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

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?

• You don't really need surjectivity of the exponential map for this argument. The inverse function theorem shows that $\phi$ is a bijection between neighborhoods of the identity. It follows that the image is an open subgroup, but $SO(3)$ is connected so it's everything. Commented Oct 15, 2021 at 17:19

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.

• Thanks, this is the answer I was waiting for. Commented Oct 16, 2021 at 10:14