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Dec 20, 2023 at 7:54 comment added Kevin Casto (oops, and scale by $\|a\|$)
Dec 20, 2023 at 7:24 comment added Kevin Casto I guess the construction would go something like: given the first vector $a$, we can view the orthogonal complement to $a$ as $T_{\hat a} S^n$. Then project $b$ onto this subspace and use the almost complex structure to rotate it 90°. Does that work?
Dec 20, 2023 at 7:20 comment added Ryan Budney It seems the more natural implication goes the other way. If you have a cross product you get the almost complex structure $J$ as rotation by $\pi/2$ using the basis $v, p \times v$ in $T_p S^n$, i.e. the linear transformation of $T_p S^n$ that sends a tangent vector $v \longmapsto p \times v$ and $p \times v \longmapsto -v$, being the identity on the orthogonal complement.
Dec 20, 2023 at 7:02 history edited Random CC BY-SA 4.0
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Dec 19, 2023 at 6:25 history asked Random CC BY-SA 4.0