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Problem 4.3 in Mathematical Methods for Physics and Engineering by M. Blennow asks us to find the symmetries of a parallelogram tiling of the plane:

Parallelogram tiling

The solutions manual lists the following types of symmetry transformations: translations along either edge direction by a multiple of the side length, and 180° rotations about a vertex, the midpoint of an edge, or about the midpoint of a parallelogram. The different types of rotations can all be related by translations, so all of the above can be generated by the transformations $T_1$, $T_2$, and $c$ illustrated below:

Generators of the given symmetries

However, I think that there should be one more. To get a non-ambiguous characterization of it, let us pick some vertex as the origin and introduce a basis as below:

Origin and basis

Then, of course, each point in the plane can be uniquely identified as $x^1 \vec e_1 + x^2 \vec e_2$, for some $x^1, x^2 \in \mathbb R$, and we can define a transformation $\sigma$ by its action $$x^1 \vec e_1 + x^2 \vec e_2 \mapsto -x^1 \vec e_1 + x^2 \vec e_2.$$ It seems to me that $\sigma$ is a symmetry of the tiling. Moreover, it cannot be constructed from the generators already listed, because it is not parity conserving (i.e., as a linear transformation it has negative determinant).

So, why is $\sigma$ not regarded as a symmetry transformation? Is there some conventional restriction on the types of transformations that we regard as symmetry transformations that forbids $\sigma$?


One commenter did not believe that the pattern is invariant under $\sigma$, so here is a short proof. In the introduced basis each horizontal line can be written as $$H_n=\{r\vec e_1+n\vec e_2:r\in\mathbb R\}$$ for some integer $n$. Similarly each "skew-vertical" line can be written $$V_n=\{n\vec e_1+r\vec e_2:r\in\mathbb R\}.$$ Hence $$\sigma:H_n\mapsto\{-r\vec e_1+n\vec e_2:r\in\mathbb R\}=H_n$$ and $$\sigma:V_n\mapsto\{-n\vec e_1+r\vec e_2:r\in\mathbb R\}=V_{-n}.$$

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    $\begingroup$ Any (reversible) variable change that helps solve the poroblem (for, example, separate variables) is useful in practice; don't stick to symmetries - they represent too narrow case of variable changes. $\endgroup$ May 3, 2021 at 17:48
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    $\begingroup$ @VladimirKalitvianski Sure, but I don't quite see how that is relevant in this case, because the problem is specifically about finding symmetries. $\endgroup$
    – ummg
    May 3, 2021 at 18:01
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    $\begingroup$ Because after your horizontal flip, the pattern is not the same as the original pattern and hence isn't a symmetry. $\endgroup$
    – Triatticus
    May 3, 2021 at 19:05
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    $\begingroup$ @HansWurst No, it is more like a reflection (but not quite since it is not orthogonal). As you say, a point $x^1\vec e_1+x^2\vec e_2$ is translated by $-2x^1\vec e_1$, but, since this is $x^1$-dependent, points on different lines are are translated by different amounts. A translation transformation on the other hand translates the whole plane equally. $\endgroup$
    – ummg
    May 3, 2021 at 21:27
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    $\begingroup$ Check en.wikipedia.org/wiki/Wallpaper_group. "The types of transformations that are relevant here are called Euclidean plane isometries". Your transformation is not an isometry because angle $\theta$ changes to $180-\theta$. Isometries must preserve length and angle. $\endgroup$
    – isometry
    May 3, 2021 at 21:35

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Check wallpaper group. "The types of transformations that are relevant here are called Euclidean plane isometries". Your transformation is not an isometry because the angle θ between $e_1$ and $e_2$ changes to 180−θ. Isometries must preserve length and angle.

Mathematically we can (and physicists often do) consider larger symmetry groups than Isometry Groups. For example transformations which preserve angle but not length form the Conformal Group; smooth reversible transformations which preserve neither angle or length form the Diffeomorphism Group.

We can also consider isometry groups which preserve length and angle with respect to a metric other than the Euclidean metric. For example in Special Relativity the appropriate isometry group is called the Lorentz Group.

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    $\begingroup$ This seems to be it. It seems a bit arbitrary though, since the reason that $\theta=\vec e_1\angle\vec e_2\mapsto\sigma(\vec e_1)\angle\sigma(\vec e_2)=\pi-\theta$ is that vectors have an orientation (a notion of backwards and forwards), and the vectors are not fundamental to the pattern, but something we imposed. The lines themselves do not have that, and the angle between two lines could be said to be conserved by the transformation because the pattern looks the same after the transformation. $\endgroup$
    – ummg
    May 4, 2021 at 13:31
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    $\begingroup$ Agree. But the reason isometries are interesting is because you can physically apply them to solid objects (directly in the case of translations and rotations; for reflections you need to flip the paper). So yes your transformation does preserve the pattern, but it does not belong to the set of such rigid actions. $\endgroup$
    – isometry
    May 4, 2021 at 14:58
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    $\begingroup$ That is a fair point! However, we cannot physically apply a reflection to a solid object either, yet we commonly consider reflective symmetry transformations. $\endgroup$
    – ummg
    May 4, 2021 at 15:03

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