Map from unit ball in quaternions to $\text{SU}(2)$ a group homomorphism?

Question

We have an embedding $\mathbb{C} \hookrightarrow \mathbb{H}$, $1 \mapsto 1$, $i \mapsto i$. Right multiplication by elements of the field $\mathbb{C} \subset \mathbb{H}$ makes $\mathbb{H}$ a $2$-dimensional vector space over $\mathbb{C}$ with basis $\{1, j\}$. Therefore, associated with any quaternion $x = x_0 + x_1i + x_2j + x_3k$, there is a $\mathbb{C}$-linear map $x': \mathbb{H} \to \mathbb{H}$, $y \mapsto xy$, which we view as a map $\mathbb{C}^2 \to \mathbb{C}^2$.

Consider the set $X := \{x \in \mathbb{H} \mid |x| = 1\}$. It is a group under multiplication.

Question. Is the map $f: X \to \text{SU}(2)$, $f(x) = x'$ a group homomorphism?

Answer 1

Yes. Let $L_q$ denote the left multiplication map on $\mathbb{H}$ given by $L_a(x)=ax$. Since quaternions are associative, we have $L_a(x\lambda)=L_a(x)\lambda$ for any other quaternion $\lambda$, in particular for complex numbers (here we are identifying $\mathbb{C}$ with a subspace of $\mathbb{H}$ in the usual way), so it is a linear transformation of $\mathbb{H}$ as a complex vector space. With $\{1,\mathbf{j}\}$ as a right $\mathbb{C}$ vector space basis, we get an identification $\mathbb{H}\cong\mathbb{C}^2$ allowing us to interpret $L_a$ as a matrix in $M_2(\mathbb{C})$.

Clearly $L:\mathbb{H}\to M_2(\mathbb{C})$ (sending $a$ to $L_a$) is an $\mathbb{R}$-linear map.

That it is an $\mathbb{R}$-algebra homomorphism means:

• $L_a\circ L_b=L_{ab}$ for all $a,b$,
• i.e. $L_a(L_b(c))=L_{ab}(c)$ for all $a,b,c$,
• i.e. $a(bc)=(ab)c$ for all $a,b,c$.

The last interpretation is just the associative property of the quaternions.

It follows that $\mathbb{H}^\times\to\mathrm{GL}_2(\mathbb{C})$ is a group homomorphism, in particular it is a group homomorphism when restricted to $\mathrm{Sp}(1)$ (the group of unit quaternions).

To check that the image of $\mathrm{Sp}(1)$ is $\mathrm{SU}(2)\subseteq\mathrm{GL}(2,\mathbb{C})$, some linear algebra is in order. First we're using $\{1,\mathbf{j}\}$ as a right $\mathbb{C}$-basis, since $\mathbb{H}=\mathbb{C}\oplus\mathbf{j}\mathbb{C}$. Since $w\mathbf{j}=\mathbf{j}\overline{w}$, left multiplication by $w$ is

$$ L_w \quad \longleftrightarrow \quad \begin{pmatrix} w & 0 \\ 0 & \overline{w} \end{pmatrix}. $$

Left multiplication by $\mathbf{j}$ has the effect of $L_{\mathbf{j}}(1)=0+\mathbf{j}1$ and $L_{\mathbf{j}}(\mathbf{j})=-1+\mathbf{j}0$, so

$$ L_{\mathbf{j}} \quad \longleftrightarrow \quad \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. $$

Since $\mathbf{q}\mapsto L_{\mathbf{q}}$ is an algebra homomorphism, we may write $\mathbf{q}=a+b\mathbf{j}$ ($a,b\in\mathbb{C}$) and get

$$ L_{a+b\mathbf{j}} \quad \longleftrightarrow \quad \begin{pmatrix} a & 0 \\ 0 & \overline{a} \end{pmatrix} + \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} b & 0 \\ 0 & \overline{b} \end{pmatrix} = \begin{pmatrix} a & -b \\ \overline{b} & ~\overline{a} \end{pmatrix}.$$

When $\mathbf{q}\in\mathrm{Sp}(1)$ is a unit quaternion, the condition $|\mathbf{q}|^2=1$ is equivalent to $|a|^2+|b|^2=1$ which makes the above any possible matrix in $\mathrm{SU}(2)$.