Lecture 15: Finite and Discrete Subgroups

Lecture 15: Finite and Discrete Subgroups, Continued 15 15.1 Finite and Discrete Subgroups Review Last time, we began studying certain subgroups of M2 . The group of isometries of R2 is precisely #» 2 M2 = {t #» b ◦ A : b ∈ R , A ∈ O2 }, where O2 is the group of orthogonal matrices. Guiding Question What are the fnite subgroups of O2 ?a a The discrete subgroups of O turn out to be the same as the fnite subgroups, either C or D (we omit the proof, as it n n 2 is in the homework.) One way in which subgroups of M2 naturally arise is with symmetries of plane fgures. Example 15.1 For the following two plane fgures, they both have discrete symmetries including translations, rotations, and glide refections. Last time, we looked at fnite subgroups of the orthogonal matrices G ⊆ O2 . We found the following theorem which greatly restricts the possibilities for such subgroups: Theorem 15.2 Any fnite subgroup G ⊆ O2 is either • G∼ = Cn = ⟨ρ2π/n ⟩, the cyclic group generated by a rotation by 2π/n; or • G∼ = Dn = ⟨ρ2π/n , r⟩ which is the group Cn with an extra refection r. The elements of the form ρ2π/n , which are rotations by 2π/n, are orientation-preserving, while elements of the form ρ2π/n r, which are refections over certain lines through the origin, are orientation-reversing. 15.2 Finite Subgroups of M2 Now that we have found the fnite and discrete subgroups of O2 , we bring our attention to fnite subgroups G ⊆ M2 . Guiding Question What are the fnite subgroups of M2 ? Do we get more subgroups now that we have more elements? In fact, there are no new fnite subgroups obtained from allowing G to be in M2 instead of O2 . Theorem 15.3 Any fnite subgroup G ⊆ M2 is also isomorphic to Cn or Dn . Lecture 15: Finite and Discrete Subgroups, Continued Proof. In order to show that G is isomorphic to Cn or Dn , it is enough to fnd s0 ∈ R2 such that g(s0 ) = s0 for all g ∈ G. Then, by changing coordinates such that s0 is the new origin53 , G fxes the origin (formerly s0 ) and so G ⊆ O2 . As a result, by applying Theorem 15.2, G must in fact be isomorphic to Cn or Dn . • Step 1. First, we fnd some fnite set S fxed by every element g: we require that gS = S for all g ∈ G. For any p ∈ R2 , let S = {g(p) ∈ R2 : g ∈ G}54 . Then, for any element s ∈ S, it is equal to s = g ′ (p) for some g ′ ∈ G, by the defnition of S. In addition, for any g ∈ G, the action of g on s is g(s) = g(g ′ (p′ )) = (gg ′ )(p) ∈ S, again by how S is defned. So gS = S. • Step 2. Intuitively, to fnd s0 , we would take the average, or the center of mass, of all the points. For example, for the set of rotations ⟨2π/3⟩, S would be 3 equidistant points, and the center of the equilateral triangle would be fxed by such rotations. From this intuition, we can apply the following averaging trick. This is where G being fnite is required, as we wouldn’t be able to take the average otherwise. Where S = {s1 , · · · , sn }, let 1 (s1 + · · · + sn ) n be the average of all the elements in S. For any isometry f = tb ◦ A,   1 (As1 + · · · + Asn ) f (s0 ) = tb n 1 = ((As1 + b) + · · · + (Asn + b)) n 1 = (f (s1 ) + · · · + f (sn )), n s0 = since A is a linear operator. As a result, for any g ∈ G, 1 (g(s1 ) + · · · + g(sn )) n 1 = (s1 + · · · + sn ) n = s0 , g(s0 ) = since g permutes the elements in S. So we see that G does fx s0 , and by changing coordinates so that s0 is the origin, G must in fact be isomorphic to Cn or Dn . 15.3 Discrete Subgroups of M2 No new fnite subgroups are obtained by taking elements in M2 instead of O2 ; what if we take discrete subgroups55 instead of fnite subgroups? Guiding Question What about discrete subgroups of M2 ? The defnition of discreteness in M2 combines the two defnitions for the rotations and translations. 53 We take t−s0 Gts0 is called the orbit of p, since it is all the points that p can reach by some transformation in G, or all the points that p orbits to. 55 We will formalize the notion of discreteness in M now! 2 54 This Lecture 15: Finite and Discrete Subgroups, Continued Defnition 15.4 A group G ⊆ M2 is discrete if there exists some ε > 0 such that any translation in G has distance ≥ ε and any rotation in G has angle ≥ ε.a a In fact, for discreteness, it would make more sense to require two diferent ε and ε for translations and rotations, just 1 2 to ensure that there are not continuously many translations and rotations. In this case, we can simply acquire the ε for this defnition by taking the minimum of the two; then any translation in G has distance ≥ ε1 ≥ ε and any rotation has angle ≥ ε2 ≥ ε. 15.3.1 Discrete Subgroups of R2 As a warmup, let’s consider the copy of the plane inside M2 , (R2 , +) ⊆ M2 , consisting of the translations tb . What are the discrete subgroups of (R2 , +)? The result and argument is similar to the discrete subgroups of (R, +) that we covered last week. Theorem 15.5 If G ⊆ R2 is discrete, then 1. G = {0}; or 2. there exists some α ⃗ ∈ R2 such that G = Zα ⃗ ; or 3. there exist linearly independent vectors ⃗a, ⃗b ∈ R2 such that G = Z⃗a + Z⃗b. This is called a lattice inside R2 . Proof. First pick any α̂ ̸= 0 ∈ G. The intersection G ∩ Rα̂ must be discrete, so there is some smallest length vector in G ∩ Rα̂; call it α. Then if G ∩ Rα̂ = G, then G ∩ Rα̂ = Zα, and we are done. Otherwise, pick β ∈ G such that β ∈ / Rα, minimizing the distance from β to Rα. There exists such a β because in any bounded region of R2 , there can only be fnitely many points of G; then we can simply pick the point in G closest to Rα. Claim: G = Zα + Zβ. If this were not true, then there would exist a point γ ∈ G that is not on the lattice formed by α and β. Thus, by shifting by α and β, the parallelogram with sides α and β would contain a point closer to Rα, violating the minimality of β. Lecture 15: Finite and Discrete Subgroups, Continued 15.3.2 Back to Discrete Subgroups of M2 ! Now that we have considered the translations in M2 , which are isomorphic to the plane R2 , we can move on to the entire M2 . Guiding Question How can we study discrete groups G ⊆ M2 ? Recall that there exists a projection π from M2 to O2 , where R2 , the group of translations, is the kernel. The projection takes π ker(π) = R2 ,→ 56 M2 − → O2 t #» b ◦ A 7→ A. The restriction of π to G takes π|G : G −→ O2 . The kernel L = ker(π|G ) consists of the translations in G. Under this map, the image of G is a subgroup G := π(G) ⊆ O2 , known as the point group of G. The projection takes π|G ker(π|G ) = L ⊆ G −−→ G. Example 15.6 For this infnite plane fgure, the group of translations L in the symmetry group G is a rectangular lattice. #» The point group G contains rotation by π around 0 and refection across ℓ; as a result, G is isomorphic to D2 . As we can see in the example, by using the projection π, each G can be decomposed into a discrete point group G isomorphic to Cn or Dn , and a discrete group L ⊆ R2 , classifed in Theorem 15.5. In fact, we can constrain the possibilities even more! The following proposition is a start. Proposition 15.7 Every A ∈ G maps L to L. Proof. Next time!