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// SPDX-License-Identifier: CC-BY-SA-4.0
// Copyright © Paweł Redman
#import "/common.typ": post
#show: post.with(
title: "Possible rotational symmetries in a Bravais lattice",
date: datetime(year: 2019, month: 4, day: 16),
)
Any 2- or 3-dimensional Bravais lattice will have *at most* the following rotational symmetries or their compositions:
#table(
columns: 3,
table.header([*Symmetry*], [*Angle of rotation*], []),
[Identity], [$0$], [0°],
[2-fold], [$pi$], [180°],
[3-fold], [$2 pi / 3$], [120°],
[4-fold], [$pi / 2$], [90°],
[6-fold], [$pi / 3$], [60°],
)
Every other rotational symmetry (e.g. 5-fold) is impossible in such a Bravais lattice, regardless of the lattice basis or the chosen rotation axis. This is also known as the #link("https://en.wikipedia.org/wiki/Crystallographic_restriction_theorem")[Crystallographic restriction theorem].
= Proof in three dimensions
Consider a Bravais lattice, a collection of points described as follows:
$
bold(x) = n_1 bold(b)_1 + n_2 bold(b)_2 + n_3 bold(b)_3
$
for $n_1, n_2, n_3 in ZZ$ and some oblique basis ${bold(b)_1, bold(b)_2, bold(b)_3}$. This can be rewritten more compactly as:
$
bold(x) = mat(delim: "[", bold(b)_1, bold(b)_2, bold(b)_3) vec(n_1, n_2, n_3) = hat(B) bold(n)
$
Rotational symmetry means that if the points are rotated around some axis by angle $theta$, they will exactly coincide with the original points. That is, they can be written in the same form as the original points. Representing the multiplication by premultiplication by the rotation matrix $hat(R)_theta$ yields:
$
hat(R)_theta bold(x) &= hat(B) bold(n)'
$
where $bold(n)'$ is some new vector with integer components. Substituting $bold(x) = hat(B) bold(n)$ and premultiplying both sides by $hat(B)^(-1)$:
$
hat(R)_theta hat(B) bold(n) &= hat(B) bold(n)' \
hat(B)^(-1) hat(R)_theta hat(B) bold(n) &= hat(B)^(-1) hat(B) bold(n)' \
&= bold(n)'
$
The matrix $hat(B)^(-1) hat(R)_theta hat(B)$ transforms one vector with integer components into another. Therefore, it must itself have only integer components and an integer trace. Furthermore, change of basis does not change the trace. Finally, the trace of any rotation matrix in three dimensions equals $1 + 2 cos theta$. Putting this all together yields:
$
op("Tr")(hat(B)^(-1) hat(R)_theta hat(B)) = op("Tr")(hat(R)_theta) = 1 + 2 cos theta in ZZ
$
Rewriting using $k in ZZ$ and knowing that $(forall x in RR)(cos x in [-1, 1])$:
$
1 + 2 cos theta &= k \
cos theta &= (1 + k) / 2 \
cos theta &in {-1, -1 / 2, 0, 1 / 2, 1}
$
Ignoring angles outside $[0, 2 pi)$ and solving for $theta$:
$
(cos theta = -1) &=> (theta = pi) \
(cos theta = -1 / 2) &=> (theta = 2 pi / 3) or (theta = 4 pi / 3) \
(cos theta = 0) &=> (theta = pi / 2) or (theta = 3 pi / 2) \
(cos theta = 1 / 2) &=> (theta = pi / 3) or (theta = 5 pi / 3) \
(cos theta = 1) &=> (theta = 0)
$
These are the angles from the table in the beginning of the article and their multiples. For any $theta$ not in this set of solutions, the trace of $hat(B)^(-1) hat(R)_theta hat(B)$ is a non-integer, the matrix itself necessarily has non-integer entries and the $bold(n)'$ vector ends up with non-integer entries. This contradicts the premise, hence the restriction on possible $theta$'s.
= Proof in two dimensions
The proof is almost identical to the case in three dimensions. The only differences are:
- There's no $n_3$ or $bold(b)_3$ (the appropriate expressions get shorter by a term).
- The trace of any 2D rotation matrix is $2 cos theta$ (rather than $1 + 2 cos theta$). The results are still the same, though.
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