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Given a planar domain $\Omega \subset \Bbb{R}^2$ bounded and open we can associate to it the spectrum of the Laplace operator with Dirichlet boundary condition. It is known that there are planar domains which are essentially different, but share the same spectrum. The known isospectral domains are not convex, and I think that there are no known isospectral convex domains.

Still, there are a few cases in which the spectrum determines (up to a rigid motion) the shape. For instance we have the circles and rectangles.

Are there other classes for which we know that the spectrum completely determines the shape?

Are there any results on "almost isospectral convex domains"? (i.e. the fact that $\lambda_k(\Omega_1)=\lambda_k(\Omega_2)$ for $k \leq n$ can say something about the resemblance of the convex sets $\Omega_1$ and $\Omega_2$)

If you know any good references on these matters, please share them.

After doing the suggested search I found the following results:

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    $\begingroup$ Google 'hearing the shape of a triangle' for a proof that a triangle is determined by its spectrum $\endgroup$
    – Otis Chodosh
    Commented Dec 22, 2013 at 18:56
  • $\begingroup$ @OtisChodosh: Thanks. I have found a few interesting articles. $\endgroup$
    – Beni Bogosel
    Commented Dec 22, 2013 at 19:52
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    $\begingroup$ Nice question by the way! Here's a related open problem which I think is quite nice (sort of related to your "almost isospectral" question for $k=1$): What is the Faber-Krahn (en.wikipedia.org/wiki/…) inequality on polygons? In other words, what $n$-gon of fixed area minimizes the first Dirichlet eigenfunction? For $n=3,4$ (triangles, quadrilaterals) it is known that the minimizer is the regular $n$-gon. Polya and Szego conjectured that this holds for $n\geq 5$ but very little progress has been made! $\endgroup$
    – Otis Chodosh
    Commented Dec 23, 2013 at 21:32
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    $\begingroup$ The following paper contains some information about it (and some related questions): projecteuclid.org/… $\endgroup$
    – Otis Chodosh
    Commented Dec 23, 2013 at 21:33
  • $\begingroup$ @OtisChodosh: I was doing some numerical computations yesterday on exactly the same thing (Faber-Krahn for polygons), obtaining regular polygons, as expected. $\endgroup$
    – Beni Bogosel
    Commented Dec 23, 2013 at 22:02

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