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Math. 512. Plane crystallographic groups
October 13, 2006
Let T be the subgroup of the group of motions M (2) of the plane P which consists of translations. The subgroup T is a normal subgroup of M (2). For each g ∈ M (2) and ta ∈ T we have
g ◦ ta = tg(a) ◦ g. (1)
Fix a point O on the plane so that any vector is equal to a vector a = OP~. thus we can identify vectors with their end points P. Then vectors form a vector space of dimension 2. If we fix a rectangular coordinate system, then any vector a has coordinates (a, b). Let O(2) be the group of motions which fix the point P. Then we have proved in class that any g ∈ O(2) is a linear transformation and its matrix in the basis i, j is an orthogonal matrix A. It is of determinant 1 if g is a rotation, and of determinant −1 if g is reflection with respect to a line ` containing O. Consider a homomorphism
φ : M → O(2)
defined as follows. For any motion g we have
t−g(O) ◦ g(O) = O.
This shows that any element m ∈ M (2) is of the form m = ta ◦ g, where g ∈ O(2). Since a = m(O), the end point of the vector a, the vector a is defined uniquely by g, hence the orthogonal part g is defined uniquely. We set φ(ta ◦ g) = g.
By (1), we have
(ta ◦ g) ◦ (tb ◦ g′) = ta ◦ (g ◦ tb) ◦ g′^ = (ta ◦ tg(b)) ◦ (g ◦ g′) = ta+g(b) ◦ g ◦ g′.
Thus φ((ta ◦ g) ◦ (tb ◦ g′)) = g ◦ g′. This checks that φ is a homomorphism. By Chasles’s Theorem any motion is either rotation ρθ(c) about an angle θ with fixed point c, or a reflection rwith the mirror line, or a translation ta, or a glide reflection ta ◦ r`. We can write
ρθ(c) = tc ◦ ρθ ◦ t−c,
r= ta ◦ r′ ◦ t−a,
where ′^ is the line parallell to and containing O, and ta() =′. It follows from the definition of φ that φ(ta) = 1, the identity map. Since φ is a homomorphism, we get
φ(ρθ(c)) = ρθ, φ(r) = r′.
It follows from the definition of φ that its kernel is T and the image is the whole O(2). Thus M (2)/T ∼= O(2).
Lemma 1. Let g = ta ◦ r. Then g = tb ◦ r′ , where the vector b is parallel to the line ′. Here b is equal to the orthogonal projection of a to, and ′^ is the line parallel to which passes through the end point of the vector 1 2 (a^ −^ b).
Proof. Choose coordinate system so that is the x-axis. Let a = (a, b). Consider the line′^ given by the equation y = b/2. For any x = (x, y) we have
t(a,0) ◦ r′^ (x) = (x, y − 2(y − b/2)) + (a, 0) = (x + a, −y + b) = ta ◦ r(x).
b
a
`
`′
A glide reflection is the motion g of the form ta ◦ r, where a is not orthogonal to the line (equivalently, its orthogonal projection to is not equal to zero). Note that if a is orthogonal to then g is the reflection with respect to a line parallel to `. The main difference between a glide reflection
Proof. If g ∈ Γ, then¯ h = ta ◦ g ∈ Γ for some translation ta ∈ T. For any x ∈ L we get
tg(x) = g ◦ tx ◦ g−^1 = (t−a ◦ h) ◦ tx ◦ (h−^1 ◦ ta)
= t−a ◦ (h ◦ tx ◦ h−^1 ) ◦ ta) = t−a ◦ th(x) ◦ ta = th(x ∈ Γ.
This shows that g(x) ∈ L.
Note that elements g in the point group Γ are not necessary in Γ. How-¯ ever, ta ◦ g ∈ Γ for some translation ta (not necessary in LΓ).
Lemma 3. The group ¯Γ is a finite group. Its intersection with O(2)+^ (the subgroup of rotations) is a cyclic group Cn of order n = 1, 2 , 3 , 4 or 6. If ¯Γ contains a reflection, then it is a group D 2 , D 3 , D 4 ,or D 6.
Proof. One proves that Γ is finite because Γ is discrete (we omit the proof).¯ Thus any g ∈ ¯Γ ∩ O(2)+^ is represented by an orthogonal matrix of finite order. It looks as
A =
cos θ − sin θ sin θ cos θ
, θ = 2π/n.
Let L = Za + Za. Since g(L) ⊂ L, we get
g(a) = m 1 a + n 1 b, g(b) = m 2 a + n 2 b,
for some integers m, n. The matrix
B =
m 1 n 1 m 2 n 2
represents g in the basis (a, b). Hence B = CAC−^1 for some matrix C (the basis change matrix). Also, this implies that the traces of A and B coincide (it is the sum of the eigenvalues of g). This gives
T r(A) = 2 cos 2πi/n = T r(B) = m 1 + n 2 ∈ Z.
Thus 2 cos 2πi/n = 0, ± 1 , ±2. This gives 2π/n = π/ 2 , 3 π/ 2 , π/ 3 , 2 π/ 3 , 0 , π. This proves the first assertion. If Γ contains a reflection, then it is a dihedral group with the cyclic subgroup of index 2 of one of Cn from the assertion.
Definition 1. A lattice L is called rectangular (resp. hexagonal) if one can choose a rectangular fundamental parallelogram (resp. parallelogram with two equal side and the angle π/ 3 between them). A lattice is called half-rectangular if it is spanned by a side of a rectangular and its middle point.
In the following, if L is rectangular (resp. hexagonal) we will choose a basis (a, b) of a lattice L which defines a fundamental parallelogram which is a rectangle (resp. has equal length sides with the angle π/3 between the sides). However, note that there are bases of the same lattice of different shape.
Assume that a reflection r∈ Γ.¯ Then, for any x ∈ L, x + r(x) ∈ L. But it is easy to see that this vector is equal to 2pr(x), the twice of the orthogonal projection of x to. Thus is spanned by a vector from the lattice L. By taking a lattice vector on of smallest length, we may assume that ` is spanned by a. Now
2 pr(b) = b + r(b) ∈ L,
hence, after replacing b with b + ma for some m ∈ Z, we may assume that pr`(b) = 0, or 12 a. In the first case we get a rectangular lattice, in the second case we get a half-rectangular lattice. Now we are ready to classify all crystallographic groups.
- Type I: No element of Γ fixes a point.
In this case Γ contains only translations and glide reflections. Also the composition of two glide reflections whose glide vectors are not parallel has a fixed poins since
(tc ◦ r) ◦ (tc′^ ◦ r′^ ) = tc ◦ tr(c′) ◦ r′^ ◦ r′^ = tc+r(c′) ◦ ρ 2 θ,
where θ is the angle between and′. Thus all glide reflections s,c have lines parallel to the same vector. Let s = s,c ∈ Γ. By changing O we may assume that O ∈. Applying (2) we see that 2c ∈ L. Let a′^ be the smallest length vector in L which lies on . By composing s = tc ◦ r with tma with m ∈ Z we may replace c with a vector equal to a′/2. We know that r`(L) ⊂ L. As we observed before this implies that L is a a rectangular lattice or a half-rectangular lattice. Let ΩΓ be spanned by a′^ and b′.
Lemma 2 to write it as tc ◦ r′^ , where′^ is parallel to and the distance between the lines and ′^ is equal to the length of the vector 12 (x − pr(x)). Write x = ma + nb for some integers m, n. Then x − pr(x) = nb, and hence the length of 12 (x − pr(x)) is equal to n||b||/2.
III 2 : L = Za + Zb is a hexagonal lattice, Γ¯ ∼= C 2 is generated by r. The mirror lines of reflections from Γ are lines parallel to with distance between each other equal to an integer multiple of
3 ||b||/2. b
O^ a
`
Similar to the previous case. We note that the vector c = −a+b belongs to L. It is perpendicular to a and its length is
3 ||a|| =
3 ||b||.
III 3 : L = Za + Zb is a square lattice, Γ¯ ∼= D 2 is generated by ρπ, r`. The mirror lines of reflections from Γ are lines parallel to a or b with distance between parallel lines equal to an integer multiple of ||a||/2. b
O (^) a
`
We use (3) to see that lines spanned by a and b are mirror lines, and then get more applying the argument from case III 1.
III 4 : L = Za + Zb is a square lattice, Γ¯ ∼= D 4 is generated by ρπ/ 2 , r`. The mirror lines of reflections from Γ are lines parallel to a, b, c = a + b with distance between parallel lines equal to an integer multiple of
3 ||a||/2. b
O (^) a
`
III 5 : L = Za + Zb is a hexagonal lattice, ¯Γ ∼= D 3 is generated by ρ 2 π/ 3 , r`. The mirror lines of reflections from Γ are lines parallel to a or bwith distance
between parallel lines equal to an integer multiple of
3 ||a||/2. b
O^ a We use (3) to see that lines spanned by a and b are mirror lines, and then get more applying the argument from case III 2.
III 6 : L = Za + Zb is a hexagonal lattice, Γ¯ ∼= D 6 is generated by ρπ/ 3 , r`. The mirror lines of reflections from Γ are lines parallel to a, b, a − b with distance between parallel lines equal to an integer multiple of
3 ||a||/2. b
O^ a
- Type IV: Γ contains glide reflections s`,c with c 6 ∈ LΓ but not of type I.
Since Γ is not of type I it must contain either rotations or reflections. In fact, if it contains a reflection then the composition with a glide reflection will be a rotation. So, we may assume that Γ contains a non-trivial rotation ρθ. We choose a fundamental parallelogram such that c = a/2 and use that
s,c ◦ ρθ = (tc ◦ r) ◦ ρθ = tc ◦ rρθ/ 2 (`), (4)
This shows that the projection of 2c to the line ρθ/ 2 (`) belongs to the lattice LΓ.
IV 1 , IV 2 : L = Za + Zb is a rectangular lattice, Γ¯ ∼= D 2 is generated by ρπ, r. Γ contains glide reflections s,c parallel to a reflections with respect to lines parallel to b. The distances between the lines parallel to a (resp.b) are integer multiple of the half-lengths of b (resp. a). In this case θ = π, hence in (4) we have c is perpendicular to ρπ/ 2 (). As we had remarked before this means that s,c ◦ ρθ is a reflection with respect to a line parallel to a line perpendicular to `. This implies that L is rectangular (case IV 1 ) or half-rectangular (case IV 2 ). The other properties had been explained before.
