Why Electrical conductivity tensor is symmetric? Or is it not always symmetric? How to show that the electrical conductivity tensor is symmetric? (or it's not always symmetric?)
 A: It is not necessarily symmetric. The conductivity tensor $\boldsymbol \sigma$ is given by:
$$\mathbf J = \boldsymbol \sigma \mathbf E$$
And its inverse $\boldsymbol \sigma^{-1}=\boldsymbol \rho$ is the resistivity tensor. If you use matrix notation you have:
$$\begin{pmatrix} J_1\\ J_2\\J_3\end{pmatrix}=\begin{pmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22} & \sigma_{23}\\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{pmatrix}\begin{pmatrix} E_1\\ E_2\\E_3\end{pmatrix}$$
As you can see here, $\sigma_{12}$ tells you how a component of the electric field in the "$2$" direction generates a current in the "$1$" direction, whereas $\sigma_{21}$ tells you how an electric field in the "$1$" direction generates a current in the "$2$" direction (you could think of the $x$ and $y$ directions, in Cartesian coordinates). These two need not be the same, and in a solid they could depend on, for example, the type of anisotropy in the material.
Wikipedia gives us a nice example: "In the Hall effect,  due to rotational invariance about the $z$-axis $\rho_{xx}=\rho_{yy}$ and $\rho_{yx}=-\rho_{xy}$". So $\boldsymbol \rho$ is not symmetric, and the inverse of a non-symmetric matrix is also not symmetric, so $\boldsymbol \sigma$ is not symmetric.
