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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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    Ah thanks a lot that's what I suspected.2012-01-15
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    Can this local isometry always be promoted to a global isometry (of course if there is no topological obstruction) ?2014-04-11
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    @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.2014-04-11
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    Yes, the isometry is given by a unit speed geodesic.2015-07-22