Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

The only rotation axis obvious to me is rotation by 360 degrees, the identity. Vertical mirror planes I've been dicing and cutting it through several planes and I still see none. Yet, the structure looks fairly symmetric and I feel like I'm missing something.

No center of inversion either.

Not too terribly sure about horizontal mirror planes. I suppose that one bisecting the plane horizontally would be there but I am not entirely sure if this is appropriate.

Right now I'm saying this structure belongs to the C1 point group.

share|cite|improve this question
Do you have any more information on the lattice vectors? Specifically, how the two sublattices are offset? – Chay Paterson May 7 '13 at 21:30
@ Chay Paterson The lattice vectors are given by a = {-1/2, -Sqrt[3]/2}; b = {1, 0}; The basis of atom red is {2/3,1/3}, basis of atom blue is {0,0}. – user17338 May 7 '13 at 21:47
I make the angle between the red atom and the blue atom at (1,0) to be acos(sqrt(4/5)). Since acos(sqrt(anything))/pi is almost always irrational and this doesn't correspond to any of the special cases, I believe you are correct and there are no nontrivial rotational angles. – Chay Paterson May 7 '13 at 22:16

I have consulted the International Tables for Crystallography (which is the authorative reference for symmetries, point groups, space groups and the like; unfortunately, it is not freely available on the web), and as drawn in the question, so the plane group is indeed p1, so no rotation axes.

However, it would be nice if you could please clarify how you have specified the basis: When you say it is $\{2/3, 1/3\}$, does that mean $2\hat{\mathbf{x}}/3 + \hat{\mathbf{y}}/3$ or $2\mathbf{a}/3 + \mathbf{b}/3$? The latter is the most frequently used method of specifying a basis, and the international tables actually list the coordinates $2\mathbf{a}/3, \mathbf{b}/3$ as a position that is compatible with a threefold rotation, and hence plane group p3m1. Could it be that you have misunderstood the problem formulation?

I'll add a technical note add the end: In two dimensions, the p1 plane group has the oblique lattice. Triclinic is a beast belonging to three dimensions (and one of its space groups is P1, note the capital P).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.