# Question about complete metric on manifolds

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!

## 2 Answers

$\textbf{Theorem 1}$ 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.

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.

• 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. – Giuseppe Negro Aug 29 '19 at 9:06
• "... and whether all manifolds have complete metrics" - that's what I answered! – Mathy Aug 29 '19 at 12:16
• Fair enough, the question was admittedly a bit unclear. +1 – Giuseppe Negro Aug 29 '19 at 12:32