Commutative Rotations

Solution 1:

In $\mathbb{R}^3$, the only sorts of rotations that commute are the following:

• Rotations about the same axis commute.

• Any pair of $180^\circ$ rotations about perpendicular axes commute.

Thus, the maximal commutative groups of rotations in $\mathbb{R}^3$ are:

• The group of rotations about any axis, and

• Four-element groups consisting of the identity plus three $180^\circ$ rotations about three perpendicular axes (isomorphic to the Klein four-group).

The latter can be thought of as the group of rotations of a rectangular box with three different side lengths

In $\mathbb{C}^3$, it depends on exactly what you mean by "rotation". There are two obvious possible meanings:

1. $3\times 3$ (complex) unitary transformations, which form the unitary group $U(3)$.

2. $6 \times 6$ (real) orthogonal rotations, which form the special orthogonal group $SO(6)$

The former act naturally on the unit sphere in $\mathbb{C}^3$, while latter is more obviously thought of as rotations of the unit sphere in $\mathbb{R}^6$. Unitary transformations preserve the complex structure, sending complex subspaces to complex subspaces, while orthogonal transformations do not.

Unitary Transformations: Every unitary transformation is diagonalizable over the complex numbers, with the eigenvalues being complex numbers of modulus one. Geometrically, the unitary transformation represents rotation in three "planes" (one-dimensional complex subspaces), which are the eigenspaces of the matrix. The angle of rotation on each plane is determined by the argument of the corresponding eigenvalue. These three eigenspaces are well-defined unless the matrix has a repeated eigenvalue.

Two unitary transformations commute if and only if they are simultaneously diagonalizable. For most pairs of transformations, this is the equivalent to having the same three eigenspaces. However, the situation gets a bit more complicated if one of the transformations has a repeated eigenvalue -- there needs to be a common decomposition of $\mathbb{C}^3$ into three eigenspaces. The most extreme case are the scalar matrices, which commute with everything.

To be precise, the centralizer of a typical element of $U(3)$ is isomorphic to $U(1)\times U(1)\times U(1)$, where $U(1)$ is the group of complex numbers of unit modulus. If a unitary matrix has a repeated eigenvalue, the centralizer is isomorphic to $U(1)\times U(2)$, or all of $U(3)$ if the matrix is scalar.

The maximal commutative subgroups of $U(3)$ are all isomorphic to $U(1)\times U(1)\times U(1)$.

Orthogonal Transformations: Orthogonal transformations are a bit more complicated. A typical orthogonal transformation of $\mathbb{R}^6$ has three perpendicular "planes of rotation". Each of these planes is two-dimensional, and is invariant under the rotation. (Unlike the unitary case, these planes are not required to be complex subspaces.) Usually each plane has a different angle of rotation between $0^\circ$ and $180^\circ$. However, if some of the angles are the same, then the planes are not uniquely determined. This is similar to the "repeated eigenvalues" case above. (Indeed, such an orthogonal transformation has repeated complex eigenvalues.) Things also get complicated when one of the angles of rotation is $0^\circ$ and one is $180^\circ$.

For a typical orthogonal transformation with three different planes of rotation and three different angles, its centralizer consists of all elements that involve the same three planes, and is isomorphic to $SO(2)\times SO(2)\times SO(2)$ (where $SO(2)$ is rotations of the unit circle, isomorphic to the group $U(1)$ mentioned above). The centralizers in other cases are more complicated.

There are also several types of maximal commutative subgroups in the orthogonal case. One is the subgroups of the form $SO(2)\times SO(2)\times SO(2)$ mentioned above. However, there are other types, e.g. the group of all diagonal matrices with $1$'s and $-1$'s on the diagonal and determinant $1$, which has $32$ elements and is isomorphic to $(\mathbb{Z}_2)^5$.

Solution 2:

If $R_1$ is a rotation (in the space $\mathbf{R}^3$) about the axis $L$, then $R_2R_1R_2^{-1}$ is a rotation about the axis $R_2(L)$. So for $R_1$ and $R_2$ to commute, $R_2$ must map $L$ to itself (in either orientation). From this it easily follows that either the two (non-trivial) rotation must be about the same axis, or both of them must be rotations by 180 degrees.

Together with the identity mapping the rotations by 180 degrees about a system of any three orthogonal axes form a group isomorphic to the Klein four group. Looks like that is the only alternative.