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?



Periodic tilings of the plane by regular polygons
I claim that the tiling <https://upload.wikimedia.org/wikipedia/commons/6/66/5-uniform_310.svg> does not admit a orthogonal period. 


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 $b\_1 = b\_2 = 0$. 


