SU(2)/<±I> is Isomorphic to SO(3)

By definition,

$SU(2) = \{ \begin{bmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha} \end{bmatrix} : \alpha, \beta \in \mathbb{C}, |\alpha|^2 + |\beta|^2 = 1 \}$ .

Alternatively, we can write each matrix $A \in SU(2)$ as

$\begin{bmatrix} a + bi & -c + di \\ c + di & a - bi \end{bmatrix}$ where $a^2 + b^2 + c^2 + d^2 = 1$ .

Thus, we have a bijective map from $S^3 \rightarrow SU(2)$ , given by:

$a + bi + cj + dk \mapsto \begin{bmatrix} a + bi & -c + di \\ c + di & a - bi \end{bmatrix}$ .

It is easy to see that this map is smooth, and so, as a manifold $SU(2) \cong S^3$ .

Claim: this map is not only a diffeomorphism between manifolds, but it is also a group isomorphism between the groups $SU(2)$ and $S^3$ .

Proof: Recall quaternion multiplication: Let $q = a + bi + cj + dk$ and $p = x + yi + zj + wk$ . Then,

$qp = (ax - by - cz - dw) + (ay + bx + cw - dz)i + (az - bw + cx + dy)j + (aw + bz - cy + dx)k$ .

The corresponding matrices for $q$ and $p$ in $SU(2)$ are given by:

$q \mapsto Q = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix}$

$p \mapsto P = \begin{bmatrix} x + yi & z + wi \\ -z + wi & x - yi \end{bmatrix}$ .

It follows that

$QP = \begin{bmatrix} (ax - by - cz - dw) + i(ay + bx + cw - dz) & (az - bw + cx + dy) + i(aw + bz - cy + dx) \\ -(az - bw + cx + dy) + i(aw + bz - cy + dx) & (ax - by - cz - dw) - i(ay + bx + cw - dz) \end{bmatrix}$ ,

which is exactly the corresponding matrix for $qp \in S^3$ in $SU(2)$ .

Hence, the map from $S^3 \rightarrow SU(2)$ is a group isomorphism. $\square$

Claim: Every unit quaternion corresponds to a rotation in $\mathbb{R}^3$ .

Proof: We identify $\mathbb{R}^3$ with the space of purely imaginary quaternions, i.e. $\text{Im} \mathbb{H} = \{ bi + cj + dk: b, c, d \in \mathbb{R} \}$ :

$(x, y, z) \leftrightarrow xi + yj + zk$

Given $q$ be a unit quaternion. We consider the map:

$R_{q} : \text{Im} \mathbb{H} \rightarrow \text{Im} \mathbb{H}$ given by

$v \mapsto qvq^{-1}$ for any $v \in \text{Im} \mathbb{H}$ .

It is easy to check that:

$qvq^{-1} \in \text{Im} \mathbb{H}$
$|qvq^{-1}| = |v|$

Thus, $R_{q}$ preserves the length of the imaginary quaternions. We now show that the angle is also preserved, i.e. this map preserves the inner product.

Let $v, w \in \text{Im} \mathbb{H} \cong \mathbb{R}^3$ . We want to show that:

$R_q(v) \cdot R_q(w) = v \cdot w$ .

Key: For $v, w \in \text{Im} \mathbb{H}$ , $vw = v \cdot w + v \times w$ . In particular, this means $\text{Re}(vw) = v \cdot w$ .

Thus, to show that $R_{q}$ preserves inner product, it suffices to show that $\text{Re}(R_{q}(v) R_{q}(w)) = \text{Re}(vw)$ .

We have $R_{q}(v) R_{q}(w) = (qvq^{-1})(qwq^{-1}) = qvwq^{-1} = R_q(vw)$ .

It is easy to check that for any $x \in \mathbb{H}$ , $\text{Re}(R_q(x)) = \text{Re}(x)$ . For instance, write $x = a + \textbf{x}$ where $\textbf{x}$ is the purely imaginary part and $a \in \mathbb{R}$ . Then,

$qxq^{-1} = q(a + \textbf{x})q^{-1} = qaq^{-1} + q\textbf{x}q^{-1} = a + q\textbf{x}q^{-1}$ .

Hence, $\text{Re}(R_{q} x) = a = \text{Re}(x)$ . Thus, every unit quaternion corresponds to an element of $SO(3)$ .

Claim: The map $\Psi: S^3 \rightarrow SO(3), \Psi(q) = R_q$ is a group homomorphism.

It suffices to show that group multiplication is preserved.

Let $p, q \in S^3$ and $v \in \mathbb{R}^3$ be arbitrary. Then,

$\Psi(pq)v = R_{pq}(v) = (pq)v(pq)^{-1} = p(qvq^{-1})p^{-1} = p(R_q(v))p^{-1} = R_p \circ R_q (v)$ .

This shows that $\Psi$ is a group homomorphism.

Claim: The kernel of $\Psi$ is {1, -1}.

It is easy to see that both 1 and -1 are both mapped to the identity element in $SO(3)$ .

Let $p$ be an element in the kernel. Then, for any $v \in \mathbb{R}^3$ , we have:

$pvp^{-1} = R_p(v) = v$ .

Recall: $v$ can be expressed as a purely imaginary quaternion. This means, $pv = vp$ for any purely imaginary quaternion. This is only possible if $p \in \mathbb{R}$ . Since $p \in S^3$ , $p \in \{-1, 1 \}$ .