Let $(M, g)$ be a non-compact smooth Riemannian manifold of dimension $n \ge 2$, and $G$ a subgroup of the isometry group of $(M,g)$, say with $G$ contained in the component of the identy.

Let $W^{1,2}_{G}(M)=\{f \in W^{1,2}(M)| \quad f\circ \phi =\phi \quad \forall \phi \in G\}$.

Is there any known result concerning the compactness of the Sobolev imbedding $W^{1,2}_{G}(M) \hookrightarrow L^p(M)$ for some subgroup $G$?

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    $\begingroup$ What do you mean by $\textrm{dim}\,G<\infty$? As I recall, the isometry group of a Riemannian manifold is finite dimensional. $\endgroup$ Commented May 3, 2011 at 21:09
  • $\begingroup$ Sorry, it's a mistake. What I meant is: $G$ contained in the component of the identity $\endgroup$ Commented May 3, 2011 at 21:29
  • $\begingroup$ You should not take the same letter for the metric and for elements of the group. $\endgroup$ Commented May 4, 2011 at 11:59
  • $\begingroup$ There is an old note by Naceur Achtaich (circa 1988) when $M\subset\mathbb R^3$ has a rotational symmetry about the $z$-axis. Very localised, but at least an exmple. $\endgroup$ Commented May 4, 2011 at 12:02

2 Answers 2


Sorry for posting this as an Answer, but I can't comment yet.

As far as I know (judging from Emmanuel Hebeys work on this subject), there are no generalised results on Sobolev embeddings on non-compact Riemannian manifolds unless they are complete.


Hi, You need additional geometric condition for the general case but considering the case of $\mathbb{R}^n$ with $G=SO(n)$, i.e. $H^1_ {radial}$, you have compact injection. You will find all the details in chapter 9 of the excellent book of Hebey Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities.

  • $\begingroup$ I'm quite new in this field, but in the new book by Hebey [Sobolev spaces on Riemannian manifolds] he only considers the case when $M$ is compact. Anyhow, I have to the reference you've just mentioned. $\endgroup$ Commented May 4, 2011 at 9:54
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    $\begingroup$ I give you the precise reference, it is also about non-compact case! you have just to read... $\endgroup$
    – Raphael
    Commented May 4, 2011 at 13:36

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