Riemannian curvature

Let $𝑀$ be a $𝐶 𝛼$ -manifold equipped with an affine connexion $\nabla$ . The Riemannian curvature $𝑅 𝑐 𝑑 𝑎 𝑏 \in T 13 (𝑀)$ is a tensor field defined so that

𝑅 𝑐 𝑑 𝑎 𝑏 𝑋 𝑎 𝑌 𝑏 𝑍 𝑑 = [\nabla 𝑋, \nabla 𝑌] 𝑍 𝑐 - \nabla [𝑋, 𝑌] 𝑍 𝑐

and thus

𝑅 𝑐 𝑑 𝑎 𝑏 𝑍 𝑑 = [\nabla 𝑎, \nabla 𝑏] 𝑍 𝑐 + 𝑇 𝑑 𝑎 𝑏 \nabla 𝑑 𝑍 𝑐

where $𝑇 𝑐 𝑎 𝑏$ is the torsion tensor. A manifold with null Riemannian curvature is said to be flat.

Proof of equality and tensoriality

Let $𝑋 𝑎, 𝑌 𝑎, 𝑍 𝑎 \in 𝔛 (𝑀)$ be vector fields. Then

\nabla 𝑋 \nabla 𝑌 𝑍 𝑐 = 𝑋 𝑎 \nabla 𝑎 𝑌 𝑏 \nabla 𝑏 𝑍 𝑐 = 𝑋 𝑎 𝑌 𝑏 \nabla 𝑎 \nabla 𝑏 𝑍 𝑐 + (𝑋 𝑎 \nabla 𝑎 𝑌 𝑏) \nabla 𝑏 𝑍 𝑐 = 𝑋 𝑎 𝑌 𝑏 \nabla 𝑎 \nabla 𝑏 𝑍 𝑐 + \nabla \nabla 𝑋 𝑌 𝑍 𝑐

and thus using the first equation

𝑅 𝑐 𝑑 𝑎 𝑏 𝑋 𝑎 𝑌 𝑏 𝑍 𝑑 = \nabla 𝑋 \nabla 𝑌 𝑍 𝑐 - \nabla 𝑌 \nabla 𝑋 𝑍 𝑐 - \nabla [𝑋, 𝑌] 𝑍 𝑐 = 𝑋 𝑎 𝑌 𝑏 \nabla 𝑎 \nabla 𝑏 𝑍 𝑐 + \nabla \nabla 𝑋 𝑌 𝑍 𝑐 - 𝑋 𝑎 𝑌 𝑏 \nabla 𝑏 \nabla 𝑎 𝑍 𝑐 - \nabla \nabla 𝑌 𝑋 𝑍 𝑐 - \nabla [𝑋, 𝑌] 𝑍 𝑐 = 𝑋 𝑎 𝑌 𝑏 \nabla 𝑎 \nabla 𝑏 𝑍 𝑐 - 𝑋 𝑎 𝑌 𝑏 \nabla 𝑏 \nabla 𝑎 𝑍 𝑑 + 𝑋 𝑎 𝑌 𝑏 𝑇 𝑑 𝑎 𝑏 \nabla 𝑑 𝑍 𝑐

for any $𝑋 𝑎, 𝑌 𝑎 \in 𝔛 (𝑀)$ , so indeed

𝑅 𝑐 𝑑 𝑎 𝑏 𝑍 𝑑 = \nabla 𝑎 \nabla 𝑏 𝑍 𝑐 - \nabla 𝑏 \nabla 𝑎 𝑍 𝑑 + 𝑇 𝑑 𝑎 𝑏 \nabla 𝑑 𝑍 𝑐

as claimed.

To show tensoriality it suffices to show that the map $𝑍 𝑑 \mapsto 𝑅 𝑐 𝑑 𝑎 𝑏 𝑍 𝑑$ is $𝐶 𝛼 (𝑀)$ -linear. To this end let $𝑓 \in 𝐶 𝛼 (𝑀)$ be a scalar field. Then

\nabla 𝑎 \nabla 𝑏 𝑓 𝑍 𝑐 = \nabla 𝑎 (𝑓 \nabla 𝑏 𝑍 𝑐 + 𝑍 𝑐 d 𝑓 𝑏) = 𝑓 \nabla 𝑎 \nabla 𝑏 𝑍 𝑐 + d 𝑓 𝑎 \nabla 𝑏 𝑍 𝑐 + d 𝑓 𝑏 \nabla 𝑎 𝑍 𝑐 + 𝑍 𝑐 \nabla 𝑎 \nabla 𝑏 𝑓

and

𝑇 𝑑 𝑎 𝑏 \nabla 𝑑 𝑓 𝑍 𝑐 = 𝑇 𝑑 𝑎 𝑏 (𝑓 \nabla 𝑑 𝑍 𝑐 + 𝑍 𝑐 d 𝑓 𝑑) = 𝑓 𝑇 𝑑 𝑎 𝑏 \nabla 𝑑 𝑍 𝑐 + 𝑇 𝑑 𝑎 𝑏 𝑍 𝑐 d 𝑓 𝑑 .

Thus

𝑅 𝑐 𝑑 𝑎 𝑏 (𝑓 𝑍) 𝑑 = 𝑓 𝑅 𝑐 𝑑 𝑎 𝑏 𝑍 𝑑 + 𝑍 𝑐 [\nabla 𝑎, \nabla 𝑏] 𝑓 + 𝑍 𝑐 𝑇 𝑑 𝑎 𝑏 \nabla 𝑑 𝑓

where the final terms cancel since $0 = [\nabla 𝑎, \nabla 𝑏] 𝑓 + 𝑇 𝑐 𝑎 𝑏 \nabla 𝑐 𝑓$ .

Conflicting conventions

The convention used here is that used by Evgeny Buchbinder and (for the most part) Wikipedia. Wald's General relativity defines the torsion-free case acting on a 1-form $𝜔 𝑎 \in Ω 1 (𝑀)$ so that

˜ 𝑅 𝑎 𝑏 𝑐 𝑑 𝜔 𝑑 = [\nabla 𝑎, \nabla 𝑏] 𝜔 𝑐

meaning the action on a vector field $𝑋 𝑎 \in 𝔛 (𝑀)$ is

˜ 𝑅 𝑎 𝑏 𝑑 𝑐 𝑋 𝑑 = - [\nabla 𝑎, \nabla 𝑏] 𝑋 𝑐

meaning $˜ 𝑅 𝑎 𝑏 𝑑 𝑐 = 𝑅 𝑐 𝑑 𝑏 𝑎$ .

Given local coördinates $𝑥 : 𝑈 \to ℝ 𝑚$ , the components of $𝑅 𝛾 𝛿 𝛼 𝛽$ can be computed explicitly in terms of the connexion coëfficients $Γ 𝛾 𝛼 𝛽$ as¹

𝑅 𝛾 𝛿 𝛼 𝛽 = 𝜕 𝛼 Γ 𝛾 𝛽 𝛿 - 𝜕 𝛽 Γ 𝛾 𝛼 𝛿 + Γ 𝜎 𝛽 𝛿 Γ 𝛼 𝛼 𝜎 - Γ 𝜎 𝛼 𝛿 Γ 𝛼 𝛽 𝜎 .

Derivation

We have

𝑅 𝑐 𝑑 𝑎 𝑏 (𝜕 𝛼) 𝑎 (𝜕 𝛽) 𝑏 (𝜕 𝛿) 𝑑 = [\nabla 𝛼, \nabla 𝛽] (𝜕 𝛿) 𝑐 - \nabla [𝜕 𝑎, 𝜕 𝛽] (𝜕 𝛿) 𝑐 = \nabla 𝛼 Γ 𝛾 𝛽 𝛿 (𝜕 𝛾) 𝑐 - \nabla 𝛽 Γ 𝛾 𝛼 𝛿 (𝜕 𝛾) 𝑐 = 𝜕 𝛼 Γ 𝛾 𝛽 𝛿 - 𝜕 𝛽 Γ 𝛾 𝛼 𝛿 + Γ 𝜎 𝛽 𝛿 Γ 𝛼 𝛼 𝜎 - Γ 𝜎 𝛼 𝛿 Γ 𝛼 𝛽 𝜎 .

as required.

Relation to parallel transport

The curvature of an (intrinsic) space can be thought of as the failure a vector to return to its initial value after parallel transport along a loop, explaining why the structure of an affine connexion is a prerequisite.

#to/complete

Properties

$𝑅 𝑐 𝑑 𝑎 𝑏 = 𝑅 𝑐 𝑑 [𝑎 𝑏]$ , i.e. $𝑅 𝑐 𝑑 𝑎 𝑏 = - 𝑅 𝑐 𝑑 𝑏 𝑎$ .

For a torsion free connexion

Assume $\nabla$ is torsion-free.

Bianchi Identity I. $12 𝑅 𝑑 [𝑐 𝑎 𝑏] = 𝑅 𝑑 𝑐 𝑎 𝑏 + 𝑅 𝑑 𝑎 𝑏 𝑐 + 𝑅 𝑑 𝑏 𝑐 𝑎 = 0$ .

Proof

#missing/proof

Computing curvature

Besides the above expression using the connexion coëfficients, see the Vielbein method for computing curvature.