MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $G$?

share|cite|improve this question
What do you mean by $\textrm{dim}\,G<\infty$? As I recall, the isometry group of a Riemannian manifold is finite dimensional. – Somnath Basu May 3 '11 at 21:09
Sorry, it's a mistake. What I meant is: $G$ contained in the component of the identity – Mercy King May 3 '11 at 21:29
You should not take the same letter for the metric and for elements of the group. – Denis Serre May 4 '11 at 11:59
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. – Denis Serre May 4 '11 at 12:02

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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. – Mercy King May 4 '11 at 9:54
I give you the precise reference, it is also about non-compact case! you have just to read... – Raphael May 4 '11 at 13:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.