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.

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