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On a noncompact Riemannian manifold $M$, the $L^2$-spectrum of the Laplace-Beltrami operator $\Delta$ sits inside $\mathbb{R}$ (by self-adjointness), either to the left or to the right of $0$ depending on sign convention. I know that under various curvature assumptions one can show that the $L^p$-spectrum is equal to the $L^2$-spectrum for $p \neq 2$.

Without making any geometric assumptions on $M$, is it possible to conclude that the $L^p$-spectrum of $\Delta$ is a subset of $\mathbb{R}$? If not, is there anything we can say about the $L^p$-spectrum without having to make any additional assumptions?

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1 Answer 1

up vote 6 down vote accepted

The answer has a lot to do with the off-diagonal decay of the resolvent, (\Delta - \lambda)^{-1}. There are two rather different cases to consider, when M is R^n and when M = H^n. The former is [0,\infty), just as for L^2. This can be computed explicitly, but in fact there is a general theorem due to Sturm which states that the L^p spectrum of the scalar Laplacian is independent of p if the Ricci curvature is bounded below and if the volume of balls grows subexponentially. On the other hand, the L^p spectrum of \Delta on hyperbolic space behaves quite differently. There is a very nice paper by Davies, Simon and Taylor which explains this, but in retrospect the answer depends in that case rather simply on the known and very simple off-diagonal asymptotics of the Green function or resolvent (and the precise exponential decay of the volume form). The recent paper http://arxiv.org/pdf/0707.2477.pdf by A. WEber, see also some earlier work of his, shows that for various locally symmetric spaces of higher rank, the same sort of phenomenon occurs.
There are various other types of results scattered around the literature on this topic, bt I am not aware if there is (or could be) a sharp necessary and sufficient geometric condition for the spectrum to be independent of p.

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this is great, thanks! –  Alex Amenta Nov 7 '13 at 5:17

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