Can you explain it using schematics? I think the reason is the symmetry that doesn't allowed to unique lattice points.

  • $\begingroup$ Let's assume you stacked two face-centered (base-centered) tetragonal lattices. At the side where they meet the atom, on the center of a surface, turns into an atom that is in the center of the combined volume of the two cells (just draw two face-centered tetragonal cells). That means by stacking you also get atoms inside the volume and not only on it's surface (face-centered). And in this case a much smaller body-centered tetragonal cell can be constructed from this arrangement. $\endgroup$ – Justanotherchemist Mar 6 '19 at 7:39
  • $\begingroup$ @Justanotherchemist You seem to have proved that face-centered and base-centered cells are the same (which they are not), and moreover, that they all are impossible. I'm not ready to agree with this. $\endgroup$ – Ivan Neretin Mar 6 '19 at 7:56
  • $\begingroup$ Well if been calculating this through with all possible orientations right now. In the cubic case, wouldn't the body centered lattice become tetragonal and thus have a lower symmetry? Because no matter how I draw it on side will be 'a' and the other two will be smaller as they are diagonal. While for the tetragonal face-centered case the resulting body-centered lattice remains tetragonal in symmetry. $\endgroup$ – Justanotherchemist Mar 6 '19 at 9:30
  • $\begingroup$ Face-centered case (that is, with positions in the centers of all faces) is not even mentioned in the question, so I see no reason to bring it up. Base-centered tetragonal cell will be equivalent to a smaller primitive cell which is still tetragonal. And of course the body-centered cubic cell is equivalent to some primitive cell which is not cubic. So you are essentially right, but I'd rather put it in somewhat different words. Lattice, for one, does not care about our choice of unit cell; lattice symmetry does not depend on it. $\endgroup$ – Ivan Neretin Mar 6 '19 at 9:57
  • $\begingroup$ @ivan sounds like an answer to me. $\endgroup$ – Oscar Lanzi Mar 6 '19 at 10:03

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