Let $(M, g)$ be a complete Riemannian manifold diffeomorphic to $\mathbb{R}^n$. Under appropriate geometric assumptions concerning the geometry near infinity, but without any curvature sign assumptions, is the spectrum of the Laplace-Beltrami operator known to be qualitatively similar to other situations? (i.e. positive, discrete, ... )
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$\begingroup$ Discrete? That's not true even in the scalar case. $\endgroup$– Willie WongCommented Dec 15, 2023 at 2:32
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$\begingroup$ But for the Laplace-Beltrami operator, if $T$ is a smooth, compactly supported tensor field, and denoting by $\langle,\rangle$ the metric pairing of tensors, then $\int_M \langle\triangle g T,T\rangle = - \int_M \langle \nabla T, \nabla T\rangle$ which should give you whatever positivity (or negativity I guess) you want. $\endgroup$– Willie WongCommented Dec 15, 2023 at 2:40
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