Can someone remind me in a nutshell why the associativity of the exterior product fails to transfer to the cross product? (It's been over a decade since I had to deal with the former back in school.)
And is there a more general lesson here, such as why the Jacobi identity (instead of associativity) is impressed upon a Lie algebra by whatever algebraic structure the exterior product and Hodge dual form? (I don't think I ever knew what the latter structure is.)
In $\mathbb{R}^3$, we can indeed more or less identify $u\times v$ with $u\wedge v$ for two vectors $u,v \in \mathbb{R}^3$ (more accurately, $u\wedge v$ is the linear form given by scalar product with $u\times v$; or another way of putting thigs it is that $u\wedge v$ gives an "axial vector" or "pseudovector", not a vector), but this no longer works for triple products: $u\wedge v\wedge w$ is the triple product $(u\times v)\cdot w$ (which is also $u\cdot(v\times w)$ and is symetric under rotations of $u,v,w$, incidentally), not the triple cross product $(u\times v)\times w$. To put things differently, what allows $\wedge$ to be associative is that we have a grading (in physicists' terms: scalars, vectors, "axial vectors" and "pseudoscalars"; more mathematically: $\bigwedge^i(\mathbb{R}^3)$ for $i=0,1,2,3$) whose structure is destroyed if we identify $u\wedge v$ with $u\times v$.