# Question about the proof of $S^3/\mathbb{Z}_2 \cong SO(3)$

I’m trying to show $S^3/\mathbb{Z}_2 \cong SO(3)$ completely rigorously.

For that purpose I considered three-sphere $S^3$ as a subspace of the ring of quaternions $\mathbb{H}$ and looked into the map $f:S^3 \to SO(3)$ defined as follows. For each $x \in S^3$ and each $y \in \mathbb{R}^3$ (again considered as a subspace of $\mathbb{H}$ spanned by $\bf{i}$,$\bf{j}$ and $\bf{k}$), $f(x)y = xyx^{-1}$ (note that here $f(x)$ is a $3\times 3$ matrix).

I’ve already checked that $f$ is a well-defined continuous group homomorphism with kernel $\mathbb{Z}_2$. So it only remains to show the ontoness of $f$, which I’m having trouble to do. I’ll appreciate any help. Thanks a lot.

#### Solutions Collecting From Web of "Question about the proof of $S^3/\mathbb{Z}_2 \cong SO(3)$"

To show that $f$ is onto, note that for all $A\in SO(3)$, $A$ has an eigenvalue 1 with respect to some eigenvalue $v_1$. Let $v_2, v_3$ be an orthonormal basis of $\mathbb R^3$. Then one can show that $A$ is the rotation along the $v_2 – v_3$ plane with angle $\theta$. Then one can check (let $v_1 = (a, b, c)$)

$$(*) \ \ \ f(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}v_1) = f(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}(ai + bj + ck)) =A$$

Indeed, you can check that if $v = ai + b j + ck$ and $w = di + ej + fk$, then

$$vw = -\langle v, w \rangle + v \times w,$$

if $v_1, v_2, v_3$ are chosen such that

$$v_1\times v_2 = v_3,\ \ v_2\times v_3 = v_1,\ \ v_3\times v_1 = v_2 ,$$

then write $\phi =\theta/2$, hence (for example)

$$f(\cos\phi+ \sin\phi v_1) v_2 = (\cos\phi + \sin\phi v_1)v_2 (\cos\phi – \sin\phi v_1)$$
$$= \big(\cos\phi v_2 + \cos\phi \sin\phi v_1\times v_2\big)(\cos\phi – \sin\phi v_1)$$
$$= \cos^2 \phi v_2 – \cos\phi\sin\phi v_2 \times v_1 + \cos\phi \sin\phi v_1 \times v_2 -\sin^2\phi v_3\times v_1$$
$$\cos(2\phi) v_2 + \sin(2\phi) v_3 = A(v_2)$$

Similar one can check for $v_1$ and $v_3$ to show (*).