3.4 Discrete groups Lemma 3.4.1. Let Λ be a subgroup of R2, and let ~a;~b be two elements of Λ that are linearly independent as vectors in R2. Suppose that the paralleogram spanned by ~a and ~b contains no other elements of Λ other than ~0;~a;~b and ~a +~b. Then Λ is the group Λ = fm~a + n~b j m; n 2 Zg: (3.4) Proof. The plane is tiled up with copies of the parallellogram P spanned by the four vectors ~0;~a;~b;~a+~b. The vertices of these parallellograms are m~a+n~b for some integers m and n. Since Λ is a group all these vertices are in Λ. If Λ is not the group fm~a + n~b j m; n 2 Zg, then there exists an element x 2 Λ not in located at any of the vertices of the parallellograms. Since these parallellograms tile the whole plane there exist m and n such that x is contained in the parallellogram m~a + n~b; (m + 1)~a + n~b; m~a + (n + 1)~b and (m + 1)~a + (n + 1)~b. The element x0 = x − m~a − n~b is in the group Λ, and in the first paralleogram P . By assumption x0 is one of the four vertices of P , and it follows that also x is an vertex of the parallellograms. Hence Λ is the group fm~a + n~b j m; n 2 Zg. Definition 3.4.2. A subgroup Λ of R2 is called a lattice if there exists an > 0 such that every non-zero element ~a 2 Λ has length j~aj > . Groups of the form (3.4) are planar lattices. Lemma 3.4.3. Let Λ be a lattice, and let B ⊂ R2 be a bounded set. Then Λ \ B is a finite set. Proof. Let > 0 be such that any non-zero element ~a in Λ has length at least . Then in particular the disc D(~a) of radius /2 and centered around ~a contains no elements from Λ other than ~a. As B is bounded there can only be a finite number of non-over lapping discs in B with a fixed radius /2. Proposition 3.4.4. Let Λ be a lattice. Then Λ is one of the following groups, 1. the additive group Λ = fm~a j m 2 Zg, for some vector ~a 2 R2, 2. the group Λ is a planar lattice. Proof. If Λ is not trivial (that is the zero vector only) then there exists a non- zero element ~a 2 Λ. Assume now that Λ is contained in the line ` spanned by ~a. Since Λ is discrete there exist two non-zero elements in Λ having minimal length. Let ~b be one of these (the other one is −~b.) Then we have that Λ = fn~b j n 2 Zg. 53 The other case is that Λ is not contained in the line `. From the above argument we have that Λ\` is a discrete group, and of the form fn~b j n 2 Zg for some vector ~b on ` and with minimal distance. Since there exists vectors in Λ not lying on the line `, we pick one ~c with minimal length. Then ~c and ~b are linearly independent. The parallellogram spanned by ~c and ~b contains only a finite set of elements in Λ. Let ~a be one such not lying on `, and having minimal distance to the line `. Then the parallellogram with vertices ~0;~a;~b;~a + ~b contain no other element from Λ, and thus by Lemma 3.4.1 generate the whole group. Definition 3.4.5. A subgroup G of isometries is called discrete if there exists an > 0 such that 1. for any non-zero translation t~a in G we have that the length j~aj > , 2. for any non-zero rotation ρθ in G we have that jθj > . Lemma 3.4.6. An isometry T is determined by an orthogonal matrix A and a vector b, where T (X) = AX + b for any vector X. Let 2 ': Isom(R ) −! O(2) be the map that sends an isometry T (X) = AX + b to the orthogonal matrix A. Then ' is a homomorphism of groups. Proof. An isometry can be written uniquely as t~aρθ or t~aρθr from where it follows that the map ' is well-defined. That the map is a homomorphism is more surprising. Take two isometries, T1 = taρθ1 r1 and T2 = tbρθ2 r2, where 2 r1 and r2 are either r or the identity, and where a and b are vectors in R . Then '(T1) = ρθ1 r1, and '(T2) = ρθ2 r2, and in particular we have that '(T1)'(T2) = ρθ1 r1ρθ2 r2; and we need to prove that this also equals '(T1T2). We will use the relations (3.3) that are present in the group of isometries. The product is associative, and with the parantheses we only indicate where we apply the relations. We have 0 T1T2 = taρθ1 (r1tb)ρθ2 r2 = taρθ1 (tb r1)ρθ2 r2 0 where b = r1(b). Furthermore, 0 00 ta(ρθ1 tb )r1ρθ2 r2 = ta(tb ρθ1 )r1ρθ2 r2 00 where b = ρθ1 . Thus we have that the unique presentation of the product 00 T1T2 = ta+b ρθ1 r1ρθ2 r2; from where it follows that '(T1T2) = '(T1)'(T2), and that the map is a homomorphism of groups. 54 Lemma 3.4.7. Let G be a discrete group. Then subset of translations in G is a normal subgroup, and this subgroup is isomorphic to the lattice 2 Λ = f~a 2 R j t~a 2 Gg: Proof. By restricting the homomorphism ' to the subgroup G we get a ho- momorphism 'jG : G −! O(2). Its kernel is a normal subgroup and are precisely the translations in G. Definition 3.4.8. Let G be a discrete group. Restricting the homomorphism (3.4.6) to G gives a homomorphism G −! O(2). The image is a group G¯ called the point group of G. Proposition 3.4.9. Any discrete subgroup of O(2) is finite. In particular the point group G¯ of a discrete group is finite. Proof. Let G be a discrete subgroup of O(2). If G contains no rotations, then G is either the group of one element, or the group of two elements G = f1; rg. We may therefore concentrate on G having rotations. Since G is discrete there exists one rotation ρθ with smallest positive angle θ > 0. Then there exists an integer n such that nθ ≥ 2π. Let n be the smallest such integer n. If we do not have equality nθ = 2π then nθ − 2π < θ. It follows that our chosen angle θ was not the smallest after all. In other words it must be that nθ = 2π. It follows that the subgroup of rotations in G are Cn, and that the group G is either Cn or Dn. The last statement follows as the point group G¯ is the image of a discrete group under a homomorphism, which is discrete. Corollary 3.4.10. The point group G¯ of a discrete group is either the cyclic group Cn of order n, or the dihedral group Dn of order 2n. Proof. This follows since these are the finite subgroups of O(2). Lemma 3.4.11. Let G be a discrete group with translation group Λ. Since Λ is a normal subgroup the group G acts on Λ by conjugation, and we have −1 gt~ag = t'(g)(~a); for any g 2 G, and ~a 2 Λ, and where ' is the homomorphism (3.4.6). Proof. Write an element g 2 G as g = tbρθr1, where r1 is either r or the identity. Then, using the relation (3.3) we have that gt~a = tbρθr1t~a = tbρθtr1(~a)r1 = tbtρθ(r1(~a))ρθr1: 55 0 −1 Note that b = ρθ(r1(~a)) is '(g)(~a). We have that g = r1ρ−θt−b, such that −1 gt~ag = tbtb0 ρθr1r1ρ−θt−b = tb0 = t'(g)(~a); proving the claim. Lemma 3.4.12. Let G be a discrete group with translation group Λ and point group G¯. For any g¯ 2 G¯, andy ~a 2 Λ we have that g¯(~a) 2 Λ. In particular we have that G¯ is naturally a subgroup of the symmetry group of Λ. Proof. By definition G¯ is a subgroup of O(2), so for any vector ~a in R2 the g¯(~a) is an element of R2. We need to show that if ~a 2 Λ theng ¯(~a) is in Λ as −1 well. This follows from Lemma 3.4.11 since Λ is normal subgroup and gt~ag is therefore in Λ, for any g 2 G and any t~a in Λ. Proposition 3.4.13 (Crystallographic Restriction). Let G be a discrete group whose translation group Λ is a plane lattice. Then the point group ¯ G is either Cn or Dn, with n = 1; 2; 3; 4 or 6. In fact, any finite group H ⊂ O(2) of symmetries of the plane lattice is one of the ten groups listed above. Proof. Let H ⊂ O(2) be a finite group of symmetries of the plane lattice Λ. It suffices to show that any rotation in H has order 1, 2, 3, 4 or 6. Let ρ be the rotation in H with smallest order θ > 0, and let ~a be a non-zero vector ~ ~ with smallest length.