Symmetries and eigenvalues of the Laplacian.

Lets consider a domain $\Omega \subseteq \mathbb R^2$ smooth enough, and the eigenvalue for the laplacian \begin{align} -\Delta u &= \lambda u &x\in\Omega\\ u &= 0 &x\in \partial \Omega \end{align} I am interested in an explicit relation between the group of symmetries of $\Omega$ and the multiplicity of the eigenvalues of $-\Delta$.The idea is that if I have a symmetry $R$ and an eigenfunction $u$, then $u(R(x))$ will be also an eigenfunction with the same eigenvalue. For instance, in the case of the square we have as symmetry group $D_4$.In this case a base of eigenfunctions is (for the square with sides of length $\pi$) $$u_{nm}(x,y) = \sin(n x)\sin(m y)$$ so for $n \not=m$ we have that $u_{nm}$ and $u_{mn}$ are two l.i. function associated to the same eigenvalue, and we can obtain one from the other with the symmetry $R:(x,y) \rightarrow (y,x)$.But not every degeneracy can be attributed to the symmetries, for instance $u_{5,5}$ and $u_{1,7}$ are associated to the same eigenvalue, but there is not symmetry between them.

On the other hand, symmetries doesn't implies degeneracy, for instance for a rectangle with aspect ratio not the square root of a rational, the spectrum is simple, but still has some symmetries $\mathbb Z_2 \times \mathbb Z_2$.

In particular, i would like to know if no symmetries implies no degeneracy of the spectrum.

Solution 1:

For the last question, you already gave an answer. You noticed that on the square, $u_{5,5}$ and $u_{1,7}$ are not linked by symmetries, but have the same eigenvalues.

Notice now that the sinuses functions take the value zero on any multiple of $\pi$, so you can build a non-symmetric domain $\Omega$ by gluing togheter adjacent squares. For example, take $$ \Omega = [-\pi,\pi]^2 \,\,\cup\,\, [-\pi,\pi]\times[\pi,3\pi]\,\,\cup\,\, [\pi,3\pi]\times[-\pi,\pi]\,\,\cup\,\, [3\pi,5\pi]\times[-\pi,\pi] $$ This domain has no symmetries, but $u_{5,5}$ and $u_{1,7}$ are still l.i. and have the same eigenvalues.