Let $O$ denote the division algebra of octonions over $\Bbb R$, and write $V$ for the 7-dimensional quotient space $O/{\Bbb R}$. The compact group $G_2:={\rm Aut}(O)$ naturally acts on $V$, and clearly the 7-dimensional representation of $G_2$ in $V$ is isomorphic to its representation in the space of pure octonions.

I know from a classification of alternating trilinear forms in dimension 7 that there exists a $G_2$-invariant alternating trilinear form $\omega\in\Lambda^3 V^*$

Question. What is a coordinate-free description of a $G_2$-invariant alternating trilinear form on $V$?

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    $\begingroup$ What about $(x,y,z)\mapsto\mathrm{Re}(x(yz)+y(zx)+z(xy)-x(zy)-y(xz)-z(yx))$? It's clearly invariant alternating. On $(i,j,k)$ its value is $-6$, so it's nonzero. $\endgroup$ – YCor Nov 30 '20 at 15:20
  • $\begingroup$ @YCor: Yes, it is clearly invariant and alternating. How did you compute the value on $(i,j,k)$? $\endgroup$ – Mikhail Borovoi Nov 30 '20 at 15:27
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    $\begingroup$ Just from the table: $i(jk)=j(ki)=k(ij)=-i(kj)=-j(ik)=-k(ji)=-1$... $\endgroup$ – YCor Nov 30 '20 at 15:40
  • $\begingroup$ @YCor: Thank you, that is an answer! $\endgroup$ – Mikhail Borovoi Nov 30 '20 at 15:46

The form $(x,y,z)\mapsto \mathrm{Re}(x(yz)+y(zx)+z(xy)−x(zy)−y(xz)−z(yx))$ is clearly invariant and alternating. It is nonzero, since its value at $(i,j,k)$ (which satisfy the quaternions relations) is $-6$.

Actually, it can be checked that the symmetrized form $\mathrm{Re}(x(yz)+y(zx)+z(xy)+x(zy)+y(xz)+z(yx))$ vanishes. So the invariant form $$(x,y,z)\mapsto\mathrm{Re}(x(yz)+y(zx)+z(xy)$$ is already alternating (and takes the value $-3$ at $(i,j,k)$: it's actually zero modulo $3$ on the basis).

  • $\begingroup$ How did you check that the symmetrized form vanishes? $\endgroup$ – Mikhail Borovoi Nov 30 '20 at 16:17
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    $\begingroup$ I did a little brute Sage program this time to check all cases (to check the vanishing on all 343 possible triples). $\endgroup$ – YCor Nov 30 '20 at 16:37
  • $\begingroup$ Excellent! Thank you! $\endgroup$ – Mikhail Borovoi Nov 30 '20 at 16:43
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    $\begingroup$ Also if I'm correct, the 3rd symmetric power, of dimension 84, decomposes in irreducibles as 77+7. In particular it doesn't contain the trivial representation, so there's no nonzero invariant symmetric trilinear form. (While the 3rd alternating power, of dimension 35, decomposes as 27+7+1.) $\endgroup$ – YCor Nov 30 '20 at 22:27
  • $\begingroup$ My calculation using Table 5 in the book by Onishchik and Vinberg shows that $V\otimes S^2 V$ does not contain the trivial representation. Thus indeed there is nonzero invariant symmetric bilinear form. $\endgroup$ – Mikhail Borovoi Dec 1 '20 at 10:38

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