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For three-dimensional cross product, the following property holds true: \begin{equation} (R\mathbf x) \times (R \mathbf y)=R(\mathbf x \times \mathbf y) \end{equation} where $R\in SO(3)$.

Is the analogous property (with $R\in SO(7)$) true for seven-dimensional cross product?

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  • $\begingroup$ Ho do you define the cross product in 7 dimensions?. $\endgroup$ – Rogelio Molina Jul 2 '15 at 6:23
  • $\begingroup$ @RogelioMolina Here's the definition: en.wikipedia.org/wiki/Seven-dimensional_cross_product $\endgroup$ – Danijel Jul 2 '15 at 6:25
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    $\begingroup$ Well, I haven't verified or thought about it, but the wiki page you linked to says it's not. Have you tried computing a few examples to try to find a counterexample? $\endgroup$ – Callus Jul 2 '15 at 6:30
  • $\begingroup$ @Callus Thanks, I didn't notice that short section in the wiki article. I guess this answers the question. $\endgroup$ – Danijel Jul 2 '15 at 6:38
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    $\begingroup$ Look at the rotations section in the wiki entry. It explicitly says that the cross product is not invariant under the group of rotations in seven dimensions, $SO(7)$. Instead, it is is invariant under the exceptional Lie group $G_2$, a subgroup of $SO(7)$. $\endgroup$ – achille hui Jul 2 '15 at 6:56
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Counterexample: $$\mathbf{x}:=(1,0,...,0)^T,\mathbf{y}:=(0,...,0,1)^T\implies \mathbf z:=\mathbf{x}\times\mathbf{y}=(0,0,0,0,0,1,0)^T\\ R_{ij}:={1\over\sqrt{2}}(\delta_{i1}\delta_{j1}+\delta_{i2}\delta_{j2}-\delta_{i1}\delta_{j2}+\delta_{i2}\delta_{j1})+\delta_{ij}(1-\delta_{i1})(1-\delta_{i2})\\ R\mathbf{x}={1\over\sqrt{2}}(1,1,0,0,0,0,0)^T\\ R\mathbf{y}={1\over\sqrt{2}}(-1,1,0,0,0,0,0)^T\\ (R\mathbf{x})\times(R\mathbf{y})=(0,0,1,0,0,0,0)^T\\ R\mathbf z=\mathbf z $$

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