Let $(M,g)$ be a $d$-dimensional smooth, riemannian manifold without boundary, which is geodesically complete, connected, has positive injectivity radius and a bounded geometry (i.e. all derivatives of the metric are bounded and $\det(g)\geq c>0$ uniformly on every local chart repsectively). Additionally we furnish this manifold with a geodesic atlas, which is at most countable and uniformly locally finite if we choose the radius of the geodesic balls small enough (and we assume this in the following), and consider the $L_2$-based Sobolev space $H^1_2(M)$ defined via the covariant derivative. For this space we still have the usual continuous Sobolev embeddings $H^1_2(M)\hookrightarrow L_q(M)$ for $\frac{1}{2}\geq \frac{1}{q}\geq \frac{1}{2}-\frac{1}{d}>0$. Of course in this generality, they cannot be compact, but I wonder if there is some "local compactness" comparable to the situation on $R^d$, where the sobolev embeddings are compact for bounded domains with Lipschitz boundary. More precisely I'm interested in the following: Let $(O,\kappa)$ be a local geodesic chart. Can $\overline{O}$ be defined as a compact smooth riemannian manifold with boundary? (For such manifolds the above embeddings $H^1_2(\overline{O})\hookrightarrow L_q(\overline{O})$ are compact)

Best regards

  • $\begingroup$ I may be missing something, but if you take a normal ball inside the neighborhood, its closure is diffeomorphic to the standard closed ball, and of course it is a Riemannian manifold with boundary. Did I get anything wrong? $\endgroup$ – Amitai Yuval Aug 25 '14 at 9:43
  • $\begingroup$ @amitai-yuval Yes, it is a manifold with boundary but I guess I should have emphasized the smooth and compact bit. However, the compactness as a metric space (!) follows from the fact, that the closure of the chart itself is diffeomorphic to a closed euclidean ball, which is compact. For the smooth structure we use the atlas of $(M,g)$ restricted to $\overline{O}$ which by the uniform local finiteness is even finite, since also $\overline{O}$ can only be nontrivially intersected with finitely many other charts of $(M,g)$. Thanks for the answer, it brought me on the right track. $\endgroup$ – somnaris Aug 25 '14 at 13:41
  • $\begingroup$ Great, have fun then! $\endgroup$ – Amitai Yuval Aug 25 '14 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.