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Every element in the symmetry group G of Rn is of the form Ta ∘ R with R ∈ O(n). If one composes two such elements, Ta ∘ R1 and Tb ∘ R2 then one gets an element whose linear part is the composite R1 ∘ R2 .
This is the motivation for the following:
Definition
The set of all linear parts of elements of a symmetry group G is called the point group P of G.
Remarks
Theorem
The point group of a symmetry group G is the factor group G/L where L is the normal subgroup of translations in G.
Proof
Given a symmetry f, we may write it as Ta∘ R with R ∈ O(n) and we may define a homomorphism θ : G O(n) by Ta∘ R ↦ R
It is easy to verify that this is a group homomorphism with kernel the lattice L of translations and image the point group. The result then follows from the first isomorphism theorem.
Although the elements of the point group P are not elements of the symmetry group G, they do act on the lattice.
Theorem
If A ∈ P ⊆ O(2) and a ∈ L then A(a) ∈ L also.
Proof
Since A is in the point group for some f ∈ G we have f = Tv ∘ A.
Then A ∘ Ta(x) = A(a + x) = A(a) + A(x) = TA(a) ∘ A(x).
Now calculate f ∘ Ta ∘ f -1 which is an element of G.
This is Tv ATa A -1T-v = Tv(TA(a)A)A -1T-v = TvTA(a)T-v = TA(a) and so A(a) is in the lattice.
This means that (notwithstanding the fact that P is not a subgroup of G) that the elements of P act on the lattice and so satisfy the crystallographic restriction. This limits the possible subgroups of O(2) which P can be.
Here are the possible choices for the groups P and the lattices on which they can act. We will see later how the final two columns can be filled in.
Point group | Lattice | #groups | Names of groups |
C1 | general | 1 | p1 |
C2 | general | 1 | p2 |
C3 | equilateral | 1 | p3 |
C4 | square | 1 | p4 |
C6 | equilateral | 1 | p6 |
D1 | rectangular rhomboid | 2 1 | pm pg cm |
D2 | rectangular rhomboid | 3 1 | pmm pgg pmg cmm |
D3 | equilateral | 2 | p3m1 p31m |
D4 | square | 2 | p4m p4g |
D6 | equilateral | 1 | p6m |
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