Not all smooth manifolds admit a complete Riemannian metric. While it is possible to assign a Riemannian metric to any smooth manifold, the differentiable structure does not guarantee the existence of a complete metric. A counterexample is the punctured Euclidean space, which cannot be given a complete Riemannian metric due to its topology. The discussion emphasizes the distinction between assigning a metric and achieving completeness in the context of smooth manifolds.
PREREQUISITESMathematicians, particularly those specializing in differential geometry, topology, and Riemannian geometry, as well as graduate students exploring advanced concepts in manifold theory.
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.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.