Can one hear orientability?

Mark Kac asked in a paper from 1966 the following question: Can one hear the shape of a drum?

The mathematically precise question is the following: Assume that \((M,g)\) and \((N,h)\) are two compact Riemannian surfaces (thought of as the heads of two drums). If the Laplace operators of \((M,g)\) and of \((N,h)\) have the same eigenvalues (for Dirichlet boundary conditions), must \((M,g)\) and \((N,h)\) be isometric?

The motivation for the question comes from the following facts (which are true for arbitrary [closed] Riemannian manifolds):

  • First, note that the Laplace operator determines the Riemannian metric \(g\). The reason for this that in local coordinates the Laplace operator is of the form \[\Delta f = -g^{ij} \partial_j \partial_i f + (\text{lower order terms}).\] This means that if we evaluate the operator on suitable functions we can recover the metric coefficients \(g^{ij}\).
  • If we only know the spectrum of the Laplace operator, then we can extract geometric information from it in the following way. If we denote by \(e(t,x,y)\) the heat kernel, i.e., the fundamental solution to the heat equation on \((M,g)\), then it satisfies \[e(t,x,y) = \sum_{\lambda \in \sigma(\Delta)} e^{-\lambda t} \phi_\lambda(x) \phi_\lambda(y),\] where \(\phi_\lambda\) is a normalized eigenfunction for the eigenvalue \(\lambda\). So especially, the quantity \(\int_M e(t,x,x) dx = \sum_{\lambda \in \sigma(\Delta)} e^{-\lambda t}\) only depends on the spectrum of \(\Delta\). On the other hand, expanding asymptotically for \(t \to 0\) we get \[\int_M e(t,x,x) dx = (4\pi t)^{-\mathrm{dim}(M)/2} \big( \mathrm{vol}(M) + t \cdot \mathrm{scal}_g / 6 + \cdots \big).\] Therefore, the spectrum of the Laplace operator determines the dimension, the volume and the total scalar curvature (and infinitely many more geometric quantities, but they are complicated expressions in the curvature tensor and its derivatives and therefore it is hard to extract geometric information from them).

First examples of closed Riemannian manifolds which have the same spectrum of the Laplacian but which are not isometric were given by Milnor (these were 16-dimensional flat tori). But on the other hand, it is known that the round sphere in dimensions up to 6 is spectrally rigid, i.e., if a Riemannian manifold \((M,g)\) has the same spectrum of the Laplacian as a round sphere of dimension at most 6, then \((M,g)\) is actually indeed a round sphere.

So one can ask which further properties of Riemannian manifolds (besides the above mentioned ones) are encoded in the spectrum of the Laplacian. Last month there was a paper put on the arXiv (arXiv:2008.12498) showing that orientability of surfaces with boundary is not encoded in the spectrum of the Laplacian (with Neumann boundary conditions). But note that this result is not knew: as the authors write in the abstract, the paper was already circulated in 1995.

Finally, what about the original question of Mark Kac? Well, in 1991 it was answered negatively by Gordon-Webb-Wolpert: they constructed non-isometric planar domains with both the same Dirichlet and Neumann spectrum. However, these domains do not have a smooth boundary; and actually, Kac’ question is still unsolved in this case.

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