Hearing the 17 planar symmetry groups

2010-12-16T21:41:32Z

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.)

Answer by Greg Kuperberg for Hearing the 17 planar symmetry groups

2011-02-05T22:58:23Z

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.

Hearing the 17 planar symmetry groups - MathOverflow

Hearing the 17 planar symmetry groups

Answer by Greg Kuperberg for Hearing the 17 planar symmetry groups