# Can you hear your location on a manifold?

Can one hear the shape of a drum? is a famous paper by Mark Kac from 1966. The precise mathematical question stated therein is the following: If the Laplace operators of the two Riemannian manifolds with boundary (M,g) and (N,h) have the same eigenvalues (for Dirichlet boundary conditions), must (M,g) and (N,h) be isometric? I blogged about this problem already twice (here and here).

In April of this year a preprint titled Can you hear your location on a manifold? was put on the arXiv by Wyman and Xi (arXiv:2304.04659). A bit more concretely, the question is this: You stand on a compact Riemannian manifold, make a single sharp “snap” sound, and then listen intently to its reverberations. If you have perfect hearing and perfect knowledge of the shape of the manifold, can you deduce your location (up to isometries) within it?

Mathematically the question means the following: Let $$(e_j)_{j \in \mathbb{N}}$$ be an orthonormal basis of eigenfunctions of the Laplace operator (with Dirichlet or Neumann boundary conditions in case the manifold has non-empty boundary) an denote by $$\lambda_j$$ the corresponding eigenvalues. For a point $$x$$ of the manifold the Weyl counting function is defined as $N_x(\lambda) := \sum_{\lambda_j \le \lambda} |e_j(x)|^2\,.$ Can one now recover (up to isometries) the point $$x$$ from the knowledge of the function $$N_x$$, i.e. if $$N_x = N_y$$ must there exist an isometry mapping $$x$$ to $$y$$?

The result of the above mentioned preprint is now the following: On any compact, smooth manifold without boundary and of dimension at least 2 a generic Riemannian metric has the property that $$N_x = N_y$$ implies $$x=y$$. Generic means here that the set of these metrics has meager complement in the $$C^\infty$$-topology on all Riemannian metrics.