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Hello,

Given a hexagonal cell and a kpoint set that doesn't have the full symmetry of the lattice I see the vasp code applying the symmetry operations of the full symmetry group to each kpoint before reducing it with the symmetry of the lattice. ( in function IBZKPT in symmetry.F). This leads to a non-equidistant kpoint set with significantly more kpoints in the irreducible zone than maximally expected from the choice of the monkhorst pack parameters.

Despite the explicit discouragement of this choice of kpoint set, I wonder why the above stated is necessary, particularly because only the lattice symmetry is used in the symmetrization of the charge density.
What consequences would not enforcing the full symmetry group have?

Thanks in advance

the consequence is a (significant) increase in computation time (depending on how large the fraction of the IBZ is of the full BZ).
especially for k-meshes of non-rectangular lattices the symmetrization of the k-mesh is sometimes not easily done by vasp if the mesh does not explicitely include Gamma (typically for even meshes, please have a look at the k-points.pdf tutorial file of the vasp-workshop). Please try if the behavior pesists with an odd k-mesh as well