Lagmental Vicfred

The Riemann Tensor Is a Commutator Corrected by the Lie Bracket by Vicfred

Last updated: Mon 14 July 2025

tags: mathematics, curvature, Gauss--Bonnet, Jacobi fields

Curvature compares two orders of covariant differentiation and subtracts the bracket correction. The example is deliberately concrete; it is a test of the statement, not a substitute for it.

Set-up

Curvature measures the failure of covariant derivatives to commute. The Riemann tensor \(R(X,Y)Z\) contracts to Ricci curvature and restricts to sectional curvature \(K(\sigma)\).

$$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z $$

The formulas should not be merged too early. The datum \(\mathsf D\), the conclusion \(\mathsf C\), and the bridge \(\Longrightarrow\) have three different logical jobs.

$$ R(X,Y)=-R(Y,X) $$

The calculation

A worked instance is useful here because it exposes every index that the compressed statement hides.

$$ R^l{}_{ijk}=\partial_i\Gamma^l_{jk}-\partial_j\Gamma^l_{ik}+\Gamma^m_{jk}\Gamma^l_{im}-\Gamma^m_{ik}\Gamma^l_{jm} $$

What survives abstraction

What survives the example is not its particular numbers but the relation encoded by the two rows below. That relation is the part worth transporting to a new setting.

$$ \begin{aligned} \mathsf{D}\;&:\quad R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,\\[5pt] \mathsf{C}\;&:\quad R(X,Y)=-R(Y,X). \end{aligned} $$

The boundary

Sign conventions for \(R\) vary by author. A sphere may receive the opposite tensor sign unless the convention is stated.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] R(X,Y)=-R(Y,X) \end{gathered}} $$

I would use the boxed line as a reference later, while returning to the full display whenever a hypothesis becomes uncertain. That division keeps compression from becoming ambiguity.

This article was posted on Sun 08 November 2020. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.