# Isomorphism between $SU(2)$ and $U(1, \mathbb H)$

Question:

Prove $SU(2)$ is isomorphic to the group of quaternions of norm $1$, that is, $U(1,\mathbb H) \simeq SU(2)$.

Attempt:

How could I start finding the isomorphism? Intuitively, a quaternion can be expressed as two complex numbers $a+bi+cj+dk=a+bi+(c+di)j$, and as an element of $SU(2)$ is $\left[ \begin{array}{ c c } \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{array} \right]$ such that $\mid{\alpha}\mid^2+\mid\beta\mid^2=1$ with $\alpha,\beta \in\mathbb{C}$ then we can map $\alpha=a +bi$ and $\beta=c +di$.

$a^2 + b^2 +c^2 +d^2 = 1$ holds for quaternions of norm $1$, hence $\mid{\alpha}\mid^2+\mid\beta\mid^2$ also holds.

• Have you done the analogous complex problem, namely, proving that $SO(2)$ is isomorphic to the group of complex numbers of norm $1$? May 3, 2015 at 15:12
• i think with the $SU(2)$ representation that way round you may need $\alpha=a -bi$ and $\beta = c-id$ May 3, 2015 at 15:40
• @Travis In that case I guess $e^{i\phi} \to \left[ \begin{array}{ c c } \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{array} \right]$
– jjjx
May 3, 2015 at 15:54
• Right, now what's the analogue of the second sentence in your question? "Intuitively, a [unit] complex number be expressed as..." May 3, 2015 at 15:58
• Have you checked if your guess works by just multiplying it out? May 3, 2015 at 16:00

Just to formalize your idea for future reference. (I came looking for an answer but it wasn't complete).

Consider the map \begin{align}\varphi: U(1, \mathbb H) &\to SU(2)\\x = z + wj &\mapsto \begin{pmatrix}z & w \\ -\overline w &\overline z\end{pmatrix}\end{align}

where $x = x^0 \textbf 1 + x^1 \textbf i + x^2 \textbf j + x^3 \textbf k$, $z = x^0 + x^1 \textbf i \in \mathbb C$ and $w = x^2 + x^3 \textbf i \in \mathbb C$.

It's clear that $$|x|^2 = |z|^2 + |w|^2 = \det \begin{pmatrix} z & w \\-\overline w & \overline z \end{pmatrix}$$

So $x \in U(1, \mathbb H)$ if, and only if, $\begin{pmatrix} z & w \\-\overline w & \overline z \end{pmatrix}$ has determinant $1$. Plus, such matrices account for all elements of $SU(2)$.

Now we must check $\varphi$ is in fact a group isomorphism.

• $\varphi$ is a group homomorphism.

A simple calculation shows that $jz = \overline z j$ and $j w = \overline w j$. Thus if $x,y \in U(1,\mathbb H)$ are given we have \begin{align} xy &= (u + vj)(z + w j)\\&= uz + uw j + v(jz) + v(jw)j\\&= (uz - v\overline w) + (uw + v\overline z)j \end{align}

Applying \begin{align}\varphi(xy) &= \begin{pmatrix}uz - v\overline w & uw + v\overline z \\-\overline{uw + v \overline z} & \overline{uz - v\overline w}\end{pmatrix} \\&=\begin{pmatrix}uz - v\overline w & uw + v\overline z \\- \overline u \,\overline w - \overline v z & \overline u \, \overline z - \overline v \,w\end{pmatrix} \\&=\begin{pmatrix}u & u \\-\overline v & \overline u\end{pmatrix} \begin{pmatrix}z & w \\-\overline w & \overline z\end{pmatrix}\\&= \varphi(x)\,\, \varphi(y)\end{align}

• $\varphi$ is clearly a bijection.

Thus we conclude that $U(1,\mathbb H) \simeq SU(2)$ as we wanted.

Remark: As $S^3 = U(1,\mathbb H) \simeq SU(2)$ we get that $SU(2)$ is simply connected.