# Hearing the 17 planar symmetry groups

Though I'm sure it's really hard to work out for myself, does anyone know a reference for the spectra of the Laplacian on the 17 flat compact orbifolds that underlie the 17 planar symmetry groups.

I'm thinking Neumann boundary conditions to model reflection lines. And I do realize that these spectra may vary according to a certain number of moduli depending on the group.

(Feel free to add more tags if appropriate.)

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Wouldn't you use rather boundary conditions on the fundamental domain imposing that the functions be compatible with the identifications? –  Mariano Suárez-Alvarez Dec 16 '10 at 22:28
I may be mistaken, but perhaps you mean Neumann boundary conditions, I believe a different guy from von Neumann. –  drbobmeister Dec 17 '10 at 0:32
Tricky. Thanks drbob! –  David Feldman Dec 17 '10 at 1:59
Carl Gottfried Neumann (May 7, 1832 - March 27, 1925) –  David Feldman Dec 17 '10 at 2:54
You probably know this, but one may compute the spectrum of the minimal torus (orbifold) cover using Fourier analysis, and then you need to analyze the invariant subspaces under the finite symmetry group. I suspect this may have been worked out classically for the wallpaper groups, but I'm not enough of an expert to know a reference. However, here's a reference where this has been carried out in 3-D: front.math.ucdavis.edu/0407.5422 You might be able to work backwards through the literature from these authors to the 2-D case. –  Ian Agol Dec 17 '10 at 22:59
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I don't think that it's so hard to work out the solution non-rigorously, nor all that onerous to do it rigorously following Ian Agol's suggestion in the comments. A reference could be nice, but I don't think that it's quite necessary, because the answer itself is not all that different from its derivation.

First, in the Euclidean plane $\mathbb{R}^2$, the plane wave $$f(\vec{x}) = \exp(i\vec{k} \cdot \vec{x})$$ is an eigenfunction of the Laplacian $-\nabla$ with eigenvalue $\vec{k} \cdot \vec{k}$. Suppose that the orbifold is expressed as $X = \mathbb{R}^2/\Gamma$, where $\Gamma$ is a discrete, cocompact group. Then if you average $f$ with respect to $\Gamma$, you will get a Laplace eigenvector on $X$ with the same eigenvalue. Since these plane waves are complete (in an analytic sense) upstairs, their averages are at least complete downstairs.

Let $A \subseteq \Gamma$ be the subgroup of translations of $\Gamma$. Then the $A$-average of $f$ is either $f$ again or it vanishes. It's $f$ precisely when $\vec{k} \in 2\pi A^*$, where $A^*$ is the dual lattice of $A$.

Then there is a rotation group $R = \Gamma/A$ which is some finite group. You can now average $f$ over any lift of $R$ and see what you get. If $R$ lifts to a finite subgroup of $\Gamma$, then that subgroup, call it $R$ again, fixes a point, say the origin. In this case you get a basis of eigenfunctions using the $R$-orbits of $f$ and $\vec{k}$. The answer is all $\vec{k} \cdot \vec{k}$, where $\vec{k}$ represents each $R$-equivalence class in $2\pi A^*$.

For example, suppose that $\mathbb{R}^2/A$ is the standard square torus and $A$ is the standard square lattice, so that $A^* = A$. Suppose that $R$ is generated by a rotation by 90 degrees at the origin. Then $\vec{k} = 2\pi(n,m)$ and the eigenvalue is $4\pi^2(n^2 + m^2)$, where you should be careful to choose the pair of integers $(n,m)$ with $n \ge 0$ and $m > 0$, or $n = m = 0$.

If $\Gamma$ is a non-split extension of $R$ by $A$, then the answer is a little more complicated, but it's not that much more complicated. In fact the basic analysis is the same for compact Euclidean orbifolds in any dimension.

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