TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30

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Coordinate differentials form the dual cotangent basis

Statement

If $(U, x)$ is a smooth chart around $p$ , then the covectors $d x_{p}^{1}, \dots, d x_{p}^{n}$ form the basis of $T_{p}^{*} M$ dual to the tangent basis $\partial_{x^{1}} ∣_{p}, \dots, \partial_{x^{n}} ∣_{p}$ .

Facts & Assumptions

Given: A smooth chart $(U, x)$ around $p$ .

[F1]

For a smooth real-valued function $f$ , the differential is the functional $d f_{p} (v) = v ([f])$ (The differential of a smooth real-valued function).

[L1]

The coordinate derivations form a basis of $T_{p} M$ (Coordinate derivations form a basis of the tangent space).

Proof

technique · direct

1.1F1given

By [F1], one has $d x_{p}^{i} (\partial_{x^{j}} ∣_{p}) = \partial_{x^{j}} ∣_{p} ([x^{i}]) = δ_{i j}$ .

2.1L1step 1.1

Thus $d x_{p}^{1}, \dots, d x_{p}^{n}$ are dual to the basis from [L1].

3.1step 2.1∎

Therefore they form a basis of $T_{p}^{*} M$ .

Coordinate differentials form the dual cotangent basis

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources