Here is an exercise problem from Modern Condensed Matter Physics (Girvin and Yang, 2019):©

Exercise 3.9. Show that five-fold rotation symmetry is inconsistent with lattice translation symmetry in 2D. Since 3D lattices can be formed by stacking 2D lattices, this conclusion holds in 3D as well.

Before I saw this problem, I had never thought about whether a plane lattice can have $m$-fold symmetry for any positive integer $m$. I was surprised at first that I cannot have a translationally symmetric lattice with 5-fold symmetry. After some thinking, I did realize that I cannot imagine a 5-fold symmetric plane lattice, so such a lattice cannot exist intuitively.

Actually, the only allowed rotational symmetries are 2-fold, 3-fold, 4-fold, and 6-fold. This result is known as the crystallographic restriction theorem. Then, how to prove it?

After jiggling around the possible structure of the symmetry group of a plane lattice, I finally proved it. I found that this proof is actually a simple and good example of how algebraic number theory can be used in physics.

Before dive into the proof, we need to first prove a simple lemma about real analysis:

Lemma 1. If $G$ is a subgroup of $(\mathbb R^2,+)$ that is discrete and spans $\mathbb R^2$, then there exist two linearly independent elements in $\mathbb R^2$ that generate $G$.

Proof. Because $G$ spans $\mathbb R^2$, there exist two linearly independent elements $g_1,g_2\in G$.

Consider the vector subspace $V_1\coloneqq g_1\mathbb R$ and the subgroup $G_1\coloneqq G\cap V_1$. Obviously, $G_1$ should be generated by some element $h_1\in G_1$ (this is because $V_1\simeq\mathbb R$, and $G_1$ as a discrete set must have a smallest positive element under that isomorphism, which must be the generator of $G_0$ because it would otherwise not be the smallest positive element). Therefore, $G_1=h_1\mathbb Z$. Also, because $h_1\ne0$, $\left\{h_1,g_2\right\}$ must span $\mathbb R^2$.

Let $T\coloneqq\left\{ah_1+bg_2\in G\,\middle|\,a\in\left[0,1\right),b\in\left[0,1\right]\right\}.$ Then, $T$ must be discrete (because $G$ is) and bounded, and contains at least the element $g_2$. Express every element in $T$ as $ah_1+bg_2$ and pick out the one element with the smallest non-zero $b$, and denote it as $h_2=a^\star h_1+b^\star g_2$. Certainly, $\left\{h_1,h_2\right\}$ span $\mathbb R^2$.

Now, for any $g\in G$, we can express it uniquely as $g=ah_1+bg_2$. Define $c_2\coloneqq\left\lfloor\frac{b}{b^\star}\right\rfloor,\quad c_1\coloneqq\left\lfloor a-a^\star c_2\right\rfloor,\quad g'\coloneqq g-c_1h_1-c_2h_2.$ Then, $g'\in T$, and if we express it as $g'=a'h_1+b'g_2$, then $b'$ is smaller than $b^\star$. By definition of $b^\star$, $b'=0$, so $g'\in G_1$. Hence, $\left\{h_1,h_2\right\}$ generates $G$. $\square$

Now, we are ready to prove our main result:

Theorem. There is a discrete subset of $\mathbb R^2$ that has both translational symmetry and $m$-fold symmetry iff $\varphi(m)\le2$, where $\varphi$ is Euler’s totient function.

Proof. For the neccessity, prove by contradiction. I instead prove that a set that has the said symmetries must not be discrete.

Denote the plane as $\mathbb C$. Assume that there is an $m$-fold symmetry around point $0$. Then, for any lattice site $z$, the point $Rz\coloneqq\alpha z$ (where $\alpha\coloneqq\mathrm e^{2\pi\mathrm i/m}$) is also a lattice site. Assume that there is a translational symmetry with translation $a$, then the point $Tz\coloneqq z+a$ is also a lattice site. Without loss of generality, we can adjust the orientation of our coordinate system and the length unit so that $a=1$.

The group $G$ generated by $\{R,T\}$ is a subgroup of the symmetry group of the lattice. Its action $S\coloneqq\left\{g0\,\middle|\,g\in G\right\}$ on the point $0$ is a subset of all the lattice sites (this is only true when $0$ is a lattice site; I will discuss later the other case). Notice that for any $z\in S,n\in\mathbb Z$, we have $T^nRz=n+\alpha z\in S$. Therefore, by expanding any polynomial with integer coefficients using Horner’s rule, we can see that $\mathbb Z[\alpha]\subseteq S$.

Because $\alpha$ is an algebraic integer of degree $\varphi(m)$ (the minimal polynomial of $\alpha$ is known as the $m$th cyclotomic polynomial), the generating set of $\mathbb Z[\alpha]$ must have at least $\varphi(m)$ elements. Therefore, according to Lemma 1, $\mathbb Z[\alpha]$ is discrete iff $\varphi(m)\le2$.

For the case where $0$ is not a lattice site, we can generate $S$ by acting $G$ on any lattice site $z_0$. We can then easily prove that $z_0+\mathbb Z[\alpha]\subseteq S$. To prove this, we just need to see that we can act $R^{-k}$ on $z_0$ before further acting $T^nR$ on it for $k$ times. All the other steps are the same and still valid.

For the sufficiency, because there are only finitely many $m$’s that satisfy $\varphi\!\left(m\right)\le2$. Therefore, we can enumerate these $m$’s and see that we can easily construct a plane lattice with both translational symmetry and $m$-fold symmetry for each $m$. $\square$

I know the original problem in the book was probably not intended to be solved in this way, but it is really amazing how some seemingly purely mathematical areas can have their applications in physics, especially in an exercise problem of a physics textbook where pure mathematics is pretty unexpected.

Unfortunately, this proof, which is based on algebraic properties of certain complex numbers, does not generalize to higher dimensions because we cannot use the complex plane to represent a high-dimensional space.