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

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Coordinate derivations form a basis of the tangent space

Statement

If $(U, x)$ is a smooth chart on an $n$ -manifold $M$ with $p \in U$ , then the coordinate derivations $\partial_{1} ∣_{p}, \dots, \partial_{n} ∣_{p}$ form a basis of $T_{p} M$ .

Facts & Assumptions

Given: A smooth chart $(U, x)$ with $p \in U$ .

[L1]

Each coordinate operator $\partial_{i} ∣_{p}$ is a derivation at $p$ (Coordinate derivations are well-defined derivations).

[L2]

Every derivation annihilates constant germs (A derivation annihilates constant germs).

[L3]

Smooth functions on a Euclidean neighbourhood admit a first-order Hadamard factorization (First-order Hadamard factorization near a point).

Proof

technique · direct

1.1L2L3given

By [L1], the coordinate operators belong to $T_{p} M$ . If $v \in T_{p} M$ and $f$ is represented in the chart by $\tilde{f}$ , then [L3] gives $\tilde{f} (u) - \tilde{f} (a) = \sum_{i} (u^{i} - a^{i}) g_{i} (u)$ near $a = x (p)$ ; applying $v$ to the corresponding germ and using [L2], one obtains $v ([f]) = \sum_{i} v ([x^{i}]) g_{i} (a) = \sum_{i} v ([x^{i}]) \partial_{i} ∣_{p} ([f])$ .

2.1step 1.1

Step 1.1 shows $v = \sum_{i} v ([x^{i}]) \partial_{i} ∣_{p}$ , so the coordinate derivations span $T_{p} M$ .

2.2step 1.1

If $\sum_{i} c_{i} \partial_{i} ∣_{p} = 0$ , apply this derivation to the coordinate germ $[x^{j}]$ ; only the $j$ th term survives, so $c_{j} = 0$ . Hence the coordinate derivations are linearly independent.

3.1step 2.1step 2.2∎

Therefore $\partial_{1} ∣_{p}, \dots, \partial_{n} ∣_{p}$ form a basis of $T_{p} M$ .

Coordinate derivations form a basis of the tangent space

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources