3
$\begingroup$

It is known that the only spheres that admit an almost complex structures are $\Bbb S^2$ and $\Bbb S^6$ (Borel and Serre, 1953). In particular, $\Bbb S^4$ cannot be given an almost complex structure (Ehresmann and Hopf). My question is the following.

Question: Let $J$ be an almost complex structure on $\Bbb S^n$. Is it true that $J$ induces a cross product on $\Bbb R^{n+1}$?

Definition: A cross product on $\Bbb R^n$ is a skew-symmetric bilinear map $\times \colon \Bbb R^n\times \Bbb R^n\to \Bbb R^n$ such that $\langle a\times b, a\rangle=0$ and $\|a\times b\|^2=\|a\|^2\|b\|^2-\langle a, b\rangle^2$ for $a,b\in \Bbb R^n$. Here, $\|\cdot\|$ and $\langle\cdot, \cdot \rangle$ are the usual norm and inner product of $\Bbb R^{n}$, respectively.

Remark: Using characteristic classes, etc., one can show that the only spheres that admit almost complex structures are $\Bbb S^2$ and $\Bbb S^6$. Also, a theorem of Hurwitz (1898) says that $\Bbb R^n$ has a normed division algebra structure if and only if $n=1,2,4,8$. So, $\Bbb R^n$ has a cross product if and only if $n=1,3,7$. But the above question is different, actually, irrespective of all these facts, and it asks whether the formula of a cross product on $\Bbb R^{n+1}$ can be written in terms of an almost complex structure $J\colon T\Bbb S^n\to T\Bbb S^n$.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Commented Dec 20, 2023 at 7:20
  • $\begingroup$ 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? $\endgroup$ Commented Dec 20, 2023 at 7:24
  • $\begingroup$ (oops, and scale by $\|a\|$) $\endgroup$ Commented Dec 20, 2023 at 7:54

0

You must log in to answer this question.