How many faces, edges and vertices are fixed when $S_4$ permutes the diagonals of a cube?

Question

Consider the action of $S_4$ on a cube, where it acts by permuting the long diagonals. The conjugacy classes of $S_4$ are denoted by $id$, (12), (123), (1234) and (12)(34). I want to know the number of faces, vertices and edges each of the elements (12), (123), (1234) and (12)(34) fix. For example, (12) and (123) fix no face and (1234) fixes 2. I think (12)(34) also doesn't fix a face. Is that so? And can someone please enumerate the number of vertices and edges each of these elements fix. I'm having trouble imagining the permuation of the diagonals.

Answer 1

Note that each permutation of $S_4$ acts as a rotation of the cube. So it's helpful to consider the kinds of rotations a cube can have. The easiest kind to visualize is rotation about the center of two opposite faces. There are 3 possible axes of rotation, and 4 rotations about each axis (with the identity, or null rotation shared by all). That's 10 rotations, of which 6 are of order 4 (and thus correspond to 4-cycles), and 3 of order 2. All of this kind of rotation fix 2 faces, but no edges or vertices. Now there are two kinds of elements of $S_4$ of order 2, transpositions and double transpositions. Transpositions leave two main diagonals fixed, whereas this kind of rotation (180 degrees around an axis through the center of opposite faces) swaps diagonals in pairs, so these 3 rotations correspond to the double transpositions.

There are two kinds of rotations that are somewhat harder to visualize: along an axis from the midpoint of opposite edges, and rotations about one of the long diagonals itself. The latter kind has order 3, and so must correspond to 3-cycles. A little experimentation with an actual cube shows such a rotation fixes 2 vertices, no faces, and no edges. This accounts for 8 more rotations.

The only thing left, then, are the rotations about the midpoints of two opposite edges, which therefore must correspond to the transpositions. These fix the pair of said edges, but no faces, nor vertices.

Answer 2

First of all, the question is not quite well specified, because there are $2^3\times 3!=48$ symmetries of the cube and only $24$ permutations of its diagonals, so each permutation corresponds to two different symmetries. So to get an action of $S_4$ you must choose a subgroup of symmetries that realises each permutation exactly once. There are in fact two such subgroups, but the most natural choice is to take the orientation-preserving symmetries, those that have determinant$~1$ (rather than$~{-}1$).

Then I think it is easiest here to change the point of view: focus on one face/edge/vertex and find the orientation-preserving symmetries that fix it, and see what permutations of the diagonals these give. If the permutations all have a different cycle type than the specified permutation, then that permutation will fix no face/edge/vertex; on the other hand if some permutation does have the right cycle type, then you have at least one fixed face/edge/vertex, and it is fairly easy to see if thee are any others.

Concretely, fixing a face, there are $4$ orientation-preserving symmetries that fix it: the identity (which has cycle type $(1,1,1,1)$), a quarter turn about the centre of the face (cycle type $(4)$ on diagonals, since every corner of the face is on a different diagonal), and its second and third powers (cycle types $(2,2)$ respectively $(4)$). So among the non-identity permutations you listed, only $(1234)$ and $(12)(34)$ fix any faces (for the latter a more natural conjugate to consider would be $(1234)^2=(13)(24)$) and both fix two opposite faces. (For $(12)(34)$ this contradicts the guess in your question.) For an edge there are only two orientation-preserving symmetries that fix it, the non-identity one of which (again rotation about an axis through its centre) gives cycle type $(2,1,1)$ on the diagonals. So here only $(12)$ fixes any edges, and it fixes two opposite ones. The vertex case is similar: the cycle type is $(3,1)$; only $(123)$ fixes any vertices and it fixes two opposite ones.

Answer 3

It is easier to study the different "types" of rotation of a cube and to check what they do to the space diagonals. Therefore let $C:=[{-1},1]^3$ be our cube, and let $D$ be the set of space diagonals. Any rotation $T$ of $C$ induces a certain permutation $\pi_T$ of $D$. The cycle structure of $\pi_T$ depends only on the conjugacy class of $T$. At the same time it is much easier to visualize the induced permutations of the vertices, edges, and faces of $C$ if $T$ is given in advance.

There are five conjugacy classes of rotations of $C$:

${\rm id}:\quad$ The Identity.

$A:\quad$ Rotation around a coordinate axis about an angle $\pm{\pi\over2}$. Such a rotation cyclically permutes the vertices of, say, the top face of $C$ and therefore permutes $D$ cyclically. The cycle structure of $\pi_T$ is $(1,2,3,4)$.

$A^2:\quad$ Rotation around a coordinate axis about an angle $\pm\pi$. The cycle structure of $\pi_T$ is then the square of the former, i.e., $(1,2)(3,4)$.

$B:\quad$ Rotation around an axis through the midpoints of two opposite edges about an angle $\pm\pi$. Such a rotation swaps the space diagonals through the end points of the chosen edges and keeps the two other space diagonals fixed (but reverses their direction). It follows that the cycle structure of $\pi_T$ is $(12)$.

$C:\quad$ Rotation around a space diagonal about an angle $\pm{2\pi\over3}$. Such a rotation cannot leave another space direction fixed; therefore $\pi_T$ necessarily has a cycle of length $3$. Its cycle structure is then $(1,2,3)$.

So it has turned out that each conjugacy class of ${\cal S}_4$ is realized in the set of permutations of $D$ induced by rotations of $C$.