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

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A vector field is flow-invariant if and only if its Lie derivative vanishes

Statement

Let $X$ have maximal flow $Φ$ . A smooth vector field $Y$ is invariant under that flow, meaning

$(Φ_{t})_{*} Y = Y$

whenever both sides are defined, if and only if $L_{X} Y = 0$ .

Facts & Assumptions

Given: Smooth vector fields $X, Y$ on $M$ and the maximal flow $Φ$ of $X$ .

[L1]

The Lie derivative is defined by $(L_{X} Y)_{p} = {\frac{d}{d t} ∣}_{t = 0} (Φ_{- t})_{*} Y_{Φ_{t} (p)} .$ (The Lie derivative of a vector field).

Proof

technique · direct

1.1

If $(Φ_{t})_{*} Y = Y$ for all admissible $t$ , then $(Φ_{- t})_{*} Y_{Φ_{t} (p)} = Y_{p}$ for every $p$ . Differentiating at $t = 0$ and using [L1] gives $(L_{X} Y)_{p} = 0$ for every $p$ .

L1given

1.2

Conversely, assume $L_{X} Y = 0$ . For fixed $p$ , define $F_{p} (t) : = (Φ_{- t})_{*} Y_{Φ_{t} (p)}$ . The same difference-quotient formula as in [L1], applied at the point $Φ_{t} (p)$ and then transported back by $Φ_{- t}$ , shows $F_{p}^{'} (t) = 0$ for every admissible $t$ . Hence $F_{p}$ is constant, so $F_{p} (t) = F_{p} (0) = Y_{p}$ .

L1given

2.1

Rewriting the identity from step 1.2 gives $(Φ_{t})_{*} Y = Y$ wherever defined. Therefore $Y$ is flow-invariant if and only if $L_{X} Y = 0$ .

step 1.1step 1.2∎

Depends on

Used by

Two vector fields commute if and only if their local flows commute Theorem

Dependency tree · two levels

7 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

Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)