Do all bivectors simplify to 2-blades in seven-dimensional space?

Question

The wedge product of two vectors $\vec{v}, \vec{w}\in\mathbb{R}^{n}$ can be defined as an anti-symmetrized tensor product. In three dimensions, there is a correspondence between the wedge product of two vectors and the cross product of two vectors. As a result, given any bivector $$ \overline{v} = v_{1}\vec{e}_{y}\wedge\vec{e}_{z} + v_{2}\vec{e}_{z}\wedge\vec{e}_{x} + v_{3}\vec{e}_{x}\wedge\vec{e}_{y}, $$ we can convert it to a vector $$ v_{1}\vec{e}_{x} + v_{2}\vec{e}_{y} + v_{3}\vec{e}_{z}, $$ find two perpendicular vectors $\vec{a}, \vec{b}$ for which $\vec{a}\times\vec{b} = \sum_{i} v_{i}\vec{e}_{i}$, and then $\vec{a}\wedge\vec{b} = \overline{v}$.

Thus, any bivector in three-dimensional space is a 2-blade. This fact is true in 1, 2, and 3 dimensional space, but it fails to hold in 4 dimensions. Given that seven-dimensional space has a cross product of the form $\mathbb{R}^{7}\times\mathbb{R}^{7}\rightarrow\mathbb{R}^{7}$ according to this post, does it stand to reason that any bivector can rewritten as a 2-blade as well in seven-dimensional space?

If yes, can we make a concrete proof? If no, is there a counterexample (e.g. $\vec{e}_{1}\wedge\vec{e}_{2}+\vec{e}_{3}\wedge\vec{e}_{4}$) with a proof that it is a counterexample? Moreover if no, why would my reasoning sketch involving the cross product in 3D fail for 7D?

Answer 1

First off, if $\vec{e}_{1}\wedge\vec{e}_{2}+\vec{e}_{3}\wedge\vec{e}_{4}$ is not simple in 4D, it is not going to be simple in any higher dimension. To see this, just note that the proofs, as given here, go through for $n$-dimensional space for any $n\ge 4$.

Second, the correspondence between bivectors and vectors I mentioned is the Hodge dual $\bigwedge^{1}\mathbb{R}^{3}\leftrightarrow\bigwedge^{2}\mathbb{R}^{3}$. In general, it is $\bigwedge^{k}\mathbb{R}^{n}\leftrightarrow\bigwedge^{n-k}\mathbb{R}^{n}$. For $n=7$, it takes 1-vectors to 6-vectors and vice-versa. There is no bijective correspondence (that will have any meaning) between 1-vectors and 2-vectors, especially given that $\mathbb{R}^{7}$ is of dimension $7$ but $\bigwedge^{2}\mathbb{R}^{7}$ is of dimension $\binom{7}{2} = 21$. In fact, because the latter has a greater dimension, it is inevitable that any attempt at a linear mapping between the two will fail to be bijective (either some bivectors will fail to have a corresponding vector or two bivectors will map onto the same vector).

This second point prevents the reasoning I used in 3D to go through in any other dimension. My reasoning crucially relied on this correspondence, which breaks in 7D.

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Now I also wanted to say that I find it a peculiarity that the cross product corresponds to the wedge product in 3D so well. In 3D we have $\vec{v}\times\vec{w} = \star(\vec{v}\wedge\vec{w})$, but this just doesn't hold in 7D.

In 3D, there are two possible cross products (assuming an inner-product is fixed): a "left-handed one" and a "right-handed one." We always choose the "right-handed one," and the Hodge star operator in the formula for 3D cross-products sneaks in a preferred orientation (an orientation and inner-product are required to define $\star$).

One reason it works in 3D is that given a plane spanned by two vectors, there's only one perpendicular axis to that plane. This isn't true in 7D, and that's somewhat in line with the fact that the cross product in 7D isn't unique even up to orientation/sign. See this post.

My understanding at the moment is that the formula $\vec{v}\times\vec{w} = \star(\vec{v}\wedge\vec{w})$ is only a coincidence that holds in 3D only. If there is any analogous formula for 7D of any sort, I'd be interesting to see one (feel free to post a comment). Other than that, I can't say much more.