PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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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 commutator $[X, Y]$ is again a derivation (The commutator of vector-field derivations is again a derivation).

[L2]

Every derivation of $C^{\infty} (M)$ comes from a unique smooth vector field (Derivations of smooth functions are exactly smooth vector fields).

[L3]

A smooth vector field is determined in a chart by its coefficient functions (Smoothness of a vector field is equivalent to smooth coordinate components).

[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

1.1

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 $W$ .

L4givenconstruct

2.1

By [L1] and [L2], $[X, Y]$ is induced by a unique smooth vector field on $M$ . Evaluating it on ${\tilde{x}}^{j}$ at $p$ and using step 1.1 gives $[X, Y] ({\tilde{x}}^{j}) (p) = X (Y^{j}) (p) - Y (X^{j}) (p) = \sum_{i} X^{i} (p) \partial_{i} Y^{j} (p) - \sum_{i} Y^{i} (p) \partial_{i} X^{j} (p) .$

L1L2step 1.1given

3.1

Because ${\tilde{x}}^{j}$ and $x^{j}$ agree near $p$ , the $j$ -th coordinate coefficient of $[X, Y]$ at $p$ is exactly $[X, Y] ({\tilde{x}}^{j}) (p)$ . Since $p$ was arbitrary, step 2.1 and [L3] give the displayed coordinate formula for $[X, Y]$ on $U$ .

L3step 2.1∎

Depends on

Used by

Two pairs of vector fields can agree at a point and still have different bracket values there Counterexample
A coordinate computation of a nonzero Lie bracket Example
FALSE: the point values Xₚ and Yₚ determine the bracket value [X,Y]ₚ False statement
Coordinate vector fields commute Proposition

Dependency tree · two levels

14 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)