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 mm-fold symmetry for any positive integer mm. 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 GG is a subgroup of (R2,+)(\mathbb R^2,+) that is discrete and spans R2\mathbb R^2, then there exist two linearly independent elements in R2\mathbb R^2 that generate GG.

Proof. Because GG spans R2\mathbb R^2, there exist two linearly independent elements g1,g2Gg_1,g_2\in G.

Consider the vector subspace V1g1RV_1\coloneqq g_1\mathbb R and the subgroup G1GV1G_1\coloneqq G\cap V_1. Obviously, G1G_1 should be generated by some element h1G1h_1\in G_1 (this is because V1RV_1\simeq\mathbb R, and G1G_1 as a discrete set must have a smallest positive element under that isomorphism, which must be the generator of G0G_0 because it would otherwise not be the smallest positive element). Therefore, G1=h1ZG_1=h_1\mathbb Z. Also, because h10h_1\ne0, {h1,g2}\left\{h_1,g_2\right\} must span R2\mathbb R^2.


T{ah1+bg2G|a[0,1),b[0,1]}.T\coloneqq\left\{ah_1+bg_2\in G\,\middle|\,a\in\left[0,1\right),b\in\left[0,1\right]\right\}.

Then, TT must be discrete (because GG is) and bounded, and contains at least the element g2g_2. Express every element in TT as ah1+bg2ah_1+bg_2 and pick out the one element with the smallest non-zero bb, and denote it as h2=ah1+bg2h_2=a^\star h_1+b^\star g_2. Certainly, {h1,h2}\left\{h_1,h_2\right\} span R2\mathbb R^2.

Now, for any gGg\in G, we can express it uniquely as g=ah1+bg2g=ah_1+bg_2. Define

c2bb,c1aac2,ggc1h1c2h2.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, gTg'\in T, and if we express it as g=ah1+bg2g'=a'h_1+b'g_2, then bb' is smaller than bb^\star. By definition of bb^\star, b=0b'=0, so gG1g'\in G_1. Hence, {h1,h2}\left\{h_1,h_2\right\} generates GG. \square

Now, we are ready to prove our main result:

Theorem. There is a discrete subset of R2\mathbb R^2 that has both translational symmetry and mm-fold symmetry iff φ(m)2\varphi(m)\le2, where φ\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 C\mathbb C. Assume that there is an mm-fold symmetry around point 00. Then, for any lattice site zz, the point RzαzRz\coloneqq\alpha z (where αe2πi/m\alpha\coloneqq\mathrm e^{2\pi\mathrm i/m}) is also a lattice site. Assume that there is a translational symmetry with translation aa, then the point Tzz+aTz\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=1a=1.

The group GG generated by {R,T}\{R,T\} is a subgroup of the symmetry group of the lattice. Its action

S{g0|gG}S\coloneqq\left\{g0\,\middle|\,g\in G\right\}

on the point 00 is a subset of all the lattice sites (this is only true when 00 is a lattice site; I will discuss later the other case). Notice that for any zS,nZz\in S,n\in\mathbb Z, we have TnRz=n+αzST^nRz=n+\alpha z\in S. Therefore, by expanding any polynomial with integer coefficients using Horner’s rule, we can see that Z[α]S\mathbb Z[\alpha]\subseteq S.

Because α\alpha is an algebraic integer of degree φ(m)\varphi(m) (the minimal polynomial of α\alpha is known as the mmth cyclotomic polynomial), the generating set of Z[α]\mathbb Z[\alpha] must have at least φ(m)\varphi(m) elements. Therefore, according to Lemma 1, Z[α]\mathbb Z[\alpha] is discrete iff φ(m)2\varphi(m)\le2.

For the case where 00 is not a lattice site, we can generate SS by acting GG on any lattice site z0z_0. We can then easily prove that z0+Z[α]Sz_0+\mathbb Z[\alpha]\subseteq S. To prove this, we just need to see that we can act RkR^{-k} on z0z_0 before further acting TnRT^nR on it for kk times. All the other steps are the same and still valid.

For the sufficiency, because there are only finitely many mm’s that satisfy φ ⁣(m)2\varphi\!\left(m\right)\le2. Therefore, we can enumerate these mm’s and see that we can easily construct a plane lattice with both translational symmetry and mm-fold symmetry for each mm. \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.