How does an almost complex structure on a sphere determine a 1-fold cross product?

Question

In the article "VECTOR CROSS PRODUCTS ON MANIFOLDS", by Alfred Gray, the author states, between theorem 2.8 and corollary 2.9, that

The existence of vector cross product of Type I [1-fold vector cross product] on spheres is the well-known problem of the existence of almost complex structures on spheres.

How can I see this? I need specifically to show that the existence of an almost complex structure on a sphere determines a 1-fold vector cross product on the same sphere. Does anyone know where I can find the answer?

Edit 1: Let $V$ be a finite-dimensional vector space over the real numbers and let $\langle \cdot,\cdot \rangle$ be a nondegenerate bilinear form on $V$. A 1-fold vector cross product is a linear map $P:V \rightarrow V$ such that \begin{equation} \langle P(v), v \rangle = 0 \end{equation} \begin{equation} \lVert P(v) \rVert^2 = \lVert v \rVert^2. \end{equation}

Let $J$ be the almost complex structure. Since $J^2 = -Id$, the idea is to associate $J$ with a rotation of $\pi/2$ radians. But $J$ is not that well behaved, and so it has to be corrected to keep the norm and be orthogonal. In any Euclidean space of even dimension it is easy to fix $J$. But in the sphere the situation is different: this fix has to vary continuously as the base point of the space tangent to the sphere varies.

It might be of some help (or not): if $J$ preserves norm, $(J+I)(v)$ is orthogonal to $v$, but $J+I$ fails to preserve norm.

Answer 1

As you noted, the definition of a cross-product requires a choice of a Riemannian metric on the given manifold. The correct reading of Gray's statement is:

Suppose that $(M,J)$ is an almost complex manifold. Then there exists a Riemannian metric $h$ on $M$ (one can take any metric Hermitian with respect to $J$), such that $J$ yields a 1-fold cross-product on $(M,h)$.

In order to prove this, first note that every almost complex structure $J$ on a manifold $M$ (the exact topology of $M$ is irrelevant here) has a compatible Hermitian metric $h$. Next, let $u$ be a nonzero vector in $T_xM$. Then $u$ and $Ju$ span a real 2-dimensional subspace $V(u)$ in $T_xM$. The condition that $J^2=-Id$ and the assumption that $J$ preserves $h$, implies that $J$ acts as an order $4$ rotation in $V(u)$. Hence, $Ju$ is orthogonal to $u$ and has the same norm as $u$. Thus, the map $$ u\mapsto Ju $$ is a 1-fold cross-product with respect to the metric $h$.