I read in David Vanderbilt's book named "Berry Phases in Electronic Structure Theory - Electric Polarization, Orbital Magnetization and Topological Insulators" the definition of Berry curvature: "Berry curvature $\Omega(\mathbf{\lambda})$ is simply defined as the Berry phase per unit area in ($\lambda_x,\,\lambda_y$) space".

Berry Curvature is defined by: \begin{equation} \Omega_{n,\mu\nu}(\mathbf{k})=\partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k})\tag{1} \end{equation}

where $A_{n\mu}(\mathbf{k})=\langle u_{n\mathbf{k}}|i\partial_{\mu}u_{n\mathbf{k}}\rangle$ and $A_{n\nu}(\mathbf{k})=\langle u_{n\mathbf{k}}|i\partial_{\nu}u_{n\mathbf{k}}\rangle$ are Berry connections.

Berry's curvature has the following property: $\Omega_{n,\mu\nu}=-\Omega_{n,\nu\mu}$.

How is this property mathematically demonstrated?

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    $\begingroup$ Didn't you prove this already? $\endgroup$
    – Cody Aldaz
    Jul 22, 2020 at 19:56
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    $\begingroup$ @CodyAldaz In this question, I only asked how the operators of spatial inversion symmetry and time reversal symmetry operate on Berry's curvature, but this property about the Berry curvature that I put now has not been demonstrated and we thought it best to ask a separate question. But both questions complement each other and thankfully you drew attention to the post because anyone interested in Berry's curvature will probably want to read both posts. $\endgroup$ Jul 22, 2020 at 20:00

1 Answer 1


You can just exchange the $\mu,\nu$ indices to verify the antisymmetry: $$ \Omega_{n,\mu\nu}(\mathbf{k})=\partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k})\\ \Rightarrow \Omega_{n,\nu\mu}(\mathbf{k})=\partial_{\nu}A_{n\mu}(\mathbf{k})-\partial_{\mu}A_{n\nu}(\mathbf{k}) = - \left( \partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k}) \right) = -\Omega_{n,\mu\nu}(\mathbf{k}). $$

  • $\begingroup$ Thanks for the quick response! I thought it would be more difficult, because I thought it involved bra and kets, but after all it was simpler! Thank you! $\endgroup$ Jul 22, 2020 at 19:57

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