Correspondence between rotations and pairs of antipodal unit quaternions

Question

I'm having some trouble understanding how rotations of $\mathbb{R}^3$ correspond to antipodal pairs of unit quaternions.

In section 1.5 of his Naive Lie Theory, John Stillwell proves the theorem that conjugation by $\cos \theta + u \sin \theta$, where $u$ is a purely imaginary unit quaternion, results in a rotation of the purely imaginary quaternions by $2 \theta$ about $u$.

He then says that

This theorem shows that every rotation of $\mathbb{R}^3$, given by an axis $u$ and angle of rotation $\alpha$, is the result of conjugation by the unit quaternion $$t = \cos \frac{\alpha}{2} + u \sin \frac{\alpha}{2}.$$ The same rotation is induced by $-t$, since $(-t)^{-1}s(-t) = t^{-1}st$. But $\pm t$ are the only unit quaternions that induce this rotation, because each unit quaternion is uniquely expressible in the form $t = \cos \frac{\alpha}{2} + u \sin \frac{\alpha}{2}$, and the rotation is uniquely determined by the two (axis, angle) pairs $(u, \alpha)$ and $(-u, -\alpha)$. The quaternions $t$ and $-t$ are said to be antipodal, because they represent diametrically opposite points on the 3-sphere of unit quaternions.

Thus the theorem says that rotations of $\mathbb{R}^3$ correspond to antipodal pairs of unit quaternions.

I don't understand what Stillwell means when he says that "each unit quaternion is uniquely expressible in the form $t = \cos \frac{\alpha}{2} + u \sin \frac{\alpha}{2}$"? Indeed, this seems to be false, since $$\cos \frac{\alpha}{2} + u \sin \frac{\alpha}{2} = \cos \frac{-\alpha}{2} + (-u) \sin \frac{-\alpha}{2}, \tag{*}$$ but $\alpha \neq -\alpha$ and $u \neq -u$.

I also don't understand what he means by "the rotation is uniquely determined by the two (axis, angle) pairs $(u, \alpha)$ and $(-u, -\alpha)$". Surely only one of these is sufficient to determine the rotation in question?

I also note that on the Wikipedia page for Quaternions and spatial rotations, it is stated that the fact each rotation is represented by two distinct unit quaternions

reflects the fact that each rotation can be represented as a rotation about some axis, or, equivalently, as a negative rotation about an axis pointing in the opposite direction (a so-called double cover).

This confuses me since this "double cover" property of rotations seems to correspond more to $(*)$ than to the fact that conjugation by $-t$ gives the same rotation as conjugation by $t$. Indeed, writing

$$ \begin{align} -t &= -\cos \frac{\alpha}{2} - u \sin \frac{\alpha}{2} \\ &= \cos(\frac{\alpha}{2} + \pi) + u \sin(\frac{\alpha}{2} + \pi) \\ &= \cos \frac{\alpha + 2 \pi}{2} + u \sin \frac{\alpha + 2 \pi}{2} \end{align}$$

it appears $-t$ corresponds to the rotation $(u, \alpha + 2 \pi)$ rather than $(-u, -\alpha)$ as intimated by this remark. (I say this somewhat metaphorically -- I am aware that as objects $(u, \alpha + 2 \pi)$ and $(-u, -\alpha)$ are in fact identical.)

How does this correspondence actually work?

Answer 1

Stillwell is being a little sloppy here. When we say that a complex number is $r e^{i\theta}$, we're being a little informal, since it's also $(-r) e ^{i (\theta + \pi)}$ ... but we all agree to use only positive $r$, and to treat $z = 0$ as a special case, and at THAT point, the $(r, \theta)$ description is unique. Something similar is going on here.

I believe that the uniqueness he's claiming is that every unit quaternion $$ q = a + bi + cj + dk $$ can be uniquely expressed as $$ q = \cos(\theta) + \sin(\theta) \mathbf u $$ for some unit quaternions $\mathbf u$ and for some $\theta$ between $0$ and $\pi$. To get this represention, we write $$ \theta = \arccos(a), $$ compute $e = \sin(\theta)$, and write $$ \mathbf u = (b/e, c/e, d/e). $$

He then says (to himself), "Oh, and to make things simpler later on, let's write $\alpha = 2 \theta$, so $\alpha$ is between $0$ and $2 \pi$.

As for "The rotation is uniquely determined the the two axis-angle pairs...", that's a bit sloppy, too. What's meant is, I think, this:

"Each of the two axis-angle pairs blah and blah determines the same rotation, and furthermore, there's no other axis-angle pair that corresponds to this rotation: the map from axis-angle pairs to rotations is 2-to-1, and surjective."

But here's the fact: the unit quaternions $q$ and $-q$ really do correspond to the same rotation. A 2-to-1 mapping like this (with a few other properties, all of which this map has!) is called a "double cover".

I'm going to go all out and illustrate this for you by looking at the unit quaternions of the form $$ a = a + 0i + 0j + bk $$ which happen to correspond to rotations of the xy-plane of 3-space (or "the ij plane of pure vector quaternion space")

Each such quaternion is expressible as $$ q = \cos(\theta) + \mathbf k \sin(\theta) $$ for some theta between $0$ and $2 \pi$. Let's look at the map $$ s \mapsto q^{-1} s q $$ on the pure vector quaternions. We can see what it does to each of $s = i, j, k$. I'll let you check that $q^{-1} k q = k$. So it's a rotation about the $k$ axis. Let's see by how much: \begin{align} q^{-1} i q &= (a - bk) i (a + bk)\\ &= (ai - bj)(a + bk)\\ &= a^2 i - ab j + ab (-j) - b^2 jk \\ &= (a^2 - b^2) i - 2ab j \end{align} where $a = \cos \theta$ and $b = \sin \theta$. By the double-angle formulas, we have $a^2 - b^2 = \cos^2 \theta - \sin^2 \theta = \cos 2\theta$ and $2ab = \sin 2 \theta$.

Summary so far: the great circle of unit quaternions in the $(1,k)$ plane each corresponds to a rotation of the pure vector quaternions in the $ij$ plane, with $\cos \theta + k \sin \theta$ corresponding to a rotation by $2 \theta$.

Each quaternion in this great circle corresponds to a rotation, but clearly, two opposite quaternions (for, say, $\theta$ and $\theta + \pi$) correspond to the same rotation of the $ij$-plane (by amount $2 \theta$, or equivalently, by $2\theta + 2 \pi$).

Does that help any?

Answer 2

You can display this from the matrix side of things by invoking Euler's theorem of rigid-body rotations [1]. It states that the orientation of a body after having undergone any sequence of rotations is equivalent to a single rotation by the unit vector $\vec{n}=(n_x,n_y,n_z)^{\top}$. So for a given rotation; \begin{align}R = \begin{bmatrix}r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \end{bmatrix} \qquad (1) \end{align} We have the equivalent representation; \begin{align} R(\vec{n},\theta) = \begin{bmatrix}\cos(\theta)+n_{x}^2(1-\cos(\theta)) & n_xn_y(1-c)-n_zs & n_xn_z(1-\cos(\theta))+n_y\sin(\theta) \\ n_yn_x(1-\cos(\theta)) + n_z\sin(\theta) & \cos(\theta)+n_y^2(1-\cos(\theta)) & n_yn_z(1-\cos(\theta)) + n_x\sin(\theta) \\ n_zn_x(1-\cos(\theta))-n_y\sin(\theta) & n_zn_y(1-\cos(\theta)) + n_x\sin(\theta) & \cos(\theta)+n_z^2(1-\cos(\theta)) \end{bmatrix} \end{align} Now what is not obvious from inspection is the following change of variables, let $\vec{e} = (e_0, e_1, e_2, e_3)$; \begin{align}e_0 &= \cos(\theta/2) \\ e_1 &= n_x\sin(\theta/2) \\ e_2 &= n_y\sin(\theta/2) \\ e_3 &= n_z\sin(\theta/2) \end{align} These are known as $\it{Euler~parameters}.$ From an algebraic point of view we have effectively formed a double covering of the space of rotations. Our equivalent representation $R(\vec{n},\theta)$ that we defined earlier can now be expressed as; \begin{align} R(e_0, e_1, e_2, e_3) = \begin{bmatrix} 2(e_0^2 + e_1^2)-1 & 2(e_1e_2-e_0e_3) & 2(e_1e_3+e_0e_2) \\ 2(e_1e_2-e_0e_3) & 2(e_0^2 + e_2^2)-1 & 2(e_1e_3-e_0e_1) \\ 2(e_1e_3-e_0e_2) & 2(e_2e_3+e_0e_1) & 2(e_0^2 + e_2^3)-1 \end{bmatrix} \qquad (2) \end{align} From inspection we now see that $\vec{e}$ and $-\vec{e}$ both represent the same rotation matrix. Computationally however we have to solve the system of non-linear equations by setting our original matrix $(1)$ equal to the matrix we arrived at $(2)$.

[1] Goldstein, Herbert, Classical mechanics, Cambridge: Addison-Wesley Press, Inc. XII, 399 pp. (1951). ZBL0043.18001.