Relation Between Cross Product and Infinitesimal Rotations, Generators, Etc Looking into the infinitesimal view of rotations from Lie, I noticed that the vector cross product can be written in terms of the generators of the rotation group $SO(3)$. For example:
$$\vec{\mathbf{A}} \times \vec{\mathbf{B}} =  (A^T \cdot J_x \cdot B)    \>\> \hat{i} + (A^T \cdot J_y \cdot B)    \>\> \hat{j} + (A^T \cdot J_z \cdot B)    \>\> \hat{k}$$
where,
$$J_x =  \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} \>\>\>;\>\>\>;J_y = \begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix} \>\>\>;\>\>\>;J_z = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} $$
I couldn't help but think I was missing some profound connection here.  Why is the vector cross product seemingly intimately related to the generators of the $SO(3)$ rotation group?
 A: The mystery dissipates when we consider how this generalizes to $D$-dimensional space. The cross product only makes sense when $D=3$, but the idea of using the antisymmetric part of $A_a B_b$ to represent an oriented element of area works for any $D$, as does the concept of a rotation in a given plane. So let's see what this looks like for aribtrary $D$.
We can express the vectors $A$ and $B$ in a canonical orthonormal basis, say
$$
\mathbf{E_1} =\left[\matrix{1\cr 0\cr 0\cr \vdots}\right]
\hskip1cm
\mathbf{E_2} =\left[\matrix{0\cr 1\cr 0\cr \vdots}\right]
\hskip1cm
\mathbf{E_3} =\left[\matrix{0\cr 0\cr 1\cr \vdots}\right]
\hskip1cm
\cdots
$$
and so on (again, this is for arbitrary $D$). I'm using boldface for a vector (represented here as a single-column matrix) and using non-boldface for its components. In this basis, we have
$$
\mathbf{A} = \sum_a A_a \mathbf{E}_a
\hskip2cm
\mathbf{B} = \sum_b B_b \mathbf{E}_b,
\tag{1}
$$
which gives
\begin{align*}
\mathbf{A}\mathbf{B}^T-
\mathbf{B}\mathbf{A}^T
&= ( \sum_a A_a \mathbf{E}_a)(\sum_b B_b \mathbf{E}_b^T)
- (\sum_b B_b \mathbf{E}_b)( \sum_a A_a \mathbf{E}_a^T)\\
&= \sum_{a,b} A_a B_b 
(\mathbf{E}_a\mathbf{E}_b^T -
\mathbf{E}_b\mathbf{E}_a^T) \\
&= \sum_{a,b} \frac{A_a B_b - B_a A_b}{2}
(\mathbf{E}_a\mathbf{E}_b^T -
\mathbf{E}_b\mathbf{E}_a^T)
\tag{2}
\end{align*}
where $T$ means transpose. The last step uses the fact that the quantity in parentheses changes sign if the indices $a$ and $b$ are exchanged, so the value of the sum is unchanged if we replace $A_a B_b$ with $-B_a A_b$. Therefore, the value of the sum is also unchanged if we replace $A_a B_b$ with half of $A_a B_b$ plus half of $-B_a A_b$. (In other words, the sum in the second-to-last line equals half of itself plus half of itself, which gives the last line.)
Now recognize that the matrix $\mathbf{E}_a\mathbf{E}_b^T - \mathbf{E}_b\mathbf{E}_a^T$ is the generator of rotations in the $a$-$b$ plane, and recognize that $A_a B_b - B_a A_b$ are the components of the "cross product" when $D=3$. Equation (2) shows how the phenomenon noted in the OP generalizes to arbitrary $D$.
The quantity $\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T$ is a matrix representation of a bivector, the oriented element of area defined by the two vectors $\mathbf{A}$ and $\mathbf{B}$. The antisymmetry isolates their "mutually orthogonal part," which is the part we need for specifying an element of area. In fact, the bivector $\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T$ is the (unnormalized) generator of a rotation in the $\mathbf{A}$-$\mathbf{B}$ plane. The corresponding rotation matrix is
$$
R(\theta)=\exp\left(\theta\frac{\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T}{N}\right)
\tag{3}
$$
where $N$ is defined by the condition 
$$
\left(\frac{\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T}{N}\right)^3
=-\frac{\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T}{N}.
\tag{4}
$$
Notice that the generator $\mathbf{E}_a\mathbf{E}_b^T - \mathbf{E}_b\mathbf{E}_a^T$ satisfies this condition with $N=1$. The minus sign is a consequence of antisymmetry, and this is also what ensures that $R$ is an orthogonal matrix: $R^TR =R R^T=I$.
When $D=3$, given any plane through the origin, there is a unique line through the origin that is orthogonal to the given plane. For this reason, we can get away with pretending that a bivector is a vector when $D=3$, and this is why we can define the "cross product" when $D=3$. (Vectors and bivectors still behave differently under reflections, though.) But for general $D$, such as $D=2$ or $D\geq 4$, that trick doesn't work, and we need to use a two-index quantity to represent a bivector. 
