# Do all compact manifolds have finite volume?

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 '18 at 20:02

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.
• Can you clarify something for me? Why can we find a bounded coordinate chart around $p$? And why does every component of metric $g=(g_{ij})$ is bounded, actually what does this line mean exactly? – clear Feb 14 at 5:43