Question: What is the geometric significance of the fact that the exterior products of the unit basis vectors in $\mathbb{R}^3$ generate a basis for the Lie algebra of the pure rotation group $SO(3)$?
Also, if applicable, what is the algebraic significance or the physical significance?
This is one of the first things I found to be true when investigating all of these subjects, but strangely I have yet to find a single source documenting this phenomenon and explaining its significance. There are no coincidences in mathematics, so seemingly this should have a fairly profound explanation, because otherwise it would be a fairly big coincidence.
My guess is that it means somehow (I'm not sure how) that $\mathbb{R}^3$ is the only Euclidean space for which rotations in a plane can be identified with rotations about an axis, since $3-2=1$. In general, rotation about an axis is not defined, while rotation in a plane always is.
Identifying a plane with rotation in the same plane seems like somewhat of a conceptual leap, however. Seemingly the two should be denoted differently somehow.
Context: Given $\mathbb{R}^n$, in what follows $n=3$.
The infinitesimal rotations are one possible basis for the 3-dimensional vector space of skew-symmetric matrices. This vector space is actually a Lie algebra $\mathfrak{so}(3)$ of the Lie group $SO(3)$ of rotation matrices with determinant $1$ (i.e. pure rotations, those which do not contain a reflection).
$$\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix} $$
Given the unit basis vectors for $\mathbb{R}^3$, $e_1, e_2, e_3$, they have three linearly independent exterior products, $e_1 \wedge e_2, e_2 \wedge e_3,$ and $e_3 \wedge e_1$. Their coordinate representations are familiar:
$$e_1 \wedge e_2 = e_1 \otimes e_2 - e_2 \otimes e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 0 \\ 1 \\0 \end{bmatrix} - \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}^T - \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} - \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$
$$e_2 \wedge e_3 = e_2 \otimes e_3 - e_3 \otimes e_2 = \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}\otimes \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} - \begin{bmatrix} 0 \\0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}^T - \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}$$
$$e_3 \wedge e_1 = e_3 \otimes e_1 - e_1 \otimes e_3 = \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}^T - \begin{bmatrix} 1 \\ 0 \\0 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$
Other almost certainly related facts:
$e_1 \wedge e_2, e_2 \wedge e_3, e_3 \wedge e_1$ are basis $2-$vectors in geometric algebra and hence represent planes. Specifically, $e_1 \wedge e_2$ represents the plane spanned by $e_1$ and $e_2$, $e_2 \wedge e_3$ represents the plane spanned by $e_2$ and $e_3$, and $e_3 \wedge e_1$ represents the plane spanned by $e_3$ and $e_1$.
If $*$ denotes the Hodge dual, then we have $$(e_1 \wedge e_2)^* = e_1 \times e_2 = e_3$$ $$(e_2 \wedge e_3)^* = e_2 \times e_3 = e_1$$ $$(e_3 \wedge e_1)^* = e_3 \times e_1 = e_2$$ These three equations are just the right hand rule, and represent the facts that:
- rotation about the $x_3$ axis is the same as rotation in the $x_1,x_2$ plane,
- rotation about the $x_1$ axis is the same as rotation in the $x_2, x_3$ plane,
- rotation about the $x_2$ axis is the same as rotation in the $x_3, x_1$ plane.
Angular momentum is the integral of motion corresponding to rotational invariance, i.e. symmetry with respect to $SO(3)$. (This means, by Noether's theorem, that if the laws of physics are unchanged under pure rotations, then angular momentum is conserved.) Angular momentum also generates the infinitesimal rotations using the Poisson bracket.
The cross-product algebra on $\mathbb{R}^3$ can be identified with the Lie algebra of pure quaternions (at least when normalizing by a factor of 2). Conjugate multiplication of quaternions can also be used to represent space rotations, leading to an identification of antipodal pairs of unit quaternions with space rotations in $\mathbb{R}^3$. (See, for example, section 4.4 or sections 1.4-1.6 of Stillwell's Naive Lie Theory.) $2-$vectors in geometric algebra function similarly.
If you consider the exterior algebra $\Lambda^{2}(\mathbb{R}^{n})$ and $\mathfrak{so}(n)$, they are in fact isomorphic as you have hinted at in the above for the case $n=3$.
If you pick $u \wedge v \in \Lambda^{2}(\mathbb{R}^{n})$, and then define the linear mapping $\phi(u\wedge v):\mathbb{R}^{n} \to \mathbb{R}^{n}$ as: $$ \phi(u \wedge v)(x) = \left( u \cdot x \right) v - \left( v \cdot x \right) u $$ Then we can think of $\phi(u \wedge v) $ as a matrix, which after a little work you can show to be skew-symmetric - so we can say that $\phi(u\wedge v) \in \mathfrak{so}(n)$.
In this sense, we have a mapping $\phi:\Lambda^{2}(\mathbb{R}^{n}) \to \mathfrak{so}(n)$. This mapping is in fact an isomorphism between $\Lambda^{2}(\mathbb{R}^{n})$ and $\mathfrak{so}(n)$.