8-fold symmetry wrt one plane implies 8-fold symmetry wrt all planes?
Let a bounded, open, connected set $A \subset \mathbb{R}^N$ (with $N \geq 3$) satisfy the following symmetry assumptions:
- $A$ is 4-fold rotational symmetric with respect to any coordinate plane $(x_i,x_j)$.
- There exists a coordinate plane $(x_k,x_l)$ such that $A$ is 8-fold rotational symmetric with respect to $(x_k,x_l)$.
Do the assumptions 1. and 2. imply that $A$ is 8-fold rotational symmetric with respect to any coordinate plane $(x_i,x_j)$?
In the 3D case, my visual intuition says that it is true, but I have no clue about higher dimensions. Maybe there is a general result about that?
P.S. I'm not closely familiar with the language of group theory, so, please, let me know if the property of being $n$-fold symmetric with respect to any coordinate plane has its own name or a common notation in the literature.
$\endgroup$1 Answer
$\begingroup$Yes. You can conjugate the rotational subgroup on $(x_k,x_\ell)$-plane to $(x_i,x_j)$ with your order 4's. Assuming $i,j,k,\ell$ are distinct (so in particular, dimension $\geq 4$). Then you can do the following:
- Rotate on $(x_i,x_k)$ plane by $\pi/2$. This effectively switches the $x_i,x_k$ modulo a sign, say $x'_i=-x_k,x'_k=x_i$.
- Rotate on $(x_j,x_\ell)$ plane by $\pi/2$. This switches $x_j,x_\ell$ say with the sign on $x'_j$.
- Now do the rotation on $(x_k,x_\ell)$.
- Undo steps 1 and 2 by a rotation by $-\pi/2$.
Or in symbols,$$ r^{-1}_{ik}r^{-1}_{j\ell}S_{k\ell}r_{ik}r_{j\ell}=S_{ij} $$where $S_{ij}$ is the rotation group of order $8$ on $(x_i,x_j)$-plane and fixing the other orthogonal directions, $r_{ij}$ is a rotation by $\pi/2$ on the $(x_i,x_j)$-plane (again fixing the other directions).
If only 3 of them, then there is a common direction, say $(x_k,x_j)$ to $(x_k,x_\ell)$. You just need to bring $j$ to $\ell$, do the rotation, and move $\ell$ back to $j$, i.e.,$$ r^{-1}_{j\ell}S_{k\ell}r_{j\ell}=S_{jk}. $$
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