# Why is the automorphism group of the octonions $G_2$ instead of $SO(7)$

I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of $$G_2$$. However, when I do the same calculation for the quaternions, I end up with the three generators of $$SO(3)$$, which basically tells me that I can rotate the set of imaginary units anyway I like.

Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of $$SO(7)$$. Instead, the calculations show, that the possible transformations are restricted to the $$G_2$$ subgroup.

Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions?

• The quaternions are generated by any two imaginary elements x and y that are orthonormal, i.e., they are spanned by 1, x, y, and xy. Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, x, y, and z that are orthonormal and z is perpendicular to xy. This means that any automorphism of the octonions that fixes three such elements is the identity. Thus, SO(7) is too large to be the automorphism group of the octonions because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions. – Robert Bryant Mar 18 at 15:50
• @Robert Can I talk you into leaving this answer as an answer? – Theo Johnson-Freyd Mar 18 at 17:08
• @TheoJohnson-Freyd: Sure. I'll be happy to do this. – Robert Bryant Mar 18 at 18:02

The quaternions are generated by any two imaginary elements $$x$$ and $$y$$ that are orthonormal, i.e., $$\bigl(1,\, x,\, y,\, xy\bigr)$$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions.

Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, $$x$$, $$y$$, and $$z$$ that are orthonormal and $$z$$ is perpendicular to $$xy$$. In fact, $$\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $$x'$$, $$y'$$, $$z'$$ are another orthonormal triple of imaginary octonions such that $$z'$$ is perpendicular to $$x'y'$$, there is a unique automorphism of the octonions that carries $$(x,y,z)$$ to $$(x',y',z')$$.

Note that the set of orthonormal triples $$(x,y,z)$$ in the imaginary octonions is the Stiefel manifold $$V_{7,3}$$, which has dimension $$6+ 5+ 4 = 15$$, and the single extra relation $$z\cdot xy = 0$$ cuts out a submanifold of dimension $$14$$. Hence the automorphism group of the octonions is a Lie group of dimension $$14$$.

Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $$\mathrm{SO}(7)$$ (which has dimension $$21$$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions.

(Sadly, when $$x$$, $$y$$, $$xy$$, and $$z$$ are imaginary orthogonal octonions, the defining property $$uv\cdot uv = (u\cdot u)(v\cdot v)$$ turns out to imply that $$(xy)z = -x(yz)$$, so the octonions are not associative. However, this is not really the reason that $$\mathrm{SO}(7)$$ is not the automorphisms of the octonions.)

• In three dimensions, a "basic" rotation fixes a $1d$-subspace, in four dimensions there can be a fixed $2d$-plane, and so on. Do I understand correctly, that "basic" rotations that leaves a $3d$-subspace invariant would not be part of the automorphim-group? Basically, only combinations of a few "basic" rotations are allowed that rotate all imaginary units more or less simultaneously? – p6majo Mar 18 at 19:59
• However, a typical generator of $G_2$ has four non-zero entries [twice as many as a generator of $SO(7)$]. It seems to me that it fixes three imaginary units. But perhaps it acts non-trivially on the additional constraint $z\cdot xy=0$? – p6majo Mar 18 at 21:50
• @p6majo: A $\mathrm{G}_2$ element can have a 1-dimensional, 3-dimensional, or 7-dimensional $+1$-eigenspace of imaginary octonions. For example, if $x$ and $y$ are orthonormal imaginary octonions, then the set of unit imaginary octonions $z$ that are perpendicular to the $3$-plane spanned by $x$, $y$, and $xy$ is a $3$-sphere, and there is a subgroup of $G_2$ isomorphic to $\mathrm{SU}(2)$ that fixes $x$ and $y$ (and hence $xy$) and acts simply transitively on that 3-sphere. (Be careful not to confuse a subspace of fixed vectors of an automorphism with a subspace that is preserved by it.) – Robert Bryant Mar 18 at 23:59