Smooth Manifold admits Riemannian Metric

Theorem

Let $M$ be a smooth manifold.

Then $M$ admits a Riemannian metric.

This article, or a section of it, needs explaining. In particular: What does this "admit" mean? You can construct a metric from the Euclidean metric and the differentiable structure on $M$ OK, now I probably understood. The statement should be then more explicit like: there exists a Riemannian metric such that 1). .. 2). .. etc. Otherwise it is not clear what is to be proved. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Proof

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 2018: John M. Lee: Introduction to Riemannian Manifolds (2nd ed.) ... (previous) ... (next): $\S 2$: Riemannian Metrics. Definitions