Is there a simple proof that the homomorphism $\phi:SU(2)\rightarrow SO(3)$ is surjective?

Question

I am wondering, is there a simple, direct proof that the double cover $$\phi:SU(2)\rightarrow SO(3)$$ is a surjective map? Explicitly, if we identify $\mathbb R^3$ with $\text{im }\mathbb H$, and then this space with a subspace of $2\times 2$ complex matrices, then we can define $\phi$ as $$A\mapsto(X\mapsto AXA^+)$$ however, it seems to me that it is really complicated to see directly that this map is surjective. The only way I see one can prove this is surjective is via studying the associated lie algebra map $$\phi^*:\mathfrak{su}(2)\rightarrow\mathfrak{so}(3)$$ which, after some tedious computation, one can check is an isomorphism. Then, one knows that the exponential map of a connected compact Lie group, like $SU(2)$ and $SO(3)$, is surjective. This fact, however, is aconsequence of the existence of maximal tori, etc., which is a highly nontrivial fact. That $\phi^*$ is an isomorphism toghether with the surjectivity of the exponential maps implies the surjectivity of $\phi$.

My question is, can we conclude directly the surjectivity of $\phi$ via a simpler argument?

Answer 1

You can just directly verify it.

To that end, I prefer to use the isomorphism $SU(2)\cong Sp(1)$ see this MSE question and answer).

In this language, a unit quaternion $q$ acts on $\mathbb{R}^3\cong \operatorname{Im}\mathbb{H}$ by conjugation: $q$ corresponds to the linear map $C_q:\mathbb{R}^3\rightarrow \mathbb{R}^3$ given by $C_q(v) = qvq^{-1} = qv\overline{q}$.

A general element $B\in SO(3)$ is given geometrically by specifying a fixed axis and an angle of rotation about that axis. So, to show the map $Sp(1)\rightarrow SO(3)$ is surjective, it suffices to show we can get any axis of rotation and any angle of rotation using a function as above.

So, suppose the axis is specified by a point $v = (v_1,v_2,v_3)\in S^2\subseteq \mathbb{R}^3$, and the angle is given by some $\theta \in [0,\pi]$.

Consider the unit quaternion $q:= \cos(\theta/2) + \sin(\theta/2)v$. Then $q\in Sp(1)$ and $C_q(v) = qvq^{-1} = v$ since real numbers and $v$ commute with themselves.

Moreover, if a unit vector $w$ is perpendicular to $v$, then \begin{align*} C_q(w) &= (\cos(\theta/2) + \sin(\theta/2)v)w(\cos(\theta/2)-\sin(\theta/2)v)\\ &= \cos^2(\theta/2) w +\cos(\theta/2)\sin(\theta/2)(vw - wv) -\sin^2(\theta/2) vwv.\end{align*}

Now, because $v$ and $w$ are perpendicular and purely imaginary, $wv = -vw$. (See, e.g., this MSE question.) Moreover, this implies that $vwv = -v^2 w = w$. Here, $v^2 = vv = -v \overline{v} = -|v|^2 = -1$, which is true for any purely imaginary unit length quaterion.)

Thus, $C_q(w) = (\cos^2(\theta/2) - \sin^2(\theta/2)) w + 2\cos(\theta/2)\sin(\theta/2)(vw) = \cos(\theta)w + \sin(\theta)(vw).$

Finally, simply notice that $vw$ is perpendicular to both $v$ and $w$. Indeed, $1\bot w$ and multiplication by $v$ is an isometry, so $v\bot vw$ and likewise, $1\bot v$ and multiplication by $w$ is an isometry, so $w\bot vw$.

Thus, we identify the formula for $C_q(w)$ as rotation of by angle $\theta$ in the plane perpendicular to $v$, so we are done.