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On a Riemannian manifold $M$, the Laplace operator $L$ is uniquely defined.

If $M$ is not compact, then $L$ admits a continuous spectrum.

Is there a way of "changing" $M$ and/or $L$ in say $M'$ and $L'$, such that the spectrum of $L'$ is now purely discrete and contains the spectrum of $L$? I am interested in examples of this type.

Feel free to restrict to homogenous varieties or any other geometric gadget, if have an example of similar nature in mind.

Motivation: More general it is true that, given an self adjoint operator $T$ on a Hilbert space, does there exists a decomposition such that $T_1 + T_2 = T$, where $T_1$ has the same point spectrum as $T$. I am not looking for this. Is there a better geometric construction?

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What if the operator $T$ has no point spectrum? (Or in the manifold case, what if $(M,L)$ is the Euclidean space?) I may be misunderstanding what you mean/want, can you clarify? –  Willie Wong Jun 24 '11 at 1:55
    
I was just talking about containment, the empty set is contained everywhere. –  plusepsilon.de Jun 24 '11 at 9:19

2 Answers 2

up vote 4 down vote accepted

Surely that would have outstanding consequences if it were true.

Let's just consider the case where $M$ is a hyperbolic Riemann surface, and assume arithmetic if non-compact. Then the main term in Weyl's laws for the discrete spectrum doesn't depend upon being closed or having cusps, just upon the volume, but the next term in the remainder does. The discrete spectrum grows faster in the presence of cusps, so one cannot hope to pass from an arithmetic surface with cusps to a compact surface with the same discrete spectrum. See for example M\"uller's paper: "Weyl's law in the theory of automorphic forms."

The work of Phillips, Sarnak and others, indicates that that discrete spectrum varies wildly under perturbations, and it would seem rather improbable (at least to me), that the phenomenon you're looking for ever occurs (or at least no more common than instances of isopectral manifolds).

Incidentally, Phillips-Sarnak suggests that for a general non-arithmetic manifold with cusps, the discrete spectrum will be finite, in contrast to the case of a compact manifold. I don't have any intuition in this case for whether such a spectrum might be contained in the spectrum of a compact manifold, but neither is it clear to me how useful knowing this would be.

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That is a perfect answer, maybe the Jacquet Langland correspondance is a suitable example in this context. –  plusepsilon.de Jun 24 '11 at 9:17

A different "adjustment" is as in Colin-de-Verdiere's 1981 treatment of meromorphic continuation of Eisenstein series (at least for rank-one?): he replaces the usual Laplacian (which is essentially self-adjoint) by a pseudo-Laplacian created as the minimal ("Friedrichs") extension of the restriction of the Laplacian to automorphic forms whose constant terms vanish above a fixed height. A variant of the argument for discreteness of cuspforms proves that this pseudo-Laplacian has a compact resolvent, hence, discrete spectrum. The cuspforms remain eigenfunctions, the constants cease to be eigenfunctions because they are not in the space, and certain truncated Eisenstein series become genuine eigenfunctions for the pseudo-Laplacian (rather than failing to be eigenfunctions for the ordinary Laplacian).

Edit: and, in regard to some aspects of @Kimball's points, specifically (looking at the details of C-de-V's discussion), the truncated Eisenstein series are essentially those whose truncated constant term is still continuous, that is, so that, at height $y_o$, in the simplest case the condition for the truncation $\wedge^{y_o} E_s$ to be a "new" eigenfunction is that $y_o^s+c_s y_o^{1-s}=0$, where $c_s=\xi(2s-1)/\xi(2s)$. Thus, standard (if non-trivial) facts about zeta (and a refined Weyl's Law for cuspforms?) still assure that the "new" eigenfunctions are asymptotically fewer, quantifiably.

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