5
$\begingroup$

A friend told me that, the same way you can represent a 3-vector as an imaginary quaternion then conjugate it by a unit quaternion with real part $\cos(\frac{\theta}{2})$ to rotate it by $\theta$ around the axis of the imaginary part, you can do the same thing to a 7-vector and a unit octonion. Can't find anything verifying this-if its true a source for the derivation for the formula would be appreciated (as would one for the quaternion formula).

$\endgroup$
  • 2
    $\begingroup$ Remember that octonions are also non-associative, so it may not be possible to compose rotations using octonions in the same way that standard rotations are composed. $\endgroup$ – Michael McGovern Dec 16 '16 at 19:39
  • 2
    $\begingroup$ See the seven dimensional cross product $\endgroup$ – Omnomnomnom Dec 16 '16 at 19:39
  • $\begingroup$ @Omnomnomnom I actually skimmed that article before posting the question-as interesting as it is I'm interested in writing programs that simulate higher dimensional rotations, so unless the cross-product has a relevance to rotation that I've forgotten $\endgroup$ – Eben Cowley Dec 16 '16 at 19:51
  • 1
    $\begingroup$ I thought that one could come up with an analog to the Rodriguez rotation formula using this cross-product, but that doesn't look to be as promising as it did. In particular, this cross-product isn't connected to the Lie algebra of $SO(7)$ in the way that the 3D cross-product is connected to $SO(3)$. You could generate some of the rotations using cross-products, though. $\endgroup$ – Omnomnomnom Dec 16 '16 at 20:18
  • 1
    $\begingroup$ The unit quaternions cover $SO(3)$ by parameterizing $SU(2)$. The latter happens because there's Lie group isomorphism which maps the quaternion ball to $SU(2)$ by sending the roots of $-1$ to the Pauli matrices. I don't actually know if $SO(7)$ has an $SU(6)$ universal covering, but this is evidently irrelevant because the Octinions are non-associative, thus shouldn't be isomorphic to $SU(6)$. $\endgroup$ – SZN Dec 17 '16 at 0:10
4
$\begingroup$

First off, some facts. Every isometry fixing the origin is an orthogonal linear transformation of space, the group of all such linear operators is called $O(V)$. The ones which preserve the orientation of space (equivalently, are path-connected to the identity map) form a subgroup called the special orthogonal group $SO(V)$, which has index two.

A hyperplane is a codimension one subspace, or equivalently the orthogonal complement of a 1D line, and a reflection is an isometry which acts as negative one on a line and as the identity on its orthogonal complement (a hyperplane). Reflections are in $O(V)$ but not $SO(V)$, and every isometry is a composition of reflections. In particular, a composition of reflections across the complements of lines $\ell_1$ and $\ell_2$ will be a rotation in the 2D plane $\mathrm{span}(\ell_1,\ell_2)$ by an angle twice that of the angle between $\ell_1$ and $\ell_2$.

A plane rotation is a map which acts as a usual rotation on a 2D subspace and acts as the identity map on the plane's orthogonal complement. Plane rotations are elements of $SO(V)$. Two plane rotations with respect to orthogonal planes will commute with each other. Every rotation, i.e. every element of $SO(V)$, is a product of plane rotations with respect to a collection of pairwise orthogonal 2D planes. If $R$ is a plane rotation with respect to the (oriented) plane $\Pi$ by an angle of $\theta$, and $S$ is any other isometry, then $SRS^{-1}$ will be a plane rotation with respect to the oriented plane $S\Pi$ by an angle of $\theta$.

In a normed division algebra $\mathbb{K}$, there is a multiplicative norm $|\cdot|$ (i.e. it satisfies $|xy|=|x||y|$), and it induces an inner product $\langle-,-\rangle$ satisfying $|x|^2=\langle x,y\rangle$. The imaginary elements are those that are perpendicular to $1$ with respect to this inner product. If $u,v$ are orthogonal imaginaries, then they anticommute, i.e. $uv=-vu$

More generally, for any imaginaries $u$ and $v$, there is a formula $uv=-\langle u,v\rangle+u\times v$. For quaternions, $u\times v$ is the usual cross product, but for octonions it is a much less symmetrical seven-dimensional cross product (which has symmetry group $G_2\subset SO(7)$).

The division algebra $\mathbb{K}$ will also be alternative. This means every pair of elements generates an associative subalgebra, even if $\mathbb{K}$ itself isn't associative. Concretely this means the algebraic expressions $aab$, $aba$ and $baa$ are unambiguous without parenthesization for all $a,b$.

The square roots of negative one are precisely the unit pure imaginaries $u$. Since $\mathbb{K}$ is alternative, $u(ux)=(uu)x=-x$ makes sense, and thus $\mathbb{O}$ becomes a left and right four-dimensional complex vector space over $\mathbb{R}[u]$ with scalar multiplication from the left and right. The two sides don't necessarily match: multiplying anything in $\mathrm{span}(1,u)$ on either side does the same thing, but $uv=-vu$ for all $v$ in the orthogonal complement of $\mathrm{span}(1,u)$.

In the quaternions $\mathbb{H}$, given any unit imaginary $u$ we can form an oriented orthonormal basis $1,u,v,w$ (satisfying the same relations as $1,i,j,k$). Then multiplying by $e^{\theta u}=\cos(\theta)+\sin(\theta)u$ on the left will rotate $\mathrm{span}(1,u)$ and $\mathrm{span}(v,w)$ each by $\theta$, whereas right multiplication will rotate $\mathrm{span}(1,u)$ and $\mathrm{span}(v,w)$ by $\theta$ and $-\theta$ respectively. If we combine the two actions using a half angle, we find that $x\mapsto e^{(\theta/2)u}xe^{-(\theta/2)u}$ acts as the identity on $\mathrm{span}(1,u)$ and a rotation by $\theta$ on $\mathrm{span}(v,w)$. If we restrict this conjugation action to the 3D subspace of imaginary quaternions, this is rotation around the oriented axis $\mathbb{R}u$ by an angle of $\theta$.

The unit quaternions are closed under multiplication and form a group called $\mathrm{Sp}(1)$ or $S^3$ (some abusively call it $\mathrm{SU}(2)$ which I dislike). Using this conjugation action, we get a group homomorphism $\mathrm{Sp}(1)\to\mathrm{SO}(3)$ with kernel $\pm1$.

The same thing happens in $\mathbb{O}$, except now $e^{(\theta/2)u}xe^{-(\theta/2)u}$ will be a rotation in three different planes within the 7D space of imaginary octonions. I don't see any obvious description of which three planes (indeed, the three planes are not an invariant - one could use different sets of three planes to describe the same rotation, such is the ambiguity of isoclinic rotations). I assume they generate $SO(7)$ but it's not clear to me how. One could presumably check using some calculation with lie algebras but that doesn't sound fun or enlightening.

For $\mathbb{O}$ we could also do the following. The effect of $x\mapsto e^{(\theta/2)u}xe^{(\theta/2)u}$ is the plane rotation by $\theta$ with respect to the 2D plane $\mathrm{span}(1,u)$. We can conjugate two of these plane rotations to get an arbitrary plane rotation of $\mathbb{O}$, which generates all of them in $SO(8)$.

Alternatively, note that $x\mapsto -\overline{x}$ (where $\overline{x}$ is conjugation - it negates the imaginary part while preserve the real part) is a reflection across the hyperplane orthogonal to the real axis, i.e. across the subspace of imaginary elements. If we conjugate this reflection by a rotation from $1$ to $q$ (e.g. left multiplication by $q$, where $q$ is a unit quaternion), we get $x\mapsto q(-\overline{qx})=-q\overline{x}q^{-1}$ which is reflection across the hyperplane orthogonal to $q$. If we do this for both $p$ and $q$, we get that $x\mapsto p(qxq^{-1})p^{-1}$ is some plane rotation.

This gives another way to see that multiplication on the left and right by unit octonions generates $SO(8)$. For quaternions, in fact we have a $2$-to-$1$ group homomorphism $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to SO(4)$.

Far less obviously, since $u^2=-|u|^2$ for imaginary octonions, we get a left action of the Clifford algebra $\mathrm{Cliff}(7)$ on $\mathbb{O}\cong\mathbb{R}^8$. From the classification of Clifford algebras, we know $\mathrm{Cliff}(7)\cong M_8(\mathbb{R})\oplus M_8(\mathbb{R})$. Since the algebra representation $\mathrm{Cliff}(7)\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$ is nontrivial, by checking dimensions and using the fact the only ideals are the two summands we see the map is surjective. Therefore, every rotation of $\mathbb{O}$ can be attained simply by left multiplying by enough pure imaginary octonions. (The same works on the right side as well.)

$\endgroup$
  • $\begingroup$ What's wrong with the group of unit quaternions being referred to as SU(2)? Do you just dislike implicitly speaking up to isomorphism? $\endgroup$ – Eben Cowley Jan 3 '18 at 16:43
  • $\begingroup$ @EbenCowley I am comfortable speaking up to isomorphisms in many contexts. Not a hundred percent sure why I don't like it for $SU(2)$ and $Sp(1)$. Partly because I consider the isomorphism between them in some sense nontrivial, partly because they have different explanations for their double covers of $SO(3)$. It's kind of like referring to the group of unit complex numbers as $SO(2)$, or the additive group of reals as $SO^+(1,1)$. $\endgroup$ – arctic tern Mar 12 '18 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.