# Periodic tilings of the plane by regular polygons

Let $A$ be a tiling of $\mathbb{R}^{2}$ using regular polygons. Assume that the tiling is edge-to-edge. Assume also that there are two directions of periodicity, so that $\mathbf{u},\mathbf{v}\in \mathbb{R}^{2}$ are linearly independent vectors, and $A+\mathbf{u}=A+\mathbf{v}=A$.

Question: Must there always exist orthogonal directions of periodicity? That is, must there always exist non-zero vectors $\mathbf{u},\mathbf{v}\in \mathbb{R}^{2}$ such that $\mathbf{u}\cdot \mathbf{v}=0$, and $A+\mathbf{u}=A+\mathbf{v}=A$?

Note that the assumption of regularity is necessary, since we can tile the plane with identical parallelograms in such a way that there are no orthogonal directions of periodicity. It is also necessary that we have an edge-to-edge tiling, since otherwise we can construct an example with identical squares that does not have orthogonal directions of periodicity.

Does this follow from some feature of the wallpaper groups?

• Does this upload.wikimedia.org/wikipedia/commons/6/6a/2-uniform_17.png have an orthogonal period? – Dongryul Kim Mar 3 '17 at 23:32
• It seems that, computationally, there is no orthogonal period, but someone who knows number theory would need to confirm this. – Dongryul Kim Mar 3 '17 at 23:46
• Actually, there are orthogonal periods. I don't know if I can post an image in a comment, so I am going to post an answer with an image that illustrates the orthogonal directions. – Dillon M Mar 4 '17 at 0:02

The basis for the period lattice is given by the two vectors $$v_1 = \begin{pmatrix} 3 + \sqrt{3} \\ -1 \end{pmatrix}, v_2 = \begin{pmatrix} 1/2 \\ (2+\sqrt{3})/2 \end{pmatrix}.$$ So if there is an orthogonal period, then there has to be integers $a_1, a_2, b_1, b_2$ such that $\langle a_1 v_1 + a_2 v_2, b_1 v_1 + b_2 v_2 \rangle = 0$ and $(a_1, a_2), (b_1, b_2) \neq (0,0)$. Expanding out, we can write this as $$(13 + 6\sqrt{3}) a_1 b_1 + \frac{1}{2} (a_1 b_2 + a_2 b_1) + (2 + \sqrt{3}) a_2 b_2 = 0.$$ Since $\sqrt{3}$ is irrational, we then obtain \begin{align*} 0 &= 6 a_1 b_1 + a_2 b_2, \\ 0 &= 26 a_1 b_1 + (a_1 b_2 + a_2 b_1) + 4 a_2 b_2. \end{align*} Multiplying $-25 + \sqrt{7}$ to the first equation and multiplying $6$ to the second equation and adding them up gives \begin{align*} 0 &=(6 + 6 \sqrt{7}) a_1 b_1 + 6(a_1 b_2 + a_2 b_1) + (-1 + \sqrt{7}) a_2 b_2 \\ &= (-1 + \sqrt{7}) ((1 + \sqrt{7}) a_1 + a_2) ((1 + \sqrt{7}) b_1 + b_2). \end{align*} Then again by irrationality of $\sqrt{7}$, either $a_1 = a_2 = 0$ or $b_1 = b_2 = 0$.