Construct the three polyhedral rotation groups

Try

The pole-count argument gives possible orders for finite spatial rotation groups. It does not construct the solids or certify their faces. Here we construct three convex hulls from coordinates and prove that their rotation groups have orders 12,24 and 60.

For a finite set V of unit vectors, its convex hull consists of all convex combinations of its vertices. A supporting plane n·x=h has every vertex on the side n·x≤h. If exactly three noncollinear vertices attain h, their triangle is a face. A rotation preserving V also preserves its convex hull and its faces.

Begin with the four vectors (1,1,1), (1,−1,−1), (−1,1,−1), (−1,−1,1), divided by √3. Compute the dot product of two different vectors. Then compare the six vectors ±e₁,±e₂,±e₃. Which pairs can be edges, and which pairs are antipodal? We use exact support planes to answer this, rather than relying on a perspective sketch.

Predict

A rotation fixing two nonparallel vectors is the identity: it also fixes their cross product, and these three vectors form a basis. Consequently, once the image of an ordered adjacent pair of vertices is prescribed, there is at most one rotation. If a polyhedron has E edges, its vertex-preserving rotation group has at most 2E elements, because there are 2E possible images of one ordered edge.

Predict the bound for a tetrahedron with six edges and an octahedron with twelve. To achieve each bound we must exhibit enough actual rotations. Counting possible target pairs alone is not an existence proof.

We reuse spatial axes and Rodrigues rotations, including the distinction between an orientation-preserving rotation and an orthogonal reflection.

Experiment

Put φ=(1+√5)/2, so φ²=φ+1. Form twelve raw vectors by cyclically permuting (0,±1,±φ), with the two signs independent. Divide them all by R=√(φ+2) to get unit vertices. We work with raw vectors for simpler arithmetic; normalization divides every dot product by φ+2 and every length by R.

Choose v=(0,1,φ). Test the following list: these are exactly its five vertices at raw distance 2. All other distinct vertices are their negatives and −v.

Their cyclic order is A,D,B,C,E. In that order neighboring raw dot products are φ and nonneighboring ones are −φ. Verify A·D=φ and A·B=−φ yourself. Projecting perpendicular to v subtracts the same component φv/(φ+2) from all five points.

Predict why rotating this pentagon by one fifth of a turn also permutes the other seven vertices of the solid. A local symmetry would not automatically extend if those other vertices were arbitrary.

Understand

Tetrahedron and octahedron certificates. The tetrahedral vectors have pairwise dot product −1/3. For any omitted vertex w, the other three satisfy w·x=−1/3 while w·w=1. Thus the plane with normal −w supports exactly that triple. These are the four faces and their six edges. For completeness, the four vectors are affinely independent, so every point in their affine space has unique barycentric coordinates of sum one; the four face inequalities are precisely nonnegativity of those coordinates. No other face can be missing from this simplex description.

For the octahedron V=❴±e₁,±e₂,±e₃❵, every sign choice s=(s₁,s₂,s₃), with each sᵢ=±1, gives the supporting plane s·x=1. Exactly s₁e₁,s₂e₂,s₃e₃ lie on it. The convex hull is exactly |x₁|+|x₂|+|x₃|≤1: a point satisfying this is a convex combination of signed coordinate vectors, using a cancelling pair to supply any remaining weight. Its eight sign inequalities therefore describe all eight triangular faces. Edges join precisely nonantipodal vertices from distinct coordinate axes, giving twelve edges.

Even coordinate sign changes (product of signs +1) combined with cyclic coordinate permutations preserve the tetrahedral set. They give 4·3=12 distinct determinant-one matrices. The ordered-edge bound gives at most 12, so these are its entire rotation group. For the octahedron every signed permutation matrix preserves V; exactly half of the 3!·2³ matrices have determinant +1, giving 24 rotations. The twelve-edge bound gives at most 24. Thus its rotation group has exactly 24 elements.

The icosahedral neighbor pentagon. Every raw vertex has squared norm φ+2. For distinct vertices the possible dot products are φ,−φ and −(φ+2); the last occurs only at antipodes. The four even sign changes and three cyclic coordinate permutations are twelve distinct rotations preserving the set. They act transitively: cyclic permutation selects the zero coordinate, and even sign changes independently select the two nonzero signs, with the unused sign fixed by their product. It suffices therefore to check neighbors at v, as in the table. Each vertex has five raw-distance-two neighbors, giving thirty candidate edges. Their unit-vector dot product is φ/(φ+2)=1/√5; this is the edge value needed in reconstruction.

For the five projected neighbors the squared radius is

ρ² = (φ+2) − φ²/(φ+2).

The cosine between consecutive projected vectors is [φ−φ²/(φ+2)]/ρ²=(φ−1)/2=cos(2π/5). Their oriented successive angles have the same positive sign: the raw triple products v·(A×D), v·(D×B), v·(B×C), v·(C×E), v·(E×A) all equal 2φ²>0. They are therefore five successive angles 2π/5, not an unordered equal-distance assertion. This proves the regular pentagon. The rotation about v through 2π/5 permutes these neighbors, fixes v and −v, and permutes their five negatives. These are all twelve vertices, so it is a symmetry of the whole set. The remaining vertices being exactly these negatives is essential.

Certify the edges and faces. The representative triangle v,A,D has all three mutual raw dot products φ. Its normal n=v+A+D=(φ,0,2φ+1) gives value h=3φ+2 on its three vertices. On the remaining nine vertices the values are φ three times, −φ three times, and −h three times. Hence n·x=h is a supporting plane containing exactly this triangle. Normalizing the vertices simply divides these values by R. Transitivity on vertices and the fifth-turn stabilizer carry this certificate to every triangle consisting of a vertex and two consecutive neighbors.

Each neighbor list is a five-cycle, so an adjacent pair has exactly two common neighbors and lies in exactly two of these triangles. There are 12·5/3=20 distinct certified triangular faces and thirty distinct edges. To see that no additional facets have been overlooked, use the elementary boundary fact that the facets of a three-dimensional convex polytope are connected through shared edges. One proof radially projects the boundary from an interior point onto a sphere and joins interior points of two facets by a path avoiding the finitely many vertices and crossing edges transversely; the crossed facets form such a chain. Here the origin is interior because the spanning vertex set is antipodally symmetric. Every edge of every certified triangle already belongs to its two certified facets, and a convex polytope edge has exactly two incident facets (intersect by a small transverse plane). Thus a chain cannot leave the certified collection. It contains every facet. This proves the convex hull has exactly the stated faces and edges, rather than guessing them from a count.

Count the rotations. The vertex-transitive twelve-element subgroup and the fifth-turn symmetry show that the full rotation group has at least 12·5=60 elements by orbit–stabilizer. It is finite since its faithful action on a spanning finite vertex set embeds it into a finite permutation group. Any rotation sends an ordered edge to an ordered edge. There are 2·30=60 such targets, and two independent vertex vectors determine a rotation uniquely. This supplies the matching upper bound, so the group order is exactly 60. The argument proves construction and order, not yet an identification with an abstract alternating group.

Apply

Replace the candidate triangle v,A,D by v,A,B. Decide whether these vertices form one of the triangular faces by checking their dot products and a supporting-plane candidate. Then reverse just the first coordinate in each tetrahedral vertex. Does this operation preserve its vertex set, and can it count toward the rotation group?

Check the failed candidates

A·B=−φ, so A and B are not consecutive neighbors of v. In particular their pair is not an edge of the certified convex hull. More directly, let m=v+A+B=(1,φ,2φ). Its value on v is 3φ+2, while its values on A and B are φ+2. Thus this sum-normal does not even place the three vertices in one supporting plane. The missing adjacency is the decisive obstruction; failing a single guessed normal by itself would not prove absence of a face.

For the tetrahedron, each raw vertex has coordinate product +1. Reversing one coordinate changes it to −1, outside the original vertex set; the matrix also has determinant −1. It fails both the set-preservation and rotation tests. Reversing two coordinates preserves the product and has determinant +1, so it is one of the twelve exhibited tetrahedral rotations.

Continue

Compare these constructions with the pole-count restrictions. The supporting planes establish the solids; explicit rotations and independent upper bounds establish their group orders. An abstract identification such as A₄, S₄ or A₅ needs a further faithful action or group-theoretic argument. Keep that next obligation separate from the exact coordinate certificates proved here.

Continue to reconstruct polyhedra from rotation orbits to prove that the exceptional pole-count cases are conjugate to these models.