Is the Lorentz 3-force a 3-vector or 3-pseudovector?

Question

Usually, we say that the Lorentz force is a vector. However, in group theory, we make a distinction between vectors and pseudovectors based on how they transform under $O(3)$. Vectors and pseudovectors transform similarly under rotations but not under reflections. For example, the magnetic field is a pseudovector, because it can be written in terms of a cross product, particularly as a curl:

$$\vec{B} = \nabla \times \vec{A}.$$

Any cross product is a pseudovector, because if both vectors reflect, then the cross product of those vectors do not, since the minus signs cancel out.

Now consider the Lorentz force:

$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}).$$

The electric field is a vector. The velocity is a vector and the magnetic field is a pseudovector, so the cross product is a pseudovector as well. Because if the velocity reflects, $B$ does not, but the cross product does since it picks up a minus sign. So is the total force then a vector or pseudovector? How does it transform under reflections?

Answer 1

I indeed made the mistake to conclude that the cross product term is a pseudovector, since it does pick up a minus sign similarly to the velocity, so it is in fact a vector.

So the Lorentz force as a whole is a vector, as expected.

Answer 2

Index notation might help here, so:

$$ B_i = ({\bf \nabla}\times {\bf A})_i = \epsilon_{ijk}\partial_j A_k $$

Then Lorentz's force law:

$$ F_i = q(E_i + \epsilon_{ijk}v_jB_k) $$

becomes:

$$ F_i = q(E_i + \epsilon_{ijk}v_j\epsilon_{klm}\partial_lB_m) $$

From there:

$$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} $$

shows that it is a polar vector (parity odd).