Why is the Berry curvature odd under time reversal? The question is: Why is the Berry curvature, defined as
$$\mathcal{F}=-\mathrm i\, \epsilon_{ij}\, \left\langle\partial_{ki}u_{n}(k)\mid \partial_{kj}u_{n}(k) \right\rangle ,$$
odd if I apply time reversal? From my understanding, the time reversal operator would simply be $\mathcal{T}=K$ the complex conjugation operator. I just do not seem to be able to see why the Berry curvature is odd under time reversal in a time reversal invariant system.
 A: This is easiest to see if we explicitly write out the wavefunctions. We have that the Berry Curvature is given by
$$
-i\epsilon_{ij}\int dr \frac{\partial}{\partial k_i}u^*(k,r)\frac{\partial}{\partial k_j}u(k,r)
$$
Here, $r$ represents all the position degrees of freedom, and $k$ is just a label that indexes our wavefunctions. Under time-reversal, the wavefunction is sent to its complex conjugate, so that the Berry Curvature is
$$
-i\epsilon_{ij}\int dr \frac{\partial}{\partial k_i}u(k,r)\frac{\partial}{\partial k_j}u^*(k,r)
$$
This is identical to the original expression, except with $i$ and $j$ switched. Since $\epsilon_{ij}$ is antisymmetric, this amounts to an overall minus sign. Thus, the Berry Curvature is time-odd.
As an aside: since the Berry Curvature integrated over the Brillouin zone gives the number of counterclockwise edge modes, this quantity must be odd. After all, 10 counterclockwise edge modes become 10 clockwise edge modes (or -10 counterclockwise edge modes) under time reversal!
