Do All Smooth Manifolds Admit a Complete Riemannian Metric?

[mtiano]

Is it true that any smooth manifold admits a complete riemannian metric? Can you prove it? If not can you give a counter example? Obviously we can always put a riemannian metric on any smmoth manifold the question is does the differentable structure allow us to find a complete one.

 

[pmb_phy]

Is it true that any smooth manifold admits a complete riemannian metric? Can you prove it? If not can you give a counter example? Obviously we can always put a riemannian metric on any smmoth manifold the question is does the differentable structure allow us to find a complete one.

Although I'm not 100 % sure on what you mean by the qualifier "complete" but any smooth manifold can be given more structure by adding a metric to the space. To see this recall that every point on a manifold looks locally like Rⁿ which can always be given a metric. So when you specify the distance relationship you wish to to all points on the manifold then you've specified a metric for that manifold.

Best wishes

Pete