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I understand that for infinite-area hyperbolic surfaces, there are no $L^2$-eigenfunctions of the Laplace-Beltrami operator but there are a lot resonances.

But I am confused about the notion of resonances for finite-area hyperbolic surfaces. Are there resonances which are not eigenvalues?

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    $\begingroup$ I don't know the answer to the question, but I'll point out that, for some infinite-area hyperbolic surfaces, there do exist L^2-eigenfunctions (for instance, if the Hausdorff dimension $\delta$ of the limiting set is large enough, there is a "base" eigenfunction with eigenvalue $\delta(1-\delta)$). I think this is explained in the paper of Lax and Phillips in Journal of Functional Analysis 46, 1982. $\endgroup$
    – Denis Chaperon de Lauzières
    Commented Oct 20, 2013 at 17:21
  • $\begingroup$ Thanks a lot, I clearly misread finite for none in Borthwick. $\endgroup$
    – Sabine Müller
    Commented Oct 21, 2013 at 18:15

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I presume by a "resonance" you mean a scattering resonance, so a pole of the determinant of the scattering matrix. On finite-area hyperbolic surfaces all scattering resonances correspond to eigenvalues, but not the other way around. The "cusp form" spectrum, which includes all embedded eigenvalues, does not give rise to scattering resonances.

See page 20 of Introduction to Spectral Theory on Hyperbolic Surfaces by David Borthwick.

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  • $\begingroup$ Thanks a lot, in particular for the reference. On the same page 20, below the "resonances = scattering poles + cusp resonances" there is this "not quite accurate". Does it means it might happen that a scattering pole is also a cusp resonance? If so, are there known examples? $\endgroup$
    – Sabine Müller
    Commented Oct 21, 2013 at 18:20
  • $\begingroup$ what is intended by this "not quite accurate", I think, is that it may happen, exceptionally, that an eigenvalue is not of the cusp form and still does not give rise to a scattering resonance. $\endgroup$
    – Carlo Beenakker
    Commented Oct 21, 2013 at 20:59

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