7D octonionic cross product versus bivector cross product Why the octonionic cross product product is not "unique" up to some permutations in the octonionic multiplication table BUT the bivector cross product in 7D is unique and isomorphic to one of those octonionic tables? That is, why the octonions see "more freedom" to select a cross product that simply $x\times y=x_iy_j-x_jy_i$?
 A: It looks like you've defined $x\wedge y:=xy^T-yx^T$, which turns two vectors into a matrix. There is no way the multiplication table can be "isomorphic" to that of the octonions, since the outputs aren't even the same kind of thing as the inputs in this operation. Moreover, the range of this is the $21$-dimensional space of real antisymmetric $7\times 7$ matrices, which is bigger than the $7$-dimensional space of the cross product.
By the universal property, every antisymmetric bilinear map $\mathbb{R}^7\times\mathbb{R}^7\to\mathbb{R}^7$ factors through the universal bivector map $\mathbb{R}^7\times\mathbb{R}^7\to \Lambda^2(\mathbb{R}^7)$ (you can think of $\Lambda^2$ as the set of antisymmetric matrices here). This means any $7$D cross product is the composition of this map with some map $\Lambda^2(\mathbb{R}^7)\to\mathbb{R}^7$. However there are stipulations that must be satisfied in order to count as a cross product, so not every linear map $\Lambda^2(\mathbb{R}^7)\to\mathbb{R}^7$ corresponds to a cross product. Nonetheless, there are still many choices that work. The universal property means (sort of) that the bivector is "agnostic" about where the two vectors end up - it is only with the choice of map out of the space of bivectors that their fate is determined.
Note also the octonionic multiplication table has more symmetries than just "permutations" and sign changes, although these are an interesting cross-section of the full symmetry group to consider. Call an orthogonal transformation $R$ of $\mathbb{R}^7$ a symmetry if $(Ru)\times(Rv)=R(u\times v)$ for all $u$ and $v$. The signed permutation matrices which are symmetries form a finite group isomorphic to $\mathbb{Z}_2^3\rtimes\mathrm{SL}_3(\mathbb{Z}_2)$, I think, so the space of all octonionic cross products using a given set of axes is $(\mathbb{Z}_2\wr S_7)/(\mathrm{Aff}(3,2)$. But the full symmetry group is actually called $G_2$, so the full space of $7$D cross products on $\mathbb{R}^7$ is $\mathrm{O}(8)/G_2$.
