Chapter 4: Octonions

Geometry

As with the quaternions, the units $i$, $j$, $k$, $k\ell$, $j\ell$, $i\ell$, and $\ell$ are by no means the only square roots of $-1$. Rather, any imaginary octonion squares to a negative number, so it is only necessary to choose its norm to be 1 in order to get a square root of $-1$. The imaginary octonions of norm 1 form a 6-sphere in the 7-dimensional space of imaginary octonions.

Any such unit imaginary octonion $\shat$ can be used to construct a complex subalgebra of $\OO$, which we will also denote by $\CC$, and which takes the form \begin{equation} \CC = \{a + b\,\shat\} \end{equation} with $a,b\in\RR$. We can again use the identity (4) of §2.4. to write \begin{equation} e^{\shat\theta} = \cos\theta + \shat\sin\theta \end{equation} so that any octonion can be written in the form \begin{equation} x = r e^{\shat\theta} \end{equation} where \begin{equation} r=|x| \end{equation} Two unit imaginary octonions $\shat$ and $\that$ which point in different directions (this excludes $\that=-\shat$) determine a quaternionic subalgebra of $\OO$, which we also denote by $\HH$, and which takes the form \begin{equation} \HH = \{a + b\,\shat + c\,\that + d\,\shat\,\that\,\} \label{Hsub} \end{equation} where $a,b,c,d\in\RR$.

An important technique when working with the octonions is to work with what we call generic octonions. Any single octonion $x$ can be assumed to lie in $\CC$, that is \begin{equation} x = x_1 + x_2 i \end{equation} A second octonion $y$ lies in $\HH$, but adds just one new direction, that is \begin{equation} y = y_1 + y_2 i + y_3 j \end{equation} while a third can be assumed to take the form \begin{equation} z = z_1 + z_2 i + z_3 j + z_4 k + z_8 \ell \end{equation} that is, a general element of $\HH$ plus one further direction. Only with 4 or more octonions is it necessary to use the full 8 dimensions of $\OO$! This approach makes it obvious that any products involving no more than 2 distinct octonions (or their conjugates) must associate — they lie in $\HH$! But this is just alternativity!

It is often useful to consider octonions as vectors in $\RR^8$. The norm of an octonion is precisely the same as its vector norm. But the dot product can be obtained from the norm by a process known as polarization, as follows. If $\vv$, $\ww$ are vectors in $\RR^n$, we have \begin{align} |\vv+\ww|^2 &= (\vv+\ww)\cdot(\vv+\ww) \nonumber\\ &= \vv\cdot\vv + 2\vv\cdot\ww + \ww\cdot\ww \nonumber\\ &= |\vv|^2 + |\ww|^2 + 2\vv\cdot\ww \end{align} which can be solved for the last term. By analogy, the dot product of two octonions is given by \begin{equation} 2 x \cdot y = |x+y|^2 - |x|^2 - |y|^2 = x\bar{y} + y\bar{x} = \bar{x}y + \bar{y}x \end{equation} since $|x|^2=x\bar{x}=\bar{x}x$. For imaginary octonions, orthogonality is equivalent to anticommutativity, that is, \begin{equation} x \cdot y = 0 \Longleftrightarrow x\,y = -y\,x \end{equation} which in turn implies that \begin{equation} \bar{xy} = \bar{y}\>\bar{x} = y\,x = -x\,y \end{equation} so that this is also equivalent to the product being pure imaginary. We can therefore use the dot product to ensure that the product $\shat\,\that$ in (\ref{Hsub}) is pure imaginary: If not, simply replace $\that$ by its orthogonal component $\that-(\that\cdot\shat)\,\shat$ (rescaled to have norm 1).

What about the cross product? The cross product of two vectors in $\RR^3$ points in the unique direction (up to sign) orthogonal to the plane spanned by the given vectors. In higher dimensions, there is in general no such preferred direction; there are many directions perpendicular to a given plane. It is therefore somewhat surprising that restricting the octonionic product to imaginary octonions yields a cross product in $\RR^7$ with the usual properties (which are essentially linearity, anticommutativity, and orthogonality to each original vector; there is also a condition on the norm of the cross product). Remarkably, such products exist only in $\RR^3$ and $\RR^7$, corresponding to imaginary quaternions and octonions, respectively.

One way to specify a unique direction in $\RR^n$ is to give $n-1$ directions orthogonal to it. One might therefore suspect that there is a generalized “cross product” in $\RR^n$ involving $n-1$ vectors. This is correct; the resulting product is most easily described in the language of differential forms. But there is precisely one further “generalized cross product” of more than two vectors, namely a product of 3 vectors in $\RR^8$, which also turns out to be related to the octonions. This triple cross product is defined by \begin{equation} x\times y\times z = {1\over2}\Big( x(\bar{y}z)-z(\bar{y}x) \Big) \end{equation} for any octonions $x$, $y$, $z$. (Note that this is not an iterated cross product, but a product directly defined on 3 factors.) The real part of this product defines the associative 3-form $\Phi$, namely \begin{equation} \Phi(x,y,z) = \Re (x \times y \times z) = {1\over2} \, \Re\Big( [x,\bar{y}] z\Big) \end{equation} which we will use later. (The last equality follows by direct computation.)

As with quaternions, a useful strategy for solving problems is to break up the octonions into complex or quaternionic pieces. We have \begin{align} x &= x_\Hone + x_\Htwo \ell \\ &= x_\Cone + x_\Ctwo i + x_\Cthree j + x_\Cfour k \end{align} where $x_\Hone,x_\Htwo\in\HH$ and $x_\Cone,x_\Ctwo,x_\Cthree,x_\Cfour\in\CC$, and where our default conventions are that $\HH$ is the quaternionic subalgebra generated by $i$, $j$, $k$, but $\CC$ is the complex subalgebra determined by $\ell$.

  • What is the result of conjugating an octonion by $\ell$?

First of all, there are no associativity issues here since there are only two octonions involved, namely $\ell$ and $x$. The easiest way to work this out is to expand $x$ as in (\ref{xdef}), noting that $\ell$ anticommutes with each term except the first and last, so that \begin{equation} \ell y \bar{\ell} = y_1 - y_2 i - y_3 j - y_4 k - y_5 k\ell - y_6 j\ell - y_7 i\ell + y_8 \ell \end{equation} As a special case, for $q\in\HH$ we have \begin{equation} \ell q \bar{\ell} = \bar{q} \label{lcom1} \end{equation} and similarly \begin{equation} \ell (q \ell) \bar{\ell} = \bar{q} \ell \label{lcom2} \end{equation} or equivalently \begin{align} \ell q &= \bar{q} \ell \\ \ell (q \ell) &= -\bar{q} \end{align} These last two expressions are pieces of the general multiplication rule, which can be expressed in the form \begin{equation} (x_\Hone + x_\Htwo \ell) (y_\Hone + y_\Htwo \ell) = (x_\Hone y_\Hone - \bar{y_\Htwo} x_\Htwo) + (y_\Htwo x_\Hone + x_\Htwo \bar{y_\Hone}) \ell \qquad\qquad \end{equation} and from which (\ref{lcom1}) and (\ref{lcom2}) could have been derived, in the form \begin{equation} \ell (y_\Hone + y_\Htwo \ell) \bar{\ell} = \bar{y_\Hone} + \bar{y_\Htwo} \ell \end{equation}

  • What is the result of conjugating an octonion by $e^{\ell\theta}$?

This follows immediately from the similar computation (10) of §3.4 over the quaternions. Write $x$ in terms of 4 complex numbers as above. Looking first at the quaternionic subalgebra containing $i$ and $\ell$, and then replacing $i$ in turn with $j$ and $k$, leads us to \begin{align} e^{\ell\theta} x e^{-\ell\theta} &= e^{\ell\theta} x_\Cone e^{-\ell\theta} + e^{\ell\theta} x_\Ctwo i e^{-\ell\theta} + e^{\ell\theta} x_\Cthree j e^{-\ell\theta} + e^{\ell\theta} x_\Cfour k e^{-\ell\theta} \nonumber\\ &= x_\Cone + x_\Ctwo e^{2\ell\theta} i + x_\Cthree e^{2\ell\theta} j + x_\Cfour e^{2\ell\theta} k \end{align} As we will see later, this corresponds to a rotation by $2\theta$ in 3 planes at once!


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