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

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: $$$$\Omega_{n,\mu\nu}(\mathbf{k})=\partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k})\tag{1}$$$$

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?

• Didn't you prove this already? – Cody Aldaz Jul 22 at 19:56
• @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. – Carmen González Jul 22 at 20:00

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