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I've recently been wondering about whether non-complete metrics on manifolds can be transformed into complete metrics on manifolds and whether all manifolds have complete metrics. After some googling I came across this link and the first comment says that any metric is actually conformal to a complete metric. I was wondering if anybody can show me a proof of this because I have had difficulty finding one. Thank you!

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$\textbf{Theorem 1}$ of The Existence of complete Riemannian Metrics is what you're looking for :

For any Riemannian metric $g$ on $M$ there exists a complete Riemannian metric which is conformal to $g$

The proof starts right on the first page.

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Another way to argue that every second countable differentiable manifold $M$ admits a complete Riemannian metric is the following: By Whitney, $M$ can be embedded into $\mathbb{R}^{2n+1}$ as a closed submanifold. The pullback metric on $M$ from $\mathbb{R}^{2n+1}$ then is complete since closed subsets of complete metric spaces are complete.

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  • $\begingroup$ Yes, but the complete metric you put in this way is unrelated to the one that was originally put on $M$. The question is whether there is a complete metric that is conformal to the given one. $\endgroup$ – Giuseppe Negro Aug 29 '19 at 9:06
  • $\begingroup$ "... and whether all manifolds have complete metrics" - that's what I answered! $\endgroup$ – Mathy Aug 29 '19 at 12:16
  • $\begingroup$ Fair enough, the question was admittedly a bit unclear. +1 $\endgroup$ – Giuseppe Negro Aug 29 '19 at 12:32

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