For vectors $x$ and $y$, is there any differentiable function $f(\cdot)$, such that $f(x, y) = -f(y, x)$? For 3d vectors, the cross-product is one such function. But what about arbitrary dimensional vectors?
More specifically I am looking for a function $f(\cdot)$ whose output is a single real number.
For any differentiable function $g(x,y)$, take $f(x,y)=g(x,y)-g(y,x)$.
Then, $f$ is differentiable and will automatically satisfy the equality $f(x,y)=-f(y,x)$.
This method works because the construction of $f$ is basically sort of averaging over the $\mathbb{Z}_2$ action on the function $g(x,y)\mapsto -g(y,x)$.