Rotational invariance of the conductivity tensor (Classical Hall Effect) In classical Hall effect, the conductivity tensor is given as
$\sigma = \frac{\sigma_{DC}}{1+\omega_B^2 \tau^2} \begin{pmatrix}
1 & -\omega_B \tau \\ \omega_B \tau & 1
\end{pmatrix}$
where the author suggests that since it is rotationally invariant, it must be in the form of
$\sigma = \frac{\sigma_{DC}}{1+\omega_B^2 \tau^2} \begin{pmatrix}
1 & -\omega_B \tau \\ \omega_B \tau & 1
\end{pmatrix} = \begin{pmatrix}
\sigma_{xx} & \sigma_{xy} \\ -\sigma_{xy} & \sigma_{xx}
\end{pmatrix}$.
My understanding of rotational invariance is that of
$\forall A \in SO(2), \quad A^T \sigma A = \sigma$ , i.e. $[\sigma, A]=0$, which would result in that form as outlined in https://math.stackexchange.com/questions/173639/properties-for-a-matrix-being-invariant-under-rotation.
However, I want to know what it really means physically; $\sigma$ is a conductivity, hence a tensor, which used to be a scalar when magnetic field is absent. So what does it mean that the conductivity is rotationally invariant?
 A: A generic $n\times n$ matrix $\sigma$ can be decomposed as
$$\sigma = \sigma_0 + \sigma_H + \sigma_S$$
where $\sigma_0$ is proportional to the identity matrix, $\sigma_H$ is antisymmetric, and $\sigma_S$ is symmetric and trace-free.  Explicitly for $2\times 2$ matrices,
$$\pmatrix{a&b\\c&d}=\frac{a+b}{2} \pmatrix{1&0\\0&1} + \frac{b-c}{2}\pmatrix{0&1\\-1&0} + \Delta \pmatrix{\cos(\varphi)&\sin(\varphi)\\ \sin(\varphi)&-\cos(\varphi)} $$
where $\Delta \equiv \sqrt{\left(\frac{a-d}{2}\right)^2+\left(\frac{b+c}{2}\right)^2} $ and $\tan(\varphi)=\frac{b+c}{a-d}$.
Writing $\mathbf E = \sigma \mathbf J$, we can examine the contributions to $\mathbf E$ from the various parts of $\sigma$.  The scalar part contributes a scalar multiple of $\mathbf J$, while the antisymmetric part contributes a component which is orthogonal to $\mathbf J$.
The symmetric part requires additional interpretation.  It defines two special directions in the plane, offset by $\varphi$ from the chosen reference axes. $\mathbf J$ is resolved into its components along these special axes, scaled by $\Delta$ in one direction and $-\Delta$ in the other, and finally added back together.  You can see this in the following GIF:

Rotational symmetry precludes the existence of these two special directions, which means that the only contribution can come from the scalar and antisymmetric parts.

In slightly more mathematical language, the scalar part of $\sigma$ transforms (unsurprisingly) like a scalar, while the antisymmetric part transforms as a pseudoscalar in $2$ dimensions.  Under proper rotations, both of these parts are invariant.  The symmetric trace-free part transforms non-trivially under rotations, and so imposing rotational symmetry rules it out.
