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

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    $\begingroup$ 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. $\endgroup$ Oct 31, 2013 at 20:34
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    $\begingroup$ @JosephO'Rourke: The word "agree" was used in a sophisticated way. $\endgroup$
    – Ben McKay
    Oct 31, 2013 at 20:56
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    $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Oct 31, 2013 at 20:58
  • $\begingroup$ 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. $\endgroup$
    – Wintermute
    Oct 31, 2013 at 21:01
  • $\begingroup$ @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. $\endgroup$ Nov 1, 2013 at 2:17

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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 $-1$.

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