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The three-dimensional cross product can be viewed as the wedge product corresponding to the exterior power $\Lambda^2(\mathbb R^3)$. An explanation that I have come up with for the scarcity of cross products is the following: we have the classical result $$\dim(\Lambda^m(\mathbb R^n))=\binom{n}{m}$$ and in particular $$\dim(\Lambda^2(\mathbb R^3))=\binom 3 2 = 3.$$ Since other $n$ don't satisfy $\binom n 2 = n$, there will be no other dimensionality in which the wedge product is a function $\mathbb R^n\to\mathbb R^n$ (in particular similar logic applies to why the triple product on $\mathbb R^3$ is a scalar: $\binom 3 3=1$.) In this case, how is my logic losing the possibility of a 7-dimensional cross product, noting that $\binom 7 2 = 21$ and thus the 7d wedge product maps into $\mathbb R^{21}$?

Note: I'm perfectly aware about quaternions and octonions providing an explanation for the existence of a 7d cross product - I'm not asking for a construction.

theage
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  • Although the map $(x, y) \to *(x \wedge y)$ doesn't generalize to a map $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ in general, not every cross-product has to be generated as such. In fact, the cross-product on $\mathbb{R}^3$ is unique modulo sign, so you could think of the construction you outline as just an effect of that uniqueness: anything that 'looks' like a cross-product (linear, antisymmetric, etc.) has to be the usual one derived from, e.g., the quaternion product. – anomaly Apr 30 '15 at 17:29
  • @anomaly So in particular, the 7-dimensional cross product is not derived directly from an exterior power at all? – theage Apr 30 '15 at 17:33
  • why did you pick $7$ and not $4$ ? – mercio Apr 30 '15 at 17:35
  • @mercio It is well-know that there is no cross product in 4d while there is one in 7d. This question is asking why the 7d one does not arise as a wedge product on $\mathbb R^7$ (and it appears to be answered now). – theage Apr 30 '15 at 17:42
  • @theage: Sort of. To simplify notation, I'll deal with tensor products rather than wedge products. (The outcome is the same; we just need to remember that we're working with antisymmetric functions throughout). The tensor (or wedge) product gives a map $V \times V \to V^{\otimes 2}$. If we have an element $v\in V^{\otimes 3}$, dualizing gives $v^*\in Hom(V^{\otimes 3}, \mathbb{R}) = Hom(V^{\otimes 2}, Hom(V, \mathbb{R})) = Hom(V^{\otimes 2}, \mathbb{R})$. That's exactly what we're looking for, although we have to check that it's a valid cross-product. For $\mathbb{R}^3$, we have a canonical – anomaly Apr 30 '15 at 17:44
  • ...choice of $v$: the element $*1 = e_1 \otimes e_2 \otimes e_3$. (Note that we actually care about wedge products here rather than tensor products, despite the sloppy notation). In fact, this construction is exactly what the Hodge star is, just in more generality. In higher dimensions, though, there's no canoncial choice of $v$. – anomaly Apr 30 '15 at 17:46
  • @anomaly I see, that all makes sense (and is quite beautiful). If you copy and paste it into an answer I'll accept it. – theage Apr 30 '15 at 17:48
  • Sure, I'd be happy to. – anomaly Apr 30 '15 at 17:48
  • @theage The tail of my answer here gives another, more specifically 3-dimensional perspective on this that you might be interested in. – Steven Stadnicki Apr 30 '15 at 18:17

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Although the map $(x,y)\to *(x\wedge y)$ doesn't generalize to a map $\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n$ in general, not every cross-product has to be generated as such. In fact, the cross-product on $\mathbb{R}^3$ is unique modulo sign, so you could think of the construction you outline as just an effect of that uniqueness: anything that 'looks' like a cross-product (linear, antisymmetric, etc.) has to be the usual one derived from, e.g., the quaternion product.

So, how do we generalize this contruction to higher dimensions? To simplify notation, I'll deal with tensor products rather than wedge products. (The outcome is the same; we just need to remember that we're working with antisymmetric functions throughout). The tensor (or wedge) product gives a map $V\times V \to V^{\otimes 2}$. If we have an element $v\in V^{\otimes 3}$, dualizing gives $$v^∗\in\operatorname{Hom}(V^{\otimes 3},\mathbb{R})=Hom(V^{\otimes 2},\operatorname{Hom}(V,\mathbb{R}))=Hom(V^{\otimes 2},V).$$ That's exactly what we're looking for, although we have to check that it's a valid cross-product. For $V = \mathbb{R}^3$, there's a canonical choice of $v$: the element $∗1=e_1 \otimes e_2 \otimes e_3$, which corresponds to a nonzero element of the $1$-dimensional space $\Lambda^3 V$. (Note that we actually care about wedge products here rather than tensor products, despite the sloppy notation). In fact, this construction is exactly what the Hodge star is, just in more generality. In higher dimensions, though, there's no canoncial choice of $v$.

anomaly
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