Why is the magnetic field $B$ a pseudo-vector? Physically speaking, "pseudo-vectors" are vectors $v\in \mathbb{R}^3$ which transform as $ v'= (\det {R})v$ if the "system were to transform as $R\in O(3)$". However, what does this mean mathematically? And in particular, why is the magnetic field $B$ a pseudo-vector?
I would imagine that by "vectors", we actually mean smooth differential forms with the isomorphism $v\mapsto \sum v_i dx_i$, and by "transforming the system as $R\in O(3)$", I would imagine that it means we are applying a pullback on 1-forms corresponding to the map $x\mapsto Rx$. Assuming that $B =*dA$ where $*$ is the hodge-star operator, how would $\det R$ be factored into this transform?
 A: I don’t know the math jargon, but the magnetic field is a pseudo vector because it is determined for a given configuration of moving charges by the right hand rule. This means that if you look at a mirror reflection of a physical setup with the B vector drawn in, it will be reversed. Angular momentum is another example of this. This property of reversing direction relative to the rest of the scenery upon mirror reflection is how I recognize pseudo vectors.
A: I thought about this a little further, and I think I can explain that $B$ is a pseudo-vector  (without using too much math jargon).
Let $R\in O(3)$ and let $\omega:\mathbb{R^3} \to \mathbb{R^3}$ be a smooth vector function such that $A$ or $B$. Let me define transforming the vector as $R$ as group action on $\omega$ defined by
$$
(R\cdot \omega)(x) = R\omega (R^{-1}x)
$$
That is, to get the new vector function $\omega'=R\cdot \omega$ at some postion $x$, we need to transform the system back as $R^{-1}x$, find the corresponding vector of $\omega$ at that value, and then rotate/reflect that vector based on $R$. One way to see that this is the canonical way of transforming the system is to think about the differential forms
$$
\omega_i(x)dx_i
$$
If we take $y=Rx$ (i.e. system transforms as $R$), then the differential form becomes
$$
(R\cdot\omega)_i(y) dy_i
$$
Then using the fact that $B=\nabla \times A$, we see that
$$
R\cdot B = (\det {R}) (\nabla\times(R\cdot A))
$$
This is where the determinant comes from. In particular, if for some special system $R\cdot A =A$ for some reflection $R$, then $R\cdot B = -B$.
A: In the Lorentz force $\vec F=q\vec E + q\vec v\times\vec B$,

the force $\vec F$, the electric field $\vec E$, and the velocity $\vec v$
are [ordinary, polar] vectors, and the electric charge $q$ is a scalar.
Since the cross-product of two ordinary vectors is a pseudo-vector,
the magnetic field $\vec B$ must be a pseudo-vector so that $\vec v\times\vec B$ is an ordinary polar vector.
(Even though $\vec E$ and $\vec B$ are "vectors", since the "magnetic field vector" is a pseudovector, it can never be added [as vectors] to an "electric field vector".)
