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The invariant exterior-derivative formula is $C^{\infty}$ -multilinear

Statement

The invariant formula defining $d ω$ is alternating and $C^{\infty} (M)$ -multilinear in $X_{0}, \dots, X_{k}$ ; hence it defines a smooth $(k + 1)$ -form.

Facts & Assumptions

Given: The manifolds, forms, vector fields, maps, and coordinates explicitly named in the statement.

[F1]

The preceding result states that For $ω \in Ω^{k} (M)$ , define the candidate $d ω$ on smooth vector fields by $d ω (X_{0}, \dots, X_{k}) = \sum_{i} (- 1)^{i} X_{i} ω (X_{0}, {\hat{X}}_{i}, \dots, X_{k}) + \sum_{i < j} (- 1)^{i + j} ω ([X_{i}, X_{j}], X_{0}, {\hat{X}}_{i}, {\hat{X}}_{j}, \dots, X_{k}) .$ The next lemma proves that this candidate is a form. (The exterior derivative by the invariant vector-field formula).

Proof

technique · direct

1.1F1given

Replace $X_{i}$ by $f X_{i}$ . For $a \neq i$ , the $a$ th derivative term contributes $(- 1)^{a} (X_{a} f) ω ({\hat{X}}_{a})$ . If $a < i$ , the $(a, i)$ bracket term contributes $(- 1)^{a + i} (X_{a} f) ω (X_{i}, X_{0}, \dots, {\hat{X}}_{a}, \dots, {\hat{X}}_{i}, \dots, X_{k}) = - (- 1)^{a} (X_{a} f) ω ({\hat{X}}_{a})$ , because moving $X_{i}$ to its usual slot takes $i - 1$ swaps. If $a > i$ , the $(i, a)$ bracket correction from $[f X_{i}, X_{a}] = f [X_{i}, X_{a}] - (X_{a} f) X_{i}$ contributes $- (- 1)^{i + a} (- 1)^{i} (X_{a} f) ω ({\hat{X}}_{a}) = - (- 1)^{a} (X_{a} f) ω ({\hat{X}}_{a})$ . Thus every derivative-of- $f$ term cancels.

2.1step 1.1∎

All remaining terms are $f$ times the original formula; alternation follows by exchanging adjacent inputs, and smooth coordinate coefficients give a smooth form.

Depends on

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed.

The invariant exterior-derivative formula is C∞C∞-multilinear