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Let $(M,g)$ be a complete Riemannian manifold and $\Delta_{g}$ the Laplace-Beltrami operator.

What geometric and topological conditions on $M$ are needed so that $\Delta_{g}$ is uniformly elliptic?

yess
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    From the wiki you linked it said the Laplacian is an example of a uniform elliptic operator. –  Jan 06 '16 at 03:57

1 Answers1

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Since the Laplace-Beltrami operator is $\Delta_g = g^{ij} \nabla_i \nabla_j$, its principal symbol is simply $\sigma_\Delta(\xi) = g^{ij} \xi_i \xi_j = \Vert \xi \Vert^2$. Thus $\Delta_g$ is uniformly elliptic with elliptic constant $1$.

Anthony Carapetis
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  • Thank for your answer. I am confused as I thought the condition of uniform ellipticity involves the euclidean norm.That is $\sigma {\Delta }(\xi)\ge \theta (\sum{i}\xi_{i}^{2})$. – yess Jan 06 '16 at 12:23
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    @yess: There is no "Euclidean norm" on an arbitrary Riemannian manifold - your expression is dependent on the coordinates you use. In this setting the only reasonable definition of uniform ellipticity uses the norm induced from the metric. – Anthony Carapetis Jan 06 '16 at 13:06
  • I was under the impression that formally the condition is a condition on each point of the cotangent bundle which is just $R^{n}$ – yess Jan 06 '16 at 20:49
  • @AnthonyCarapetis on the wiki page for the definition of ellipticity, the introduction part simply says that ellipticity means the coefficients of the highest order derivatives are positive, which seems to be very weird because I saw somewhere in the literature that says $-\nabla^2$ is also elliptic – M. Zeng Jan 15 '16 at 06:09
  • @AnthonyCarapetis also, regarding your answer, is it still true if the metric is position dependent? – M. Zeng Jan 15 '16 at 06:10
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    @M.Zeng: some authors use the opposite sign for the Laplacian. And yes - I'd say all metrics are "position dependent", but there's no assumption of flatness if that's what you're asking. – Anthony Carapetis Jan 15 '16 at 06:14