Quaternions and spherical trigonometry

What's new 2024-12-19

Hamilton’s quaternion number system ${\mathbb{H}}$ is a non-commutative extension of the complex numbers, consisting of numbers of the form ${t + xi + yj + zk}$ where ${t,x,y,z}$ are real numbers, and ${i,j,k}$ are anti-commuting square roots of ${-1}$ with ${ij=k}$ , ${jk=i}$ , ${ki=j}$ . While they are non-commutative, they do keep many other properties of the complex numbers:

Being non-commutative, the quaternions do not form a field. However, they are still a skew field (or division ring): multiplication is associative, and every non-zero quaternion has a unique multiplicative inverse.
Like the complex numbers, the quaternions have a conjugation
$\displaystyle \overline{t+xi+yj+zk} := t-xi-yj-zk,$
although this is now an antihomomorphism rather than a homomorphism: ${\overline{qr} = \overline{r}\ \overline{q}}$ . One can then split up a quaternion ${t + xi + yj + zk}$ into its real part ${t}$ and imaginary part ${xi+yj+zk}$ by the familiar formulae
$\displaystyle \mathrm{Re} q := \frac{q + \overline{q}}{2}; \quad \mathrm{Im} q := \frac{q - \overline{q}}{2}$
(though we now leave the imaginary part purely imaginary, as opposed to dividing by ${i}$ in the complex case).
The inner product
$\displaystyle \langle q, r \rangle := \mathrm{Re} q \overline{r}$
is symmetric and positive definite (with ${1,i,j,k}$ forming an orthonormal basis). Also, for any ${q}$ , ${q \overline{q}}$ is real, hence equal to ${\langle q, q \rangle}$ . Thus we have a norm
$\displaystyle |q| = \sqrt{q\overline{q}} = \sqrt{\langle q,q \rangle} = \sqrt{t^2 + x^2 + y^2 + z^2}.$
Since the real numbers commute with all quaternions, we have the multiplicative property ${|qr| = |q| |r|}$ . In particular, the unit quaternions ${U(1,\mathbb{H}) := \{ q \in \mathbb{H}: |q|=1\}}$ (also known as ${SU(2)}$ , ${Sp(1)}$ , or ${Spin(3)}$ ) form a compact group.
We have the cyclic trace property
$\displaystyle \mathrm{Re}(qr) = \mathrm{Re}(rq)$
which allows one to take adjoints of left and right multiplication:
$\displaystyle \langle qr, s \rangle = \langle q, s\overline{r}\rangle; \quad \langle rq, s \rangle = \langle q, \overline{r}s \rangle$
As ${i,j,k}$ are square roots of ${-1}$ , we have the usual Euler formulae
$\displaystyle e^{i\theta} = \cos \theta + i \sin \theta, e^{j\theta} = \cos \theta + j \sin \theta, e^{k\theta} = \cos \theta + k \sin \theta$
for real ${\theta}$ , together with other familiar formulae such as ${\overline{e^{i\theta}} = e^{-i\theta}}$ , ${e^{i(\alpha+\beta)} = e^{i\alpha} e^{i\beta}}$ , ${|e^{i\theta}| = 1}$ , etc.

We will use these sorts of algebraic manipulations in the sequel without further comment.

The unit quaternions ${U(1,\mathbb{H}) = \{ q \in \mathbb{H}: |q|=1\}}$ act on the imaginary quaternions ${\{ xi + yj + zk: x,y,z \in {\bf R}\} \equiv {\bf R}^3}$ by conjugation:

$\displaystyle v \mapsto q v \overline{q}.$

This action is by orientation-preserving isometries, hence by rotations. It is not quite faithful, since conjugation by the unit quaternion ${-1}$ is the identity, but one can show that this is the only loss of faithfulness, reflecting the well known fact that ${U(1,\mathbb{H}) \equiv SU(2)}$ is a double cover of ${SO(3)}$ .

For instance, for any real ${\theta}$ , conjugation by ${e^{i\theta/2} = \cos(\theta/2) + i \sin(\theta/2)}$ is a rotation by ${\theta}$ around ${i}$ :

$\displaystyle e^{i\theta/2} i e^{-i\theta/2} = i \ \ \ \ \ (1)$

$\displaystyle e^{i\theta/2} j e^{-i\theta/2} = \cos(\theta) j - \sin(\theta) k \ \ \ \ \ (2)$

$\displaystyle e^{i\theta/2} k e^{-i\theta/2} = \cos(\theta) k + \sin(\theta) j. \ \ \ \ \ (3)$

Similarly for cyclic permutations of ${i,j,k}$ . The doubling of the angle here can be explained from the Lie algebra fact that ${[i,j]=ij-ji}$ is ${2k}$ rather than ${k}$ ; it also closely related to the aforementioned double cover. We also of course have ${U(1,\mathbb{H})\equiv Spin(3)}$ acting on ${\mathbb{H}}$ by left multiplication; this is known as the spinor representation, but will not be utilized much in this post. (Giving ${\mathbb{H}}$ the right action of ${{\bf C}}$ makes it a copy of ${{\bf C}^2}$ , and the spinor representation then also becomes the standard representation of ${SU(2)}$ on ${{\bf C}^2}$ .)

Given how quaternions relate to three-dimensional rotations, it is not surprising that one can also be used to recover the basic laws of spherical trigonometry – the study of spherical triangles on the unit sphere. This is fairly well known, but it took a little effort for me to locate the required arguments, so I am recording the calculations here.

The first observation is that every unit quaternion ${q}$ induces a unit tangent vector ${qj\overline{q}}$ on the unit sphere ${S^2 \subset {\bf R}^3}$ , located at ${qi\overline{q} \in S^2}$ ; the third unit vector ${qk\overline{q}}$ is then another tangent vector orthogonal to the first two (and oriented to the left of the original tangent vector), and can be viewed as the cross product of ${qi\overline{q} \in S^2}$ and ${qj\overline{q} \in S^2}$ . Right multplication of this quaternion then corresponds to various natural operations on this unit tangent vector:

Right multiplying ${q}$ by ${e^{i\theta/2}}$ does not affect the location ${qi\overline{q}}$ of the tangent vector, but rotates the tangent vector ${qj\overline{q}}$ anticlockwise by ${\theta}$ in the direction of the orthogonal tangent vector ${qk\overline{q}}$ , as it replaces ${qj\overline{q}}$ by ${\cos(\theta) qj\overline{q} + \sin(\theta) qk\overline{q}}$ .
Right multiplying ${q}$ by ${e^{k\theta/2}}$ advances the tangent vector by geodesic flow by angle ${\theta}$ , as it replaces ${qi\overline{q}}$ by ${\cos(\theta) qi\overline{q} + \sin(\theta) qj\overline{q}}$ , and replaces ${qj\overline{q}}$ by ${\cos(\theta) qj\overline{q} - \sin(\theta) qi\overline{q}}$ .

Now suppose one has a spherical triangle with vertices ${A,B,C}$ , with the spherical arcs ${AB, BC, CA}$ subtending angles ${c, a, b}$ respectively, and the vertices ${A,B,C}$ subtending angles ${\alpha,\beta,\gamma}$ respectively; suppose also that ${ABC}$ is oriented in an anti-clockwise direction for sake of discussion. Observe that if one starts at ${A}$ with a tangent vector oriented towards ${B}$ , advances that vector by ${c}$ , and then rotates by ${\pi - \beta}$ , the tangent vector now at ${B}$ and pointing towards ${C}$ . If one advances by ${a}$ and rotates by ${\pi - \gamma}$ , one is now at ${C}$ pointing towards ${A}$ ; and if one then advances by ${b}$ and rotates by ${\pi - \alpha}$ , one is back at ${A}$ pointing towards ${B}$ . This gives the fundamental relation

$\displaystyle e^{kc/2} e^{i(\pi-\beta)/2} e^{ka/2} e^{i(\pi-\gamma)/2} e^{kb/2} e^{i(\pi-\alpha)/2} = 1 \ \ \ \ \ (4)$

relating the three sides and three equations of this triangle. (A priori, due to the lack of faithfulness of the ${U(1,\mathbb{H})}$ action, the right-hand side could conceivably have been ${-1}$ rather than ${1}$ ; but for extremely small triangles the right-hand side is clearly ${1}$ , and so by continuity it must be ${1}$ for all triangles.) Indeed, a moments thought will reveal that the condition (4) is necessary and sufficient for the data ${a,b,c,\alpha,\beta,\gamma}$ to be associated with a spherical triangle. Thus one can view (4) as a “master equation” for spherical trigonometry: in principle, it can be used to derive all the other laws of this subject.

Remark 1 The law (4) has an evident symmetry ${(a,b,c,\alpha,\beta,\gamma) \mapsto (\pi-\alpha,\pi-\beta,\pi-\gamma,\pi-a,\pi-b,\pi-c)}$ , which corresponds to the operation of replacing a spherical triangle with its dual triangle. Also, there is nothing particularly special about the choice of imaginaries ${i,k}$ in (4); one can conjugate (4) by various quaternions and replace ${i,k}$ here by any other orthogonal pair of unit quaternions.

Remark 2 If we work in the small scale regime, replacing ${a,b,c}$ by ${\varepsilon a, \varepsilon b, \varepsilon c}$ for some small ${\varepsilon>0}$ , then we expect spherical triangles to behave like Euclidean triangles. Indeed, (4) to zeroth order becomes
$\displaystyle e^{i(\pi-\beta)/2} e^{i(\pi-\gamma)/2} e^{i(\pi-\alpha)/2} = 1$
which reflects the classical fact that the sum of angles of a Euclidean triangle is equal to ${\pi}$ . To first order, one obtains
$\displaystyle c + a e^{i(\pi-\gamma)/2} e^{i(\pi-\alpha)/2} + b e^{i(\pi-\alpha)/2} = 0$
which reflects the evident fact that the vector sum of the sides of a Euclidean triangle sum to zero. (Geometrically, this correspondence reflects the fact that the action of the (projective) quaternion group on the unit sphere converges to the action of the special Euclidean group ${SE(2)}$ on the plane, in a suitable asymptotic limit.)

The identity (4) is an identity of two unit quaternions; as the unit quaternion group ${U(1,\mathbb{H})}$ is three-dimensional, this thus imposes three independent constraints on the six real parameters ${a,b,c,\alpha,\beta,\gamma}$ of the spherical triangle. One can manipulate this constraint in various ways to obtain various trigonometric identities involving some subsets of these six parameters. For instance, one can rearrange (4) to get

$\displaystyle e^{i(\pi-\beta)/2} e^{ka/2} e^{i(\pi-\gamma)/2} = e^{-kc/2} e^{-i(\pi-\alpha)/2} e^{-kb/2}. \ \ \ \ \ (5)$

Conjugating by ${i}$ to reverse the sign of ${k}$ , we also have

$\displaystyle e^{i(\pi-\beta)/2} e^{-ka/2} e^{i(\pi-\gamma)/2} = e^{kc/2} e^{-i(\pi-\alpha)/2} e^{kb/2}.$

Taking the inner product of both sides of these identities, we conclude that

$\displaystyle \langle e^{i(\pi-\beta)/2} e^{ka/2} e^{i(\pi-\gamma)/2}, e^{i(\pi-\beta)/2} e^{-ka/2} e^{i(\pi-\gamma)/2} \rangle$

is equal to

$\displaystyle \langle e^{-kc/2} e^{-i(\pi-\alpha)/2} e^{-kb/2}, e^{kc/2} e^{-i(\pi-\alpha)/2} e^{kb/2} \rangle.$

Using the various properties of inner product, the former expression simplifies to ${\mathrm{Re} e^{ka} = \cos a}$ , while the latter simplifies to

$\displaystyle \mathrm{Re} \langle e^{-i(\pi-\alpha)/2} e^{-kb} e^{i(\pi-\alpha)/2}, e^{kc} \rangle.$

We can write ${e^{kc} = \cos c + (\sin c) k}$ and

$\displaystyle e^{-i(\pi-\alpha)/2} e^{-kb} e^{i(\pi-\alpha)/2} = \cos b - (\sin b) (\cos(\pi-\alpha) k + \sin(\pi-\alpha) j)$

so on substituting and simplifying we obtain

$\displaystyle \cos b \cos c + \sin b \sin c \cos \alpha = \cos a$

which is the spherical cosine rule. Note in the infinitesimal limit (replacing ${a,b,c}$ by ${\varepsilon a, \varepsilon b, \varepsilon c}$ ) this rule becomes the familiar Euclidean cosine rule

$\displaystyle c^2 = a^2 + b^2 - 2ab \cos \alpha.$

In a similar fashion, from (5) we see that the quantity

$\displaystyle \langle e^{i(\pi-\beta)/2} e^{ka/2} e^{i(\pi-\gamma)/2} i e^{-i(\pi-\gamma)/2} e^{-ka/2} e^{-i(\pi-\beta)/2}, k \rangle$

is equal to

$\displaystyle \langle e^{-kc/2} e^{-i(\pi-\alpha)/2} e^{-kb/2} i e^{kb/2} e^{i(\pi-\alpha)/2} e^{kc/2}, k \rangle.$

The first expression simplifies by (1) and properties of the inner product to

$\displaystyle \langle e^{ka/2} i e^{-ka/2}, e^{-i(\pi-\beta)/2} k e^{i(\pi-\beta)/2} \rangle,$

which by (2), (3) simplifies further to ${-\sin a \sin \beta}$ . Similarly, the second expression simplifies to

$\displaystyle \langle e^{-kb/2} i e^{kb/2} , e^{i(\pi-\alpha)/2} k e^{-i(\pi-\alpha)/2}\rangle,$

which by (2), (3) simplifies to ${-\sin b \sin \alpha}$ . Equating the two and rearranging, we obtain

$\displaystyle \frac{\sin \alpha}{\sin a} = \frac{\sin \beta}{\sin b}$

which is the spherical sine rule. Again, in the infinitesimal limit we obtain the familiar Euclidean sine rule

$\displaystyle \frac{\sin \alpha}{a} = \frac{\sin \beta}{b}.$

As a variant of the above analysis, we have from (5) again that

$\displaystyle \langle e^{i(\pi-\beta)/2} e^{ka/2} e^{i(\pi-\gamma)/2} i e^{-i(\pi-\gamma)/2} e^{-ka/2} e^{-i(\pi-\beta)/2}, j \rangle$

is equal to

$\displaystyle \langle e^{-kc/2} e^{-i(\pi-\alpha)/2} e^{-kb/2} i e^{kb/2} e^{i(\pi-\alpha)/2} e^{kc/2}, j \rangle.$

As before, the first expression simplifies to

$\displaystyle \langle e^{ka/2} i e^{-ka/2}, e^{-i(\pi-\beta)/2} j e^{i(\pi-\beta)/2} \rangle$

which equals ${\sin a \cos \beta}$ . Meanwhile, the second expression can be rearranged as

$\displaystyle \langle e^{-i(\pi-\alpha)/2} e^{-kb/2} i e^{kb/2} e^{i(\pi-\alpha)/2}, e^{kc/2} j e^{-kc/2} \rangle.$

By (2), (3) we can simplify to

$\displaystyle e^{-i(\pi-\alpha)/2} e^{-kb/2} i e^{kb/2} e^{i(\pi-\alpha)/2} = (\cos b) i - (\sin b) \cos(\pi-\alpha) j + (\sin b) \sin(\pi-\alpha) k$

and so the inner product is ${\cos b \sin c - \cos b \sin c \cos \alpha}$ , leading to the “five part rule”

$\displaystyle \cos b \sin c - \sin b \cos c \cos \alpha = \sin a \cos \beta.$

In the case of a right-angled triangle ${\beta=\pi/2}$ , this simplifies to one of Napier’s rules

$\displaystyle \cos \alpha = \frac{\tan c}{\tan b}, \ \ \ \ \ (6)$

which in the infinitesimal limit is the familiar ${\cos \alpha = \frac{c}{b}}$ . The other rules of Napier can be derived in a similar fashion.

Example 3 One application of Napier’s rule (6) is to determine the sunrise equation for when the sun rises and sets at a given location on the Earth, and a given time of year. For sake of argument let us work in summer, in which the declination ${\delta}$ of the Sun is positive (due to axial tilt, it reaches a maximum of ${23.5^\circ}$ at the summer solstice). Then the Sun subtends an angle of ${\pi/2-\delta}$ from the pole star (Polaris in the northern hemisphere, Sigma Octantis in the southern hemisphere), and appears to rotate around that pole star once every ${24}$ hours. On the other hand, if one is at a latitude ${\phi}$ , then the pole star an elevation of ${\phi}$ above the horizon. At extremely high latitudes ${\phi > \pi/2-\delta}$ , the sun will never set (a phenomenon known as “midnight sun“); but in all other cases, at sunrise or sunset, the sun, pole star, and horizon point below the pole star will form a right-angled spherical triangle, with hypotenuse subtending an angle ${\pi/2-\delta}$ and vertical side subtending an angle ${\phi}$ . The angle subtended by the pole star in this triangle is ${\pi-\omega}$ , where ${\omega}$ is the solar hour angle ${\omega}$ – the angle that the sun deviates from its noon position. Equation (6) then gives the sunrise equation
$\displaystyle \cos(\pi-\omega) = \frac{\tan \phi}{\tan(\pi/2-\delta)}$
or equivalently
$\displaystyle \cos \omega = - \tan \phi \tan \delta.$
A similar rule determines the time of sunset. In particular, the number of daylight hours in summer (assuming one is not in the midnight sun scenario ${\phi > \pi/2 -\delta}$ ) is given by
$\displaystyle 24 - \frac{24}{\pi} \mathrm{arccos}(\tan \phi \tan \delta).$
The situation in winter is similar, except that ${\delta}$ is now negative, and polar night (no sunrise) occurs when ${\phi > \pi/2+\delta}$ .