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Linear algebra part 2
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Linear algebra part 2
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18.701 October 2003Plane crystallographic groups with point groups C4 or D4
This describes the discrete subgroups G of M whose translation group L is a lattice and whose point group
is either the cyclic group C4 or the dihedral group D4 .
The cyclic group C4 is generated by the rotation through the angle π/2 about the origin, which we denote by
ρ. In its standard representation, the dihedral group D4 contains ρ and the reflection r about the horizontal
axis. The elements of D4 are the four rotations ρi and the four reflections ρi r, where i = 0, 1, 2, 3.
Every element of the point group, in particular ρ, carries the lattice L to itself, so L is a square lattice. We
may choose coordinates in V ≈ R2 so that L becomes the lattice of vectors with integer coordinates. We
recall that
L = {a ∈ V |ta ∈ G}.
We first consider the case of a discrete subgroup H whose point group is C4 . The next proposition shows
that there is just one type of group in this case. Since ρ is in the point group H, H contains an element
of the form m = tv ρ. This element is a rotation through angle π/2 about some point of the plane P . So
H contains a rotation through angle π/2. We translate the coordinates in P so that the origin becomes the
center of the rotation m. In this new coordinate system, m = ρ, so when we write the isometries using these
coordinates, ρ ∈ H.
Proposition. Let H be a discrete subgroup of M whose translation group L is a lattice and whose point
group H is the cyclic group C4 . Then with coordinates chosen as above, H consists of the elements ta ρi ,
where a ∈ L (i.e., a is an integer vector) and i = 0, 1, 2, 3.
Proof. Let S denote the set of vectors ta ρi with a ∈ L. By definition, ta ∈ H when a ∈ L, and by our choice
of coordinates, ρ ∈ H. Therefore the products ta ρi are in H, which shows that S ⊂ H. Conversely, let h be
an element of H. Since the point group of H is cyclic, every element of H preserves orientation. So m has
the form tv ρθ for some v ∈ V and some θ. The image of m in the point group is ρθ , so θ is a multiple of
π/2, and ρθ = ρi for some i. Since H is a group, the product mρ−i = tv is in H, so v ∈ L. Therefore m ∈ S.
This shows that S ⊃ H, hence that S = H. �
We now consider the case of a discrete group G whose point group is D4 . To begin with, we note that the
orientation-preserving elements of G form a subgroup H of G of index 2. Moreover H is a discrete subgroup
of M whose translation group is L, and whose point group is C4 . With suitable choices of coordinates, H is
the group described by the previous proposition. In particular, ρ ∈ G.
Proposition. Let G be a discrete subgroup of M with point group D4 . We choose coordinates so that the
origin is a point of rotation of order four, and so that the lattice L is the integer lattice. Let U denote the
translated lattice c + L in V , where c = ( 12 , 12 )t . There are two possibilities G1 and G2 :
(a) The elements of G1 are the products ta p where a ∈ L and p ∈ D4 , or
(b) The elements of G2 are the products tv p, such that p ∈ D4 ,
v ∈ L if p is a rotation, and v ∈ U if p is a reflection.
Proof. Since the point group G contains the reflection r, G contains an element of the form m = tu r. If u
is in L, then the product t−u m = r is also in G. Because ρ ∈ G and G is a group, G contains the elements
of G1 , and an argument similar to that used in the proof of the previous proposition shows that G = G1 .
Suppose that u ∈ L. Let u = ρu. Since ρ ∈ G,
m = ρm = ρtu r = tu� r
1
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