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3.5 Finite rotation groups
Let g ∈ SO(3) be a non-trivial rotation. Then g is the rotation around a unique line in R^3. The two intersection points of and the unit sphere S = {x^2 + y^2 + z^2 = 1} ⊂ R^3 we refer to as poles.
Lemma 3.5.1. Let G be a non-trivial subgroup of SO(3), and let X be the set of poles arising from g ∈ G, with g 6 = id. Then G acts on the set X.
Proof. Let P be a pole, and g ∈ G a group element having P as a pole. Let h be an element of G. Then the action is natural, and we have that h.P = h(P ) is an element in R^3 , and in fact on the unit sphere. We need only to show that h(P ) is a pole, that is we need to see that h(P ) is fixed by a non-trivial element of G. Consider the element hgh−^1. Then hgh−^1 (h(P )) = h(P ), and since g 6 = id we have that hgh−^1 is a non-trivial element.
Proposition 3.5.2. Let G be a non-trivial, finite subgroup of SO(3). Let m be the number orbits |X/G|, where X is the set of poles. Let ri denote the order of the stabilizer group of an element Pi in one orbit Bi (where i = 1,... , m). Then we have the formula
|G|
∑^ m
i=
ri
Moreover, the number of orbits m ≤ 3 is atmost three.
Proof. Let P be a pole. The stabilizer group of that point P is all rotations g ∈ G around the line defining P. This is a cyclic group of order r. The opposite pole has the same stabilizator group. Moreover, each non- trivial element g ∈ G is a rotation about one unique line, and produce two poles. By removing the identity element from that observation we obtain the equality (^) ∑
P ∈X
(rp − 1) = 2(|G| − 1).
Let the orbits of X be B 1 ,... , Bm. The cardinality of each orbit |Bi| = ni. The order r of the stabilizator group Cr of a point P is independent of the chosen point in orbit containing P. Therefore the above equality becomes
∑^ m
i=
(ri − 1)ni = 2|G| − 2.
We have from the Orbit Formula (2.1.18) that niri = |G|, for each i = 1 ,... , m. Using this when dividing the above formula with |G| establishes the famous formula of the proposition.
The last stament follows as the stabilizer GP contains at least another element different from the identity, so the summand to the right is at least 1/2 whereas the sum to the left is strictly less than 2.
Corollary 3.5.3. Let r = (r 1 ,... , rm) be the order of the stabilizer groups where r 1 ≤ · · · ≤ rm, where m is the number of orbits X/G. Then the only possibilites that can occur are
- Two orbits, and r = (N, N ), where |G| = N ≥ 2.
- Three orbits, and r = (2, 2 , N ), where |G| = 2N ≥ 8.
- Three orbits, and r = (2, 3 , 3), where |G| = 12.
- Three orbits, and r = (2, 3 , 4), where |G| = 24.
- Three orbits, and r = (2, 3 , 5), where |G| = 60.
Proof. Let |G| = N. One orbit m = 1 is impossible since 2 − 2 /N ≥ 1 whereas 1 − 1 /r 1 < 1. With m = 2 the formula can be written as
2 /N = 1/r 1 + 1/r 2 ,
which is only true with r 1 = r 2 = N , proving the first case. With three orbits m = 3 we rewrite the formula as
2 /N = 1/r 1 + 1/r 2 + 1/r 3 − 1.
If r 1 ≥ 3 then the right hand side will not sum to a strictly positive integer as the left hand side is. Thus r 1 = 2. If also r 2 = 2 then r 3 = N/2, and since each ri ≥ 2 we have proven the second case. If r 1 = 2 then we note that r 2 ≥ 4 together with r 3 ≥ 4 is impossible since the right hand side sum needs to be strictly positive. Similarily, as 1/2 + 1/3 + 1/ 6 − 1 = 0, we are only left with three possibilities with
r = (2, 3 , 3), r = (2, 3 , 4) and r = (2, 3 , 5).
The group order N = |G| is in this three cases 12, 24 and 60, respectively.
Lemma 3.5.4. Let G be a finite subgroup of SO(3). Let Q be a pole, let H be the stabilizator group of Q. Let P be a pole defined by another line that defines Q. Then the orbit of P under the action of H consists of |H| poles.
Proof. By assumption the line defining P is different from the line′^ defining Q. Hence the intersection of the stabilizator groups GP ∩ H is the trivial group. The result then follows from the Orbit Formula 2.1.18.
In the case with (2, 3 , 5). The twelve points forming the orbit B 3 have all a regular pentagon of points in B 2 attached to it. The 20 points forming orbit B 2 have all regular triangles attached to it. The points in B 2 form the vertices of a dodecahedron, and the 12 points in B 3 form the vertices of an icosahedron. In these cases we get two platonic solids inscribed in the unit sphere. The symmetry group of one object equals the symmetry group of the other.
Corollary 3.5.6. The only regular polyhedra in three space are the Platonic solids; the tetraheder, the cube, the octaheder, the dodecaheder, and the icosa- heder.
Exercises. Finite rotation groups
- The twelve points (± 1 , ±α, 0), (0, ± 1 , ±α), (±α, 0 , ±1) fomr the ver- tices of a regular icosahedron if α is suitably chosen.
(a) Verify this and determine α. (b) Determine the matrix of rotation with angle 2π/5, about the line containing the point (1, α, 0).
- Prove Corollary 3.5.6 directly, with elementary methods.
