5
$\begingroup$

Consider the octonions $\mathbb O$ and in particular their imaginary part $\operatorname{Im}\mathbb O$. Let $(-,-)$ be the scalar product induced by the identification of the imaginary octonions with $\Bbb R^7$. Furthermore, define the cross product of (any) octonions by $x\times y:= \frac{1}{2}(xy-yx)$. For imaginary octonions $x,y$, there is the nice identity $x\times y=(x,y)+x\cdot y$, where the dot denotes octonion multiplication.

Define the linear map $L_{x,y}:\operatorname{Im}\mathbb O\to \operatorname{Im}\mathbb O$ by $L_{x,y}(v)=x\times (y\times v)$, where $x,y$ are also imaginary quaternions. It is easy to prove by explicit calculation (plugging in a basis) that there is the following identity:

$$(x,y)=-\frac{1}{6}\operatorname{tr}L \qquad \qquad x,y\in \operatorname{Im}\mathbb O$$

This identity is useful because it shows that octonion multiplication of imaginary octonions only depends on the cross product, and thereby yields the equivalence of two common definitions of the exceptional Lie group $G_2$. I don't like the brute force proof, so I am left wondering whether there is a better way to see that this identity holds. Is anyone aware of a slick(er) proof?

EDIT: It was just pointed out to me by Ted Shifrin that $\operatorname{tr}L$ is the would-be Killing form on $\Bbb R^7$ seen as an (almost-but-not-quite) Lie algebra (equipped with the cross product). This might point the way towards a nice proof (?)

$\endgroup$
2
$\begingroup$

Since $L_{x,y}$ is linear in $x$ and $y$, it suffices to compute the trace in two cases: when $x$ and $y$ are parallel and when they are perpendicular (assuming $x$ and $y$ have norm $1$ as well).

In the first case, it's just the square of $L_x=x\times-$, which annihilates $x$ and equals multiplication by $x$ on its orthogonal complement, so $L_{x,x}$ will be multiplication by negative one on $x$'s orthogonal complement; the trace is obviously $-6$.

In the second case, $L_{x,y}$ annihilates $y$ and $xy$, sends $x$ to $y$, and $z\perp xy \Rightarrow x\perp yz$, implying

$$ \langle L_{x,y}z,z\rangle=\langle x(yz),z\rangle=\langle yz,\bar{x}z\rangle=\langle y,\bar{x}\rangle|z|^2=0$$

when $z\perp \{x,y,xy\}$, thus $L_{x,y}$ is skew-hermitian on $\{x,y,xy\}^\perp$ hence has trace zero.

$\endgroup$
  • $\begingroup$ What a slick way to "brute force" it. $\endgroup$ – Danu Apr 30 '17 at 0:19

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.