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Consider a smooth, compact Riemannian surface $\mathcal{S}$ in $\mathbb{R}^3$ and suppose we are given the complete set of eigenfunctions $\{\phi_i\}$ of the associated Laplace-Beltrami operator.

Is this information sufficient to fully reconstruct the surface?

[As far as I know, one can infer the Riemannian metric from the Laplace-Beltrami eigenfunctions and thus fundamental geometrical properties like curvature. But I don't have enough information to actually draw the surface in its embedding space, correct? With what (minimal) information can I supplement knowing the set $\{\phi_i\}$ such that I am able to draw the surface?]

EDIT: The above statement is wrong (thanks for the comments).

I still think the topic is interesting: To what extend and in what way are the Riemannian metric and the Laplace-Beltrami eigenvalue problem related and determined by each other?

madison54
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    How do we infer the metric from the eigenfunctions,exactly? – Robert Lewis Feb 13 '14 at 09:31
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    For a fixed Riemannian metric $g$ on $S$, there might be different isometric embeddings to $R^3$, while eigenfunctions and spectrum are intrinsic properties of $g$; therefore, you cannot reconstruct an embedding from the eigen-data. (I am also not sure if you can reconstruct $g$ either, but this is another story.) – Moishe Kohan Feb 13 '14 at 15:57
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    You cannot reconstruct a metric from the spectrum alone (the eigenvalues). You say you can do it with the eigenfunctions... that's interesting (if true). How? As for actually embedding it (isometrically), it's more difficult, as this is a global question (locally you can deform a surface isometrically, think of a cylinder vs cone). Sounds like a good question for Mathoverflow, to get people like Robert Bryant to answer. +1. – Gil Bor Feb 13 '14 at 17:06
  • I had the following paper in mind (Zeng et al., "Discrete Heat Kernel Determines Discrete Riemannian Metric"), which actually deals with a discrete case, but states that "...[the heat kernel] fully determines the Riemannian metric (uniquely up to a scaling)" and make another statement of this relation in their Theorem 2.2. The heat kernel itself is determined with the eigenvalue problem. But it seems that I misinterpreted their statement and claimed too much. Now that you ask, I am not able to write down the metric given the eigenfunctions. Thanks for pointing it out! I've edited my question. – madison54 Feb 13 '14 at 17:54
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    A version of this question has been asked before, and answered. It goes under the title made famous by Kac in 1966, "Can One Hear the Shape of a Drum?." Hear refers to the eigenvalues of the Laplacian. The answer is No, although the problem in 2D remained open into the 1990's, when it was shown to be No in the plane as well. – Joseph O'Rourke Feb 13 '14 at 18:36

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The Kac paper is about reconstructions from eigenvalues, and the question seems to be about eigenvectors. What do you mean 'eigenvectors are given' Eigenvectors are defined on the surface, so surface must be given before giving eigenvectors. Perhaps the OP meant eigenvalues? Coming back to the reconstruction from eigenvalues, from Kac paper mentioned above, we know that reconstruction of arbitrary surfaces is not possible. But if we know that the surface has spherical topology, the reconstruction is possible. I don't know the mechanism though.

AnandJ
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