How can i express the scalar curvature for a one - dimensional Riemannian manifold (M, g) in terms of the metric g ?
scalar curvature on one - dimensional Riemannian Manifold
3
$\begingroup$
riemannian-geometry
1 Answers
7
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 $R_{ijkl}$.
-
0Ah thanks a lot that's what I suspected. – 2012-01-15
-
0Can this local isometry always be promoted to a global isometry (of course if there is no topological obstruction) ? – 2014-04-11
-
1@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
-
1Yes, the isometry is given by a unit speed geodesic. – 2015-07-22