Complete riemannian metric

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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.
 

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mtiano said:
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
 

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