Possible rotational symmetries in a Bravais lattice

Any 2- or 3-dimensional Bravais lattice will have at most the following rotational symmetries or their compositions:

Symmetry Angle of rotation
Identity $0$
2-fold $\pi$ 180°
3-fold $\frac{2\pi}{3}$ 120°
4-fold $\frac{\pi}{2}$ 90°
6-fold $\frac{\pi}{3}$ 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 Crystallographic restriction theorem.

Proof in three dimensions

Consider a Bravais lattice, a collection of points described as follows:

$$ \mathbf{x} = n_1 \mathbf{b}_1 + n_2 \mathbf{b}_2 + n_3 \mathbf{b}_3 $$

for $ n_1, n_2, n_3 \in \mathbb{Z} $ and some oblique basis $ \{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3\} $. This can be rewritten more compactly as:

$$ \mathbf{x} = \begin{bmatrix} \mathbf{b_1} & \mathbf{b_2} & \mathbf{b_3} \end{bmatrix} \begin{bmatrix} n_1 \\ n_2 \\ n_3 \end{bmatrix} = \hat{B}\mathbf{n} $$

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:

$$ \hat{R}_\theta\mathbf{x} &= \hat{B}\mathbf{n}' $$

where $\mathbf{n}'$ is some new vector with integer components. Substituting $\mathbf{x} = \hat{B}\mathbf{n}$ and premultiplying both sides by $\hat{B}^{-1}$:

$$ \begin{align}
\hat{R}_\theta\hat{B}\mathbf{n} &= \hat{B}\mathbf{n}' \\
\hat{B}^{-1}\hat{R}_\theta\hat{B}\mathbf{n} &= \hat{B}^{-1}\hat{B}\mathbf{n}' \\
&= \mathbf{n}'
\end{align} $$

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:

$$ \textup{Tr}(\hat{B}^{-1}\hat{R}_\theta\hat{B}) = \textup{Tr}(\hat{R}_\theta) = 1 + 2\cos{\theta} \in \mathbb{Z} $$

Rewriting using $k \in \mathbb{Z}$ and knowing that $(\forall x \in \mathbb{R})(\cos{x} \in [-1, 1])$:

$$ \begin{align}
1 + 2\cos{\theta} &= k \\
\cos{\theta} &= \frac{1 + k}{2} \\
\cos{\theta} &\in \left\{-1, -\frac{1}{2}, 0, \frac{1}{2}, 1\right\}
\end{align} $$

Ignoring angles outside $[0, 2\pi)$ and solving for $\theta$:

$$ \begin{align}
(\cos{\theta} = -1) &\implies (\theta = \pi) \\
\left(\cos{\theta} = -\frac{1}{2}\right) &\implies \left(\theta = \frac{2\pi}{3}\right) \lor \left(\theta = \frac{4\pi}{3}\right) \\
(\cos{\theta} = 0) &\implies \left(\theta = \frac{\pi}{2}\right) \lor \left(\theta = \frac{3\pi}{2}\right)\\
\left(\cos{\theta} = \frac{1}{2}\right) &\implies \left(\theta = \frac{\pi}{3}\right) \lor \left(\theta = \frac{5\pi}{3}\right) \\
(\cos{\theta} = 1) &\implies (\theta = 0) \\
\end{align} $$

These are the angles from the table in the beginning of the article and their multiplies. 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 $\mathbf{n}'$ 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:

  1. There's no $n_3$ or $\mathbf{b}_3$ (the appropriate expressions get shorter by a term).
  2. The trace of any 2D rotation matrix is $2\cos{\theta}$ (rather than $1 + 2\cos{\theta}$). The results are still the same, though.