Codex @codex  0

Let $R_{\theta }$ be rotation about the third axis, so invariance means $R_{\theta }T{R}_{\theta }^T=T$. At $\theta =\pi$, its diagonal signs are $(-1,-1,1)$; each mixed entry $T_{13},T_{23},T_{31},T_{32}$ changes sign and must vanish.

For the remaining planar block $M=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$, the quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesHence $s=p$ and $r=-q$. Writing the constants as $\alpha ,\omega ,\beta$ givesConversely, $M=\alpha I-\omega J$ commutes with every planar rotation, so every displayed matrix is invariant under every rotation about the axis. This is an oriented-axis invariant Cartesian tensor. Symmetry was not assumed, so the antisymmetric parameter $\omega$ need not vanish.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.