Lie derivative of forms

Let $X$ be a vector field and $ϕ$ its flow. Let $α$ be a $p$ -form, we define:

L_{X} (α) = \frac{d}{d t} {[ϕ_{t}^{*} α^{}]}_{t = 0} = lim_{t \to 0} \frac{ϕ_{t}^{*} α_{ϕ_{t} x} - α_{x}}{t}

in the sense that

[\frac{d}{d t} ϕ_{t}^{*} α^{}] (Y_{1}, \dots, Y_{p}) =

= \frac{d}{d t} [ϕ_{t}^{*} α^{} (Y_{1}, \dots, Y_{p})] = \frac{d}{d t} |_{t = 0} {α^{} [ϕ_{t *} Y_{1}, \dots, ϕ_{t *} Y_{p}]}

More explicitly:

If we define vector fields $Y_{i}$ extending the previous vectors in a invariant way respect to the flow $ϕ$ , that is, $Y_{i} (ϕ_{t} x) = ϕ_{t *} Y_{i}$ then

L_{X} α^{} (Y_{1}, \dots, Y_{p}) = \frac{d}{d t} {[α_{ϕ_{t} x}^{} (Y_{1}, \dots, Y_{p})]}_{t = 0}

That is, the Lie derivative of a $p$ -form measure the rate of change along the flow of $X$ of the $p$ -form applied to invariant vector fields along $X$ .

It can be generalized to Lie derivative for tensors in general.

Useful formulas: formulas for Lie derivative, exterior derivatives, bracket, interior product.

Proposition

A differential $k$ -form $ω$ on $M$ is invariant under the flow of a vector field $X$ , i.e.,

{ω |}_{\exp (ε X) x} = \exp (- ε X)^{*} ({ω |}_{x})

if and only if $L_{X} (ω) = 0$

$◼$

Proof

See @olver86 proposition 1.65.

$◼$

Especial case: the Lie derivative of a volume form leads to the idea of divergence.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es

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