Find rotation axes and constrain finite spatial groups

Try

A spatial rotation can fix a line while moving every point off that line. Work in oriented Euclidean space ℝ³ with the standard right-handed basis. Consider

R = [[0,−1,0],[1,0,0],[0,0,1]].

Apply R to (1,0,0), (0,1,0), and (0,0,1). The first two vectors turn through a positive quarter-turn in the horizontal plane; the third stays fixed. Predict the image of (2,−1,3), then check directly: it is (1,2,3). The vertical coordinate is unchanged, but this does not mean the entire map is the identity.

We will prove that every nonidentity orientation-preserving orthogonal map in three dimensions has this axis–plane description. Then we will count the special points fixed by rotations in a finite group. Counting will constrain possible groups; it will not by itself construct them or prove their uniqueness.

Predict

Write SO(3) for the real matrices R satisfying RᵀR=I and det R=1. Orthogonality preserves lengths and inner products; the determinant condition preserves orientation. A pole of a nonidentity rotation is one of its two fixed points on the unit sphere. An axis is an unoriented line through the origin, so its two unit vectors name the same axis.

Predict whether a half-turn has one fixed unit vector or two. Also predict whether the identity has a unique axis. These boundary cases matter: the identity fixes the entire sphere, whereas a half-turn fixes exactly the two poles of its axis.

For a finite group G contained in SO(3), a point stabilizer Gₓ consists of all rotations in G fixing a specified unit vector x. Distinguish this from preserving the line through x: a rotation might exchange x and −x instead of fixing either one.

Experiment

For a unit vector u and an angle θ, separate a vector x into its parallel part (u·x)u and its perpendicular part x−(u·x)u. The dot denotes the Euclidean inner product. Our cross-product convention is e₁×e₂=e₃; the positive quarter-turn on the plane perpendicular to u sends v to u×v.

Use these operations to calculate the following images before checking the table. Every row uses the same oriented coordinate system.

For the third row, the parallel part is (1/2,1/2,0). A half-turn keeps that part and negates the perpendicular part, giving 2(u·x)u−x. The resulting matrix is [[0,1,0],[1,0,0],[0,0,−1]]. Its determinant is 1, despite the negative last entry.

Change the axis and check your calculation

A half-turn about (1,0,1)/√2 has matrix [[0,0,1],[0,−1,0],[1,0,0]]. It sends (1,2,3) to (3,−2,1). Its eigenvalue 1 space is exactly the line spanned by (1,0,1), and its perpendicular plane has eigenvalue −1. Do not identify the axis by looking only for a coordinate that stays unchanged.

Understand

An axis exists and is unique except for the identity

A real cubic characteristic polynomial has a real root, so a real 3-by-3 matrix has a real eigenvalue. If R is orthogonal, a real eigenvalue must be 1 or −1, because an eigenvector keeps its length. Nonreal eigenvalues occur in conjugate pairs and have absolute value 1: on the complexified space, the same real orthogonal matrix is unitary. Such a pair has product 1. Since det R=1, the remaining eigenvalue is 1. If instead all eigenvalues are real, their product is 1 and each is ±1, so at least one is 1. Thus choose a unit vector u with Ru=u.

The plane u⊥ is invariant: for x perpendicular to u, (Rx)·u=x·Rᵀu=x·u=0. In an orthonormal basis of that plane its restriction is an orthogonal 2-by-2 matrix with determinant 1, hence a planar rotation. If R fixes two independent vectors, it fixes their span; its orthogonal one-dimensional complement is invariant and its determinant there must also be 1. It therefore fixes the complement and is the identity. Consequently a nonidentity R has a one-dimensional fixed space, and exactly two poles. This argument includes angle π.

Deriving Rodrigues’ formula with a declared orientation

Choose a unit v perpendicular to u and put w=u×v. Then (v,w,u) is positively oriented and (v,w) is an orthonormal basis of u⊥. For some θ modulo 2π, Rv=cos θ v+sin θ w and Rw=−sin θ v+cos θ w. The linear map x↦u×x has the same positive-quarter-turn action on this plane and is zero on u. Apply the planar formula to the perpendicular part and leave the parallel part fixed. The result is

R x = cos θ x + (1−cos θ)(u·x)u + sin θ (u×x).

This is Rodrigues’ formula, derived from the two-dimensional action rather than assumed. Conversely this formula fixes u and is a planar rotation on u⊥, so it defines an element of SO(3). The identity admits every axis with θ=0 modulo 2π. A nonidentity rotation has a unique unoriented axis, but replacing u by −u replaces θ by −θ. For a half-turn these angles agree modulo 2π, so the formula does not distinguish the two orientations of its axis.

Let Kᵤx=u×x. The vector triple-product identity gives Kᵤ²x=(u·x)u−x and Kᵤ³=−Kᵤ. Grouping the absolutely convergent exponential series yields exp(θKᵤ)=I+sin θ Kᵤ+(1−cos θ)Kᵤ², the same formula. This is the three-dimensional specialization of the existing real skew-adjoint rotation-block proof, not a new assumption about matrix exponentials.

Point stabilizers are cyclic

If every element of a subgroup fixes a unit x, restriction to x⊥ embeds it in SO(2): an element restricting to the identity fixes all of ℝ³. Every finite subgroup of planar rotations is cyclic. To see this, for a nontrivial subgroup choose its smallest positive angle α in [0,2π). Dividing any other angle by α leaves a remainder in [0,α); closure under composition and inverse puts that remainder in the group, so minimality makes it zero. Dividing 2π by α gives the same conclusion because the negative of a multiple of α is also a group angle. Hence α=2π/m and all angles are its multiples. The trivial subgroup is cyclic as well.

Count poles in two ways

Now assume G is finite and nontrivial, and put N=|G|≥2. Let P be the finite union of the poles of all its nonidentity elements. It is G-invariant: if gx=x, then hgh⁻¹ fixes hx. Write its distinct G-orbits as O₁,…,Oₖ, choose a representative xᵢ in each, and put mᵢ=|Gₓᵢ|. Since xᵢ is a pole, mᵢ≥2. Orbit–stabilizer gives |Oᵢ|=N/mᵢ.

Count pairs (g,x) with g≠identity and gx=x. Each g contributes two poles, giving 2(N−1). Each x in Oᵢ contributes mᵢ−1 elements, giving

Σᵢ (N/mᵢ)(mᵢ−1)=2(N−1), hence Σᵢ(1−1/mᵢ)=2−2/N.

No pole of the identity was included. For the trivial group P is empty and k=0; treat that case separately instead of imposing mᵢ≥2 on an invented orbit.

Exhaust the numerical possibilities

Each summand is at least 1/2 and the right side is less than 2, so k≤3. The set P is nonempty. The case k=1 is impossible: its left side is less than 1, while its right side is at least 1. Thus k=2 or k=3.

If k=2, the equation becomes 1/m₁+1/m₂=2/N. Since mᵢ≤N, each reciprocal is at least 1/N, so both mᵢ=N. Both poles are fixed by all of G. Thus G embeds in their planar stabilizer and is cyclic; its nonidentity rotations share exactly these two poles.

If k=3, order the integers m₁≤m₂≤m₃. Their reciprocals sum to 1+2/N>1. If m₁≥3, the sum is at most 1, impossible. Therefore m₁=2. If m₂≥4 the sum is again at most 1, so m₂=2 or 3. For m₂=2, write m₃=n≥2; the equation yields N=2n. For m₂=3, the remaining reciprocal must exceed 1/6, so m₃ is 3, 4 or 5. Substitution gives the table.

These are necessary possibilities. The count alone neither constructs the polyhedral groups nor proves conjugacy or uniqueness for these data. The rotational dihedral example has n rotations around a vertical axis and n half-turns around horizontal axes. All are spatial orientation-preserving rotations. By contrast, the planar dihedral group contains reflections of determinant −1 on its two-dimensional plane. The abstract groups can be isomorphic while their geometric actions and determinants differ.

Apply

Suppose a noncyclic finite rotation group has a pole with stabilizer order 5. Use the complete list to determine the possible group orders, and explain why choosing only the row (2,3,5) would be a mistake. Separately, decide whether order 18 can be excluded by pole counting.

Check what the count does and does not decide

The stabilizer order 5 occurs in (2,2,5) and (2,3,5), giving orders 10 and 60. In the first case the orbit sizes are (5,5,2); in the second they are (30,20,12). A noncyclic group of order 18 must have triple (2,2,9), so it is not excluded. Spatial rotations generated by a vertical turn through 2π/9 and a horizontal half-turn give such an example; the half-turn conjugates the vertical turn to its inverse. This construction verifies existence for that example, but the general counting argument alone remains a necessary-condition argument.

Finally reconstruct a rotation from a fixed unit axis and the images of one perpendicular unit vector. State the angle convention, check the remaining perpendicular direction using orientation, and distinguish the half-turn from the identity by its action on that plane.

Continue

Return to plane orthogonal maps and finite isometries for the SO(2) calculation and the distinction between a reflection and a rotation. Continue to unit quaternions and the rotation quotient for a different way to encode spatial rotations, including why two parameters can represent the same rotation.