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

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

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

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

  • $\begingroup$ Are your $k$ indices on your $u_k$ the same as your $k$ indices on your $\nabla_k$? $\endgroup$
    – hft
    Mar 8 at 23:40
  • $\begingroup$ 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 $\endgroup$ Mar 8 at 23:43
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    $\begingroup$ Well, have you considered what happens to the divergence of $A$ under a gauge transformation? I guess this will answer your question. $\endgroup$ Mar 8 at 23:50
  • $\begingroup$ 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? $\endgroup$ Mar 9 at 14:22

1 Answer 1


(I don't have the reputation to add this as a comment, apologies.)

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

Generally, constructing a topological invariant from a metric dependent object is non-trivial (you would need some further symmetry). Therefore, I don't think it is generally possible.

  • $\begingroup$ Thanks for your input. Is the Berry curvature metric-independent then? $\endgroup$ Mar 17 at 13:36
  • $\begingroup$ Defined as the exterior derivative of the connection, yes. $\endgroup$ Mar 17 at 15:37
  • $\begingroup$ However, there are subtleties involved in this definition in topologically nontrivial spaces, and you need to bring fibre bundles into the game. I recommend skimming through topology in physics by Nakahara, to get a better idea of these concepts. $\endgroup$ Mar 17 at 15:49
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    $\begingroup$ In particular, see section 10.6.2 of the 2nd edition of the book. $\endgroup$ Mar 17 at 15:55
  • $\begingroup$ Thanks! I'll take a look $\endgroup$ Mar 17 at 21:52

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