PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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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.

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The coordinate formula for the Lie derivative of a covariant tensor

Statement

For a covariant $k$ -tensor $T = T_{i_{1} \dots i_{k}} d x^{i_{1}} \otimes \dots \otimes d x^{i_{k}}$ , $(L_{X} T)_{i_{1} \dots i_{k}} = X^{j} \partial_{j} T_{i_{1} \dots i_{k}} + \sum_{a} (\partial_{i_{a}} X^{j}) T_{i_{1} \dots j \dots i_{k}} .$

Facts & Assumptions

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

[F1]

The preceding result states that The Lie derivative obeys $L_{X} (S \otimes T) = (L_{X} S) \otimes T + S \otimes (L_{X} T)$ and commutes with every natural contraction. (The Lie derivative is a derivation of the tensor algebra).

Proof

technique · direct

1.1

Evaluate the tensor derivation formula on $\partial_{i_{1}}, \dots, \partial_{i_{k}}$ and differentiate the component function along $X$ .

F1given

2.1

Since $L_{X} \partial_{i} = [X, \partial_{i}] = - (\partial_{i} X^{j}) \partial_{j}$ , moving the input terms to the other side gives the displayed plus-sign formula.

step 1.1∎

Depends on

Used by

Lie derivative of the Euclidean metric under dilations Example

Dependency tree · two levels

8 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)