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How can i express the scalar curvature for a one - dimensional Riemannian manifold (M, g) in terms of the metric g ?

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A one-dimensional Riemannian manifold does not have any intrinsic curvature at all. It is always locally isometric to a straight, "flat" line.

Formally, the Riemann curvature tensor has but a single component $R_{1111}$, but this element is required to be 0 due to (for example) the skew-symmetry of $R_{ijkl}$.

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  • $\begingroup$ Ah thanks a lot that's what I suspected. $\endgroup$ – harlekin Jan 15 '12 at 20:55
  • $\begingroup$ Can this local isometry always be promoted to a global isometry (of course if there is no topological obstruction) ? $\endgroup$ – Lor Apr 11 '14 at 17:23
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    $\begingroup$ @Lor: Good question. I think every Riemannian metric on $\mathbb R$ is isometric to an open subset of $\mathbb R$, but I cannot rattle off a proof. $\endgroup$ – Henning Makholm Apr 11 '14 at 18:53
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    $\begingroup$ Yes, the isometry is given by a unit speed geodesic. $\endgroup$ – Maxwell Stolarski Jul 22 '15 at 14:20

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