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What is the connection between these two statements:

  1. the berry curvature change sign under time-reversal operation

  2. If the system has the time-reversal symmetry, then berry curvature is odd in k.

This answer is basically supporting statement 1.

I feel statement 1 doesn't involve any symmetry of the system, it is just an operation. Just as you can rotate an object of any shape 90 degrees, while that object don't necessarily have any rotation symmetry.

So what is the logic connection between these two statements ? After explanation, another concrete example will be welcomed.

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1) Time reversal takes ${\bf k}\to -{\bf k}$. 2) Time reversal takes the Berry curvature ${\bf \Omega}\to -{\bf \Omega}$. Therefore $\Omega(-{\bf k})=-\Omega({\bf k})$. I think it really is this simple....

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