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If an $n$-gon $P$ is isospectral to a regular $n$-gon $Q$, what could we say about the shape of the $P$. Otherwise, what could we say about $Q$? In fact, some hints or simply some ideas would be appreciated.

Clarification : I talk about the spectrum of the Laplacian on the interior of the polygon, acting on the space of functions vanishing on the boundary.

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    $\begingroup$ Are you familiar with the examples from the 1990s (e.g. Gordon-Webb-Wolpert, Conway, others) constructed of isospectral planar polygons? If you wish to restrict to convex $n$_gons, I believe "isospectral $\Rightarrow$ isometric" is still open for convex domains in general. If $n=3$, it is known (Durso, Hillairet, Grieser-Maronna) that isospectral $\Rightarrow$ isometric. $\endgroup$
    – Neal
    Commented Jul 27, 2016 at 13:34
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    $\begingroup$ @Neal Thanks for you answer. In fact, the problem isn't open since November 2015 (see Theorem 4, The Sound of Symmetry). I'm trying to prove that problem by myself in using something simpler; that why I wanted some clues. (P.S. Yes, I know the problem of Gordon-Webb-Wolpert. It is convenient to solve that problem before launching into a such problem.) $\endgroup$
    – David Labrecque
    Commented Jul 27, 2016 at 13:53
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    $\begingroup$ Wasn't there a theorem that both area and perimeter are spectral invariants, whence $P \cong Q$ by the isoperimetric inequality? $\endgroup$
    – Noam D. Elkies
    Commented Jul 27, 2016 at 16:30
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    $\begingroup$ @NoamD.Elkies Yes, area and perimeter are (derivable from) coefficients in the heat trace. (One can also get a value derived from angles, see Mazzeo-Rowlette arxiv.org/abs/0901.0019) $\endgroup$
    – Neal
    Commented Jul 27, 2016 at 17:57
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    $\begingroup$ A websearch for "the sound of symmetry" came up with an EP by the metalcore band Sky Eats Airplane, but also with jstor.org/stable/10.4169/amer.math.monthly.122.9.815 (Lu and Rowlett, Amer Math Monthly 122 (November 2015) 815-835), which, I presume, is what OP had in mind. It might have been better for all, had OP told us what he knew when he first posted the question. $\endgroup$
    – Gerry Myerson
    Commented Jul 27, 2016 at 23:07

1 Answer 1

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Rowlett is hosting The Sound of Symmetry here. The proof of Theorem 4 is exactly as Noam Elkies suggests: Via the Dirichlet heat trace's asymptotic expansion, both area and perimeter are determined by the spectrum, and so for any $n$-gon $\Omega$ the isoperimetric ratio $|\Omega|/|\partial\Omega|^2$ is determined by the spectrum. The content of the proof is that this ratio is globally maximized among $n$-gons at the regular $n$-gon. This determines the regular $n$-gon and implies that any $n$-gon isospectral to the regular $n$-gon is in fact isometric to it. Unless there's a point in the proof where you're confused, I think this settles the question.

(I have a suspicion that the third coefficient of the heat trace in convex polygons $$ \frac{1}{24}\sum_{\mbox{angles }\alpha_i} \bigg(\frac{\pi}{\alpha_i} - \frac{\alpha_i}{\pi}\bigg) $$ is also extremal at the regular $n$-gon. This may be another way to show a version of the desired result for convex $n$-gons.)

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    $\begingroup$ Your "suspicion" is correct, and can be proved in various standard ways (convexity, AM-HM inequality, Cauchy-Schwarz, . . .). Note that the $\sum_i -\alpha_i/\pi$ part of the sum is constant. $\endgroup$
    – Noam D. Elkies
    Commented Jul 28, 2016 at 2:27
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    $\begingroup$ @NoamD.Elkies Thank you for the addendum. Glad my instinct is on point here -- looks like I have a nice little exercise for the train ride in the morning. :) $\endgroup$
    – Neal
    Commented Jul 28, 2016 at 2:42

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