Rotations in $2$ dimensional space can be given a nice representation with complex numbers.

The group $(\mathbb{C}, \times)$ under complex multiplication has two one-parameter subgroups: the real line $r + 0i$ and the unit circle $\{a+bi\ \ |\ \ a^2 + b^2 =1\}$. The real line "stretches" and the unit circle rotates.

In three dimension we can represent rotations with quaternions. Now there are four one-parameters subgroups. In general, to rotate and "stretch" we should require $1+ \begin{pmatrix} n \\ 2 \end{pmatrix}$ paremeters, the $"1"$ is for the stretching and the $\begin{pmatrix} n \\ 2 \end{pmatrix}$ is the number of simple planes we can specify.

For 2d space, the complex numbers, we have $1 + \begin{pmatrix} 2 \\ 2 \end{pmatrix} = 2$ parameters.

For 3d space, the quaternions, we have $1 + \begin{pmatrix} 3 \\ 2 \end{pmatrix} = 4$ parameters.

I would have thought:

For 4d space, the octonions, we have $1 + \begin{pmatrix} 4 \\ 2 \end{pmatrix} = 7$ parameters.

But the octonions have 8 parameters. Am I just mistaken in thinking that the octonions are the generalization to representing rotations in $4d$ space?

  • 2
    $\begingroup$ It might be worth noting that the two cases you present are often called "exceptional isomorphisms" precisely because they are not part of a larger pattern, and, to my limited knowledge, there is no such connection between orthogonal groups and Cayley-Dickson algebras. For a pattern which extends to higher dimension, it may be worthwhile to look into Clifford algebras and spinors instead. $\endgroup$
    – Kajelad
    Apr 23, 2021 at 2:30

1 Answer 1


Am I just mistaken in thinking that the octonions are the generalization to representing rotations in 4d space?

Yes, this assumption is wrong. There is more to the story here than just numerology.

Consider how you're using the reals $\mathbb{R}$ and complex numbers $\mathbb{C}$ to model transformations. You're using the regular representation of an algebra $A$, which involves taking elements $a\in A$ and then considering "multiplication-by-$a$" as transformations of $A$ as a vector space. For the two algebras $A=\mathbb{R}$ and $A=\mathbb{C}$, we get similarity transformations (rotations and homotheties aka dilations) of $\mathbb{R}$ and $\mathbb{C}\cong\mathbb{R}^2$. But in the former case, we also get reflection (multiplication by $-1$, which is orientation-reversing), which is not present in the latter case. So already the "pattern" (of real normed division algebras modeling rotations) shows a crack.

Next is quaternions, but now $\mathbb{H}$ is noncommutative so there is a left regular representation and a right regular representation, depending on if we get transformations by multiplying by a constant quaternion on the left or on the right. But neither of these give 3D rotations! It is in some sense the "difference" or "quotient" of these two that give 3D rotations.

Let's be concrete. All quaternions are the sum of a scalar and a 3D vector, or a real part and a purely imaginary part. All nonzero quaternions have polar forms $r\exp(\theta\mathbf{u})=r(\cos\theta+\sin\theta\,\mathbf{u})$, where $r>0$ is the magnitude, $\theta$ is a convex angle, and $\mathbf{u}$ is a 3D unit vector (which is unique except for purely real quaternions). If $p=\exp(\theta\mathbf{u})$ is a phasor (unit quaternion) and $\mathbf{v}$ a vector then the conjugate $p\mathbf{v}p^{-1}$ is the 3D vector obtained by rotating $\mathbf{v}$ around $\mathbf{u}$ by $2\theta$. In addition to not using just a regular representation as with $\mathbb{R}$ and $\mathbb{C}$ in the "pattern," this now also introduces redundancy in the form of a kernel: the two different quaternions $\pm1$ both yield the same 3D rotation - this is unlike the 1D and 2D cases.

4D rotations are modeled by pairs of quaternions, which uses both the left and right regular representation independently. That is, we view $\mathbb{H}\cong\mathbb{R}^4$ as a four-dimensional real inner product space (with inner product $\langle p,q\rangle=\mathrm{Re}(\overline{p}q)$), then every 4D rotation is expressible as $x\mapsto px\overline{q}$ for some pair $(p,q)$ of phasors. Once again there is redundancy, as $(-1,-1)$ yields the same rotation as $(1,1)$.

The octonions $\mathbb{O}$ have a 7D subspace of pure imaginary octonions. The octonions are no longer associative, although they are alternative, which allows conjugation to makes sense. We can conjugate imaginary octonions by unit octonions to get 7D rotations, however (a) not all 7D rotations arise in this way (the group of 7D rotations is 21-dimensional, whereas the space of unit octonions is 6-dimensional), and (b) octonion multiplication no longer corresponds to composing rotations. That is, $p(qxq^{-1})p$ does not generally equal $(pq)x(pq)^{-1}$. I believe these transformations do generate all 7D rotations by composition, though. And the expression $pxq^{-1}$ doesn't even make sense, since $p(xq^{-1})\ne(px)q^{-1}$ in general, but I think left and right multiplications by unit octonions generate all 8D rotations.

The way to generalize how quaternions model rotations to higher dimensions require us to augment the model a little bit - using unit vectors to represent reflections instead of rotations. The keyword here is Clifford algebras (which I discuss a little here).


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