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It's known that there are pairs of closed Riemannian manifolds which are isospectral but not isometric. Is it known if there are closed Riemannian manifolds which are isospectral but not homeomorphic? (By isospectral, I mean that the Laplace-Beltrami operator on functions has the same spectrum on both manifolds.)

Thanks,

Dmitri

flag	asked May 16 2012 at 3:58 Dmitri Gekhtman 245●6

Henry Cohn · Accepted Answer · 2012-05-16 04:30:30Z UTC

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There's an example due to Doyle and Rossetti ("Tetra and Didi, the cosmic spectral twins"; http://arxiv.org/abs/math.DG/0407422) of 3-manifolds that are isospectral but not even homeomorphic. I don't know if this was the first such example.

answered May 16 2012 at 4:30

Henry Cohn
6,749●1●23●40

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There's older examples, e.g. of Vigneras: ams.org/mathscinet-getitem?mr=584073 – Agol May 16 2012 at 4:40

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Sunada (ams.org/mathscinet-getitem?mr=782558) gave a general framework in which the results of Vigneras fit. – Jan Jitse Venselaar May 16 2012 at 9:43

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Today I learned that you can't hear the topology of a drum. – Steve Huntsman May 16 2012 at 15:27

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Agol, are you sure, that Vigneras constructs isospectral varieties that are not homeomorphic, as opposed as only not isometric? I haven't read the paper in details, but the mathscinet reviews, the introduction of the paper, and the last theorem (theorem 8) says "isometric". – Joël May 16 2012 at 16:14

@Joel: For hyperbolic manifolds of dimension $>2$ homonorphism implies isometry (Mostow Rigidity). – Misha Jan 2 at 23:16

Are isospectral manifolds necessarily homeomorphic?

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