# Divergence of the Berry connection

Given the Berry connection $$$$\boldsymbol{\mathcal{A}}(\mathbf{R}) = i \langle u(\mathbf{R}) | \nabla_\mathbf{R} | u(\mathbf{R}) \rangle,$$$$

the Berry curvature can be written as its curl, $$$$\boldsymbol{\Omega}(\mathbf{R}) = \nabla_\mathbf{R} \times \boldsymbol{\mathcal{A}}(\mathbf{R}).$$$$

The Chern number measures the ammount of curl inside a region $$$$C = \frac{1}{2 \pi} \oint_S \boldsymbol{\Omega} \cdot d\mathbf{S}.$$$$

What about the divergence of the Berry connection, $$$$\nabla_\mathbf{R} \cdot \boldsymbol{\mathcal{A}} (\mathbf{R}) = ~?$$$$ Does it correspond to a meaningful object? Is it associated with a topological charge?

• Are your $k$ indices on your $u_k$ the same as your $k$ indices on your $\nabla_k$?
– hft
Commented Mar 8 at 23:40
• They are not exacly indices, but rather specify that the derivatives are in k-space. I will try to improve the notation, thanks for the headsup Commented Mar 8 at 23:43
• Well, have you considered what happens to the divergence of $A$ under a gauge transformation? I guess this will answer your question. Commented Mar 8 at 23:50
• As I understand we have $\boldsymbol{\mathcal{A}} \rightarrow \boldsymbol{\mathcal{A}} + \nabla f$, meaning $\boldsymbol{\mathcal{A}}$ is not gauge invariant. However, the Chern-Simons form is also not gauge invariant, but it can still be used to construct a topological invariant in odd dimensions (doi.org/10.48550/arXiv.1505.03535). Is there some fundamental reason why we cannot do a similar thing with objects making use of the divergence of the Berry connection? Commented Mar 9 at 14:22

In differential form notation, the divergence is written as 'd$$*$$', i.e., it involves the metric, because of the Hodge star.