Rotation Double Cover — Examples

For a curve on a closed interval $[0,T]$, this item uses "$C^1$ on $[0,T]$" to mean continuous on $[0,T]$ and $C^1$ on $(0,T)$; no endpoint derivatives are required.

Fix ${\omega }_0\in \mathbb{R}$ and a unit vector $\widehat{n}\in {\mathbb{R}}^3$ (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$), put $\omega :={\omega }_0\widehat{n}$, and consider

$q:\mathbb{R}⟶S^3,q(t):=\cos ({\omega }_{0}t\mathrm{/}2)+\phi (\widehat{n})\sin ({\omega }_{0}t\mathrm{/}2),$

with $\phi$ and $S^3$ as in The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts. Then:

1. $q$ is of class $C^1$ with $q(0)=1$, and $q$ solves the lift equation $q^{\mathrm{\prime }}(t)=\frac{1}{2}q(t)\phi (\omega )$ on $\mathbb{R}$.
2. The curve $R:=\rho \circ q:\mathbb{R}\to SO(3)$ is of class $C^1$ with $R^{\mathrm{\prime }}(t)=R(t){\mathrm{\Omega }}_{\omega }$ for all $t$; that is, $\omega$ is the constant body angular velocity of $R$. Explicitly, $R(t)$ is the rotation about $\widehat{n}$ through the angle ${\omega }_{0}t$:

$R(t)\phi (v)=\phi (\cos ({\omega }_{0}t)v+(1-\cos ({\omega }_{0}t))⟨\widehat{n},v{⟩}_2\widehat{n}+\sin ({\omega }_{0}t)(\widehat{n}\times v))(v\in {\mathbb{R}}^3)\mathrm{.}$

3. If ${\omega }_0\mathrm{\ne }0$, set $T:=2\pi \mathrm{/}\mathrm{\mid }{\omega }_0\mathrm{\mid }>0$. Then $R(0)=R(T)=I_3$ while $q(T)=-1\mathrm{\ne }q(0)$, and the closed loop $t\mapsto R(t)$ on $[0,T]$ has no closed lift through $1$ with the closed-interval regularity just defined.

• The example realizes the half-angle relation explicitly: at time $t$ the quaternion angle from $1$ is ${\omega }_{0}t\mathrm{/}2$, while the rotation angle is ${\omega }_{0}t$; the factor of two is the same factor that appears in the lift equation and in the definition of the body angular velocity.

• Clause 3 is the explicit closed-loop instance referred to by Scope of the quaternion lift: the global sign ambiguity, closed loops, and the absence of a physical spin claim: the loop returns to the identity rotation, but its lift through the identity quaternion ends at the antipode. The example does not claim that no other lift is closed for some other initial value, and it makes no statement about any physical rotation.

Let $I\subseteq \mathbb{R}$ be a nondegenerate open interval with $0\in I$, let $\theta :I\to \mathbb{R}$ be of class $C^1$, let $\widehat{n}\in {\mathbb{R}}^3$ be a unit vector, let $q_0\in S^3$, and define the normalized phase $\alpha (t):=\theta (t)-\theta (0)$ and

$e(t):=\cos (\alpha (t)\mathrm{/}2)+\phi (\widehat{n})\sin (\alpha (t)\mathrm{/}2),q(t):=q_{0}e(t),$

with $\phi ,S^3,\rho$ as in The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts. Then:

1. $e$ and $q$ are of class $C^1$, they take values in $S^3$, and $q(0)=q_0$.
2. The curve $q$ solves $q^{\mathrm{\prime }}(t)=\frac{1}{2}q(t)\phi (\omega (t))$ with $\omega (t):={\theta }^{\mathrm{\prime }}(t)\widehat{n}$, on all of $I$.
3. Consequently $R:=\rho \circ q$ is a $C^1$ rotation path whose body angular velocity is $\omega (t)={\theta }^{\mathrm{\prime }}(t)\widehat{n}$, and $R(t)=\rho (q_0)\rho (e(t))$, where $\rho (e(t))$ is the rotation about $\widehat{n}$ through the angle $\alpha (t)=\theta (t)-\theta (0)$.

• The verification is a finite computation: it checks that the half-angle curve solves the lift equation and then invokes the lift-and-reconstruction items; no claim is made about the existence or uniqueness of $q$ beyond what those items supply, and none about the physical meaning of $\theta$ or $t$.

• The phase is normalized by subtracting $\theta (0)$, so every allowed $C^1$ function $\theta$ gives $e(0)=1$ and $q(0)=q_0$; the angular velocity remains ${\theta }^{\mathrm{\prime }}(t)\widehat{n}$. The initial value $q_0$ may be any unit quaternion, so left multiplication by $q_0$ produces either initial lift sign.

On $I=\mathbb{R}$ define

$u(t):=\cos (t\mathrm{/}2)+\phi (e_1)\sin (t\mathrm{/}2),v(t):=\cos (t\mathrm{/}2)+\phi (e_2)\sin (t\mathrm{/}2),q(t):=u(t)v(t),$

where $e_1,e_2,e_3$ is the standard basis of ${\mathbb{R}}^3$, $\phi$ is the identification of The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts, and the product is the quaternion product. Then:

1. $q:\mathbb{R}\to S^3$ is of class $C^1$.
2. $\bar{u}{u}^{\mathrm{\prime }}=\frac{1}{2}\phi (e_1)$ and $\bar{v}{v}^{\mathrm{\prime }}=\frac{1}{2}\phi (e_2)$, and

$q(t{)}^{-1}{q}^{\mathrm{\prime }}(t)=\bar{v}(t)(\bar{u}(t)u^{\mathrm{\prime }}(t))v(t)+\bar{v}(t)v^{\mathrm{\prime }}(t)=\frac{1}{2}\phi (\cos t{e}_1+e_2+\sin t{e}_3)\mathrm{.}$

3. Consequently $R=\rho \circ q$ is a $C^1$ rotation path whose body angular velocity is $\omega (t)=\cos t{e}_1+e_2+\sin t{e}_3$, that is, $\omega (t)=(\cos t,1,\sin t)$ in coordinates; this $\omega$ is neither constant nor along a fixed axis.

• The example checks the reconstruction formula on a product of two one-parameter rotations about different axes; the resulting $\omega$ is nonconstant, so the example is not a disguised fixed-axis case. The computation is finite and uses only the product form, the multiplicativity, and one evaluation of the axis-angle formula.

• No dynamical input is used: $q$ is an explicitly given curve and no equation of motion is solved. The example asserts nothing about which curves arise from any mechanical system.

Let $e_1,e_2,e_3$ be the standard basis of ${\mathbb{R}}^3$ and identify ${\mathbb{R}}^3$ with $\mathrm{Im}\mathbb{H}$ through $\phi$ as in The quaternion rotation map $\rho$: conjugation on the imaginary quaternions, the unit sphere $S^3$, and quaternion lifts; matrices of linear maps are taken in the ordered basis $(i,j,k)$. Then:

1. $\rho (1)$ is the identity matrix and $\rho (-1)=\rho (1)$; moreover, if $q\in S^3$ satisfies $\rho (q)=I_3$, then $q=\pm 1$.
2. $\rho (i),\rho (j),\rho (k)$ are the half-turns about the coordinate axes,

$\rho (i)=\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right),\rho (j)=\left(\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right),\rho (k)=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right),$

and $\rho (i)\rho (j)=\rho (k)=\rho (ij)$: the homomorphism identity is verified on the generating pair $i,j$ by finite matrix multiplication. 3. The exact preimages of $\rho (k)$ in $S^3$ are $\pm k$: $\rho (\pm k)=\rho (k)$, and every $q\in S^3$ with $\rho (q)=\rho (k)$ satisfies $q=\pm k$.