Classification of finite subgroups of SO(3,R)

Statement

This article explains the complete classification of finite subgroups of the Special orthogonal group (?) of order three over the field of real numbers. In other words, it completely classifies, upto conjugacy in the orthogonal group, all the finite subgroups of $SO(3,\mathbb{R})$.

Here is the complete classification:

• For any natural number $n$, the cyclic group of order $n$ acts as rotations in a plane, fixing the axis perpendicular to that plane. This can be realized as the von Dyck group with parameters $(n,n,1)$.
• For any natural number $n$, the dihedral group of order $2n$, acting as the dihedral group on a plane. The rotations in this group fix the perpendicular axis, while the reflections in this group also flip the perpendicular axis, and are effectively achieved through rotations, hence are still elements of the special orthogonal group. This can be realized as the von Dyck group with parameters $(n,2,2)$.
• There are three other groups: the alternating group of degree four is realized as the group of symmetries of the tetrahedron, the symmetric group of degree four is realized as the group of symmetries of the cube and also of the octahedron, and the alternating group of degree five is realized as the group of symmetries of the icosahedron and also of the dodecahedron. These are realized as von Dyck groups with parameters $(3,3,2)$ (for the alternating group of degree four), $(4,3,2)$ (for the symmetric group of degree four), and $(5,3,2)$ (for the alternating group of degree five).

Facts used

1. Euler's theorem: Every element of $SO(3,\mathbb{R})$ has an eigenvalue of $1$. In particular, it has two antipodal fixed points in the induced action on the sphere $S^2$. Moreover, if it is a non-identity element, it has no more fixed points, and it acts as a rotation on the perpendicular plane.
2. Fundamental theorem of group actions

Proof

Action on the finite set of fixed points of non-identity elements

Suppose $G$ is a finite subgroup of $SO(3,\mathbb{R})$. Let $S^2$ be the unit sphere in ${\mathbb{R}}^3$, i.e., the sphere of radius one centered at the origin. By fact (1), every non-identity element of $G$ has exactly two antipodal fixed points. Let $P$ be the set of points in $S^2$ that arise as fixed points of some non-identity element of $G$.

We claim that the action of $G$ sends $P$ to itself. This follows from the fact that if $p$ is fixed by a non-identity element $x\in G$, then $g\cdot p$ is fixed by the non-identity element $gx{g}^{-1}\in G$.

Decomposing the group

Using Euler's theorem (fact (1)) every non-identity element of the group fixes exactly one pair of antipodal points. Thus, we get:

$\left|G\right|-1=\frac{1}{2}{\sum }_{p\in P}({\mathrm{S}}_{\mathrm{t}}(p)-1)$.

The $1/2$ is to correct for double-counting antipodal points. We now use fact (2) to rewrite:

${\mathrm{S}}_{\mathrm{t}}(p)=\frac{\left|G\right|}{\left|{\mathcal{O}}_p\right|}$,

where ${\mathcal{O}}_p$ is the orbit of the point $p$.

We get:

$\left|G\right|-1=\frac{1}{2}{\sum }_{p\in P}(\frac{\left|G\right|}{\left|{\mathcal{O}}_p\right|}-1)$.

Rearranging to sum over orbits $\mathcal{O}\subseteq P$, we get:

$\left|G\right|-1=\frac{1}{2}{\sum }_{\mathcal{O}}(|G|-|\mathcal{O}|)$.

Rearranging and dividing by $\left|G\right|$:

$2-\frac{2}{\left|G\right|}={\sum }_{\mathcal{O}}(1-\frac{1}{|{\mathrm{S}}_{\mathrm{t}}(p)|})$,

where $p$ is any element of $\mathcal{O}$. Suppose $a_1,a_2,\dots ,a_r$ are the sizes of the stabilizers for the distinct orbits. Then we get:

$2-\frac{2}{\left|G\right|}=\sum _{i=1}^r(1-\frac{1}{a_i})$.

with the additional constraint that $a_i$ divides $\left|G\right|$, and $a_i$ and $\left|G\right|$ are positive integers. Without loss of generality, we can arrange $a_i$ in descending order.

Solving the equation

We easily obtain that $r\le 3$. Here are the solutions for various values of $r$:

1. $r=1$: In this case, $a_1=1$, and $\left|G\right|=1$. This is the trivial group.
2. $r=2$: We have two orbits. The above equation forces $a_1=a_2=\left|G\right|$. Call this number $a$. Since the stabilizer has size equal to the whole group, so both points of $P$ are fixed by all group elements. Thus, we have a pair of antipodal fixed points. Thus, $G$ acts as orientation-preserving orthogonal transformations in the perpendicular plane to the fixed points. In particular, $G$ is a cyclic group acting by rotations.
3. $r=3$, $a_1=a_2=a_3=2$, and $\left|G\right|=4$: $P$ has six points, with three orbits of size two. Since every point and its antipode have the same orbit size, there is at least one orbit comprising a point and its antipode. A little work shows that the group is the Klein four-group, behaving as the dihedral group on a regular $2$-gon.
4. $r=3$, with $a_2=a_3=2$ and $a_1>2$: In this case, if $a_1=n$, then $\left|G\right|=2n$. The orbit corresponding to $a_1$ has size two and the other two orbits have size $n$. Since the size of the orbit of a point and its antipode are equal, the orbit of size two contains two antipodal points. Its stabilizer is a subgroup of order $n$ acting as orientation-preserving orthogonal transformations on the perpendicular plane. Thus, this is a cyclic subgroup of order $n$. The element that flips the two antipodal points restricts to that plane as a reflection in some axis in that plane. It follows that the group is a dihedral group of order $2n$, with the usual action on that plane.
5. $r=3$, with $(a_1,a_2,a_3)=(3,3,2)$: In this case, $\left|G\right|=12$, and there are $14$ elements in $P$, with orbits of sizes $4,4,6$. We can show that either of the orbits of size $4$ can be chosen as the vertices of a regular tetrahedron and the group is the full group of orientation-preserving symmetries of the tetrahedron.
6. $r=3$, with $(a_1,a_2,a_3)=(4,3,2)$: In this case, $\left|G\right|=24$, and there are $26$ elements in $P$, with orbits of sizes $6,8,12$. We can show that the orbit of size $8$ can be chosen as the vertices of a cube and the group is the full group of orientation-preserving symmetries of the cube. The orbits of size $6$ can be chosen as the vertices of a regular octahedron, and the group is the full group of orientation-preserving symmetries of the octahedron.
7. $r=3$, with $(a_1,a_2,a_3)=(5,3,2)$: In this case, $\left|G\right|=60$. and there are $62$ elements in $P$, with orbits of size $12,20,30$. We can show that the orbit of size $12$ can be chosen as the vertices of a regular dodecahedron, and the group is the full group of orientation-preserving symmetries of the dodecahedron. The orbits of size $20$ can be chosen as the vertices of a regular octahedron, and the group is the full group of orientation-preserving symmetries of the octahedron.

Note that there are no other solutions for $a_1,a_2,a_3$, because we need $1/a_1+1/a_2+1/a_3>1$.

Geometric interpretation of the proof in terms of von Dyck groups

The $(a_1,a_2,a_3)$ parameters appearing above are the same as the parameters for the von Dyck presentation. This follows geometrically, as follows.

Let $p_1$ be a representative for the first orbit, with stabilizer of size $a_1$. Since the stabilizer of each point $p_1$ is of order $a_1$, and since each stabilizer is a cyclic group (on account of being a subgroup of the orientation-preserving orthogonal transformations of the perpendicular plane), the stabilizer of $p_i$ is generated by a rotation about $p_i$ by the angle $2\pi /a_1$. Let $p_2$ be any point in the second orbit. Then, the image of $p_2$ under this rotation cannot lie either in the $p_1$-orbit or in the $p_2$-orbit, hence it lies in the third orbit. Call this image $p_3$. It is now easy to see that the group $G$ is the von Dyck group for the spherical triangle with vertices $p_1,p_2,p_3$, and we get the desired presentation.

(Note that the fact that the presentation works follows from the fact that the sphere is simply connected, so there are no additional relations that arise when we descend from the cover where the action is well-defined).