Rotation Double Cover

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.

Let $I\subseteq \mathbb{R}$ be a nondegenerate open interval, let $\mathbb{H}$ be the quaternions, and let $\phi :{\mathbb{R}}^3\to \mathrm{Im}\mathbb{H}$ be the identification of The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts.

1. Product form and product rule. For $p=s+\phi (u)$ and $q=t+\phi (w)$ with $s,t\in \mathbb{R}$ and $u,w\in {\mathbb{R}}^3$ one has $pq=(st-⟨u,w{⟩}_2)+\phi (sw+tu+u\times w)$. If moreover $p,q:I\to \mathbb{H}$ are of class $C^1$ (componentwise, $C^k$ Euclidean maps and diffeomorphisms), then $pq:I\to \mathbb{H}$ is $C^1$ and $(pq{)}^{\mathrm{\prime }}=p^{\mathrm{\prime }}q+p{q}^{\mathrm{\prime }}$. Likewise, if $A,B:I\to M_3(\mathbb{R})$ are of class $C^1$, then $AB$ is $C^1$ and $(AB{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}B+A{B}^{\mathrm{\prime }}$.
2. Conjugation and norm. For all $p,q\in \mathbb{H}$ one has $\overline{pq}=\bar{q}\bar{p}$ and $N(pq)=N(p)N(q)$, and $\mathrm{Re}(p\bar{q})=⟨p,q{⟩}_2$ for $\mathrm{Re}(x):=x_0$ and the Euclidean inner product of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$. For a $C^1$ curve $q:I\to \mathbb{H}$, the curves $\bar{q}$ and $N\circ q$ are $C^1$ with ${\bar{q}}^{\mathrm{\prime }}=\overline{q^{\mathrm{\prime }}}$ and

$(N\circ q{)}^{\mathrm{\prime }}=2\mathrm{Re}(q^{\mathrm{\prime }}\bar{q})=2⟨q^{\mathrm{\prime }},q{⟩}_2\mathrm{.}$

3. Conjugation formula. If $q=a+\phi (u)$ with $a\in \mathbb{R}$ and $u\in {\mathbb{R}}^3$, and $v\in {\mathbb{R}}^3$, then

$q\phi (v)\bar{q}=\phi ((a^2-\mathrm{\parallel }u{\mathrm{\parallel }}_2^2)v+2⟨u,v{⟩}_{2}u+2a(u\times v));$

in particular $q\phi (v)\bar{q}$ is imaginary for every $q\in \mathbb{H}$, that is, its real part is $0$. 4. Axis-angle form. If $N(q)=1$ and $q=\cos (\theta \mathrm{/}2)+\phi (\widehat{n})\sin (\theta \mathrm{/}2)$ with $\theta \in \mathbb{R}$ and $\widehat{n}\in {\mathbb{R}}^3$, $\mathrm{\parallel }\widehat{n}{\mathrm{\parallel }}_2=1$, then for every $v\in {\mathbb{R}}^3$

$\rho (q)(\phi (v))=\phi (\cos \theta v+(1-\cos \theta )⟨\widehat{n},v{⟩}_2\widehat{n}+\sin \theta (\widehat{n}\times v))\mathrm{.}$

• Clause 3 is stated for arbitrary $q$ and is a purely algebraic identity; only in clause 4 and in the definition of $\rho$ is the unit condition used, through $q^{-1}=\bar{q}$. No division by $\mathrm{\parallel }u{\mathrm{\parallel }}_2$ or by the vector part of $q$ occurs anywhere, so $q=\pm 1$ is included.

• The identity of clause 4 is the Rodrigues axis-angle formula in quaternion form: for $N(q)=1$, the linear map $\rho (q)$ fixes $\phi (\widehat{n})$ and rotates the plane ${\widehat{n}}^{\perp }$ by the angle $\theta$. This is a computation about the map $\rho$; no statement about any physical rotation, its rate, or a choice of orientation is made by the lemma itself.

• The multiplicativity $N(pq)=N(p)N(q)$ is used in the following lemma to estimate the quaternion angular-velocity equation, and the product rule is used to differentiate curves of rotations and of unit quaternions. Both are recorded here once in the form in which they are consumed.

Let $I\subseteq \mathbb{R}$ be a nondegenerate open interval, let $SO(3)$ be as in The orthogonal group $O(3)$ and the rotation group $SO(3)$, and let $R:I\to SO(3)$ be of class $C^1$, that is, all nine entries of $R$ have continuous first derivatives; write $R^{\mathrm{\prime }}(t)$ for the entrywise derivative and put

$\mathrm{\Omega }(t):=R(t{)}^{\mathsf{T}}{R}^{\mathrm{\prime }}(t)\in M_3(\mathbb{R})$

for the product of the transpose of $R(t)$ with $R^{\mathrm{\prime }}(t)$ (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes, The vector space $M_{m\times n}(F):=F^{m\times n}$ of $m$ by $n$ matrices over a field, with entrywise operations). Then:

1. $\mathrm{\Omega }(t)$ is skew-symmetric for every $t\in I$, that is $\mathrm{\Omega }(t{)}^{\mathsf{T}}=-\mathrm{\Omega }(t)$.
2. There is exactly one map $\omega :I\to {\mathbb{R}}^3$ with

$R(t{)}^{\mathsf{T}}{R}^{\mathrm{\prime }}(t)z=\omega (t)\times z(z\in {\mathbb{R}}^3,t\in I),$

namely $\omega (t)=(-{\mathrm{\Omega }}_{23}(t),{\mathrm{\Omega }}_{13}(t),-{\mathrm{\Omega }}_{12}(t))$, and this $\omega$ is continuous. It is called the body angular velocity of $R$. 3. For every $t\in I$, writing ${\mathrm{\Omega }}_{\omega (t)}$ for the matrix $z\mapsto \omega (t)\times z$, one has $R^{\mathrm{\prime }}(t)=R(t){\mathrm{\Omega }}_{\omega (t)}$.

• The convention here is the body-fixed one: $\mathrm{\Omega }=R^{\mathsf{T}}{R}^{\mathrm{\prime }}$ and $R^{\mathrm{\prime }}=R\mathrm{\Omega }$. The spatial angular velocity of the same path is the vector attached to $R^{\mathrm{\prime }}{R}^{\mathsf{T}}$, and the two differ by the rotation $R$; consumers must state which convention they use. For $C^2$ paths this $\omega$ is the same vector that appears in the rotating-frame acceleration identity of Angular velocity of a moving frame and the rotating-frame acceleration identity, whose clause 2 records it under a twice-differentiability hypothesis; the present lemma covers the $C^1$ case and proves only what is used here.

• The lemma is a statement about curves of matrices. No physical object, frame or time scale is introduced; an interpretation of $\omega$ as the angular velocity of a rigid body belongs to the physical items that adopt a rigid-body model.

Let $I\subseteq \mathbb{R}$ be a nondegenerate open interval, let $\omega :I\to {\mathbb{R}}^3$ be continuous, and consider on the open time-state domain $D:=I\times \mathbb{H}\subseteq \mathbb{R}\times \mathbb{H}$ the initial value problem

$q^{\mathrm{\prime }}(t)=\frac{1}{2}q(t)\phi (\omega (t)),q(t_0)=q_0,$

read as the first-order system of First-order systems, initial value problems, and solutions on intervals with time-state domain $D$, state space $\mathbb{H}$, and right-hand side $F:D\to \mathbb{H}$, $F(t,q)=\frac{1}{2}q\phi (\omega (t))$, where $\phi$, $S^3$ and $\rho$ are those of The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts.

1. If $q:I\to \mathbb{H}$ is a $C^1$ solution of $q^{\mathrm{\prime }}=\frac{1}{2}q\phi (\omega (t))$, then $N\circ q$ is constant on $I$; if $q(t_0)\in S^3$ for one $t_0\in I$, then $q(I)\subseteq S^3$ and the curve

$R:=\rho \circ q:I⟶SO(3),R(t):=\rho (q(t)),$

is of class $C^1$ with $R^{\mathrm{\prime }}(t)=R(t){\mathrm{\Omega }}_{\omega (t)}$ for every $t\in I$; consequently $\omega$ is the body angular velocity of $R$ in the sense of The body angular velocity of a $C^1$ rotation path: $\mathrm{\Omega }=R^{\mathsf{T}}{R}^{\mathrm{\prime }}$ is skew, $\mathrm{\Omega }z=\omega \times z$ determines a unique continuous $\omega$, and $R^{\mathrm{\prime }}=R\mathrm{\Omega }$. 2. For every $t_0\in I$ and $q_0\in S^3$ the problem has exactly one maximal solution, and its domain is all of $I$; that is, there is exactly one $C^1$ curve $q:I\to S^3$ with $q(t_0)=q_0$ and $q^{\mathrm{\prime }}=\frac{1}{2}q\phi (\omega (t))$ on $I$, and clause 1 applies to it.