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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?

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  • $\begingroup$ Does this upload.wikimedia.org/wikipedia/commons/6/6a/2-uniform_17.png have an orthogonal period? $\endgroup$ – Dongryul Kim Mar 3 '17 at 23:32
  • $\begingroup$ It seems that, computationally, there is no orthogonal period, but someone who knows number theory would need to confirm this. $\endgroup$ – Dongryul Kim Mar 3 '17 at 23:46
  • $\begingroup$ 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. $\endgroup$ – Dillon M Mar 4 '17 at 0:02
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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 $b_1 = b_2 = 0$.

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enter image description here

An image of a tiling, with its orthogonal directions of periodicity, posted in response to an earlier comment.

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If you tile the plane with equilateral triangles so that the triangles share sides, they will naturally form rows, and the two directions of periodicity will be horizontal and vertical. You can shift these rows to produce a non-vertical direction. Pick a row and shift the row above it by d, then shift the row above that by 2d, the one above that by 3d, etc. Use negative values for the rows below.

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  • $\begingroup$ Thanks for the response. There is something I am not getting, however. What is d in your construction? If d is (a multiple of) the edge-length, then the tiling will be the regular tiling with equilateral triangles. If d is some other length, then the tiling will not be edge-to-edge. Is there something I am missing? $\endgroup$ – Dillon M Mar 3 '17 at 23:21
  • $\begingroup$ The tiling won't be edge to edge anymore, but it will still be periodic. $\endgroup$ – BallpointBen Mar 3 '17 at 23:31
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    $\begingroup$ Yes, I agree with that. I think the question is only interesting if you restrict to edge-to-edge tilings with regular polygons, because of the example you give and other similar ones. $\endgroup$ – Dillon M Mar 3 '17 at 23:44

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