The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts

Let $\mathbb{H}$ be the quaternion algebra with its product, conjugation $x\mapsto \bar{x}$, real-valued norm $N:\mathbb{H}\to \mathbb{R}$, and real embedding $r\mapsto \widehat{r}=(r,0,0,0)$, in the notation of The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$. Thus $x=x_0+x_{1}i+x_{2}j+x_{3}k$ with the multiplication formula of The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$, and

$N(x)=x_0^2+x_1^2+x_2^2+x_3^2,\widehat{N(x)}=x\bar{x}=\bar{x}x\mathrm{.}$

In particular, $N(x)>0$ for $x\mathrm{\ne }0$, and the imported division-ring theorem gives the typed inverse formula $x^{-1}=\widehat{N(x{)}^{-1}}\bar{x}$ ($\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$). Under the identification of $\mathbb{H}$ with ${\mathbb{R}}^4$ by the coordinates $x_0,\dots ,x_3$ one has $N(x)=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ for the Euclidean norm of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$.

Imaginary quaternions and the vector identification. Put

$\mathrm{Im}\mathbb{H}:=\{x\in \mathbb{H}:\bar{x}=-x\}=\mathbb{R}i\oplus \mathbb{R}j\oplus \mathbb{R}k,\phi :{\mathbb{R}}^3\to \mathrm{Im}\mathbb{H},\phi (v_1,v_2,v_3):=v_{1}i+v_{2}j+v_{3}k\mathrm{.}$

Then $\phi$ is a linear isomorphism with $N(\phi (v))=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$ for every $v\in {\mathbb{R}}^3$. Throughout the collection, $\phi$ translates between vectors of ${\mathbb{R}}^3$ (their inner product $⟨\cdot ,\cdot {⟩}_2$, norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ and cross product) and imaginary quaternions inside quaternion products.

The unit sphere. Define

$S^3:=\{q\in \mathbb{H}:N(q)=1\},$

the unit sphere of $\mathbb{H}\cong {\mathbb{R}}^4$. The identification of $S^3$ with $SU(2)$ is the one of the published covering example ex-su-two-to-so-three-as-a-covering-homomorphism, which is also the source of the covering statement recorded in the second remark below; that example is linked from the page summary of this pair.

Conjugation and the rotation map. For $q\in \mathbb{H}$ with $q\mathrm{\ne }0$ let $c_q:\mathbb{H}\to \mathbb{H}$ be

$c_q(x):=qx{q}^{-1}=qx\widehat{N(q{)}^{-1}}\bar{q},$

where the real scalar is embedded by $r\mapsto \widehat{r}$, and define

$\rho (q):=c_q{\mathrm{\mid }}_{\mathrm{Im}\mathbb{H}}:\mathrm{Im}\mathbb{H}⟶\mathrm{Im}\mathbb{H}\mathrm{.}$

Both the $\mathbb{R}$-linear map $\rho (q)$ and its matrix in the ordered basis $(i,j,k)$ of $\mathrm{Im}\mathbb{H}$ are written $\rho (q)$; the equality $\rho (q)\in SO(3)$ for $q\in S^3$ is part of the cited covering statement, and the matrix is the one meant whenever $\rho (q)$ is composed with the identifications above. The map $\rho$ itself is never linear in $q$ and no such structure is used.

Quaternion lifts. Let $I\subseteq \mathbb{R}$ be an interval and let $R:I\to SO(3)$ be a map, with $SO(3)$ as in The orthogonal group $O(3)$ and the rotation group $SO(3)$. A quaternion lift of $R$ is a map $q:I\to S^3$ with $\rho (q(t))=R(t)$ for every $t\in I$.

• Well-definedness of $\rho$. For $q\mathrm{\ne }0$ the map $c_q$ is a real algebra automorphism of $\mathbb{H}$: it is $\mathbb{R}$-linear, fixes $1$, and satisfies $c_q(xy)=qxy{q}^{-1}=(qx{q}^{-1})(qy{q}^{-1})=c_q(x)c_q(y)$, with inverse $c_{q^{-1}}$. It also commutes with conjugation. Since $\bar{\widehat{r}}=\widehat{r}$ for every real $r$, the anti-automorphism law and the typed inverse formula give $\overline{q^{-1}}=\widehat{N(q{)}^{-1}}q$; since embedded real scalars commute with quaternions,

$\overline{c_q(x)}=\overline{q^{-1}}\bar{x}\bar{q}=\widehat{N(q{)}^{-1}}q\bar{x}\bar{q}=q\bar{x}\widehat{N(q{)}^{-1}}\bar{q}=c_q(\bar{x})\mathrm{.}$

Therefore $x\in \mathrm{Im}\mathbb{H}$ (that is, $\bar{x}=-x$) gives $c_q(x)=-\overline{c_q(x)}$, so $c_q(x)\in \mathrm{Im}\mathbb{H}$. Thus the displayed codomain of $\rho (q)$ is correct, and $\rho (1)$ is the identity of $\mathrm{Im}\mathbb{H}$.

• The cited double cover. By the published covering example ex-su-two-to-so-three-as-a-covering-homomorphism — whose statement assumes the countable axiom of choice ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) — the assignment $q\mapsto \rho (q)$ on $S^3$ is a surjective two-sheeted covering homomorphism $S^3\to SO(3)$ of Lie groups with kernel $\{1,-1\}$. Equivalently: $\rho (q_1{q}_2)=\rho (q_1)\rho (q_2)$ for $q_1,q_2\in S^3$, every $R\in SO(3)$ has exactly two preimages $q$ and $-q$ in $S^3$, and $\rho (-q)=\rho (q)$. This double-cover conclusion is supplied by that item and is not reproved here; the pages that follow retain only the differential-equation and reconstruction content, and every later item that uses the kernel $\{1,-1\}$ or the two-preimage statement does so under this same countable-choice hypothesis, which is recorded again in its contract. The item is a published canonical root item; the pair's page summary carries the wikilink to it.

• Scaling. For $\lambda \in \mathbb{R}\setminus \{0\}$ and $q\mathrm{\ne }0$ one has $c_{\widehat{\lambda }q}=c_q$, because $(\widehat{\lambda }q)x(\widehat{\lambda }q{)}^{-1}=qx{q}^{-1}$; hence $\rho (\widehat{\lambda }q)=\rho (q)$. Normalizing $q$ by the real scalar $\widehat{N(q{)}^{-1\mathrm{/}2}}$ gives a unit quaternion, so each rotation is represented in $S^3$ by exactly the antipodal pair $\pm \widehat{N(q{)}^{-1\mathrm{/}2}}q$; $\rho (-q)=\rho (q)$ is the case $\lambda =-1$.

• Terminology. "Lift" is used only in the displayed sense; no topological path-lifting theorem is assumed anywhere on this page. Existence, uniqueness and the differential equation of lifts are proved in the items that follow.