Property related with Berry curvature: $\Omega_{n,\mu\nu}=-\Omega_{n,\nu\mu}$

Question

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?

Answer 1

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