# scalar curvature on one - dimensional Riemannian Manifold

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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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 $R_{ijkl}$.
@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. – Henning Makholm Apr 11 '14 at 18:53