Relation between symmetry of a reciprocal vector and lattice planes perpendicular to it

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


[![c2v](https://i.stack.imgur.com/kteBal.png)](https://i.stack.imgur.com/kteBal.png)


[![d2h](https://i.stack.imgur.com/p8hGNl.png)](https://i.stack.imgur.com/p8hGNl.png)





> 
> What is the relation the symmetry of a reciprocal vector and lattice planes perpendicular to it? Are the two symmetries equivalent?
> 
> 
> 


The meaning of symmetry in reciprocal space is unclear. If you are just talking about the location of lattice points in reciprocal space, every point on this lattice has the same symmetry. If you are also talking about the intensities (like a diffraction image), all symmetry elements include the origin, so other points lack point group symmetry.


However, you can ask the question whether a point k in reciprocal space lies on a symmetry element, i.e. whether applying the symmetry elements results in different k-vectors or the same.



> 
> Let $K\_m = (h, k, l)$ denotes a high-symmetry point in the first Brillouin zone [...]
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> 
> 


The Brillouin zone refers to a part of reciprocal space that only includes a single reciprocal lattice point (the origin). Conventionally, $h, k, l$ are used as indices for reciprocal space points, i.e. whole numbers. Here whole numbers would not make sense. If the numbers are not whole, no lattice planes in real space are associated with the point in reciprocal space, so the definition of $R\_n$ does not make sense, and $N\_1$ would not be an integer.



> 
> Is my proof correct?
> 
> 
> 


As I don't understand the first part of the proof, I find it difficult to tell.


This [paper](https://scholar.google.com/scholar?cluster=8988225890669585173&hl=en&as_sdt=0,22) introduces the concepts underlying the data of the Bilboa database that are now referenced in the question.


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