Rotations in three dimensions are determined by giving the axis of rotation and the angle of rotation about that axis. But how do you specify a rotation in higher dimensions? Specifying the axis of rotation is just a way of specifying the plane in which the rotation takes place. In higher dimensions, one must specify the plane of the rotation; there is more than one “axis” perpendicular to any plane.

The key property of rotations is that they preserve length! Rotations take the unit sphere to itself, and hence take any unit vector to some other unit vector. In other words, the norm $|\vv|$ of a vector is unchanged by the rotation. However, rotations are not the only transformations with this property; there are also reflections. The difference is that rotations are orientation-preserving, whereas reflections are orientation-reversing. (These are the only possibilities; any linear transformation must result in either an even or an odd permutation of the relative positions of the axes.)

The group of length-preserving linear transformations is called $O(n)$, the orthogonal group in $n$ dimensions; those transformations which are also orientation-preserving make up the rotation group in $n$ dimensions, $SO(n)$. These are groups, because a sequence of such transformations yields another.

In the appropriate dimensions, orthogonal transformations can be expressed in terms of division algebras. For instance, since an octonion $x\in\OO$ can be thought of as a vector in $\RR^8$, orthogonal transformations are simply those which preserve the norm $|x|$. Octonionic conjugation, $x\longmapsto\bar{x}$, is an example of a reflection.

What about rotations? Consider first conjugation 1) by a fixed element \hbox{$q\in\OO$}, that is, $x\longmapsto qxq^{-1}$. As we have already seen, there are no associativity issues here, and we can also assume without loss of generality that $|q|=1$. We have \begin{equation} |qxq^{-1}| = |qx\bar{q}| = |q| |x| |\bar{q}| = |x| \end{equation} Conjugation by $q$ is therefore an orthogonal transformation; it is not that hard to show that it is also orientation-preserving, and in fact a rotation. (It is not possible to implement octonionic conjugation through multiplication.)

So we have some rotations; which ones? First of all, the real part of $x$ is unchanged by this transformation; only the imaginary directions are affected. So we are talking about rotations of $\RR^7$. The question is, do we have all of them? We begin by considering several examples.

We showed in §2 that conjugation by $\shat$ yields a reflection of imaginary quaternions about the $\shat$-axis; this transformation will henceforth be referred to as a flip. Flips yield reflections not only when applied to quaternions, but also when applied to octonions, since the expression $qx\bar{q}$ involves only two directions, and hence lies in a quaternionic subalgebra of $\OO$.

But any rotation can be constructed from two flips. For instance, to rotate the $xy$-plane, first pick any line (through the origin) in that plane. Now reflect all points through that line, that is, imagine the line to be a mirror, and put each point where its image is. For example, if the chosen line is the $x$-axis, then the $x$-coordinate of a point is unaffected, while its $y$- and $z$-coordinates are multiplied by $-1$. Now pick another line in the $xy$-plane, at an angle $\theta$ from the first line, and reflect all points through this new line. Any point not in the $xy$-plane is taken back to its starting point. Any point in the $xy$-plane winds up being rotated by $2\theta$! (This is easiest to see for points along the $x$-axis.)

We note first of all that it doesn't matter which two lines in the $xy$-plane we choose, so long as they are separated by $\theta$ (with the correct orientation). And we have described this procedure as though it were taking place in 3 dimensions, but in fact it works in any number of dimensions; there can be any number of “$z$-coordinates”.

To rotate counterclockwise by an angle $2\theta$ in the $ij$-plane, we therefore begin by conjugating with $i$, thus reflecting about the $i$-axis. To complete the rotation, we need to reflect about the line in the $ij$-plane which makes an angle $\theta$ with the $i$-axis. This is accomplished by conjugating by a unit octonion $\shat$ pointing along the line, which is easily seen to be \begin{equation} \shat = i \cos\theta + j \sin\theta \label{sflip} \end{equation} Finally, note that the octonionic conjugate of any imaginary octonion is just minus itself. Putting this all together, a rotation by $2\theta$ in the $ij$-plane is given by \begin{equation} x \longmapsto (i\cos\theta+j\sin\theta)(ixi)(i\cos\theta+j\sin\theta) \label{ijflip} \end{equation} for any octonion $x$. (We have removed two minus signs.)

Over the octonions, we can not simplify this further. (Further parentheses are not needed; the remaining products are associative!) But over the quaternions, we can combine this into a single conjugation operation \begin{equation} x \longmapsto qx\bar{q} \end{equation} with \begin{equation} q = - (i\cos\theta+j\sin\theta) i = \cos\theta+k\sin\theta = e^{k\theta} \end{equation} where we have added the minus sign for convenience. We considered in §2 the result of conjugating by $e^{i\theta}$; by analogy, we have 2) \begin{align} e^{k\theta} (x_1 + x_2 i + x_3 j + x_4 k) e^{-k\theta} &= x_1+e^{2k\theta}(x_2 i + x_3 j)+x_4 k \nonumber\\ &= x_1 + (x_2\cos2\theta - x_3\sin2\theta) i \\ & \quad + (x_2\sin2\theta + x_3 \cos2\theta) j + x_4 k \nonumber \end{align} This is therefore the expression for rotating a quaternion counterclockwise by $2\theta$ in the $ij$-plane.

If desired, this transformation can be expressed in matrix form. We suppress two dimensions to write \begin{equation} \begin{pmatrix}x_2\cr x_3\cr\end{pmatrix} \longmapsto \begin{pmatrix} \cos2\theta& -\sin2\theta\cr \sin2\theta&   \cos2\theta\cr \end{pmatrix} \begin{pmatrix}x_2\cr x_3\cr\end{pmatrix} \end{equation} which admits obvious generalizations to higher dimensions (by adding ones along the diagonal and zeros elsewhere). The rotation group $SO(n)$ can therefore be thought of as the set of all $n\times n$ matrices of this form. These matrices are orthogonal 3) and have determinant 1; these two properties are often used to define the matrix group “$SO(n)$”. It is important to realize that this is merely one of several representations of the rotation group.

We summarize what we have done so far. Over $\CC$, conjugation by $e^{i\theta}$ is just the identity transformation! Not so over $\HH$; we have seen that such a transformation is a rotation by $2\theta$ in the $jk$-plane. What about over $\OO$?

For later convenience, we will consider conjugation by $e^{\ell\theta}$. If we look at any quaternionic subalgebra containing $\ell$, we must get the quaternionic result, namely a rotation by $2\theta$ in the “orthogonal” plane. But there are many such subalgebras! For instance, {$i$,$\ell$,$i\ell$} determine a quaternionic subalgebra, in which conjugation by $e^{\ell\theta}$ rotates the plane containing $\ell$ and $i\ell$. The same is true if we replace $i$ with $j$ or $k$. We conclude that conjugation by $e^{\ell\theta}$ rotates all 3 “orthogonal” planes. 4)

Even over the quaternions, however, conjugation by a phase doesn't give us all the rotations: The identity element, $1$, is left invariant by any conjugation! On the other hand, flips are special cases of conjugation, with $\theta={\pi\over2}$. But we have already seen how to construct a rotation in any imaginary plane as a sequence of 2 flips. We conclude conjugation, via flips, generates the rotation group on the space of imaginary elements.

The imaginary quaternions form a 3-dimensional space; rotations in this space are just ordinary rotations of the sphere. The rotation group is $SO(3)$, which is therefore generated by quaternionic flips. Analogously, the imaginary octonions form a 7-dimensional space, the rotations correspond to rotations of the $7$-sphere, and the rotation group is $SO(7)$, which is generated by octonionic flips.

The rotation groups $SO(n)$ are Lie groups, because they depend on continuous parameters. How many parameters does it take to specify a rotation of the sphere in $\RR^3$? Two angles to specify the axis, one more to specify the rotation about that axis, for a total of 3. Equivalently, this is the number of independent axes of rotation — really the number of independent planes. Such rotations generate the Lie group, in the sense that any rotation can be expressed as a sequence of generators. We say that $SO(3)$ is a 3-parameter Lie group. Similarly, in $\RR^7$, it takes 6 parameters to specify the new location of the first axis, 5 for the next, and so forth, for a total of 21 parameters — again the number of coordinate planes.

However, there is a big difference between these two cases when we look at the algebraic structure as well. An automorphism of an algebra is a map which preserves multiplication. Over the quaternions, conjugation does this, since \begin{equation} (qx\bar{q}) (qy\bar{q}) = q(xy)\bar{q} \label{auto} \end{equation} (since $\bar{q}=q^{-1}$ for elements of norm 1). Note in particular that the identity element is mapped to itself, as it must be for any automorphism. The automorphism group of the quaternions is therefore $SO(3)$.

But ($\ref{auto}$) fails over the octonions, since the parentheses cannot be moved in general. We therefore ask whether there are special octonions $q$ for which the parentheses can be moved, and which therefore result in automorphisms. The answer turns out to be yes: ($\ref{auto}$) yields an automorphism precisely when \begin{equation} q = e^{\shat\pi/3} \end{equation} As shown in [ 19 ], the automorphism group of the octonions is generated by these transformations. The resulting group is an exceptional Lie group known as $G_2$, about which we will hear more later. How big is $G_2$? We can count parameters as before. It takes 6 parameters to specify where $i$ goes, and $5$ more to specify where $j$ goes, but the new location of $k$ is then fixed by ($\ref{auto}$). There are now only 4 imaginary directions left, so it takes 3 parameters to specify the new location of $\ell$ (on the 3-sphere in this 4-dimensional space). Everything else is now fixed by ($\ref{auto}$); there are thus 6+5+3=14 parameters for $G_2$; the automorphism group of the octonions is not $SO(7)$.

Finally, what about the remaining rotations, between the real direction and the imaginary directions? We must clearly abandon conjugation, which fixes the real direction. We first try symmetric multiplication, that is \begin{equation} x \longmapsto qxq \end{equation} and we assume $|q|=1$, so that the mapping preserves norm — as it must in order to be a rotation. Over the complex numbers, symmetric multiplication by $q=e^{i\theta}$ yields a rotation by $2\theta$, thus generating $SO(2)$, the rotation group in 2 dimensions. It it easy to see that these transformations indeed generate the full rotation group, which is $SO(4)$ for quaternions and $SO(8)$ for octonions. First of all, rotations generated by flips are still here — ($\ref{ijflip}$) can be reinterpreted as symmetric multiplication! This gets us all of the rotations in the imaginary directions. As for the rest, noting that \begin{equation} e^{\ell\theta} i e^{\ell\theta} = i \end{equation} we see that symmetric multiplication by $e^{\ell\theta}$ only affects the complex plane (defined by $\ell$), where we already know that it yields a rotation by $2\theta$. This is sufficient to ensure that the full rotation group can be generated by symmetric multiplication [ 19 ].

But we can also generate rotations in 2 dimensions simply by multiplying by $e^{i\theta}$; there is no need for symmetric multiplication. Is this true for one-sided multiplication over the quaternions or octonions?

Remarkably, this fails over the quaternions [ 19 ]: One-sided multiplication by unit quaternions on either the left or the right generates a (non-obvious) copy of $SO(3)$; both are needed to generate $SO(4)$. This reflects the fact that \begin{equation} SO(4) \approx SO(3) \times SO(3) \end{equation}

Even more remarkably, the symmetry between one-sided multiplication and symmetric multiplication is restored over the octonions! It turns out [ 19 ] that one-sided multiplication by unit octonions generates all of SO(8). How can this be? Because the lack of associativity gives you more freedom, not less, namely the freedom to obtain a new transformation by nesting several transformations. (Associativity would collapse this new transformation to a simpler one.) This equivalence of the 3 representations of SO(8) given by multiplication on the left, multiplication on the right, and symmetric multiplication constitutes an important property of SO(8) known as triality.