpreted as restrictions of planar translations acting on $\mathbb{R}^3$ (remember we're identifying $\mathbb{R}^3$ and $\mathbb{C}\times\mathbb{R}$).
This combo of rotations, translations and scalings is the full menu of transformations seen in the Möbius Transformations Revealed video, in case anybody wondered.