Let $M$ be a compact $k$-dimensional manifold embedded in $\mathbb{R}^d$. Does $M$ being compact imply that the $k$-dimensional volume of $\operatorname{vol}_{k}(M)$ is finite?

  • Yes, use the Hopf-Rinow theorem to bound $\operatorname{vol}_k(M)$ by an integral of a bounded continuous function on a compact subset of a tangent space. – user10354138 Nov 8 at 20:02
up vote 4 down vote accepted

Yes: any compact manifold $M$ has finite volume with respect to any Riemannian metric. Given any point $p\in M$, we can find a bounded coordinate chart around $p$ on which each component of the metric is bounded, so that coordinate chart will have finite volume. Since $M$ is compact, it is covered by finitely many of these finite-volume coordinate charts, and so $M$ has finite volume.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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