SU(2) to SO(3) as a covering homomorphism

Write $q=a+bi+cj+dk$. A coordinate check from [F1] gives $\overline{qr}=\bar{r}\bar{q}$ and hence $N(qr)=N(q)N(r)$. Thus the unit sphere $S^3\subset \mathbb{H}$ is a group, with inverse $q\mapsto \bar{q}$. It is a smooth three-manifold by [F2], because $N^{-1}(1)$ is regular: $d{N}_q(w)=2⟨q,w⟩$ is nonzero on every unit $q$. Multiplication and inversion are polynomial and linear respectively, so this is a Lie group.

The determinant-nonzero matrices form an open subset $G{L}_3(\mathbb{R})$ of the nine-dimensional matrix space. Matrix multiplication is polynomial and inversion is the smooth adjugate-over-determinant formula, so this open manifold is a Lie group. Inside it, the equations $R^{T}R=I$ and $\det R=1$ define a closed subgroup; [F2] therefore gives it the embedded Lie-group structure denoted $SO(3)$. This is the exact use of ${\mathrm{A}\mathrm{C}}_{\omega }$ in the construction.

For a unit $q=a+r$, with $r=bi+cj+dk$, and an imaginary quaternion $v$, direct multiplication gives the vector formula $qv{q}^{-1}=(a^2-\mathrm{\mid }r{\mathrm{\mid }}^2)v+2⟨r,v⟩r+2a(r\times v)$. It also gives zero real part. The identities $⟨r\times v,r⟩=⟨r\times v,v⟩=0$ and $\mathrm{\mid }r\times v{\mathrm{\mid }}^2=\mathrm{\mid }r{\mathrm{\mid }}^2\mathrm{\mid }v{\mathrm{\mid }}^2-⟨r,v{⟩}^2$ show from this formula that $\mathrm{\mid }qv{q}^{-1}\mathrm{\mid }=\mathrm{\mid }v\mathrm{\mid }$. Hence conjugation defines an orthogonal transformation of $\mathrm{Im}\mathbb{H}$.

The polynomial map $a+bi+cj+dk\mapsto \left(\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right)$ preserves products by the quaternion table. Its image consists exactly of the matrices in $SU(2)$, since their columns have the displayed form and the unitary and determinant-one equations reduce to $a^2+b^2+c^2+d^2=1$. It and its coordinate inverse are smooth, so it identifies the Lie group of step 1.1 with $SU(2)$.

The unit sphere is path connected: if $q\mathrm{\ne }-1$, normalize the nonzero segment $(1-t)q+t$, while $-1$ is joined to $1$ by $t\mapsto \cos (\pi t)+i\sin (\pi t)$. Consequently the determinant of the orthogonal map in step 1.3, a continuous function with values in $\{1,-1\}$, equals its value $1$ at $q=1$. Thus $\rho (q)\in SO(3)$. Associativity gives $\rho (q_1{q}_2)=\rho (q_1)\rho (q_2)$, and the coordinate formula in step 1.3 makes $\rho$ smooth.

Identify $T_{1}SU(2)$ with $\mathrm{Im}\mathbb{H}$. Differentiating conjugation along $q(t)=1+tr+O(t^2)$ gives $d{\rho }_1(r)(v)=rv-vr=2r\times v$. Differentiating $R^{T}R=I$ at $I$ shows that $T_{I}SO(3)$ is contained in the three-dimensional space of skew-symmetric endomorphisms. The displayed cross-product map takes values in that space and is injective: if $r\times v=0$ for every $v$, take a basis vector not parallel to nonzero $r$ to get a contradiction. Its three-dimensional image lies in $T_{I}SO(3)$, so the inclusions force $T_{I}SO(3)$ to be the full skew-symmetric space and $d{\rho }_1$ to be an isomorphism. By [F3], there are neighborhoods $U_0$ of $1$ and $W_0$ of $I$ such that $\rho {\mathrm{\mid }}_{U_0}:U_0\to W_0$ is a diffeomorphism. Choose a smaller open neighborhood $U\subseteq U_0$ with $U\cap (-U)=\mathrm{\varnothing }$ and put $W=\rho (U)$; the restriction $\rho {\mathrm{\mid }}_U:U\to W$ remains a diffeomorphism and $W$ is open.

The map $\rho$ is onto. For $R\in SO(3)$, $\det (R-I)=\det (R^T-I)=\det (R^{-1}-I)=\det (I-R)=-\det (R-I)$, so $\det (R-I)=0$. Choose a unit vector $u$ fixed by $R$. Its perpendicular plane is invariant, and the restriction of $R$ there is an orientation-preserving planar orthogonal map. By [F4], in a positively oriented orthonormal basis it is rotation through some angle $\theta$. Substituting $q=\cos (\theta \mathrm{/}2)+u\sin (\theta \mathrm{/}2)$ into the formula of step 1.3 gives Rodrigues' formula, so $\rho (q)=R$. Only this one finite-dimensional choice of an axis and basis is made.

If $\rho (q)$ is the identity, then $q$ commutes with $i,j,k$. Comparing $qi$ with $iq$ first gives $c=d=0$, and comparing $(a+bi)j$ with $j(a+bi)$ then gives $b=0$. Since $q$ is unit, $a=1$ or $a=-1$. Conversely both real unit quaternions act trivially. Therefore $\ker \rho =\{1,-1\}$, and every fibre is exactly $\{q,-q\}$.

Step 3.2 now gives ${\rho }^{-1}(W)=U\bigsqcup (-U)$, and both restrictions are diffeomorphisms onto $W$. For arbitrary $R_0\in SO(3)$ choose one $q_0$ above it using step 3.1; then $R_{0}W$ is evenly covered by the two translated sheets $q_{0}U$ and $-q_{0}U$. Hence $\rho$ is a covering homomorphism in the sense of Covering homomorphisms of Lie groups, with exactly two sheets. The groups are nonempty and three-dimensional; no zero-dimensional, endpoint, degenerate, or biconditional case is hidden. The proof itself makes only finitely many choices, while ${\mathrm{A}\mathrm{C}}_{\omega }$ is used through the closed-subgroup construction of $SO(3)$ in [F2].