Two global forms with the same simple Lie algebra

Write $p=a+u$ and $q=b+v$ with $a,b\in \mathbb{R}$, $u=u_{1}i+u_{2}j+u_{3}k$, and $v=v_{1}i+v_{2}j+v_{3}k$. From the stated quaternion multiplication, $uv=-u\cdot v+u\times v$; in coordinates, $u\times v=(u_2{v}_3-u_3{v}_2)i+(u_3{v}_1-u_1{v}_3)j+(u_1{v}_2-u_2{v}_1)k$. Hence $u\cdot (u\times v)=v\cdot (u\times v)=0$ and $\mathrm{\mid }u\times v{\mathrm{\mid }}^2=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }v{\mathrm{\mid }}^2-(u\cdot v{)}^2$, by expansion of these coordinates. Therefore $pq=(ab-u\cdot v)+(av+bu+u\times v)$, and $\mathrm{\mid }pq{\mathrm{\mid }}^2=(ab-u\cdot v{)}^2+\mathrm{\mid }av+bu+u\times v{\mathrm{\mid }}^2=(a^2+\mathrm{\mid }u{\mathrm{\mid }}^2)(b^2+\mathrm{\mid }v{\mathrm{\mid }}^2)=\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{\mid }q{\mathrm{\mid }}^2$. The same multiplication table gives $q\bar{q}=\bar{q}q=\mathrm{\mid }q{\mathrm{\mid }}^2$, so a unit quaternion has inverse $q^{-1}=\bar{q}$ and $\mathrm{\mid }{q}^{-1}\mathrm{\mid }=1$.

If $\pi (q)$ is the identity, then $q$ commutes with every imaginary quaternion. Commuting with both $i$ and $j$ forces $q$ to be real, so the unit condition gives $q=\pm 1$; hence $\ker \pi =\{1,-1\}$. The differential at the identity sends $v\in \mathrm{Im}\mathbb{H}$ to the linear map $x\mapsto [v,x]=2v\times x$. It is injective because $v\times x=0$ for every $x$ forces $v=0$. The tangent space of $SU(2)\cong S^3$ has dimension three, and the tangent space of $SO(3)$ at the identity consists of the skew-symmetric $3\times 3$ matrices, also dimension three; therefore the differential is an isomorphism. It preserves brackets because it is the differential of a Lie-group homomorphism. The cross-product bracket is simple: if an ideal contains $v\mathrm{\ne }0$, then it contains $w\times v$ for every $w$, which spans $v^{\perp }$; together with $v$ this is all of ${\mathbb{R}}^3$. Therefore both Lie algebras are nonabelian and simple.

For $q=a+u$ unit and $x\in \mathrm{Im}\mathbb{H}$, the scalar part of $qx{q}^{-1}$ is $-a(u\cdot x)+(ax+u\times x)\cdot u=0$, so conjugation preserves $\mathrm{Im}\mathbb{H}$. It preserves the norm because $\mathrm{\mid }qx{q}^{-1}\mathrm{\mid }=\mathrm{\mid }q\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }{q}^{-1}\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }$ by step 1.1. The map $q\mapsto \pi (q)$ is a homomorphism by associativity. Its determinant is $+1$: it depends continuously on $q$, takes values in $\{-1,+1\}$, and equals $+1$ at $q=1$; the unit quaternions form the connected sphere $S^3$. For any $R\in SO(3)$, its eigenvalues have modulus one and nonreal eigenvalues occur in conjugate pairs. If there is a nonreal pair, the remaining real eigenvalue is $+1$ because the determinant is $+1$; if all eigenvalues are real, they are $\pm 1$ and their product $+1$ forces at least one eigenvalue $+1$. Thus $R$ has a unit fixed axis $n$. Its restriction to $n^{\perp }$ preserves orientation, since its determinant there equals $\det R=1$, and so it is a planar rotation through some angle $\theta$. Expanding with $nx=-n\cdot x+n\times x$ gives $qx{q}^{-1}=\cos (\theta )x+\sin (\theta )(n\times x)+(1-\cos (\theta ))(n\cdot x)n$ for $q=\cos (\theta \mathrm{/}2)+n\sin (\theta \mathrm{/}2)$, which is precisely that rotation. Thus $\pi$ is onto.

The center of $SU(2)$ is $\{\pm 1\}$, since an element commuting with every imaginary quaternion must be real. If $R$ is in the center of $SO(3)$, it commutes with every half-turn about a unit axis $n$; each half-turn has $\mathbb{R}n$ as its $+1$ eigenspace, so $R$ preserves every line. In particular, in the basis $i,j,k$ it sends each basis vector to itself or its negative; preserving the lines spanned by $i+j$ and $j+k$ forces the three signs equal. Thus $R=\pm I$, and only $I$ has determinant $+1$. The centers differ, so the groups are not isomorphic. The unit quaternions form the compact connected sphere $S^3$, and $SO(3)$ is its continuous surjective image under $\pi$, hence is compact and connected as well. The differential is an isomorphism by step 1.2, so these are distinct compact global forms with isomorphic simple Lie algebras.

The universal group quantifier includes both of these distinct groups. A proof with the fixed value $G=SU(2)$ establishes only that instance; it contains no argument for the separate value $G=SO(3)$. Hence it does not by itself establish the universally quantified claim.