Hearing the 17 planar symmetry groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:49:13Z http://mathoverflow.net/feeds/question/49685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49685/hearing-the-17-planar-symmetry-groups Hearing the 17 planar symmetry groups David Feldman 2010-12-16T21:41:32Z 2011-02-05T22:58:23Z <p>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. </p> <p>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.</p> <p>(Feel free to add more tags if appropriate.)</p> http://mathoverflow.net/questions/49685/hearing-the-17-planar-symmetry-groups/54472#54472 Answer by Greg Kuperberg for Hearing the 17 planar symmetry groups Greg Kuperberg 2011-02-05T22:58:23Z 2011-02-05T22:58:23Z <p>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.</p> <p>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.</p> <p>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 <code>$\vec{k} \in 2\pi A^*$</code>, where <code>$A^*$</code> is the dual lattice of $A$.</p> <p>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 <code>$2\pi A^*$</code>.</p> <p>For example, suppose that $\mathbb{R}^2/A$ is the standard square torus and $A$ is the standard square lattice, so that <code>$A^* = A$</code>. 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 $n = m = 0$.</p> <p>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.</p>