Lecture 16: Discrete Groups

Lecture 16: Discrete Groups 16 Discrete Groups 16.1 Review Last time, we looked at discrete57 subgroups G ≤ M2 . Then, we looked at a projection π: π ker(π) = (R2 , +) ⊂ M2 − → O2 , tb ◦ A 7→ A; essentially, it gets rid of the translation part of an isometry. We can restrict π to G to get a mapping π|G G −−→ O2 , and we call the kernel, L := ker(π|G ), and it consists of all the translations in G. The image of G in O2 , denoted G := π(G), is called the point group of G. For some element g ∈ G, its image g := π(g) ∈ G only "remembers" the angle of rotation or the slope of the refection line. If G is discrete, it is either Cn or Dn , which we proved earlier. If L ⊆ R2 is discrete, then we obtained three possible cases. (i) L = {0}; (ii) L = Zα where α ̸= 0; (iii) L = Zα + Zβ, where α, β are linearly independent.58 Examples for L and G 16.2 For a given plane fgure, it is actually not difcult to see what L and G are! For the translation subgroup L, since it must either be the identity translation, Zα, or a lattice, it is possible to simply eyeball which translations preserve the fgure. Let’s consider the following plane fgures. Later in this lecture, we will discuss the possibilities for G; it consists of the (untranslated) rotations and refections preserving a fgure. 57 The translations and rotations that cannot be arbitrarily small you look at two vectors and everything you generate from them, it’s called a lattice. 58 When Lecture 16: Discrete Groups Example 16.1 (A) For this frst fgure, say fgure A, the translation subgroup L is a rectangular lattice generated by two translation vectors, to the right and upward. Also, G is D2 , since it contains a refection as well as rotation by π. Example 16.2 (B) For fgure B, the translation subgroup is trivial, consisting of 0. Also, G is C3 , since there cannot be any refections but rotation by 2π/3 or 4π/3 around the center both preserve the fgure. Example 16.3 (C) For fgure C, the translation subgroup is generated by one vector, so L = Zα where α = (1, 0). Also, G is D1 , since there is a refection (corresponding to a glide refection in G) and no rotations possible. Lecture 16: Discrete Groups Example 16.4 (D) For fgure D, the translation subgroup is a triangular lattice generated by two vectors at an angle of π/3 to each other.a The point group is G = D6 , since rotation at a lattice point by any multiple of π/3 preserves the fgure, as well as refection. a Or 16.3 two vectors at an angle of 2π/3. Crystallographic Restriction Now that we have decomposed studying G into studying groups we understand better, L, a subgroup of translations, and G ⊆, the point group, we can actually constrain G further! Recall that • The translation subgroup L ⊆ (R2 , +)59 must be one of three possibilities, which we get from studying discrete subgroups of R2 ; • G must be Cn or Dn , which we get from studying discrete subgroups of O2 . Now that we understand the components L and G separately, we want to use this knowledge to understand G better. Guiding Question How do L and G interact with each other? Example 16.5 Consider our earlier example 16.4. In this case, any element of the point group D6 preserved the triangular lattice. In fact, G acts on L for any discrete group G ⊆ M2 ; this is a very strong constraint on how G and L interact. Theorem 16.6 For the point group G ≤ O2 of some discrete subgroup G of M2 , and the translation subgroup L ⊂ R2 , the group G must map L to itself. For any element A ∈ G and b ∈ L, the image of b under the action of A is b 7→ Ab ∈ L. We already know that O2 and thus G acts on the plane R2 and therefore L. The surprising part is that under the action of any element of G, an element of L is actually mapped to another element in L! Proof. Since A ∈ G, it is the image of an element of G, say t⃗c ◦ A ∈ G for some ⃗c ∈ R2 . Then, ⃗b ∈ L, so t⃗b ∈ G. The key observation in this proof is that L = ker(π|G ) is the kernel of a homomorphism! Thus, the subgroup L ⊴ G is actually normal, so conjugating an element of L by anything in G stays in L. Then for t⃗b ∈ L, (t⃗c ◦ A) · t⃗b · (t⃗c ◦ A)−1 ∈ L also. As isometries in M2 , we know how to manipulate these products, and so expanding out this expression gives us −1 t⃗c · A · t⃗b · A−1 · t⃗−1 · t−⃗c c tA⃗b · A · A c = t⃗ = t⃗c tA⃗b t−⃗c = tA⃗b ∈ L. 59 The translation subgroup L is sometimes written ambiguously in one of two equivalent ways; an element of L can either be the translation t⃗b ∈ L considered as an element in G, or simply the vector ⃗b ∈ L considered as an element in R2 . So L could be considered either as a subgroup of G or of R2 . Lecture 16: Discrete Groups Thus, conjugating t⃗b ∈ L by t⃗c ◦ A gives tA⃗b ∈ L. Using the identifcation of L with R2 , A⃗b ∈ L ⊂ R2 , and so every A ∈ G takes vectors ⃗b in L to other vectors in L, preserving the translation subgroup. Student Question. We’re studying discrete groups, which are groups with the requirement that the translations or rotations can’t be arbitrarily small. Are we also requiring that they have to be groups preserving a given diagram, or can they be any discrete groups of isometries? Answer. Earlier on in this lecture, we saw some examples of discrete groups G that came from the symmetry groups of certain diagrams, but what we are actually doing is simply looking at groups G with the condition that the rotations and translations must be arbitarily small60 , and classifying them; mathematically, there is no requirement that they come from pictures. However, the way that these discrete groups actually show up and the way that we fnd them is by drawing these kinds of pictures; this is one of the main reasons why we care about them! In fact, for every discrete subgroup G ⊆ M2 , there will be some picture that produces the group G as its symmetry group. The pictures in this lecture are mainly so that there are concrete examples to look at and think about. In Section 16.2, each of the examples has a symmetry group G consisting of the isometries of the plane sending the picture to itself.61 For example, in Example 16.2, rotation by 120 degrees preserves the "triangle," while 5 degrees does not, so ρ2π/3 ∈ G, whereas ρπ/36 ∈ / G. Theorem 16.6 states that the point group G, which is a diferent group from G, actually preserves L ⊆ R2 , the translation group. In Example 16.4, L is generated by Z(1, 0)t + Z(1/2, 3/2)t , the two sides of an equilateral triangle, and the point group is D6 . Any element of D6 will send an element of L to a diferent element in L. In fact, when L is a lattice, preservation by some point group G is a strong constraint on the possible angles that show up in the lattice; only certain angles are allowed. Given G, most lattices are not preserved by every element. Thus, the theorem constrains G and L together — not on each of them separately, but on how they interact. The groups that show up this way are often called crystallographic groups. there are only fnitely many. 62 They are well-studied; in fact, Theorem 16.7 (Crystallographic Restriction) Let L ̸= {0}. Then G = Cn or Dn , where n = 1, 2, 3, 4, or 6. a a The theorem name comes from the fact that it restricts the possible crystallographic groups. Although we could imagine that there are lots of possibilities for G and L, the fact that G preserves L constrains the possible point groups to fnitely many, and there are also only certain choices of L allowed for a given n. Proof. The group G is a discrete subgroup of O2 , and so it is Cn or Dn for some integer n. Since L is discrete, there is a (non-unique) shortest nonzero vector α ̸= 0. Consider a rotation ρ = ρθ ∈ G. The result of rotating α by θ is another vector in L, and since rotations are length-preserving, ρθ α is also a vector of shortest length. Since both vectors are in L, ρα − α is also in L.63 If θ is too small, ρα − α will have a shorter length, and there will be a contradiction. In particular, if θ < 2π/6, ρα − α is shorter than α, so θ ≥ 2π/6. Since Cn and Dn contain ρ2π/n , it must be the case that n ≤ 6. 60 These are called discrete groups each point individually is sent to itself; the picture as a whole is sent to an identical copy of itself. 62 Especially when L is a lattice, and there are two diferent directions to translate. 63 Since L is a subgroup, it is closed under addition/subtraction. 61 Not Lecture 16: Discrete Groups A similar argument holds to rule out n = 5. The vector α + ρ4π/5 α will be shorter than any α, which is also a contradiction.64 So n = 1, 2, 3, 4, or 6. Actually, for Cn or Dn where n = 1, 2, 3, 4, or 6, it is possible to constrain the translation subgroups L that can simultaneously show up. For instance, when L is a lattice,65 there are only 17 possible symmetry groups G that can occur. When L is 0, G can be Cn or Dn for any arbitrary n, but allowing nontrivial translations constrains G signifcantly. Student Question. How much does constraining G and L constrain the actual symmetry group G itself ? Answer. Finding G from G and L is precisely the same as fguring out the 17 plane symmetry groups,66 and is precisely the last step! We will do one example now. Let’s consider a specifc group G and try to fgure out what the actual symmetry group G can be! 64 This question is equivalent to the feasibility of tiling the plane with a regular pentagon, and in fact that is not possible! L is a lattice, it is two-dimensional, and it is Z⃗a + Z⃗b for generating vectors ⃗a and ⃗b. It is also possible for L to be Z⃗a, which is one-dimensional. 66 These are called wallpaper groups, since wallpapers are 2-dimensional patterns that usually have nontrivial symmetry groups. 65 When Lecture 16: Discrete Groups Example 16.8 (C4 ) π|G Suppose G = C4 .a Then L ⊂ G −−→ C4 , and the index [G : L] = 4.b Also, ρ = ρπ/2 ∈ G is a generator of G. Where α is some shortest-length vector in L, it’s possible to showc that ρα and α do generate L. Thus, L = Zα + Z(ρα), a square lattice. Also, there exists some rotation ρ ∈ G giving π(ρ) = ρ. Then ρ is in fact a rotation by π/2 around some other point, which we will call the origin.d The group G contains L, of index 4, as well as some rotation by π/2, ρ.e Thus, G is "generated" by L and ρ, and must consist of everything of the form G = {tv ◦ ρi : ⃗v ∈ L, i = 0, 1, 2, 3}. Also, ρtv = tρv ◦ ρ, so the group multiplication can be written down, and G is completely determined by knowing that G was C4 ; this is 1 out of the 17 wallpaper groups! a Rotations by 90 degrees, but no refections. index [G : ker(π|G )] = [G : L] is equal to the size of the image under π|G , which is G = C4 . c There is a more involved argument there, but it is not super relevant here. d In the discussion of the four kinds of isometries in M , the elements which were mapped to rotations were in fact rotations 2 around some point. e The rotation ρ is ρ, lifted to be in G, and it is an element of G not in L which generates the quotient, C . 4 b The Student Question. Can you explain where ρ came from? Why is it a rotation? Answer. By defnition, G is the image of G under π : M2 → O2 taking tb ◦ A 7→ A. Then there are four possibilities for elements in M2 : translation, rotations, refections, and glide refections. The frst two are orientation-preserving, and the last two are orientation-reversing. Refections and glide refections map to refections in O2 67 under π, translations will map to the identity, and rotations will map to rotations (around the origin). So ρ has an image of ρ, which is a rotation, and thus ρ is a rotation around some point. If ρ, the element in G, were a refection instead of a rotation, the preimage in G could have been either a refection or a glide refection, so when the point group G = Dn , one of the dihedral groups, instead of Cn , the analysis is more subtle. In fact, there might not be any refections in G at all. (In Example 16.3, there were no refections, only glide refections.) Example 16.9 If r = π(r) where r, then r = π(tb ◦ rℓ ), where b is some zeroa or nonzerob vector parallel to the line ℓ. Does this mean there are uncountably many possibilities for b and therefore r? In fact, b is constrained a bit more: tb rℓ tb rℓ = t2b , so 2b ∈ L. Thus, there are two possible situations: • The vector is in the lattice: b ∈ L; • The vector b is halfway between two lattice points, as in Example 16.1. a refection b glide refection From these two examples, we see that given some G, of which there are fnitely many, and working through the information that is present, there aren’t too many possibilities for G, and in fact there are fnitely many — 17 67 Refections across lines through the origin Lecture 16: Discrete Groups in total.