TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Curvature is a type (1,3) tensor

Statement

Let $\nabla$ be an affine connection on a smooth manifold $M$ . For $p \in M$ and $u, v, w \in T_{p} M$ , choose smooth local extensions $X, Y, Z$ and set

$R_{p} (u, v) w : = (R (X, Y) Z)_{p} .$

This is independent of the extensions, is trilinear in $(u, v, w)$ , and varies smoothly with $p$ . Equivalently,

$R_{p} (α, u, v, w) : = α (R_{p} (u, v) w)$

is a smooth type $(1, 3)$ tensor field. We use $R$ for both this tensor and its vector-valued representative.

Facts & Assumptions

[F1]

Curvature is $C^{\infty} (M)$ -linear separately in its three vector-field slots. Curvature is C-infinity-linear in all three vector fields.

[F2]

A smooth type $(1, 3)$ tensor field is a smooth section of $T_{3}^{1} M$ . A smooth tensor field.

Proof

Given: A point $p \in M$ , tangent vectors $u, v, w \in T_{p} M$ , and smooth local extensions $X, Y, Z$ on a common neighborhood of $p$ .

1.1

If $X^{'}$ is another extension of $u$ , take a coordinate frame $E_{1}, \dots, E_{n}$ near $p$ and write $X - X^{'} = \sum_{a} f^{a} E_{a}$ . Since every $f^{a} (p) = 0$ , [F1] gives $(R (X - X^{'}, Y) Z)_{p} = \sum_{a} f^{a} (p) (R (E_{a}, Y) Z)_{p} = 0$ . Repeating this argument in the second and third slots proves independence of all three extensions. The same identities, evaluated at $p$ , prove real trilinearity of $(u, v, w) \mapsto R_{p} (u, v) w$ .

F1algebra

2.1

On a coordinate neighborhood with frame $E_{i}$ and dual coframe $ε^{l}$ , put $R^{l}_{i j k} = ε^{l} (R (E_{i}, E_{j}) E_{k})$ . Each coefficient is smooth because the defining curvature expression applies the connection and Lie bracket to smooth fields. The identity $R_{p} (u, v) w = R^{l}_{i j k} (p) u^{i} v^{j} w^{k} E_{l} ∣_{p}$ , obtained from [F1], therefore makes the vector-valued representative smooth. Pairing its output with a covector gives the smooth fibrewise multilinear map $R_{p}$ above, hence a smooth section of $T_{3}^{1} M$ by [F2].

F1F2step 1.1algebra∎

Depends on

Used by

Riemann curvature four-tensor Definition
Curvature of a Riemannian product Example
Hyperbolic space has negative constant sectional curvature Example

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)
John M. Lee, Riemannian Manifolds: An Introduction to Curvature (standard reference, not scraped)