Let $Sp(1)$ denote the unit quaternions.
By the identification of $Sp(1)$ with $SU(2)$, there is an irreducible representation on $\mathbb{C}^2$.
Is it possible to see this action directly from the unit quaternions, via some "geometrically" flavored action?
(It would give a map $S^3 \to U(2)$, and maybe one could see that it was faithful. Then perhaps one could also see that it was onto $SU(2)$. I would prefer this over a formula...)
Thoughts:
Presumably it can come from the right action of $SP(1)$ on $\mathbb{H}$. $\mathbb{C}$ then acts on $\mathbb{H}$ from the left, so the right action of $Sp(1)$ is complex linear.
There are no fixed vectors under this action, so the representation cannot decompose into $3 + 1$. It decomposes into $4 + 0$ or $2 + 2$. ... I was being dumb, this is obviously a 2 dimensional complex representation.(Like in Jyrki's answer below.)
Not geometric, but without formulas.
Let $\Bbb{H}$ be the space of quaternions. We can view it as a 1-dimensional (left) module over itself with multiplication from the left. We can also view is as a 2-dimensional vector space over the complex numbers $\Bbb{C}=\Bbb{R}[i]$ with multiplication by scalars from the right.
Because the ring $\Bbb{H}$ is associative, the two actions commute: for all $q,q'\in\Bbb{H}$ and $z\in\Bbb{C}$ $$ q'(qz)=(q'q)z. $$ what this means is that left multiplication by a fixed element $q'\in\Bbb{H}$ is a $\Bbb{C}$-linear transformation. This means that we have a representation of any subgroup $G\le\Bbb{H}^*$ by $2\times 2$ complex matrices by restricting $q'$ to $G$. Yours is the case when $G$ consists of quaternions of unit modulus.