# Symmetry arguments on the Berry connection and the polarization charge

Consider the Berry connection $$A_n(\mathbf{k})=i \langle n(\mathbf{k})|\nabla_{\mathbf{k}}|n(\mathbf{k})\rangle$$ and the polarization charge $$\mathbf{P}=-\frac{1}{4\pi^2} \int_\mathrm{B.Z.}\mathrm dk_x \, \mathrm dk_y \mathrm{Tr}(\mathbf{A})$$ of a crystalline system (with discrete translation symmetry). If the unit cell has both time-reversal symmetry and parity symmetry (such as mirror symmetry about the x-axis), then should $$\mathbf{A}$$ satisfy $$\mathbf{A}(\mathbf{k}) = -\mathbf{A}(-\mathbf{k})$$?

Then, the polarization charge $$\mathbf{P}$$ is $$0$$ under any gauge. Is there any problem in this argument?

• Thanks! But I am more interested in a topological corner state which relies on the polarization charge. And the unit cell has both parity and time-reversal symmetries, so the Berry curvature is $0$ everywhere. – LearnerAL Mar 12 '19 at 3:41