Reconstruct cyclic and dihedral rotation groups

Try

A group of six spatial rotations is generated by a turn R through 2π/3 about the vertical axis and a half-turn H about the horizontal x-axis. Their matrices are

R = [[−1/2,−√3/2,0],[√3/2,−1/2,0],[0,0,1]], H=diag(1,−1,−1).

Predict whether H is a reflection of three-dimensional space. It fixes the x-axis but reverses both perpendicular directions, so det H=1: it is a rotation. Restricted to the horizontal plane it does act as a reflection. These two statements concern different spaces.

The preceding pole-count lesson found numerical possibilities for finite subgroups of SO(3). This lesson completes the geometric conjugacy proof for the cyclic case and the triples (2,2,n). It does not classify the three remaining polyhedral triples.

Predict

Throughout, G is a finite subgroup of SO(3), acting on the unit sphere. A pole is a fixed unit vector of a nonidentity group element. The stabilizer Gₚ fixes the vector p, not merely its unoriented line. A nonidentity rotation has exactly the two poles on its axis.

In the triple (2,2,n), the pole-count equation gives |G|=2n and orbit sizes n,n,2. Predict why an orbit consisting of two unit vectors p,q must be antipodal. If their sum were nonzero, what would every element of G do to it?

For n=2 all three orbit sizes are two. Do not select “the unique two-point orbit”: none is distinguished by its size. We must be able to start with any one of the three.

Experiment

For the six-element worked example, use R³=I, H²=I and HRH=R⁻¹. Predict the fixed axes of RH and R²H before checking the table. Products act from right to left.

Each horizontal-axis element is a half-turn. Its two poles have polar angles jπ/3 and jπ/3+π. All six horizontal poles together split into the even and odd multiples of π/3, each orbit of size three. The vertical poles form one orbit of size two, since H exchanges them. Their stabilizers have orders 2,2,3, respectively.

Check the incidence count: the two vertical poles each have two nonidentity stabilizer elements, while the six horizontal poles each have one. The total is 2·2+6·1=10=2(6−1). This confirms the table for this example. The general conjugacy result still needs a proof.

Understand

The cyclic case gives a conjugacy class, not just an abstract type

For a nontrivial finite rotation group with two pole orbits, the earlier count proves that both orbits are singletons: every element fixes each of their points. Choose a pole p. Restriction to p⊥ embeds G in SO(2); a map trivial on both this plane and p is the identity. The finite-circle theorem proves that the restricted subgroup consists exactly of the rotations by multiples of 2π/m for m=|G|.

Choose a unit vector u perpendicular to p and put v=p×u. The frame (u,v,p) is positively oriented and orthonormal. Its matrix Q, with those vectors as columns, lies in SO(3). Thus Q⁻¹GQ is precisely the standard cyclic group generated by the vertical rotation through 2π/m. This proves conjugacy by a spatial rotation. The trivial group is the separate m=1 case. Conversely every such cyclic group has exactly the two common poles when m≥2, so it does belong to this case.

A two-point orbit forces an invariant axis

Now suppose the pole data are (2,2,n), n≥2. The group is not cyclic, since a nontrivial cyclic rotation group has only the two singleton pole orbits just described. Choose an orbit O=❴p,q❵ of size two. For n>2 this is the orbit belonging to stabilizer order n. For n=2 choose any of the three two-point orbits.

Every g permutes p,q, so it fixes p+q. If p+q≠0, every element would fix the unit vector in its direction. The faithful restriction to its perpendicular plane would then make G cyclic, a contradiction. Therefore q=−p. The line ℓ=ℝp is invariant under the whole group, even though its two directions are exchanged by some elements.

Let K=Gₚ. Since gp is either p or −p, the action on O gives a homomorphism G→❴±1❵. It is onto because O has two elements; its kernel is K. Consequently K is normal of index two and |K|=n. Restriction to p⊥ identifies K with the cyclic rotations of order n about ℓ. In an oriented frame it has a generator R that turns through 2π/n about p.

Every element outside the kernel is a perpendicular half-turn

Choose H∈G outside K. It sends p to −p and preserves p⊥. Since its full determinant is 1, the determinant of its restriction to p⊥ is −1. The complete O(2) forms show that this restriction is a reflection. It fixes a unit vector u in p⊥ and negates the perpendicular direction v=p×u. Together with Hp=−p this gives

H u=u, H v=−v, H p=−p.

Thus H is exactly a half-turn about the horizontal axis ℝu, and H²=I. This is a consequence of orientation and the invariant axis, not an assumed presentation relation.

In the positively oriented frame (u,v,p), R and H become

R = [[cos(2π/n),−sin(2π/n),0],[sin(2π/n),cos(2π/n),0],[0,0,1]],

H = diag(1,−1,−1).

Conjugation by H reverses the plane rotation, so HRH=R⁻¹. Every group element lies in K or KH, hence has one of the forms Rʲ or RʲH with 0≤j<n. Each family has n distinct elements; the families are disjoint because they act differently on p. Their product law is (j,e)(k,d)=(j+(−1)ᵉk,e+d), with exponents taken modulo n and 2. These facts establish the entire standard rotational dihedral group of order 2n and its conjugacy to G. They also construct it for every n≥2.

Recover every pole of the model

Nonidentity powers of R have only the vertical poles. The restriction of RʲH to the horizontal plane is a reflection across the line at angle jπ/n; its full spatial action negates the vertical direction. Thus RʲH is a half-turn about that horizontal line. Its poles are precisely the vectors

pₖ=(cos(kπ/n),sin(kπ/n),0), 0≤k<2n.

The generators send pₖ to pₖ₊₂ and p₋ₖ, with indices modulo 2n. Parity is preserved, and repeated addition of 2 reaches every index of the same parity. Therefore there are exactly two horizontal pole orbits, each of size n. Each horizontal pole is fixed only by the identity and its own half-turn: nonidentity vertical rotations fix no horizontal vector, and different horizontal axes have different poles. Its stabilizer therefore has order two. The two vertical poles form one orbit with stabilizer K of order n. Since all nonidentity elements have been listed, this exhausts the poles and proves the claimed data.

For n=2 the model is ❴I,diag(1,−1,−1),diag(−1,1,−1),diag(−1,−1,1)❵. Its three axes are mutually perpendicular and its three pole orbits are the three antipodal coordinate pairs. Each can serve as ℓ in the proof. The resulting group is C₂×C₂; choosing an axis does not assert that the group intrinsically singles it out. For n>2 the orbit of size two is unique among the three pole orbits, so that particular way of selecting ℓ is unambiguous.

The formal n=1 version of the model would consist of identity and one half-turn, hence would be cyclic C₂. Its actual pole data are two singleton orbits, not a triple containing stabilizer order one. This is why the (2,2,n) family starts at n=2.

Distinguish the plane and spatial actions

Restriction of the standard spatial model to its invariant horizontal plane is faithful and gives the ordinary planar dihedral group. Conversely a planar orthogonal matrix A lifts to diag(A,det A) in SO(3). A planar reflection lifts to a half-turn: it reverses the vertical direction as well as one horizontal direction. The abstract groups are isomorphic, but the planar determinant and full spatial determinant answer different questions.

Apply

Change the main axis to p=(1,1,0)/√2 and choose the perpendicular half-turn axis u=(0,0,1). Let R turn through π/2 about p, and let H be the half-turn about u. Build the oriented frame, write both matrices in standard coordinates, and prove the group order. Identify all three pole-orbit sizes without guessing them from a drawing.

Check the tilted-axis group

Take v=p×u=(1,−1,0)/√2 and Q=[u v p]. Its columns are positively oriented orthonormal vectors. Rodrigues’ formula gives R with rows (1/2,1/2,1/√2), (1/2,1/2,−1/√2), (−1/√2,1/√2,0). The other matrix is H=diag(−1,−1,1). Directly, Ru=v, Rv=−u and Rp=p, while H fixes u and negates v,p. Therefore Q⁻¹RQ is the standard vertical quarter-turn and Q⁻¹HQ=diag(1,−1,−1).

The group has eight distinct forms RʲHᵉ, j=0,1,2,3 and e=0,1. Its main poles are ±p with stabilizer order four. The eight horizontal poles are the images under Q of the standard angles kπ/4, and split into two orbits of size four with stabilizers of order two. The pole-orbit sizes are therefore (4,4,2). Tilting the frame changes coordinates but not the conjugacy class or the group order.

Continue

Compare the normal-form proof with dihedral symmetry and the two-to-one lifts in unit quaternions. We have now proved existence and conjugacy classification for the cyclic and (2,2,n) cases, including their small-order boundaries. The triples (2,3,3), (2,3,4) and (2,3,5) still require their own geometric reconstruction and identification arguments; a numerical pole table alone does not supply them.