Lecture 18: Stabilizer

Lecture 18: Geometric Application of Stabilizer 18 18.1 Stabilizer Review A group action is when a group G acts on a set S by G×S −→ S and sends (g, s) 7−→ gs. The orbit of an element in s is all the elements it gets mapped to, Os = {gs ∈ S : g ∈ G}, and the stabilizer is StabG (s) := {g ∈ G : gs = s} ≤ G, 18.2 Counting Formula Figure 1: The partition of S into orbits and theeir corresponding bijections with G/ Stab(si ) One of the facts we learned about orbits is that they partition S. We further learned that there is a bijection between the left cosets of the stabilizer with the orbit. We were then able to write the size of S as X X |S| = |Osi | = [G : Stab(si )]. 18.3 Stabilizers of Products Given a group G acting on a set S and an element s ∈ S, we know that |Os | = [G : Stab(s)]. This also means that we expect that if we take the stabilizer of two elements in the same orbit, then they should be the same size. Specifcally, we are asking what can we say about the stabilizer of a product. If s′ = as for a ∈ G, if g ∈ Stab(s), then gs = s. Then aga−1 (s′ ) = aga−1 (as) = ag(s) = as = s′ . In other words, if g stabilizes s, then aga−1 stabilizes s′ . So the upshot is that StabG (s′ ) = a StabG (s)a−1 . If StabG (s′ ) was normal, then the two stabilizers would be the same, but this doesn’t have to be the case. We’ve provided a nice bijection to see that the sizes of the two stabilizers must be the same size. Lecture 18: Geometric Application of Stabilizer 18.4 Statement Today, we will look at a consequence of these counting formulae. Recall that we were able to study and classify fnite and discrete subgroups of isometries in the plane. The special orthogonal group SO3 is the group of #» rotations ρ(u,θ) in R3 fxing 0 . What are the fnite subgroups G ≤ SO3 ? In fact, there are not so many! Let’s start with the theorem. Theorem 18.1 If G ≤ SO3 , then • G∼ = Cn = ⟨ρu,2π/n ⟩, or • G∼ = Dn = ⟨ρu,2π/n ⟩, or • G is the group of rotational symmetries of a regular polyhedron. Figure 2: The regular polyhedra Although there are 5 regular polyhedra, there are only 3 distinct subgroups of symmetries. The dodecahedron and icosahedron have the same symmetries, which we denote as I. The cube and octahedron have the same symmetries as well, which we denote by O. Finally, the tetrahedron is partnered with itself and we denote its symmetries with T . As an example to see why some of the symmetries are the same, consider the symmetries of the octahedron. We can draw a point on the center of every face in the octahedron. Connecting these points lead to a cube, and thus any rotational symmetry of the cube will give a symmetry of the octahedron, and vice versa. A similar argument can be applied to the dodecahedron and icosahedron. Trying the argument for a tetrahedron just maps to the tetrahedron itself. On Wednesday, we worked out that the group of symmetries of a cube has size 24. Similarly, the tetrahedron has |T | = 12, and |I| = 60. Lecture 18: Geometric Application of Stabilizer We will prove this by studying orbits! An non-identity element g ̸= I ∈ G71 is a rotation and thus fxes two unit vectors, which are exactly the positive and negative unit vectors on the rotation axis of g. These are called the poles of g. Let [ P= {poles of g}. g̸=I Lemma 18.2 If p ∈ P and some g ∈ G, then gp ∈ P also. As a result, we learn that G acts on P. Proof. For p ∈ P, then there exists h = ̸ I ∈ G such that hp = p. If p′ = g(p). Then ghg −1 (p′ ) = p′ by earlier −1 −1 reasoning. Then ghg ∈ G, and ghg = ̸ I since h = ̸ I. thus p′ ∈ P also. Example 18.3 Let G = Cn . Then P = {p, −p} since every rotation will have the same poles. Example 18.4 Let G = O.a Then P = {pole for each vertex, edge, or face}. a The group of symmetries of an octahedron. Now, what can we say about stabilizers of these subgroups. let |G| = N. Let’s decompose P into orbits. Then P = O1 ∪ O2 ∪ · · · ∪ Ok . Then |Oi | = ni , and Oi = Opi for some pole pi . Then by our relations about the number of index of the stabilizer N | Stab(pi )| = ri = . ni Note that the stabilizer group will be a cyclic group. Geometrically, it will just contain the rotations around the axis pi . 18.5 Finding the subgroups Let’s write down an auxiliary set. It is the set of poles and group elements paired together. Let S := {(g, p), g = ̸ I, p is a pole for g}. Then we can count the order of S in two diferent ways. The order of S is |S| = X 2 = 2(N − 1), g∈G g̸=I since there are two poles for every non-identity element of G. Additionally, since we have k orbits, |S| = X | Stab(p)| − 1 = k X ni (ri − 1) = i=1 p∈P k X N i=1 ri (ri − 1). Every pole in the same orbit has the same stabilizer size, so we can group them together. Now, we have that k X N i=1 Dividing by N, 71 We use I for the identity here ri (ri − 1) = 2(N − 1).       1 1 1 2 1− + 1− + ··· + 1 − =2− . r1 r2 rk N Lecture 18: Geometric Application of Stabilizer Each of these quantities 1 − 
So 1 − r11 ∈ 12 , 1 . 1 r1 is between 1/2 and 1, since r1 must be at least two (by the defnition of a pole.) In addition 2 − N2 ∈ [1, 2). So k = 2 or 3, since this is the only way the counting formula works out from our bounds. In fact, this works out for examples 18.3 and 18.4. For G = Cn , we have two orbits, and for G = O we have three orbits, even though it is much more complicated. For k = 2, when there are two orbits, 1− and so 1 1 2 +1− =2− , r2 N r1 1 1 2 + = . r2 N r1 But since r1 , r2 ≤ N, 2 1 1 2 ≤ + = . r2 N N r1 We have r1 = r2 = N, and n1 = n2 = 1. Each of the poles is fxed by the entire group. Then G is a fnite subgroup of SO2 , and thus G = CN , a cyclic subgroup. For the three-orbit case, the numerics of the problem is also extremely constraining. When k = 3, the equation is 1 1 1 2 1− +1− +1− =2− , r2 r3 N r1 and equivalently 1 1 1 2 + + =1+ . r2 r3 N r1 Without loss of generality, let r1 ≤ r2 ≤ r3 . It is necessary for r1 = 2, or else the LHS72 would be ≤ 1. If r2 ≥ 4, then r3 ≥ 4 as well, and again the LHS would be ≤ 1. So r2 = 2 or 3. Finally, if r2 = 3, then r3 cannot be ≥ 6, again from the numerics of the problem. So r3 = 3, 4, or 5. In total, the cases are • Case 1. (2, 2, r) : r = N/2. In this case, we still have an infnite family. • Case 2. (2, 3, 3). We can solve for N to get N = 12. This corresponds to the tetrahedral group T. • Case 3: (2, 3, 4). N = 24. This corresponds to the octahedral group O. • Case 4: (2, 3, 5). N = 60. This corresponds to the icosahedral group I. We can really strongly limit the possibilities of a group G. We got this by counting a set S in two diferent ways, and playing around with the numbers and fractions. We aren’t done; we still have to show that these cases actually correspond to the groups. In each of these cases, we have three orbits, and those correspond to edges, faces, and vertices of these regular polyhedra. The pole for any vertex can be rotated to the pole for any other vertex. We’ll do the argument for the octahedral group and it will be similar for the remaining polyhedra. 18.6 72 left The Octahedral Group hand side Lecture 18: Geometric Application of Stabilizer Example 18.5 (Octahedral group) Let’s take (2, 3, 4) and argue that this group must be symmetries of a cube or an octahedron. For 4, n3 = 6. So the stabilizer has size 4 and the orbit has size 6. Let’s try to fgure out what the orbit looks like. It contains six vectors inside R3 , so we can simply see the possible confgurations. Drawing the frst vector is easy; we just pick wherever we want for p. Then −p has to be in the same orbit as well because StabG (−p) has the same order as StabG (p), but we only have one orbit with stabilizer size 4. What does the stabilizer group look like? Geometrically, it’s rotations around p and must have size 4, so we have Stabg (p) = C4 . Now, let’s try to fgure out the other 4 poles in our orbit. Suppose we draw q such that p and q are not perpendicular. For any q in the orbit, −q is also in the orbit. We have determined that rotations by π/2 around p are in our group, so those rotations of q should also be in our orbit. However, this gives us 4 vectors for rotations of q and 4 vectors for rotations of −q, and this is too many. This picture was wrong because q was not drawn perpendicular to p. If we drew q perpendicular, then rotating q by 90 degrees gives us an orbit of size 6. Whatever the group G is, it fxes the collection of vectors that we drew. In particular, it fxes the octahedron obtained by taking the convex hull of the vectors and thus G ≤ O. But the octahedral group has size 24a and this group also has size 24, so |G| = |O| = 24 implies that G = O. a we worked this out in class on Wednesday We can repeat this for all the cases to fnish the proof, but it isn’t too important. The key lesson is how we were able to use the counting formulas and orbit decomposition of the set to really constrain the possibilities in these ways. A lot of the challenge was just fnding the right set to act on; in this case it was the poles.