What is the relation the symmetry of a high-symmetry point in the first Brillouin zone and lattice planes perpendicular to it? Are the two symmetries equivalent?
I have this question because I want to "visualize" the symmetry operations of a high-symmetry point (a vector w.r.t. $\Gamma$ point) in the first Brillouin zone in direct space. I think the point group of a wave vector is equivalent to the point group that leaves the lattice planes perpendicular to the wave vector invariant. But I haven't find a solid proof for this hypothesis.
Below is a proof by myself:
Let $K_m = (h, k, l)$ denotes a high-symmetry point in the first Brillouin zone, $P_\alpha$ a symmetry operation of the point group that leaves $K_m$ invariant, $R_n$ a lattice vector with its end point on a $(hkl)$ plane. By definition of reciprocal lattice, we have
$$(P_\alpha K_m) Rn = 2\pi N_1$$
where $N_1$ is an integer. Since a point group operation on a constant is still a constant, we act $P^{-1}_\alpha$ on each element of the above equation, and we get
$$P^{-1}_\alpha (P_\alpha K_m) (P^{-1}_\alpha R_n) = 2\pi N_1 = K_m (P^{-1}_\alpha R_n)$$
Therefore, $P^{-1}_\alpha R_n$ is also a lattice point on the same $(hkl)$ plane. Since the same proof can run in the other direction, the point group of $K_m = (h, k, l)$ is equivalent to the point group which leaves the $(hkl)$ lattice plane invariant.
Is my proof correct? I doubt it since only the magnitude of $K_m$ is not considered, while $K_m$ in the same direction while of different magnitudes can be of different point groups. For instance, in rutile TiO2, the point group of high-symmetry point X (0.5, 0, 0) is $D_{2h}$, while the point group of (0.25, 0, 0) is $C_{2v}$.
The point groups (little co-group) of the two high-symmetry points were queried from Bilbao Crystallographic Server
