Showing that 4D rank-2 anti-symmetric tensor always contains a polar and axial vector In my special relativity course the lecture notes say that in four dimensions a rank-2 anti-symmetric tensor has six independent non-zero elements which can always be written as components of 2 3-dimensional vectors, one polar and one axial.
For instance in the angular momentum tensor $L^{ab} = X^aP^b -X^bP^a$ the top row $L^{0i}=ct\vec{p}-(E/c)\vec{x}$ which is obviously polar (as $\vec{x}$ and $\vec{p}$ are polar vectors) while the spatial-spatial section contains the usual 3D angular momentum components which obviously represent the axial angular momentum $\vec{L}$ vector. (And the first column is just -1 times the polar vector due to the anti symmetry of the tensor).
The notes only explain this as ‘these components transforming in identical ways to polar and axial vectors’. I would like to know how to show this, possibly from the co-ordinate transformation rule for a 4D rank 2 contravariant tensor by showing it has equivalent effects of transforming these vector components.
Specifically the notes say ‘it works because those elements do transform as a vector under rotations’. I’m also confused as to why rotations specifically as a transformation are mentioned here.
 A: *

*OP is asking for the branching rules for
$$\begin{align}H~:=~&O(3)\cr
~\cong~&\begin{pmatrix} O(3)&\cr &1\end{pmatrix}_{4\times 4}~~\cr
 ~\subseteq~& O(3,1)~=:~G.\end{align}\tag{1}$$


*The 4-vector representation decomposes as
$${\bf 4}~\cong~\underbrace{\bf 3}_{\text{vector}}\oplus \underbrace{\bf 1}_{\text{scalar}}.\tag{2}$$


*Therefore the tensor product representation becomes
$$\begin{align} {\bf 16} ~\cong~&
{\bf 4}\otimes{\bf 4}\cr 
~\cong~&({\bf 3}\oplus {\bf 1})\otimes({\bf 3}\oplus {\bf 1})  \cr
~\cong~&{\bf 3}\otimes{\bf 3}\oplus 
\overbrace{\underbrace{{\bf 3}\otimes{\bf 1} \oplus{\bf 1}\otimes{\bf 3}}_{~\cong~{\bf 3}_S ~ \oplus~ {\bf 3}_A}}^{\text{off-diagonal blocks}} \oplus {\bf 1}\otimes{\bf 1}.\tag{3}
\end{align}$$
Here ${\bf 3}_S$ and ${\bf 3}_A$ denote the symmetric and antisymmetric combination of the off-diagonal blocks, respectively.


*The symmetric part of the tensor product ${\bf 4}\otimes{\bf 4}$ reads
$${\bf 10}~\cong~ {\bf 4}\odot{\bf 4} 
~\cong~\underbrace{{\bf 3}\odot {\bf 3}}_{~\cong~{\bf 5} ~ \oplus~ {\bf 1}}\oplus 
{\bf 3}_S  \oplus {\bf 1},\tag{4}
$$
while the antisymmetric part is

$$ {\bf 6}~\cong~{\bf 4}\wedge{\bf 4} 
~\cong~\underbrace{{\bf 3}\wedge {\bf 3}}_{\text{axial vector}}\oplus 
\underbrace{{\bf 3}_A}_{\text{vector}} ,\tag{5}
$$

cf. OP's title question. In eq. (5) we used Hodge duality in 3D, cf. e.g. this Phys.SE post.
A: Qmechanic's answer is beautiful. I'll clarify one non-obvious detail, namely why the $\textbf{3}\wedge \textbf{3}$ transforms as a vector under the identity component of the rotation group. (It doesn't transform as a vector under reflections, which is why we call it an axial vector.)
Let $F_{ab}$ be an antisymmetric tensor in 4d spacetime, and use $0$ for the "time" index and $\{1,2,3\}$ for the "space" indices. When Lorentz transformations are restricted to rotations, the components $F_{jk}$ with $j,k\in\{1,2,3\}$ do not mix with the component $F_{0k}=-F_{k0}$, so we can consider only the components $F_{jk}$. These are the components of the $\textbf{3}\wedge \textbf{3}$ in Qmechanic's answer.

For the rest of this answer, all indices (including $a,b,c$) are restricted to the spatial values $\{1,2,3\}$.
The antisymmetry condition, $F_{jk}=-F_{kj}$, implies that this has only $3$ independent components, which is the correct number of components for a vector, but something doesn't seem quite right: Under rotations, the transformation rule for a vector only uses one rotation matrix, but the transformation rule for $F_{jk}$ uses two rotation matrices — one for each index. How can these possibly be equivalent to each other? Of course they're not equivalent to each other for rotations with determinant $-1$, which is why we call it an axial vector, but they are equivalent to each other for rotations with determinant $+1$, and the purpose of this answer is to explain why that's true.
Let $R_{jk}$ be the components of a rotation matrix whose determinant is $+1$. This condition means
$$
 \sum_{j,k,m}\epsilon_{jkl}R_{1j}R_{2k}R_{3m} = 1,
\tag{1}
$$
which can also be written
$$
 \epsilon_{abc} = 
 \sum_{j,k,m}\epsilon_{jkm}R_{aj}R_{bk}R_{cm}.
\tag{2}
$$
The fact that $R$ is a rotation matrix also implies
$$
 \sum_c R_{cm}R_{cn}=\delta_{mn},
\tag{3}
$$
which the component version of the matrix equation $R^TR=1$. Contract (2) with $R_{cn}$ and then use (3) to get
$$
 \sum_c\epsilon_{abc}R_{cn} = 
 \sum_{j,k}\epsilon_{jkn}R_{aj}R_{bk}.
\tag{4}
$$
Equation (4) is the key. The effect of a rotation on $F_{jk}$ is
$$
 F_{jk}\to \sum_{a,b}R_{aj}R_{bk}F_{ab},
\tag{5}
$$
with one rotation matrix for each index. Since $F_{ab}$ is antisymmetric, we can represent it using only three components like this:
$$
 v_m\equiv\sum_{j,k}\epsilon_{jkm}F_{jk}
\tag{6}
$$
The question is, how does $v$ transform under a rotation whose determinant is $+1$? To answer this, use (5) to get
$$
 v_m\to 
 v_m'=\sum_{j,k}\epsilon_{jkm}\sum_{a,b}R_{aj}R_{bk}F_{ab}
\tag{7}
$$
and then use (4) to get
$$
 v_m' =\sum_{a,b,c}
  \sum_c\epsilon_{abc}R_{cm}F_{ab}
 =\sum_c R_{cm} v_m.
\tag{8}
$$
This shows that $v$ transforms like a vector under rotations whose determinant is $+1$.
For rotations whose determinant is $-1$ (reflections), the right-hand side of equation (1) is replaced by $-1$, which introduces a minus sign in equation (4), which ends up putting a minus sign in equation (8). That's why we call $v$ an axial vector instead of just a vector.

More generally, in $N$-dimensional space:

*

*Pseudovector and axial vector are synonymous with "completely antisymmetric tensor of rank $N-1$." Intuitively, an ordinary (polar) vector has only one index, and a pseudovector/axial vector is missing only one index. As a result, they both transform the same way under rotations, but only under rotations. They transform differently in other respects, including relfections and dilations.


*Under an arbitrary coordinate transform, a (polar) vector transforms as $v_{j}\to \Lambda^a_j v_{a}$.


*Under an arbitrary coordinate transform, a rank-2 tensor transforms as $F_{jk}\to \Lambda^a_j\Lambda^b_k F_{ab}$. (The components of $\Lambda$ are the partial derivatives of one coordinate system's coordinates with respect to the other's. Sums over repeated indices are implied.)


*If $N\neq 3$, then angular momentum is an antisymmetric rank-2 tensor (also called a bivector), not an axial vector. A bivector has 2 indices, but an axial vector has $N-1$ indices.


*To illustrate the different transformation laws for (polar) vectors and bivectors, consider a dilation (also called dilatation) that multiplies the spatial coordinates by a constant factor $\kappa$. Then each factor of $\Lambda$ contributes one factor of $\kappa$, so $F_{jk}\to\kappa^2 F_{jk}$, but a vector goes like $v_j\to \kappa v_j$.
Axial vectors and bivectors are the same in 3d space, but they are not really vectors at all, even though they both happen to have 3 components in 3d space. If we only consider rotations (with determinant $+1$), then they might as well be vectors, but even that's only true in 3d space, not in other-dimensional spaces.
