How are these two ways of thinking about the cross product related? I was always bothered by the definition of the cross product given in e.g. a calculus course because it's never made clear how one would go about defining the cross product in a coordinate-free manner.  I now know, not one, but two ways of doing this, and I can't quite see how they're related:


*

*The cross product is the Lie bracket in the Lie algebra of $\text{SO}(3)$.

*The cross product is the Hodge star map $\Lambda^2(V) \to V$ where $V$ is an oriented $3$-dimensional real inner product space.


Okay, so there's one obvious relation here: $V$ has automorphism group $\text{SO}(3)$.  But for some reason I can't figure out where to go from here.  A good starting point would be to exhibit a canonical isomorphism between an oriented $3$-dimensional inner product space $V$ and the Lie algebra of $\text{Aut}(V)$.  Maybe this is obvious.  In any case, I would appreciate some clarification.
 A: Geometric (Clifford) algebra provides yet another way to understand the cross product.  If a vector represents a 1D subspace containing the origin, the wedge (outer) product is a bilinear form on 2 vectors ($a \wedge b$) that represents the subspace containing both vectors.  The vector has basis $e_1, e_2, ..., e_n$ ($n$ elements), so the wedge product has basis $e_1\wedge e_2, e_1\wedge e_3, \ldots, e_{n-1}\wedge e_n$ ($\binom{n}{2}$ elements).  The cross product is a way to represent this subspace (using its dual, the normal vector).  This works only for 3D, where the basis sizes match, using $e_1 e_2 \to e_3,\, e_2 e_3 \to e_1,\, e_3 e_1 \to e_2$.
A: To expand on Victork Protsak's comment, if $V$ is an $n$-dimensional real vector space with inner-product, the inner-product gives an isomorphism $V\to V^*$ and hence $V\otimes V \to \mathrm{End}(V)$. Under this isomorphism, $\Lambda^2(V)$ is identified with skew-adjoint endomorphisms of $V$, which is precisely the Lie algebra $\mathfrak{so}(V)$.
In the case $\dim V =3,$ the Hodge star gives an isomorphism $\Lambda^2(V) \to V$ and so in total we see that $V$ is canonically isomorphic to $\mathfrak{so}(V)$. A more direct way to see this isomorphism is to send the vector $v \in V$ to the generator of the right-handed rotation about the axis in the direction of $v$ with speed $|v|$. 
The use of the phrase "right-handed" makes it clear that in order to identify $V$ and $\mathfrak{so}(V)$ we have used an orientation on $V$; indeed, you need that for the Hodge star. What is interesting is that if you reverse the orientation on $V$, the map to $\mathfrak{so}(V)$ changes sign. This means that what ever orientation you chose on $V$, the push-forward to $\mathfrak{so}(V)$ is the same. Conclusion: $\mathfrak{so}(3)$ is naturally oriented. This is analogous to the natural orientation on $\mathbb{C}$. A more prosaic way to describe the orientation is to pick two independent elements $x,y \in \mathfrak{so}(3)$ and then use $[x,y]$ to complete them to an oriented basis. (Of course, you then need to check that this doesn't depend on your choice of $x,y$.)
A: Let $\varepsilon( )$ be the volume form in $\mathbb R^3$. For given vectors ${\bf p}$ and ${\bf q}$ the function $f:{\bf x}\mapsto\varepsilon({\bf p},{\bf q},{\bf x})$ is a linear functional and so is represented by a vector ${\bf r}\in\mathbb R^3$, i.e., one has  $f({\bf x})=\langle{\bf r},{\bf x}\rangle$. This vector ${\bf r}$ depends in a skew bilinear way from ${\bf p}$ and ${\bf q}$ and is called the $vector\ product$ of ${\bf p}$ and ${\bf q}$.
