# Why is SO(8) generated by left multiplications by octonions?

I have read that $SO(8)\simeq SO(\mathbb{O})$ is generated by the set $\{L_a \,|\, a\in S^7\}$, where $L_a:\mathbb{O}\to\mathbb{O}, \, x\mapsto ax$ is the left translation.

Since for $a\in\mathbb{O}$, the orthogonal reflection in the hyperplane $a^\bot$ is given by $x\mapsto -a\bar{x}a$ and elements of $SO(8)$ are an even number of such reflections, I understand that $SO(8)$ is generated by $\{B_a \,|\, a\in S^7\}$ where $B_a: x\mapsto axa$. I also read that there is a "Triality" that link $L_a$, $R_a$ and $B_a$, or the three irreducible representations of $Spin(8)$.

But I can't figure out why $SO(8)$ is generated by $\{L_a \,|\, a\in S^7\}$ in simple terms. I have found an article that does it by hand here but I would like a more abstract answer involving triality.

An other answer than triality would also be appreciated!

Thanks,

Paul

Given an algebra $A$, an automorphism $\phi$ is a linear operator satisfying $\phi(xy)=\phi(x)\phi(y)$. There is a more general notion called an isotopy $(\alpha,\beta,\gamma)$ such that $xy=z\Rightarrow\alpha(x)\beta(y)=\gamma(z)$. We will call this $\mathrm{Iso}_1(\mathbb{O})$ in the case of the octonions $\mathbb{O}$, and a priori restrict to $\alpha,\beta,\gamma\in\mathrm{SO}(8)$. (If I recall correctly, they are all in $\mathrm{SO}(8)$ up to some scalars $\lambda,\mu,\eta$ such that $\lambda\mu\eta=1$ I guess I'll leave this as an exercise for the enterprising). Observe this is a subgroup of $\mathrm{SO}(8)^3$, because given any pair $(\alpha_1,\beta_1,\gamma_1)$ and $(\alpha_2,\beta_2,\gamma_2)$ we have

$$\begin{array}{ll} (\gamma_1\gamma_2)(xy) & =\gamma_1(\gamma_2(xy)) \\ & =\gamma_1(\alpha_2(x)\beta_2(y)) \\ & =\alpha_1(\alpha_1(x))\beta_1(\beta_2(y)) \\ & =(\alpha_1\alpha_2)(x)\,(\beta_1\beta_2)(y) \end{array}$$

and thus $(\alpha_1\alpha_2,\beta_1\beta_2,\gamma_1\gamma_2)\in\mathrm{Iso}_1(\mathbb{O})$ as well. (I will let you check inverses.)

Replacing $z$ with $\bar{z}$ in the equation $xy=z$, we see the result is equivalent to both $(xy)z=1$ and $x(yz)=1$ so we may unambiguously write $xyz=1$. An equivalent form of isotopies is then the group of all $(\alpha,\beta,\gamma)\subset\mathrm{SO}(8)^3$ such that $xyz=1\Rightarrow\alpha(x)\beta(y)\gamma(z)=1$. Call it $\mathrm{Iso}_2(\mathbb{O})$.

These are isomorphic $\mathrm{Iso}_1(\mathbb{O})\cong\mathrm{Iso}_2(\mathbb{O})$ via $(\alpha,\beta,\gamma)\leftrightarrow(\alpha,\beta,{}^C\gamma)$, where $C(z):=\overline{z}$ is octonion conjugation and ${}^C\gamma:=C\circ\gamma\circ C^{-1}$ is group-theoretic conjugation, because $xyz=1\Leftrightarrow xy=\bar{z}$ and $\alpha(x)\beta(y)\gamma(z)=1\Leftrightarrow\alpha(x)\beta(y)=\overline{\gamma(z)}$ (replace $z$ with $\overline{z}$ in latter of each iff).

There is also a third manifestation. For real vector spaces, a duality is a nondegenerate bilinear pairing $V_1\times V_2\to\mathbb{R}$. A triality is a trilinear mapping $V_1\times V_2\times V_3\to\mathbb{R}$ for which fixing one argument yields a duality of the other two vector spaces. Without loss of generality, $V_1,V_2,V_3$ are the same inner product space and dualizing yields an equivalent multiplication operation $V\otimes V\to V$; the corresponding condition is that it makes $V$ a division algebra.

In the case of $\mathbb{O}$, multiplication gives the triality $t(x,y,z)=\langle xy,z\rangle$ (up to conjugation on $z$ this is the same as the more symmetric form $\langle xy,\bar{z}\rangle=\mathrm{Re}(xyz)$). The symmetries of the triality $t$ are all $(\alpha,\beta,\gamma)\in\mathrm{SO}(8)^3$ such that $t(\alpha(x),\beta(y),\gamma(z))=t(x,y,z)$. Call this $\mathrm{Iso}_3(\mathbb{O})$. Again I don't recall offhand why, but this is isomorphic to the other two isotopy groups. (Another exercise for readers.)

In any case, let's just look at $\mathrm{Iso}_1(\mathbb{O})$. One of the Moufang identities, $(ax)(ya)=a(xy)a$, is the same as saying $(L_a,R_a,B_a)\in\mathrm{Iso}_1(\mathbb{O})$. Since $B_a$s generate $\mathrm{SO}(8)$, the group homomorphism $\mathrm{Iso}_1(\mathbb{O})\to\mathrm{SO}(8)$ given by $(\alpha,\beta,\gamma)\mapsto\gamma$ is onto. The kernel is $\mathbb{Z}_2$ with nontrivial element $(-I,-I,I)$ (another exercise for readers), so the map is $2$-to-$1$. With some work one can show it is connected and conclude $\mathrm{Iso}_1(\mathbb{O})=\mathrm{Spin}(8)$, but that is actually not necessary for your goal.

There is an action of $S_3$ on $\mathrm{Iso}_2(\mathbb{O})$ by automorphisms (in fact this gives all of $\mathrm{Out}(\mathrm{Spin}(8))\cong S_3$, but again you don't need this). One of the $3$-cycles acts by $(\alpha,\beta,\gamma)\mapsto(\gamma,\alpha,\beta)$, and one of the $2$-cycles acts by $(\alpha,\beta,\gamma)\mapsto({}^C\beta,{}^C\alpha,{}^C\gamma)$. This is because $xyz=1\Leftrightarrow zxy=1$ (apply $z$ on left and $\bar{z}$ on right), and conjugating $xyz=1\Leftrightarrow \alpha(x)\beta(y)\gamma(z)=1$ yields $$\bar{z}\bar{y}\bar{z}=1\Leftrightarrow\overline{\gamma(z)}\,\,\overline{\beta(y)}\,\,\overline{\alpha(x)}=1$$

(then replace $\bar{z},\bar{y},\bar{x}$ with $x,y,z$ respectively). Thus, $S_3$ acts by permuting the coordinates, and conjugating each coordinate by $C$ in the case of odd permutations (i.e. transpositions). This fact about outer automorphisms being $S_3$ is also called triality.

The set of all $L_a$s under one such automorphism gets turned into the set of all $B_a$s, and therefore the $L_a$s are just as much a generating set as the $B_a$s are. Interestingly, if you restrict to pure imaginary $a$s, then $L_a^2=-1$ yields a representation of $\mathcal{C}\ell(\mathrm{Im}(\mathbb{O}))$ on $\mathbb{O}$, hence an action of $\mathrm{Spin}(7)$ on $\mathbb{R}^8$. This doesn't match the usual action of it on $\mathbb{R}\oplus\mathbb{R}^7$ (corresponding to the $B_a$s since they generate reflections which preserve the real axis), this new action acts transitively on $S^7\subset\mathbb{O}$ (with point-stabilizer $G_2=\mathrm{Aut}(\mathbb{O})$). Using $R_a$s instead of $L_a$s gives yet a third representation of $\mathrm{Spin}(7)$ on $\mathbb{R}^8$.

If we call these three representations $V_1,V_2,V_3$ then the aforementioned Moufang identity (dualized) tells us that the triality $V_1\otimes V_2\otimes V_3\to\mathbb{R}$ is a morphism of $\mathrm{Spin}(7)$-reps, where $\mathbb{R}$ is the trivial rep. (Note that dualizing a map $A\otimes B\to C$ means having a corresponding map $A\to B^{\ast}\otimes C$ or vice-versa, and similarly for more spaces being tensored. An inner product is equivalent to an isomorphism $B\cong B^{\ast}$.)

Finally, since I always wondered, here is a way to think about how to deduce the Moufang identities from scratch. Suppose we've already established alternativity (which isn't too hard using the alternator) and that octonions are not associative. These cover all possible identities between three variables $a,b,c$ in which each variable only shows up once. But what about the next step, where there are four terms and one of the variables shows up twice? Write out all possible nonassociative monomials with two $a$s, a $b$ and a $c$.

With some quick detective work we can rule some out (from $\mathbb{O}$ being nonassociative), prove others (from $\mathbb{O}$ being alternative), and conclude many of them are equivalent (by conjugating equations and replacing variables with their conjugates). The resulting identities left to check are the Moufang identities (IIRC).

To check e.g. $(ab)(ca)=a(bc)a$, it's useful to invoke a series of reductions to simplify the problem: wlog $b$ is a unit imaginary, wlog $c$ is a unit imaginary, wlog $b$ and $c$ are orthogonal, decompose $a=a_{\|}+a_{\perp}$ with respect to the quaternionic subalgebra $\mathrm{span}\{1,b,c,bc\}$, wlog $a_{\perp}\ne0$, $a$ may be normalized so $|a_{\perp}|=1$ and therefore $\{b,c,a_{\perp}\}$ act like $\{i,j,\ell\}$ and a single check of $\mathbb{O}$s multiplication table (following from the Cayley-Dixson laws) is enough to finish.

Some sources off the top of my head,

• Very nice answer. Thank you very much:) – paeolo Aug 11 '17 at 12:20
• You mentionned the representation $\mathrm{Cliff}(7)\to\mathrm{End}_\mathbb{R}(\mathbb{O})$ which is a surjective map. According to the end of math.stackexchange.com/questions/2061606/… : "Therefore, every rotation of $\mathbb{O}$ can be attained simply by left multiplying by enough pure imaginary octonions." Is this right? I missed something since $\mathrm{Cliff}(7)$ is generated by $\mathrm{Im}(\mathbb{O})$ as an algebra, not as a monoïd.. – paeolo Aug 11 '17 at 12:29
• @paeolo That is actually another one of my accounts. I should have phrased that more clearly - by "attaining" rotations of $\Bbb O$ from left-multiplications-by-imaginaries I was including the option of linearly combining such operators, since as you say they generate ${\cal C}\ell(7)$ as an algebra. – anon Aug 11 '17 at 18:44
• Okay!. Thanks again for this very nice answer. – paeolo Aug 12 '17 at 21:14