Solving for the rotor in quaternion rotation

Question

If we have 2 vectors $v_1,v_2$ which have been rotated into $v'_1,v'_2$ by the following operations:

$v'_1 = e^{θ\hat{n}/2}v_1e^{-θ\hat{n}/2}$

$v'_2 = e^{θ\hat{n}/2}v_2e^{-θ\hat{n}/2}$

Where $\hat{n}$ is a unit vector representing the axis of rotation, and $θ$ is a scalar representing the rotation angle.

If we knew $v_1,v_2,v'_1,v'_2$ how would we solve for either the rotor $e^{θ\hat{n}/2}$ or equivalently it's axis/angle $\hat{n},θ$?

$v_1,v_2$ can be assumed to be linearly independent and of course everything here is interpreted as a quaternion.

Answer 1

There's a straightforward formula when $\theta \ne \pi$. First get a third vector with the cross product: $$ v_3 = v_1\times v_2,\quad v_3' = v_1'\times v_2'. $$ Now we need to form the reciprocal basis of the prime basis. Let $$ V' = v_1'\wedge v_2'\wedge v_3',\quad |V'| = |v_1'\cdot(v_2'\times v_3')|. $$ Then $$ v^{\prime 1} = v_2'\wedge v_3'\,(V')^{-1} = \frac{v_2'\times v_3'}{|V'|}, $$$$ v^{\prime 2} = -v_1'\wedge v_3'\,(V')^{-1} = \frac{v_3'\times v_1'}{|V'|},, $$$$ v^{\prime 3} = v_1'\wedge v_2'\,(V')^{-1} = \frac{v_1'\times v_2'}{|V'|}. $$ The rotor $R$ we want is then $$ R \propto 1 + \sum_iv^{\prime i}v_i. $$

---

If $v_i$ and $v'_i$ are expressed as column vectors, then we can express all of this with neat matrix computations. Let $V = (v_1, v_2, v_3)$ and $W = (v_1', v_2', v_3')$ be the matrices with those columns. The reciprocal basis matrix is the inverse transpose $W^{-T}$; calculate the matrix of inner products $\Gamma = W^{-1}V$. Then $$ R \propto 1+\sum_iv^{\prime i}v_i = 1 + \mathrm{Tr}(\Gamma) + \sum_{i<j}(\Gamma_{ij}-\Gamma_{ji})e_ie_j $$ where $e_i$ is the standard basis.

---

Proof of formula:

We have $$ v'_i = Rv_i\widetilde R,\quad v^{\prime i} = Rv^i\widetilde R. $$ Now we astutely notice that $$ \sum_iv^{\prime i}v_i = \sum_iRv^i\widetilde Rv_i = R\dot\nabla\widetilde R\dot x. $$ In 3D our rotor is a sum $R = s + B$ of a scalar and a bivector; standard geometric calculus identities give us $$ \sum_iv^{\prime i}v_i = R(3s + (3-4)\widetilde B) = R(3s - \widetilde R + s) = 4sR - 1. $$ When $s \ne 0$ it follows that $$ R \propto 1 + \sum_iv^{\prime i}v_i. $$

The same idea will work in any number of dimensions so long as $R$ is a rotation in a single plane, yielding $$ R \propto 4-n + \sum_iv^{\prime i}v_i, $$ but not all rotors are of this form: consider that in 4D $$ e^{e_1e_2 + e_3e_4} = e^{e_1e_2}e^{e_3e_4} $$ has a pseudoscalar component. I don't know if there is a general formula; for 4D and so long as $\langle R\rangle_0 \ne 0$ I was able to work out $$ R \propto 4\sum_iv^{\prime i}v_i + \sum_{i,j}v^{\prime i}v^{\prime j}v_iv_j $$ using a method similar to the 3D case.

$s=0$ in 3D

When $s=0$, i.e. when the above rotor has no scalar part and is a bivector $B$, the above method fails. However, recovering $B$ is still straightforward.

The matrix $\Gamma' = WV^{-1}$ is precisely the representation of our rotation in the standard basis. Expressed in the $v'_i$ basis this is $$ W^{-1}\Gamma'W = V^{-1}W = \Gamma^{-1}. $$ So find the eigenvectors of $u^+, u^-_1, u^-_2$ of $\Gamma$ corresponding to eigenvalues $1, -1, -1$ respectively. Then $B$ is proportional to the dual of $u^+$ via a pseudoscalar $I$ or is the wedge of $u^-_1$ and $u^-_2$ (after going back to the standard basis using $W$): $$ B\propto [Wu^+]I \quad\text{or}\quad B\propto [Wu^-_1]\wedge [Wu^-_2]. $$ Also notice the following for any eigenvector $u$ of $\Gamma$: $$ \lambda u = \Gamma u \iff \lambda Wu = Vu \iff (V - \lambda W)u = 0 $$ So it suffices to determine the nullspace of $V - W$ to get $u^+$ or the nullspace of $V+W$ to get $u^-_1, u^-_2$.

Answer 2

I think you can do this in a more "quaternionic" way if you want:

Quick review of quaternion and $\mathrm{SO}_3(\mathbb R)$:

We write $\mathbb H$ for the quaternions, and $h \mapsto \bar{h}$ for the quaternionic conjugation map, that is, if $h = h_0+h_1 \mathbf i + h_2\mathbf j+h_3\mathbf k\in\mathbb H$ then $\bar{h} = h_0-h_1 \mathbf i - h_2\mathbf j-h_3\mathbf k$. Let us also set $\Re(h) = h_0$ and $\Im(h) = h-h_0$. It is well-known that $\mathbb H$ is a normed division algebra over $\mathbb R$ with norm $$ N(h) = h.\bar{h} = h_0^2+h_1^2+h_2^2+h_3^2. $$ Thus in particular $\mathbb H$ is a 4-dimensional $\mathbb R$-vector space with inner product given by $\langle h,h' \rangle = \Re(h\overline{h'})$ for which the basis $\{1,\mathbf i,\mathbf j,\mathbf k\}$ is orthonormal. Let $\mathbb U = \{h \in \mathbb H: N(h)=h\bar{h}=1\}$ be the group of unit quaternions and let $\mathbb I = \{h \in \mathbb H: \bar{h}=-h\} = \text{span}_{\mathbb R}\{\mathbf i,\mathbf j,\mathbf k\}$ be the subspace of purely imaginary quaternions. We identify $\mathbb R^3$ with the standard "dot product" with $\mathbb I$.

It is easy to check that $\mathbb U$ acts on $\mathbb I$ by conjugation, that is, if $u \in \mathbb U, h\in \mathbb I$ then $uhu^{-1} = uh\bar{u} \in \mathbb I$ and that this yields a homomorphism $\pi\colon\mathbb U \to \mathrm{O}(\mathbb I)$. Since $\mathbb U$ is connected, $\pi(\mathbb U)\subseteq \mathrm{SO}(\mathbb I) \cong \mathrm{SO}_3(\mathbb R)$. Now if $u = \cos(\phi)+\sin(\phi)u_{im}$ where $u_{im}\in \mathbb U\cap \mathbb I$, then $\cos(\phi) \in Z(\mathbb H)$, hence $u_{im}$ and $u$ commute with each other, and hence $u.u_{im}u^{-1}=u_{im}$. Thus $\pi(u) \in \mathrm{SO}_3(\mathbb R)$ is a rotation which preserves the line $\mathbb R.u_{im}$. It is then easy to check that $\pi(u)$ is a rotation by $2\phi$ about $u_{im}$.

Finally, if $a,b \in \mathbb I$, then $$ a.b = \Re(a.b) + \Im(a.b) = -\langle a,b\rangle + \Im(a.b). $$ since $\langle h_1,h_2\rangle = \Re(h_1\overline{h_2}) = -\Re(h_1h_2)$ if $h_2 \in \mathbb I$. Moreover, $a^2 = -a(-a) = -a.\bar{a} = -N(a)$ is real, hence $\Im(a.a)=0$. Thus $\Im(a.b)$ is an alternating bilinear map $\mathbb I\times \mathbb I\to \mathbb I$, which becomes the vector cross product via the standard identification.

Obtaining the rotator: If we are given $v_1,v_2$ linearly independent vectors together with $v_1' = R(v_1)$ and $v_2' = R(v_2)$, where we view all of these as elements of $\mathbb I$.

Let $w_1= v_1-v_1'=(I-R)(v_1)$ and $w_2 = (I-R)(v_2)$. If $w_1,w_2$ are linearly dependent, say $w_2=\lambda w_1$, then $\lambda v_1-v_2 = R(\lambda v_1-v_2)$ and hence $\mathbb R.(\lambda v_1 -v_2)$ is the axis of the rotation $R$. Otherwise $\text{Im}(I-R) = \text{ker}(I-R)^{\perp}$ is spanned by $\{w_1,w_2\}$, and $\Im(w_1w_2) = a$ spans the axis of rotation of $R$.

Once we have a basis $\{a\}$ of the axis of rotation, it is straight-forward to find the angle of rotaiton: set $u = a/\|a\|$ so that $\{u\}$ is an orthonormal basis of the axis of rotation and let $c_1 = \langle v_1 ,u\rangle$. Then if $p_1 = v_1 - c_1u$ and $p_1' = v_1' - c_1 u$ we have $p_1,p'_1=R(p_1) \in \mathbb R. u^{\perp}$ and if $$ \cos(\alpha) = \langle p_1,p_1'\rangle/\|p_1\|^2 $$ then $R$ is the rotation by $\alpha$ about $u$ and so the quaternion rotator is then $\cos(\alpha/2)+\sin(\alpha/2).u$.

Answer 3

There is a unique solution to this problem but it is easier to express it in the language of rotation matrices and vectors. Quaternions don't seem to provide a significant simplification to this scheme.

Note that for unit 3-vectors $v_1,v_1', v_2, v_2'$ related by:

$$v_1'=R v_1~~,~~ v_2'=Rv_2$$

it holds true that

$$v_1'\times v_2'=R(v_1\times v_2)$$

which can be proven using the properties of a rotation matrix and its determinant. The assumption that $v_1,v_2$ are linearly independent ensures that the set $B=\{v_1, v_2, v_1\times v_2\}$ forms a basis for $\mathbb{R}^3$, and since the action of the matrix $R$ on the basis is known, the matrix can be fully reconstructed. The reconstruction can be facilitated by constructing an orthonormal basis out of $B$ given by the vectors

$$q=v_1\\r=\frac{v_2-(v_1\cdot v_2)v_1}{\|v_1\times v_2\|}\\s=\frac{v_1\times v_2}{\|v_1\times v_2\|} $$

Then all one needs to do is express $q'=Rq,r'=Rr, s'=Rs$ in terms of $q,r,s$:

$$q'=aq+br+cs\\r'=dq+er+fs\\s'=gq+hr+ks$$

and the matrix in the $qrs$ basis is given by

$$R(qrs)=\begin{pmatrix}q'\cdot q&q'\cdot r&q'\cdot s\\r'\cdot q&r'\cdot r&r'\cdot s\\s'\cdot q&s'\cdot r&s'\cdot s\end{pmatrix}$$

Here you can calculate the angle from the trace:

$$1+2\cos\theta=q'\cdot q+r'\cdot r+s'\cdot s$$

The axis can be found as the eigenvector corresponding to the eigenvalue $1$ and then expressed in the original coordinates if neccessary.