# Can all pseudotensors be expressed as components of higher rank proper tensors?

Axial vectors like angular momentum $$L^k$$ are not proper vectors and thus do not follow the vector transformation law: $$\bar v^i = \Lambda^i_j v^j$$. Instead they are 'pseudovectors' which follow the vector trasnformation law except for a -1 sign under reflections/parity inversions.

These more complicated transformation rules for angular momentum can be made simpler by expressing it as a rank-2 antisymmetric tensor : $$L^{ij} = \begin{bmatrix}0&L_z&-L_y\\-L_z&0&L_x\\L_y&-L_x&0\end{bmatrix}$$.

In this form the angular momentum vector components under a transformation can be calculated using the usual rank-2 tensor transformation law, namely: $$\bar L^{ij} = \Lambda^i_a \Lambda^j_b L^{ab}$$.

I was wondering, is a similar procedure true for any pseudotensor. Can we always express it's components as the components of a higher rank new proper tensor?

• Commented Jan 2, 2021 at 13:58
• Could you elaborate, does this mean it is true that we can always do this? Commented Jan 2, 2021 at 13:59
• If you want to understand the object(s) you're working with here you could read about differential forms, angular momentum is an example. Technically the cross product on $\Bbb R^3$ does not produce another vector in $\Bbb R^3$, what you have actually used is the exterior product of two vectors: $\vec x\wedge \vec p=\vec x\otimes \vec p-\vec p\otimes \vec x$. This is an element of the exterior algebra, $\bigwedge^2 \Bbb R^3$. Commented Jan 2, 2021 at 20:48

I will offer an alternative way of looking at this. You care about transformations, right? More often than not these transformations form groups. For example rotation group is $$\text{SO}(3)$$, parity and time-reversal group is $$Z_2$$. Groups have representations over vector spaces, so different vectors simply are from different representation spaces for these groups. There is only one representation of the rotation group $$\text{SO}(3)$$ for 3d vectors. But you can have two representations of the Parity group - this gives you axial and polar vectors. You also have two representations of the time-reversal group - two more representations. So if you combine time-reversal, parity and rotations, you can have four different kinds of 3d vectors.
1. I shamelessly conflated irreducible representations with reducible ones. $$Z_2$$, being an Abelian group only has 1d irreducible representations, but $$\mbox{SO}\left(3\right)$$ has a single 3d irreducible representation. Combining time-reversal with parity, with rotations gives a group $$Z_2\times Z_2\times\mbox{SO}\left(3\right)$$. Now that has 4 different irreducible representations over 3d vector space.