Course MT3818 Topics in Geometry
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(I: Point group is cyclic) 		Contents Next page
(III: Other point groups)

II: Point group is D1

The group D1 is generated by a reflection. The result for this section is:

Theorem
There are three equivalence classes of symmetry group with point group D1.

Proof
The group D1 is generated by a reflection R in a line l.
Choose a minimal length vector v of the lattice L lying on l. (See the example of shift vectors considered above.)
Then choose a vector w in the lattice perpendicular to L and of minimal length.
Note that for any u ∈ L the vector u - R(u) is perpendicular to l since R(u - R(u)) = -(u - R(u)).

Then there are three possibilities.

  1. For some t ∈ L we have t + R(t) = u and we can take t = 1/2 u + 1/2 w and we can take the pair t and u as the basis of a centred or isosceles lattice.
    This is the case cm.


  2. The vectors v, w can be chosen as a basis for the lattice and for some u ∈ L we have Tu ∘ R ∈ G and so u + R(u) is the shift vector 0.
    This is the case pm.


  3. The vectors v, w can be chosen as a basis for the lattice and for some u ∈ R2 we have u + R(u) is the shift vector 0.
    This is the case pg.


Given another group which leads to the same case as one of the above, we can define an isomorphism of the lattice by mapping generators to generaors and then, as before, split the group up into cosets L and LR which we can use to define an isomorphism.
Patterns with these groups as symmetries are:

cm

pm

pg

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(I: Point group is cyclic) 		Contents Next page
(III: Other point groups)
JOC March 2003