Hall conductivity versus isotropy

It is known that the Hall conductivity of a two-dimensional system in magnetic field is antisymmetric: $$\sigma_{xy}=-\sigma_{yx}.$$ However the conductivity tensor, being a current response function, should obey some general symmetry properties. G. Giuliani and G. Vignale in "Quantum theory of the electron liquid" (2005) write on page 145:

First of all, the homogeneity and isotropy of the system ensure that the tensor $\chi^J_{\alpha\beta}(\vec{q},\omega)$ can be decomposed into longitudinal and transverse components, relative to the direction of $\vec{q}$, each depending only on the magnitude of $\vec{q}$. In other words, the current induced by a purely longitudinal vector potential is purely longitudinal, and that induced by a purely transverse vector potential is purely transverse. More formally, one can write $$\chi^J_{\alpha\beta}(\vec{q},\omega)=\chi_L(q,\omega)\frac{q_\alpha q_\beta}{q^2}+\chi_T(q,\omega)\left(\delta_{\alpha\beta}-\frac{q_\alpha q_\beta}{q^2}\right),$$ where $\chi_L$ and $\chi_T$ are the current–current response functions in the longitudinal and transverse channels respectively

Indeed, this property is consistent with the isotropy of space: if we perform a rotation of a coordinate system using the matrix $$\Lambda=\left(\begin{array}{rr}\cos\varphi&\sin\varphi\\-\sin\varphi&\cos\varphi\end{array}\right),$$ then the two-component vectors transform as $\mathbf{j}'=\Lambda \mathbf{j}$, $\mathbf{E}'=\Lambda \mathbf{E}$, $\mathbf{q}'=\Lambda \mathbf{q}$. Thus from the definitions of the conductivity tensors $\mathbf{j}=\hat\sigma(\mathbf{q})\mathbf{E}$ and $\mathbf{j}'=\hat\sigma(\mathbf{q}')\mathbf{E}'$ we find the property $$\hat\sigma(\Lambda\mathbf{q})=\Lambda\hat\sigma(\mathbf{q})\Lambda^+,$$ which is consistent with the longitudinal-transverse decomposition.

This decomposition implies a symmetric nondiagonal conductivity $$\sigma_{xy}=\sigma_{yx}=[\sigma_L(q)-\sigma_T(q)]\frac{q_xq_y}{q^2}.$$ The Hall conductivity obviously does not satisfy this requirement: it is antisymmetric and does not have an angular dependence $\propto q_xq_y/q^2$ at $q\rightarrow0$.

So why does the Hall conductivity disobey the isotropy relation? Does a magnetic field break isotropy of space?

• Does a magnetic field break isotropy of space? -> Obviously yes. – FraSchelle Nov 5 '17 at 16:44
• @FraSchelle In my question, I'm considering a 2D system in the $(x,y)$ plane with the magnetic field in the $z$ direction, so how does this field break isotropy in the $(x,y)$ plane? – Alexey Sokolik Nov 6 '17 at 15:00