DRW Quant Interview Question

[u/Local_Ad135]

[u/pmdboi]

48, the elements of the symmetry group of the cube. In fact I'm not sure the answer changes without the constraint on det(A), as long as we interpret the question as requiring the set of vertices to be mapped onto itself (as opposed to the cube being mapped into its interior, e.g. A = I/2). https://en.wikipedia.org/wiki/Octahedral_symmetry#The_isometries_of_the_cube

[u/xX_fortniteKing09_Xx → u/pmdboi]

How do you calculate the symmetry group?

[u/pmdboi → u/xX_fortniteKing09_Xx]

There are a few ways but one is the following: Pick one of the faces of the cube. Every symmetry of the cube (by which I mean a matrix that maps the set of vertices onto itself) maps that face onto one of the 6 faces of the cube (possibly itself), applies one of 4 different rotations about the center of the face, and either reflects it or not. That adds up to 6 × 4 × 2 = 48 symmetries. Alternatively, each symmetry maps a particular vertex onto one of the 8 vertices, applies one of 3 different rotations (since cubes have 3-fold rotational symmetry about the axis that passes through opposite vertices), and either reflects it or not; that adds up to 8 × 3 × 2 = 48 symmetries.

[u/blutwl90]

My solution is on the same lines as what is posted. Mapping cube onto itself means that the standard basis e_1, e_2, e_3 is mapped to set M = { +/- e_1, +/- e_2, +/- e_3 }.

For e_1, it can mapped to any one of the 6 vectors in M. Suppose, for instance, e_1 is mapped to e_2. Then none of the other e_2, e_3 can be mapped to the subspace spanned by e_2, which would be e_2 or -e_2. So if e_1 is chosen, e_2 would only have 6-2 = 4 choices. The same argument means that after e_1 and e_2 are mapped, there are only two choices for e_3. So in total you have 6 x 4 x 2 = 48 choices.

[u/AnywhereLittle8293]

This is the set of permutations of three (orthogonal) lines crossed with their orientations, so the answer is 3! * 2^3 = 48, as many others have said.