PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Coordinate formula for the Lie bracket

Statement

In a chart $(U, x^{1}, \dots, x^{n})$ , if

$X = \sum_{i = 1}^{n} X^{i} \frac{\partial}{\partial x^{i}}, Y = \sum_{i = 1}^{n} Y^{i} \frac{\partial}{\partial x^{i}},$

then on $U$

$[X, Y] = \sum_{j = 1}^{n} (\sum_{i = 1}^{n} X^{i} \partial_{i} Y^{j} - Y^{i} \partial_{i} X^{j}) \frac{\partial}{\partial x^{j}} .$

Facts & Assumptions

Given: Smooth vector fields $X$ and $Y$ written in a chart as above.

[L1]

The bracket acts on functions as $[X, Y] f = X (Y f) - Y (X f)$ (The Lie bracket of smooth vector fields).

[L2]

Smooth coordinate coefficient functions define a smooth local vector field (Smoothness of a vector field is equivalent to smooth coordinate components).

[L3]

Mixed second partial derivatives of a smooth function commute (Clairaut--Schwarz theorem for continuous second partial derivatives).

[L4]

For a point inside an open set there is a smooth bump function equal to $1$ on a neighbourhood of that point and supported in the open set (A manifold bump for a compact set inside an open set).

Proof

technique · direct

1.1L4givenconstruct

Fix $p \in U$ . By [L4], choose a smooth function $χ : M \to [0, 1]$ that is $1$ on a neighbourhood $W$ of $p$ and has support contained in $U$ . For each $j$ , let ${\tilde{x}}^{j}$ be the global smooth function that equals $χ x^{j}$ on $U$ and $0$ outside $U$ . Then ${\tilde{x}}^{j} = x^{j}$ on $W$ , and because $χ$ is constant on $W$ , one also has $X ({\tilde{x}}^{j}) = X^{j}$ and $Y ({\tilde{x}}^{j}) = Y^{j}$ on $W$ .

2.1L1step 1.1given

Apply [L1] to ${\tilde{x}}^{j}$ and use step 1.1. At $p$ this gives $[X, Y] ({\tilde{x}}^{j}) (p) = X (Y^{j}) (p) - Y (X^{j}) (p) = \sum_{i} X^{i} (p) \partial_{i} Y^{j} (p) - \sum_{i} Y^{i} (p) \partial_{i} X^{j} (p) .$ Each resulting coefficient is smooth on $U$ .

3.1L1L2L3step 2.1∎

On $U$ , the ordinary product rule gives the same coefficient formula when [L1] acts on any smooth test function: the mixed second derivatives cancel by [L3]. Thus the smooth local field with coefficients from step 2.1 acts as the bracket. On overlaps, two such fields agree on all local test functions, so they agree pointwise; the local fields therefore glue uniquely. This proves the formula without selecting a family of charts or local fields.

Coordinate formula for the Lie bracket

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources