Crystallographic restriction

Statement

Suppose $\Lambda$ is a lattice in the Euclidean plane, and suppose there exists an angle $\alpha$ that is not a multiple of $\pi$ and such that rotation by $\alpha$ sends $\Lambda$ to itself. Then, there exist two vectors $v,w$ in $\Lambda$ of shortest length, such that $v,w$ span $\Lambda$, and the angle between $v$ and $w$ is either $\pi /2$ or $2\pi /3$.

In the statement, the center of rotation need not be a lattice point.

Proof

Proof when the center of rotation is a lattice point

We first give the proof assuming that the center of rotation is a lattice point. In this case, we can assume that the center of rotation is the origin.

Let $v$ be a nonzero vector of shortest length in $\Lambda$, i.e., a point closest to the origin. Let $\theta$ be the smallest positive angle of rotation sending $\Lambda$ to itself. Let $w$ be the image of $v$ under $\theta$. Then, $w$ has the same length as $v$.

1. $v$ and $w$ span $\Lambda$: Given any point in the Euclidean plane, we can subtract an element in the sublattice spanned by $v$ and $w$ to arrive in the fundamental parallelogram bounded by the origin, $v,w,v+w$. It is easy to see that the maximum of the distances between this new point and the corners of the parallelogram is less than the length of $v$ or $w$. Thus, if there exists a lattice point not in the span of $v$ and $w$, we can obtain a lattice vector of shorter length. Thus, $v$ and $w$ span $\Lambda$.
2. The angle between $v$ and $w$ divides $2\pi$: Suppose the angle is $\theta$. If $\theta$ does not divide $2\pi$, the smallest multiple of $\theta$ that is bigger than $2\pi$ gives an angle smaller than $\theta$ that can be realized as an angle between $v$ and one of its images under a rotation of the lattice -- in particular, this contradicts minimality of $\theta$.
3. $v$ and $w$ have an angle of $\pi /2$ or $\pi /3$. Since $v-w$ is in $\lambda$, it has to be at least as large as $v$. If the angle between $v$ and $w$ is $\theta$, $\|v-w\|=2\sin(\theta /2)$, and thus, $\theta \geq \pi /3$. Thus, $\theta \in \{\pi /3,2\pi /5,\pi /2,2\pi /3\}$. $\theta =2\pi /3$ is not possible because if $v$ and $w$ make an angle of $2\pi /3$, we can choose the spanning set $\{v,v+w\}$, which make an angle of $\pi /3$, with a rotational symmetry of $\pi /3$. The angle $2\pi /5$ is not possible, because rotating by twice that angle gives a vector that makes an angle of $\pi /5$ with $v$. Thus, the angle is $\pi /2$ or $\pi /3$.

Proof when the center of rotation is not a lattice point

Suppose the center of rotation is $u$ and $v$ is any lattice point. Let $\theta$ be the smallest permissible positive angle of rotation about $u$. As in the previous case, $\theta$ divides $2\pi$, and the orbit of $v$ under rotation by multiples of $\theta$ forms the vertices of a regular polygon centered at $u$.

Let $v_{1},v_{2},\dots ,v_{r}$ be the other points in this orbit. Then, $(v_{i}-v)\in \Lambda$ for each $i$, and thus, $\sum (v_{i}-v)\in \Lambda$. But an elementary trigonometric computation, or a use of complex roots of unity, shows that this sum is an integer multiple of $u-v$. In particular, some nonzero integer multiple of $u-v$ is in $\Lambda$. So, a nonzero integer multiple of $u$ is in $\Lambda$.

From this, we obtain that the span of $\Lambda$ and $u$ is also a lattice containing $u$. Moreover, rotation by $\theta$ sends this new lattice to itself, so we are reduced to the previous case of the center of rotation being a lattice point. Thus, the new lattice is generated by two vectors at an angle of either $\pi /2$ or $2\pi /3$. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

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