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Symmetry-reduced or irreducible BZ #33
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Hmm, I'm not completely sure, I admit. By
I imagine you also mean that there is a hope if there is N points in the "full" BZ sampling set, that there would be N/|G| (with |G| denoting the order of the rotation group) unique points in the reduced set. If so, I think that is harder, since that puts some requirements on the initial sampling of the full BZ (just mapping a point to the irreducible BZ sector wouldn't necessarily imply that many points map to each other; they might simply map to close by points). Probably the better thing to do is to first determine the irreducible BZ, and then try to sample that. I think that can be done by taking intersections of half-spaces and the original BZ. There's an implementation in SymmetryReduceBZ.jl. The key technical challenge is to find a good way to take intersections of the polyhedron returned constructed in Brillouin with a half-space; something like Polyhedra.jl probably makes this easy, but it's a big dependency. |
Actually, the intersection of a half-space and a convex polygon is not too involved. E.g., chatGPT gives a nice, clear summary of how to adapt Sutherland-Hodgman clipping to this when given the prompt below:
So, this could be implemented directly here. The "stages" required would be:
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Yeah, sampling IBZ first is definitely a better idea. I will try this out. Thanks! |
I'll keep this open as a reminder to one day implement a method to construct the symmetry-reduced or irreducible BZ. |
How hard would it be to make a function that maps a point in the first BZ to its symmetrically identical point in the irreducible BZ? I am trying to integrate/sum over the first BZ and it would be nice to only have to go over the irreducible points and map things back out after the fact. Happy to help write code, but the symmetry operations are not something I understand amazingly well at this point.
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