Connecting the regular representation of $\mathfrak{so}(3)$ and the exterior algebra of $\mathbb{R}^3$ It is well known that the regular representation of $\mathfrak{so}(3)$ is the so-called "cross product" matrix $A(x)$ which follows $A(x)y = x\times y$, and $x,y\in\mathbb{R}^3$, while the cross product is also connected to the exterior product in $\mathbb{R}^3$. This begs the following two questions:
(1) How are $\mathfrak{so}(3)$ and $\Lambda^2(\mathbb{R}^3)$ related?
(2) Is there a general theory connecting exterior algebras and Lie algebras, or is this just a strange coincidence?
 A: I don't know what you mean by the regular representation of a Lie algebra, but the connection is the following. It's a bit confusing because there are a bunch of $3$-dimensional vector spaces that all get identified.
Let $V$ be a real inner product space. The inner product induces a canonical isomorphism $V \cong V^{\ast}$ giving an isomorphism
$$\mathfrak{gl}(V) \cong V \otimes V^{\ast} \cong V \otimes V.$$
The subalgebra $\mathfrak{so}(V) \subset \mathfrak{gl}(V)$ gets identified with the antisymmetric tensors inside $V \otimes V$, which I'll write as $\Lambda^2(V)$ even though this usually refers to a quotient of $V \otimes V$ and not a subspace. At this level of generality the Lie bracket has nothing to do with wedge products, as far as I know. 
However, if $V$ is furthermore $3$-dimensional and equipped with an orientation, then together with the inner product we get a trivialization $\Lambda^3(V) \cong \mathbb{R}$, and hence the wedge product gives a nondegenerate bilinear form
$$V \otimes \Lambda^2(V) \to \Lambda^3(V) \cong \mathbb{R}$$
inducing an isomorphism
$$\Lambda^2(V) \cong V^{\ast} \cong V.$$
(So, the $3$-dimensional vector spaces getting identified in this story: $V$, its dual, $\Lambda^2(V)$, its dual, and $\mathfrak{so}(V)$.) Up to maybe some factors of $2$, these isomorphisms intertwine the cross product, the wedge product, and the Lie bracket on $\mathfrak{so}(V)$. 
