Orthogonal and special orthogonal groups

 O(2,R) and O(3,R) have as normal subgroups of index 2 the special orthogonal groups, denoted SO(2,R) and SO(3,R), consisting of the linear transformations whose determinant is 1. We have seen that the symmetry groups of the regular polygons are the dihedral groups, represented by finite subgroups of O(2,R) and the rigid motion groups are finite cyclic groups represented by the corresponding subgroups of SO(2,R). We also saw that the symmetry groups of the square, cube and octahedron are represented by finite subgroups of O(3,R), with the rigid motion groups corresponding to the corresponding subgroups of SO(3,R). Of course the icosohedral group fits into this pattern too, but we haven't seen the proof yet.

 We shall now characterise all the finite subgroups of the orthogonal and special orthogonal groups.

 Let A in O(2,R) = 

If det A = 1 then a² + c² = 1, while if det A = -1 then
a² - c² = 1.

Case 1: det A = 1 Since -1 < or = a < or = 1, let a = arccos a. Then a = cos a, c = sin a and hence A =
 

and A is rotation anti-clockwise by a. If a = 1, then A is the identity, while if a = -1 then A is rotation by p, which is just reflection in the origin.

 Thus every element of SO(2,R) is a rotation, so every finite subgroup of SO(2,R) is cyclic, generated by a rotation by 2p/n for some n, and hence is the group of rigid motions of a regular polygon.

 Case 2: det A = -1 Let a = arccos a; then
A =

Then A is reflection in the line through the origin of angle a/2, and this line is fixed by A. If a =1, the line of reflection is the X-axis, if a = 0 it is the line y = x, and if a = -1 it is the Y-axis.

 Note that the product of two such reflections is a rotation, the product of a rotation and a reflection, in either order, is a reflection. Thus the finite subgroups of O(2,R) that are not cyclic are the reflections of order 2 and the dihedral groups D_n of order 2n, so symmetry groups of regular polygons.

 We now analyse O(3,R) the same way. Let 
     A =

be orthogonal. Look for eigenvalues. Since 
det(x I - A) is a cubic in x, it always has a real root l. So R³ has a vector v with orthogonal complement U with basis {u, w} such that with respect to basis {v,u,w}, A has matrix
 

Since A is orthogonal, |l| = 1 and the lower corner is an orthonormal 2x2 matrix with |det| = 1. In particular, vA = v or -v.

Case 1 (a) det A = 1 and l = 1. Then the lower corner is a rotation in the plane U, so A is a rotation with axis v.

 Case 1 (b) det A = 1 and l = -1. Then the lower corner has det -1, so is a reflection in a line in the plane U. Choose a basis {u,w} for U so that u is the unit vector along the line of reflection, which is fixed by A. Then {v,u,w} is a basis for R³ with respect to which A is the diagonal matrix with eigenvalues {-1,1,-1}, so A is rotation by p with axis u.

 To summarise Case 1: if det A = 1, then A is rotation about a line through the origin in R³, and every such rotation is a symmetry of R³.

 Case 2 (a) det A = -1 and l = 1, so the lower corner has determinant -1. Then A fixes v and is reflection about a line in U. Just as in Case 1 (b), we find a basis with respect to which A is the diagonal matrix with eigenvalues {1,1,-1}, i.e. a reflection in the plane spanned by the first two vectors.

 Case 2 (b) det A = -1 and l = -1. Then the lower corner is orthogonal 2x2 with determinant 1. So A is reflection in the plane U followed by a rotation in this plane. Such a symmetry is called a rotatory reflection.

 To summarise Case 2, if detA = -1, then A is a reflection or a rotatory reflection in a plane, and every such map is an indirect symmetry of R³.

 An unexpected and interesting consequence of this is that the product of two reflections in possibly different planes, as well as the product of two rotations around different axes is a rotation about an axis.
 

Finite subgroups of O(2,R) and O(3,R)

 If G is a finite subgroup of O(2,R) which is not SO(2,R), then G contains a subgroup of SO(2,R) of index 2, so has even order say 2n. Thus G contains all the reflections of the regular n-gon and hence is the dihedral group D_n or just a reflection if n = 1.

 Now let G be a subgroup of order N of SO(3,R). Then G is a group of rotations (about possibly different axes) that we may consider as acting on the unit sphere. Each rotation in G has a pair {p,-p} of poles. The set G_p of elements of G which fix p is a group of rotations of the equator with normal p, so is cyclic of order m_p say.

 Let a be a generator of G_p, and let f be any element of G. Then p f is a pole of the rotation f^-1a f Thus G acts on the set of poles of its rotations. Hence by  Lemma 4.3, the orbit of any pole p is a set of poles in 1-1 correspondence with the cosets [G:G_p]. Thus the number of poles in this orbit is N/m_p.

 Now every pole p is the pole of exactly m_p - 1 non-identity elements of G, namely the powers of a. Also, if q is in the orbit of p, say q = p f, then G_q = f^-1G_pf, which is isomorphic to G_p by Lemma 6.3, so we have m_q = m_p for all q in the orbit of p.

 So let us count the number K of pairs (a,p) consisting of a a non-identity element of G and p a pole of a. We count this set in two ways: 

1. There are N-1 possibilities for a, and each has two fixed points, so K = 2(N-1).
2. For each pole p, G has m_p - 1 elements fixing p, so the contribution to K from the orbit of p is 
3. (m_p - 1) N/m_p.

Case 1: 2 orbits The equation may be written as 
2/N = 1/m + 1/m', where each of m and m' divides N. Hence the only possibility is m = m' = N. There is only one pole in each orbit and m and m' are a pair of antipodal points and G is cyclic of order N.

Case 2: 3 orbits The equation may be written as 1+2/N = 1/m +1/m'+1/m'', and we can assume 
m <= m'<= m''. If m is at least 3, we have 
1+2/N <= 1, a contradiction, so m=2. Then the equation can be written 1/2 + 2/N = 1/m' + 1/m''. If m' = 2, this equation is satisfied by m'' = k and N = 2k for all k > 1. If m' > 2, then m' = 3, otherwise we get 2/N < 1, a contradiction. So we finally get 1/m'' = 1/2 - 1/3 + 2/N, so m''< 6. The possibilities are 3, 4 and 5. Now we work out the corresponding values of N, giving the following table of possibilities in the 3-orbit case> Suppose the size of the orbits are p, q and r:
 
 

We already know that rotation groups with these parameters exist:

 (a) dihedral group D_k
 (b) tetrahedral group A₄
 (c) octahedral group O₃
 (d) icosohedral group Icos .

 Of course it is still conceivable that there is more than one subgroup of SO (3,R) with the same parameters. This is not the case, but it is too boring to prove here. See for example Neumann, Stoy and Thompson p180. 
 

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 Last update Nov 24 1999

Author: Phill Schultz, schultz@maths.uwa.edu.au