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If two Riemannian manifolds $(M,g)$ and $(N,h)$ have the same constant curvature are they isometric?

Wintermute
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    A related (not identical) question: Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature. The answer there was: Two surfaces in $\mathbb{R}^3$ are related by an isometry of $\mathbb{R}^3$ if and only if their first and second fundamental forms agree. – Joseph O'Rourke Oct 31 '13 at 20:34
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    @JosephO'Rourke: The word "agree" was used in a sophisticated way. – Ben McKay Oct 31 '13 at 20:56
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    They are locally isometric. One proof of this uses the Frobenius theorem on the product of their unit tangent bundles, trying the match their soldering 1-forms and their Levi-Civita connections 1-forms. Globally, it is pretty clear that a flat disk is not isometric to the plane, but is locally, because you can cut a flat disk out of a plane with a band saw. – Ben McKay Oct 31 '13 at 20:58
  • What I have is two non-compact surfaces with the same metric, constant negative scalar, and constant negative Ricci curvatures. I want to find isometries between them. – Wintermute Oct 31 '13 at 21:01
  • @mtiano certainly unless you mean "sectional curvature" what you want is false, even in a local sense. For example many manifolds admit negative Einstein metrics but are not locally isometric to hyperbolic space. – Otis Chodosh Nov 01 '13 at 02:17
  • Constant curvature usually means sectional curvature. – Matthias Ludewig Nov 01 '13 at 06:40
  • @mtiano: When you describe your manifolds as surfaces, do you mean that they are 2-dimensional manifolds? It seems strange to then discuss both their scalar and their Ricci curvatures. What can you tell us about the dimensions of your manifolds? Still, clearly you can cut a small open set out of any manifold, and the open set will not be isometric to the manifold, since it is smaller, but it will be locally isometric, because it is an open subset, with metric given by the same expression. – Ben McKay Nov 01 '13 at 08:38
  • If two manifolds have constant curvature metrics with the same constant, and both metrics and complete, then the manifolds share a common covering space, so that the metrics pullback to be identical on the covering space. I think this is what you are looking for. Look at the fundamental groups, and how they act on that covering space. In particular, you can always take that covering space to be the universal covering space of one and hence both of the manifolds. But first check for metric completeness. – Ben McKay Nov 01 '13 at 08:41

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No. There are plenty of non-isometric compact surfaces of zero constant curvature (tori). There are even more surfaces of constant curvature $-1$.

Alexandre Eremenko
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