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[verstraete@pcpm.ucl.ac.be]

To my knowledge there is no such routine: the IBZ is not calculated 
explicitly or geometrically. From the full kpoint grid, we take kpt #1 and 
eliminate all its images, then go on to kpt #2 and so on. The irreducible 
kpoints are inequivalent, but can be anywhere in the BZ. If you tell me 
what you are trying to do more specifically, it may help.

Bye

Matthieu



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Matthieu Verstraete             mailto:verstraete@pcpm.ucl.ac.be
PCPM, Boltzmann, pl. Croix du Sud, 1        tel: +32 10 47 33 59
B-1348 Louvain-la-Neuve  Belgium            fax: +32 10 47 34 52