General Relativity: An Introduction for Physicists · Problem 7.15
The equivalence principle and spacetime curvature — Hobson, Efstathiou & Lasenby
Physics · curvature-tensor-two-dimensional-manifold
Problem
Show that for any two-dimensional manifold the covariant curvature tensor has the form where the scalar may be a function of the coordinates. Why does this result not generalise to arbitrary manifolds of higher dimension?Intuitive explanation
In 2D, the Riemann curvature tensor has only one independent component at each point—the Gaussian curvature —because the number of independent components is determined by the dimension. The key insight is that all curvature information must be expressible in terms of the metric and a single scalar function; the formula you're asked to prove shows that any valid curvature tensor must decompose into the unique antisymmetric structure . In higher dimensions, the curvature tensor gains additional independent components (the Weyl tensor captures traceless curvature), so you can't squeeze all geometric information into a single scalar—the manifold's shape becomes genuinely richer.
Intuitive explanation
In 2D, the Riemann curvature tensor has only one independent component at each point—the Gaussian curvature —because the number of independent components is determined by the dimension. The key insight is that all curvature information must be expressible in terms of the metric and a single scalar function; the formula you're asked to prove shows that any valid curvature tensor must decompose into the unique antisymmetric structure . In higher dimensions, the curvature tensor gains additional independent components (the Weyl tensor captures traceless curvature), so you can't squeeze all geometric information into a single scalar—the manifold's shape becomes genuinely richer.