generalized vector multiplication without identity element Is it possible to define (ideally infinitely) differentiable functions $f_n : \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $g_n : \mathbb{R}^n \times (\mathbb{R}^n \setminus \{0\}) \rightarrow \mathbb{R}^n$ where $f_n(x, g_n(y, x)) = y$ and $f_n(x, y) = 0 \iff (x = 0 \vee y = 0)$?
It's possible for $n = 1$ using $f_1(x, y) = xy$ and $g_1(x, y) = \frac{x}{y}$.
By interpretation of $\mathbb{R}^2$ as complex numbers, it's also possible for $n = 2$ using $f_2(x, y) = \left(
\begin{array}{c}
 Re (x_1 + x_2 i) (y_1 + y_2 i) \\
 Im (x_1 + x_2 i) (y_1 + y_2 i) \\
\end{array}
\right)$ and $g_2(x, y) = \left(
\begin{array}{c}
 Re \frac{x_1 + x_2 i}{y_1 + y_2 i} \\
 Im \frac{x_1 + x_2 i}{y_1 + y_2 i} \\
\end{array}
\right)$.
$n = 4$ should also work using a similar approach using quaternions. But is it possible for the general case? I'm not sure whether this is related to this ([1]) question since it might not be necessary to have full algebraic characteristics in order to define $f_n$ and $g_n$.
[1] Is there a third dimension of numbers?
 A: Assuming only that $f_n$ and $g_n$ are continuous, and also assuming the existence of a special vector $e$ satisfying $f_n(x, e) = f_n(e, x) = x$, the existence of such functions would imply the existence of an $H$-space structure on $\mathbb{R}^n \setminus \{ 0 \}$, which is homotopy equivalent to the sphere $S^{n-1}$.
It's a known but nontrivial result in homotopy theory, due to Adams, that $S^{n-1}$ has an $H$-space structure iff $n = 1, 2, 4, 8$, corresponding to the multiplication on the unit real numbers, the unit complex numbers, the unit quaternions, and the unit octonions respectively. So these are the only values of $n$ for which this is possible.
This is the most general and most difficult version of an easier result, namely the classification of real division algebras. In the simplest version of this result the multiplication is required to be bilinear (over $\mathbb{R}$) and associative (which excludes the octonions); next we drop associativity; and here we drop bilinearity.
I don't know what happens if you drop the requirement that $e$ exists. We do have $f_n(x, g_n(x, x)) = x$ so it may be possible to perform some kind of rescaling but it's tricky without either bilinearity or associativity.
