It is known that flat manifolds can be characterized as follows

If a pseudo-Riemannian manifold $M$ of signature $(s,t)$ has zero Riemann curvature tensor everywhere on $M$, then the manifold is locally isometric to $\mathbb{R}^{s,t}$.

My question is this $-$ what other properties should a pseudo-Riemannian manifold have, beyond zero curvature, for it to be globally isometric to $\mathbb{R}^{s,t}$?

My guess is that non-compactness should be at least one of the conditions, but I haven't been able to show that.

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    $\begingroup$ Geodesically complete. $\endgroup$ Jun 10, 2023 at 7:11
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    $\begingroup$ @ValterMoretti Are you sure? What if we have a $(2, 0)$-signature on a flat torus? Isn't that geodesically complete but not globally isometric to $\mathbb{R}^{(2, 0)}$? Or am I making a mistake? $\endgroup$ Jun 10, 2023 at 7:41
  • $\begingroup$ It is only necessary, as non-compactness. If you want a necessary and sufficient condition you must add that the exponential map is everywhere injective, in addition to geodesic completeness $\endgroup$ Jun 10, 2023 at 7:44
  • $\begingroup$ I think that geodesical complete + simply connected should be equivalent to global flatness, but I never tried to prove it. $\endgroup$ Jun 10, 2023 at 7:53
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    $\begingroup$ I'm able to give an answer for Riemannian manifolds (not pseudo-Riemannian). If a Riemannian manifold is geodesically complete and flat (i.e. of zero curvature everywhere), the Cartan-Hadamard theorem says that its universal cover is diffeomorphic to $\mathbb{R}^{n}$. If the manifold is further simply-connected, then it is its own universal cover and you get the desired result. $\endgroup$ Jun 10, 2023 at 8:05

2 Answers 2


Exercise 8 of Chapter 8 in Barrett O'Neill's Semi-Riemannian Geometry With Applications to Relativity (1983) asks,

(a) Let $M$ be a flat connected semi-Riemannian manifold complete at a $o\in M$ (that is, $\exp_{o}$ is defined on all of $T_{o}(M)$). Prove that $\exp_{o}: T_{o}(M) \rightarrow M$ is a semi-Riemannian covering map. (Hint: Use Proposition 6.) (b) Give an example of a connected semi-Riemannian manifold that is complete at one point but not complete.

I haven't done the exercise, but based on this, if $M$ is a pseudo-Riemannian manifold of signature $(s, t)$, then a necessary and sufficient condition that $M$ is globally isometric to $\mathbb{R}^{s, t}$ is that $M$ is

  • flat,
  • simply-connected (and connected), and
  • geodesically complete.

If $M$ is a flat pseudo-Riemannian manifold such that it is connected and geodesically complete, then by the exercise we know $T_{o}M\cong\mathbb{R}^{s, t}$ is a pseudo-Riemannian covering space of $M$. If $M$ is simply-connected, then it is globally isometric to its covering space.


Spacetimes that are of zero Riemann curvature are all either subsets or quotients of Minkowski space. A few example of such manifolds would be Minkowski space minus some closed set $\mathbb{R}^{n-1, 1} \setminus U$, identifications of Minkowski space up to some translations such as cylinders and torus, identification of Minkowski space up to some boost (like Misner space), or more complex cases like the non-time orientable Minkowski space (Minkowski cylinder plus some identification up to some involution $(\theta, t) \to (\theta + \pi, -t)$) or some spacetime made by constructing a cone with the spatial section.

From those examples, we can see that there are indeed a lot of problems with simple connectedness and geodesic completeness for those examples. As mentionned by Maximal Ideal, that is indeed sufficient conditions for Minkowski space.

It is also enough that it be two-point homogeneous. Any flat spacetime is already maximally symmetric (in the sense that it has the full Poincaré group as Killing vector fields), but in addition the flow of those Killing vectors must be complete. Alternatively it is that for any points $p, q$, there is an isometry transporting $p$ to $q$ : $\phi(p) = q$, and any four points $p, q, r, s$ having the same geodesic distance, $d(p,q) = d(r,s)$, there is an isometry transporting $p$ and $q$ to $r$ and $s$

Alternatively, it can be homogeneous and isotropic : only single points need to be transported by isometries, and there is an isomorphism mapping any two unit vectors to each other : $\phi^* v = w$.

There is probably some other conditions that one could come up with, such as the causal structure having to obey the Minkowski space causal structure and the spatial slices having to obey Euclidian axioms (in particular the ones that are violated in those counter examples, such as the prolongation of lines or the number of incident points between two lines)


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