Is each vector $y \in \{x\}^{\bot}$ a $r$-fold cross product of $x$ and some $x_1,...,x_{r-1}$? Assume that on a $n$-dimensional real linear space $V$ with inner product $\langle\cdot,\cdot\rangle$ we have defined a non-zero $r$-fold cross product, that is, a $r$-linear skew symmetric mapping
$$P : \underbrace{V \times \cdots \times V}_r \rightarrow V$$
with the following properties:

*

*$\langle P(v_1,...,v_r),v_i \rangle = 0 \textrm{ for all $v_1,\dots,v_r \in V$ and each $i \in \{1,...,r\}$},$

*$\|P(v_1,...,v_r)\|^2 = \det [\langle v_i,v_j \rangle] \ \ (=\textrm{the Gram determinant of } v_1,...,v_r).
$
My question is:

Assuming that such $P$ exists and $3 \leq r \leq n-1$, is it true that for each non-zero $x \in V$ and for each nonzero $y \in \{x\}^{\bot}$ there exist  $x_1,...,x_{r-1}$ such that $P(x,x_1,...,x_{r-1})=y$?

 A: Yes, provided that $P$ exists.
Let $x_1=x$ and $x_n=y$. Extend $\{x_1,x_n\}$ to an orthogonal basis $\{x_1,x_2,\ldots,x_n\}$ of $V$ and let
$$
W=\operatorname{span}\{x_r,x_{r+1},\ldots,x_n\}
=\left(\operatorname{span}\{x_1,x_2,\ldots,x_{r-1}\}\right)^\perp.
$$
For every $w\in W$, define $L(w)=P(x,x_2,\ldots,x_{r-1},w)$. Then

*

*$L(W)\subseteq W$ because $L(w)=P(x_1,x_2,\ldots,x_{r-1},w)\perp x_1,x_2,\ldots,x_{r-1}$;

*$L$ is linear because $w\mapsto P(x_1,x_2,\ldots,x_{r-1},w)$ is linear;

*$L$ is injective because $\|L(w)\|^2=\|P(x_1,x_2,\ldots,x_{r-1},w)\|^2=\left(\prod_{i=1}^{r-1}\|x_1\|^2\right)\|w\|^2>0$ whenever $w\ne0$.

Therefore $L:W\to W$ is a nonsingular linear operator. Now we are done, because
$$
y=L(L^{-1}(y))=P(x,x_2,\ldots,x_{r-1},L^{-1}(y)).
$$
Remark. However, $P$ does not always exist. E.g. when $n-r$ is even, our previous discussion shows that $\dim W=n-r+1$ is odd. Therefore $L$ has an eigenvector $w$. However, as $L(w)\perp w$, we must have $L(w)=0$, which is a contradiction to the invectiveness of $L$ we inferred in the above.
