# The Lie derivative of an exterior form

The Lie derivative $${L_{v}}$$ can be applied to a $${k}$$-form $${\varphi}$$ by using the pullback of $${\varphi}$$ by the diffeomorphism $${\Phi}$$ associated with the flow of $${v}$$, i.e. applied to $${k}$$ vectors $${w_{I}}$$ we define

$$\displaystyle L_{v}\varphi\left(w_{I}\right)\equiv\underset{\varepsilon\rightarrow0}{\textrm{lim}}\frac{1}{\varepsilon}\left[\varphi\left(\mathrm{d}\Phi_{\varepsilon}\left(w_{I}\right)\right)-\varphi\left(w_{I}\right)\right].$$

$${L_{v}\varphi}$$ thus measures the change in $${\varphi}$$ as its arguments are transported by the local flow of $${v}$$. In the case of a 0-form $${f}$$, this is just the differential or directional derivative $${L_{v}f=v(f)=\mathrm{d}f(v)}$$. The above illustrates the Lie derivative for a 1-form $${\varphi}$$ with $${\varepsilon=1}$$. $${L_{v}\varphi}$$ is “the difference between $${\varphi}$$ applied to $${w}$$ and $${\varphi}$$ applied to $${w}$$ transported by the local flow of $${v}$$,” so above we have $${L_{v}\varphi(w)=2-1=1}$$ (valid in the limit $${\varepsilon\rightarrow0}$$ if $${\varphi}$$ changes linearly in the range shown).

 ◊ Here and in future figures, we represent a 1-form $${\varphi}$$ as a “receptacle” $${\varphi^{\Uparrow}\equiv\varphi^{\sharp}/\left\Vert \varphi^{\sharp}\right\Vert ^{2}}$$ which when applied to a vector “arrow” argument $${v}$$ yields the number of receptacles covered by the projection of $${v}$$ onto $${\varphi^{\sharp}}$$, which is the value of $${\varphi(v)}$$. This can be seen by recalling that $${\varphi(v)/\left\Vert \varphi^{\sharp}\right\Vert}$$ is the length of the projection of $${v}$$ onto $${\varphi^{\sharp}}$$, so that this projection divided by the length of the receptacle $${\left\Vert \varphi^{\Uparrow}\right\Vert =1/\left\Vert \varphi^{\sharp}\right\Vert }$$ recovers the value $${\varphi(v)}$$. The advantage of this approach is that values can be calculated from the figure absent a length scale. Another common graphical device is to represent 1-forms as “surfaces” which are “pierced” by the arrows.
 Δ The common practice of depicting a 1-form $${\varphi}$$ in terms of the associated vector $${\varphi^{\Uparrow}}$$ as above has consequences that can be non-intuitive. For example, doubling the value of the 1-form means halving its length in the illustration, i.e. the value of the 1-form can be viewed as the “density” of receptacles. Thus, when depicting $${\varphi}$$ as changing linearly, the length $${L}$$ of the 1-form representation changes like $${L\mapsto L/(1+r\varepsilon)}$$ for some scaling factor $${r}$$, which doesn’t appear linear as a vector representation would, whose length changes like $${L\mapsto L(1+r\varepsilon)}$$.

By using the above definitions of the Lie derivative applied to vectors and 1-forms, and noting that we can derive a Leibniz rule over contraction $${L_{v}(\varphi\left(w\right))=\left(L_{v}\varphi\right)(w)+\varphi\left(L_{v}w\right)}$$, we arrive at an expression for the Lie derivative applied to general tensors, viewed as real-valued mappings on vectors and 1-forms:

\begin{aligned}L_{v}T(\varphi_{1},\ldots,\varphi_{m},w_{1},\ldots,w_{n}) & =v\left(T(\varphi_{1},\ldots,\varphi_{m},w_{1},\ldots,w_{n})\right)\\
& -\sum_{j=1}^{m}T\left(\varphi_{1},\ldots,L_{v}\varphi_{j},\ldots,\varphi_{m},w_{1},\ldots,w_{n}\right)\\
& -\sum_{j=1}^{n}T\left(\varphi_{1},\ldots,\varphi_{m},w_{1},\ldots,L_{v}w_{j},\ldots,w_{n}\right)
\end{aligned}

In a holonomic frame, this yields an expression for the Lie derivative of a tensor in terms of coordinates

\begin{aligned}L_{v}T^{\mu_{1}\dots\mu_{m}}{}_{\sigma_{1}\dots\sigma_{n}} & =v^{\lambda}\frac{\partial}{\partial x^{\lambda}}T^{\mu_{1}\dots\mu_{m}}{}_{\sigma_{1}\dots\sigma_{n}}\\
& -\sum_{j=1}^{m}\left(\frac{\partial v^{\mu_{j}}}{\partial x^{\lambda}}\right)T^{\mu_{1}\dots\mu_{j-1}\lambda\mu_{j+1}\dots\mu_{m}}{}_{\sigma_{1}\dots\sigma_{n}}\\
& +\sum_{j=1}^{n}\left(\frac{\partial v^{\lambda}}{\partial x^{\sigma_{j}}}\right)T^{\mu_{1}\dots\mu_{m}}{}_{\sigma_{1}\dots\sigma_{j-1}\lambda\sigma_{j+1}\dots\sigma_{n}}.
\end{aligned}

From this we can confirm that the Lie derivative satisfies the Leibniz rule over the tensor product, and therefore is a derivation of degree 0 on both the tensor algebra and the exterior algebra.