# Explicit fractional linear transformations which rotate the Riemann sphere about the $x$-axis

Background: Let $S^2$ denote the unit sphere in $\mathbb{R}^3$. By "stereographic projection", I mean the mapping from $S^2$ (remove the north pole) to the complex plane which sends \begin{align*} \begin{bmatrix} x \\ y \\ t \end{bmatrix} \in S^2 \mapsto z = \frac{x+iy}{1-t} \in \mathbb{C}. \end{align*} The inverse mapping, from the complex plane to the sphere, is then given by \begin{align*}z = x+iy \in \mathbb{C} \mapsto \frac{2}{|z|^2+1} \cdot \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} + \left(1 - \frac{2}{|z|^2 + 1}\right) \cdot \begin{bmatrix} 0 \\0 \\ 1 \end{bmatrix} \in S^2. \end{align*}

Using the above correspondences, we can view a transformation of $\mathbb{C}$ as a transformation of $S^2$, or vice versa. I was especially interested to learn from this question that rotations of the $2$-sphere, i.e. the transformations corresponding to matrices in $SO(3)$, actually correspond to a subset of the fractional linear transformations $z \mapsto \frac{az + b}{cz + d}$. Precisely, $z \mapsto \frac{az + b}{cz + d}$ corresponds to a rotation of $S^2$ if and only if $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$ belongs (up to scalar multiple, I guess) to $U(2)$, the group of $2 \times 2$ unitary matrices. In particular, $z \mapsto \frac{1}{z}$ corresponds to rotating $S^2$ by $180$ degrees about the $x$-axis.

Upon learning the above fact, I wondered whether I could write down a unitary matrix whose fractional linear transformation corresponds to rotation by a given angle about the $x$-axis. After a while I was able to convince myself that the fractional linear transformation corresponding to $$U_\theta = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \\ \end{bmatrix} \in SU(2)$$ i.e. the mapping $$f_\theta(z) = \frac{\cos \theta z + i \sin \theta}{i \sin \theta z + \cos \theta}$$ corresponds to rotation of $S^2$ through an angle of $\theta$ degrees about the $x$-axis, i.e. to the transformation given by the matrix $$R_{2 \theta} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(2 \theta) & - \sin( 2 \theta) \\ 0 & \sin(2 \theta) & \cos(2 \theta) \\ \end{bmatrix}.$$

The fact that the sphere spins around twice as $\theta : 0 \to 2 \pi$ I guess has something to do with the fact that $SU(2)$ is supposed to double-cover $SO(3)$.

Question: How can I efficiently prove that $f_\theta$ and $R_{2 \theta}$ really are the same transformation in different representations? To be sure, one can take a generic point $(x,y,t) \in S^2$ and check that the result of applying stereographic projection and then $f_\theta$ agrees with result of applying $R_{2 \theta}$ and then stereographic projection. However, doing this for specific points $(x,y,t)$ is more or less how I came up with the above formulae, and even then the algebra seemed to get pretty involved. Can somebody provide a more enlightening proof?

• You can first find a rotation $L$ along $y$-axis, taking the $x$-axis to the $z$-axis. Then the rotation you want is $L\Theta L^{-1}$, where $\Theta$ is a rotation along the $z$-axis, which is easier to write down: ($\Theta (z)= e^{i\theta}z)$. – user99914 Jun 8 '15 at 2:55
• @John: That is a very good idea! – Mike F Jun 8 '15 at 3:31

John's comment above is really an answer, modulo ploughing through the actual calculations. In this post, I'll apply his observation to arrive at all sorts of crazy formulae.

Notations/Conventions:

• We denote the unit sphere in $\mathbb{R}^3$ by $S^2$ and the standard unit vectors by $e_1,e_2,e_3$.
• Given a point $e \in S^2$ and an angle of rotation $\omega \in \mathbb{R}$, let $R(e,\omega) \in SO(3)$ denote counterclockwise rotation by $\omega$ about $e$. Here, "counterclockwise" is defined with respect to an observer who looks down at $e$ from outer space (so the observer faces in the $-e$ direction). Explicitly, $$R(e,\omega) = A \begin{bmatrix} \cos \omega & -\sin \omega & 0 \\ \sin \omega & \cos \omega & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} A^{-1}$$ for any $A \in SO(3)$ with $Ae_3=e$. In particular: \begin{align*} R(e_1,\omega) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \omega & -\sin \omega \\ 0 & \sin \omega & \cos \omega \\ \end{bmatrix} \\ R(e_2,\omega) = \begin{bmatrix} \cos \omega & 0 & \cos \omega \\ 0 & 1 & 0 \\ -\sin \omega & 0 & \sin \omega \\ \end{bmatrix} \\ R(e_3,\omega) = \begin{bmatrix} \cos \omega & -\sin \omega & 0 \\ \sin \omega & \cos \omega & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}. \end{align*}
• Given an azimuthal angle $\theta \in \mathbb{R}$ and a polar angle $\phi \in [0,\pi]$, the point $e$ with spherical coordinate $(\theta,\phi)$ is defined, as usual, so that $\theta$ is an argument for the projection of $e$ into the $xy$-plane, and $\phi$ is the smallest, nonnegative angle between $e$ and the north-pointing vector $e_3$. Note we can get $e$ by applying two rotations to $e_3$ as follows: $$e = R(e_3,\theta ) \ R(e_2,\phi) \ e_3.$$
• Denote the action of $GL(2,\mathbb{C})$ on $\mathbb{C} \cup \{\infty\}$ by fractional linear transformations by juxtaposition. So, $$A z = \frac{az+b}{cz+d}$$ when $A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \in GL(2,\mathbb{C})$, $z \in \mathbb{C} \cup \{\infty\}$.
• Let $P : S^2 \to \mathbb{C} \cup \{ \infty\}$ denote stereographic projection. In Cartesian coordinates, $$P (x,y,t) = \begin{cases} \frac{x+iy}{1-t} & \text{ if } (x,y,t) \neq e_3 \\ \infty & \text{ if } (x,y,t) = e_3 \end{cases}.$$ In spherical coordinates, $$P (\theta,\phi) = \begin{cases} e^{i \theta} \cot(\frac{\phi}{2}) & \text{ if } \phi \neq 0 \\ \infty & \text{ if } \phi = 0 \\ \end{cases}.$$

It is convenient to represent an element of $SO(3)$ in the form $R(e,\theta)$ described above because this makes conjugation easy to understand.

Easy Fact: If $e \in S^2,\omega \in \mathbb{R}$ and $A \in SO(3)$, then $$A \ R(e,\omega) \ A^{-1} = R(Ae,\omega).$$ That is, when one conjugates, the angle of rotation does not change, whereas the vector about which one is rotating is transformed by the conjugating matrix.

Now I'll use John's idea to calculate, for each 3d rotation $R \in SO(3)$, a unitary $U \in SU(2)$ such that the diagram $$\begin{matrix} S^2 & \overset{ P }{\longrightarrow} & \mathbb{C} \cup \{\infty\} \\ \downarrow^R & & \downarrow^U \\ S^2 & \overset{ P }{\longrightarrow} & \mathbb{C} \cup \{\infty\} \\ \end{matrix}$$ The point here is that once one learns the fact

"If $S^2$ and $\mathbb{C} \cup \{ \infty\}$ are identified via stereographic projection, then transformations in $SO(3)$ become identified with a subset of the Mobius transformations of $\mathbb{C} \cup \{ \infty\}$, namely the ones implemented by a matrix in $SU(2)$,"

one naturally wants to be able to explicitly write down a unitary corresponding to a specific rotation of the sphere.

Main Result: Fix an angle of rotation $\omega$. Fix an azimuthal angle $\theta \in \mathbb{R}$, a polar angle $\phi \in [0,\pi]$ and let $e \in S^2$ be the point with spherical coordinate $(\theta,\phi)$. Then, the unitary $$U(e, \omega) = \begin{bmatrix} \cos(\frac{\omega}{2}) + i \cos(\phi) \sin(\frac{\omega}{2}) & ie^{i \theta} \sin(\phi) \sin (\frac{\omega}{2}) \\ i e^{-i \theta} \sin (\phi) \sin(\frac{\omega}{2}) & \cos(\frac{\omega}{2}) -i\cos(\phi) \sin (\frac{\omega}{2}) \\ \end{bmatrix}$$ is such that the diagram $$\begin{matrix} S^2 & \overset{ P }{\longrightarrow} & \mathbb{C} \cup \{\infty\} \\ \downarrow^{R(e,\omega)} & & \downarrow^{U(e,\omega)} \\ S^2 & \overset{ P }{\longrightarrow} & \mathbb{C} \cup \{\infty\} \\ \end{matrix}$$ is commutative. Some special cases:

1. Putting $\theta =0$, $\phi = \frac{\pi}{2}$ so $e=e_1$, we get $$U(e_1, \omega) = \begin{bmatrix} \cos(\frac{\omega}{2}) & i \sin(\frac{\omega}{2}) \\ i \sin(\frac{\omega}{2}) & \cos(\frac{\omega}{2}) \\ \end{bmatrix}.$$
2. Putting $\theta = \phi = \frac{\pi}{2}$ so $e=e_2$, we get $$U(e_2, \omega) = \begin{bmatrix} \cos(\frac{\omega}{2}) & - \sin(\frac{\omega}{2}) \\ \sin(\frac{\omega}{2}) & \cos(\frac{\omega}{2}) \\ \end{bmatrix}.$$
3. Putting $\phi = 0$ so $e=e_3$, we get $$U(e_3, \omega) = \begin{bmatrix} e^\frac{i \omega}{2} & 0 \\ 0 & e^{- \frac{i \omega}{2}} \\ \end{bmatrix}.$$

Proof Sketch:

• First we check the claim when $e=e_3$ and $\omega$ is arbitrary. We should chase an arbitrary $(x,y,t) \in S^2$ both ways around the diagram. Applying $P \ R(e_3,\omega)$, we get \begin{align*} P \ \begin{bmatrix} \cos \omega & -\sin \omega & 0 \\ \sin \omega & \cos \omega & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ t \end{bmatrix}= P \ \begin{bmatrix} \cos(\omega)x -\sin (\omega) y \\ \sin (\omega) x + \cos (\omega) y \\ t \\ \end{bmatrix} = \frac{e^{i \omega}(x + iy)}{1-t} \end{align*} Applying $U(e_3, \omega) \ P$, we get $$\begin{bmatrix} e^\frac{i \omega}{2} & 0 \\ 0 & e^{- \frac{i \omega}{2}} \\ \end{bmatrix} \frac{x+iy}{1-t} = \frac{e^\frac{i\omega}{2}\frac{x+iy}{1-t} }{e^{- \frac{i \omega}{2}}} = \frac{e^{i \omega}(x + iy)}{1-t},$$ so the result holds in this case.
• Next we check the claim when $e = e_1$ and $\omega = - \frac{\pi}{2}$. In this case: \begin{align*} U(e_1,-\frac{\pi}{2}) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & -i \\ -i & 1 \\ \end{bmatrix} && R(e_1,-\frac{\pi}{2}) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1\\ 0 & -1 & 0 \\ \end{bmatrix} \end{align*} Applying $P \ R(e_1,-\frac{\pi}{2})$ to $(x,y,t)$, we get \begin{align*} P (x,t,-y) = \frac{x+it}{1+y} \end{align*} Applying $U(e_1,-\frac{\pi}{2}) \ P$, we get $$\begin{bmatrix} 1 & -i \\ -i & 1 \\ \end{bmatrix} \frac{x+iy}{1-t} = \frac{\frac{x+iy}{1-t}-i}{-i\frac{x+iy}{1-t}+1} = \frac{x+iy-i(1-t)}{-ix+y + 1-t}.$$ Thus, the equation to be established is $$\frac{x+it}{1+y} = \frac{x+iy-i(1-t)}{-ix+y + 1-t}.$$ Cross multiplying and cleaning things up a bit, we see the latter is equivalent to $$x(1+y) + i(-x^2 - t^2 + t(y+1)) = x(1+y) + i (y^2 - 1 + t(y+1))$$ which, since $x^2+y^2+t^2=1$, is a true equation.
• Next, we check the claim when $e=e_2$ and $\omega$ is arbitrary. Since $R(e_1,-\frac{\pi}{2}) e_3=e_2$, we can write $$R(e_2,\omega) = R(e_1,-\frac{\pi}{2}) \ R(e_3,\omega) \ {R(e_1,-\frac{\pi}{2})}^{-1}.$$ By direct calculation, \begin{align*} U(e_2,\omega) &= U(e_1,-\frac{\pi}{2}) \ U(e_3,\omega) \ {U(e_1,-\frac{\pi}{2})}^{-1} \\ &= \begin{bmatrix} 1 & -i \\ -i & 1 \\ \end{bmatrix} \begin{bmatrix} e^\frac{i \omega}{2} & 0 \\ 0 & e^{- \frac{i \omega}{2}} \\ \end{bmatrix} \begin{bmatrix} 1 & -i \\ -i & 1 \\ \end{bmatrix}^{-1} \end{align*} and so, since the claim is already established for the matrices on the RHS, it follows for the matrix on the LHS.
• Lastly, we establish the claim for a general point $e$ with spherical coordinate $(\theta,\phi)$ and a general angle of rotation $\omega$. Since $$e = R(e_3,\theta) \ R(e_2,\phi) \ e_3,$$ we have that $$R(e,\omega) = [R(e_3,\theta) \ R(e_2,\phi)] \ R(e_3,\omega) \ [R(e_3,\theta) \ R(e_2,\phi)]^{-1}.$$ A lengthy, but straightforward, calculation shows that $$U(e,\omega) = [U(e_3,\theta) \ U(e_2,\phi)] \ U(e_3,\omega) \ [U(e_3,\theta) \ U(e_2,\phi)]^{-1}.$$ Since the claim is already established for the matrices on the RHS, it follows for the matrix on the LHS.