Course MT3818 Topics in Geometry
III: Other point groups
There are nine equivalence classes corresponding to the cases where the point group is D2 (4 of them), D3 (2 of them), D4 (2 of them) and D6 (one only).
In all these cases the point group is generated two reflections R1 and R2 in lines l and m whose product R1∘ R2 is a rotation generating the cyclic subgroup Cn of Dn. So the lines l and m are at an angle of 2π/n and the reflections R1 and R2 have shift vectors a = v + R1v on l and b = w + R2w on m.
We will find that there are 9 possible combinations of point groups and shift vectors.
Let R, s be the shortest vectors in L on l and m respectively.
- If n = 2 (so that l and m are perpendicular) then we can argue as in Stage 2 above:
If 1/2 R + 1/2 s ∈ L then both R1 and R2 give the cm situation (i) above and we get the group cmm.
If R and s form a basis for the lattice L then we get three possibilities.
- The shift vectors of R1 and R2 are both 0 and we get the group pmm
- The shift vectors of R1 and R2 are both non-0 and we get the group pgg
- The shift vectors of R1 is 0 and that of R2 is non-0 and we get the group pmg
- If n = 3 then the axes of reflection l and m are at 120° or 2π/3.
Choose R and s as shown.
In this case all the shift vectors will be 0.
We get one of the following cases
- The point 1/3 R + 1/3 s ∈ L
In this case, the some of the centres of 3-fold rotation are not on reflective axes.
The group is called p31m
- The vectors R and s are a basis for the lattice.
In this case, the some of the centres of 3-fold rotation are all on reflective axes.
The group is called p3m1
- If n = 4 then R and s chosen as above will always be a basis for L and we get two cases.
- Both shift vectors are 0 ⇒ group is p4m (it should really be called p4mm)
- One shift vector is 0 and one is not ⇒ group is p4g (it should really be called p4mg)
- If n = 4 then R and s will be a basis for L and all shift vectors are 0 giving one possibility: the group called p6m.
JOC March 2003