# If a metric on $\Bbb R^n$ is flat, is the space globally isometric to Euclidean space?

Let $g$ be a metric of arbitrary signature on $\Bbb R^n$. If the Riemann tensor of $g$ vanishes identically, is there a diffeomorphism $f:\Bbb R^n\to\Bbb R^n$ with $g=f^*e$, where $e$ is the standard metric with the same signature as $g$?

In the Riemannian case, assuming $(\Bbb R^n,g)$ is complete, this is well known from the classification of space forms. I was unable to show that $(\Bbb R^n,g)$ is always complete in the Riemannian case, and I don't know if it's true or not. I also know that this is always true in a local sense, but I am looking for a global isometry.

• Pick any diffeomorphism $\phi$ from $\Bbb R^n$ to the open unit ball $\Bbb B^n$. Then, the pullback $\phi^*(e \vert_{\Bbb B^n})$ is a flat but incomplete metric on $\Bbb R^n$, and in particular it is not globally isometric to $(\Bbb R^n, e)$. – Travis Willse Apr 11 '17 at 13:50
• @Travis That's a good point. – Ryan Unger Apr 11 '17 at 14:15
• (I believe that) my comment fully addresses the question, so I've promoted it to a proper answer. – Travis Willse Apr 11 '17 at 14:23
• Bearing Travis' insight in mind, doesn't it make sense to add the assumption that $g$ is complete? – Amitai Yuval Apr 11 '17 at 15:30
• @AmitaiYuval In that case the answer is in the affirmative, cf. O'Neill Corollary 8.24. For the non-Riemannian case one must specify "geodesic completeness." – Ryan Unger Apr 11 '17 at 18:14

Pick any diffeomorphism $\phi$ from $\Bbb R^n$ to a proper (necessarily simply connected) open subset $U \subset \Bbb R^n$. (For concreteness, we could take $$U := (-1, 1)^n, \qquad \phi : (x_1, \ldots, x_n) \mapsto (\tanh x_1, \ldots, \tanh x_n)\textrm{.)}$$ Then, the pullback $\phi^* (e \vert_U)$ is a flat but incomplete metric on $\Bbb R^n$, and in particular it is not globally isometric to $(\Bbb R^n, e)$.