# Lie Derivative in Coordinates

Here we will study the coordinate dependent realization of the Lie derivative. Let $$f$$ be a scalar (zero-form), $$X$$ a vector field, and $$\alpha$$ be a one-form, where

\begin{split}\begin{aligned} X=X^{i}e_{i}=X^{i}\frac{\partial}{\partial x^{i}}\\ \alpha=\alpha_{i}e^{i}=\alpha_{i}dx^{i}\end{aligned}\end{split}

$$\{e_{1},\dots,e_{n}\}=\{\frac{\partial}{\partial x^{1}},\dots,\frac{\partial}{\partial x^{n}}\}$$ form the basis that span the space of vector fields, and $$\{e^{1},\dots,e^{n}\}=\{dx^{1},\dots,dx^{n}\}$$ are the basis elements for the space of one-forms.

Using Cartan’s magic formula the Lie derivative is given algebraically by

$$$\mathfrak{L}_{X}\alpha=\mathbf{d}\mathbf{i}_{X}\alpha+\mathbf{i}_{X}\mathbf{d}\alpha\label{eq:cartan}$$$

In coordinate form the Lie derivative of a scalar, a one-form and a vector field become

$\mathfrak{L}_{X}f=\mathbf{i}_{X}\mathbf{d}f=X^{i}\frac{\partial f}{\partial x^{i}}$
$\mathfrak{L}_{X}Y=[X,Y]=X^{i}\frac{\partial Y^{j}}{\partial x^{i}}\frac{\partial}{\partial x^{j}}-Y^{i}\frac{\partial X^{j}}{\partial x^{i}}\frac{\partial}{\partial x^{j}}$

To compute the Lie derivative of a one form we use

\begin{aligned} \mathbf{d}\alpha & =\frac{\partial\alpha_{i}}{\partial x^{j}}\,dx^{j}\wedge dx^{i}\end{aligned}
\begin{split}\begin{aligned} \mathbf{i}_{X}\mathbf{d}\alpha & =X^{k}\mathbf{i}_{\frac{\partial}{\partial x^{k}}}\mathbf{d}\alpha\\ & =X^{k}\frac{\partial\alpha_{i}}{\partial x^{j}}\mathbf{i}_{\frac{\partial}{\partial x^{k}}}\left(dx^{j}\wedge dx^{i}\right)\\ & =X^{k}\frac{\partial\alpha_{i}}{\partial x^{j}}\left(\left(\mathbf{i}_{\frac{\partial}{\partial x^{k}}}dx^{j}\right)\wedge dx^{i}-\left(dx^{j}\wedge\mathbf{i}_{\frac{\partial}{\partial x^{k}}}dx^{i}\right)\right)\\ & =X^{k}\frac{\partial\alpha_{i}}{\partial x^{j}}\left(\delta_{k}^{j}dx^{i}-\delta_{k}^{i}dx^{j}\right)\\ & =X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}dx^{i}-X^{i}\frac{\partial\alpha_{i}}{\partial x^{j}}dx^{j}\end{aligned}\end{split}
\begin{split}\begin{aligned} \mathbf{i}_{X}\alpha & =X^{i}\mathbf{i}_{\frac{\partial}{\partial x^{i}}}\alpha\\ & =X^{i}\alpha_{j}\mathbf{i}_{\frac{\partial}{\partial x^{i}}}dx^{j}\\ & =X^{i}\alpha_{j}\delta_{i}^{j}\\ & =X^{i}\alpha_{i}\end{aligned}\end{split}
\begin{split}\begin{aligned} \mathbf{d}\mathbf{i}_{X}\alpha & =d\mathbf{d}\left(X^{i}\alpha_{i}\right)\\ & =\left(\frac{\partial X^{i}}{\partial x^{j}}\alpha_{i}+X^{i}\frac{\partial\alpha_{i}}{\partial x^{j}}\right)dx^{j}\end{aligned}\end{split}

Combining the terms above we obtain the results for the Lie derivative of a one-form

$\mathfrak{L}_{X}\alpha=X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}dx^{i}+\frac{\partial X^{i}}{\partial x^{j}}\alpha_{i}dx^{j}$
$\mathfrak{L}_{X}\alpha=\left(X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}+\frac{\partial X^{j}}{\partial x^{i}}\alpha_{j}\right)dx^{i}$

The Lie derivative is a derivation and so it must satisfy a product rule over the pairing of forms and vector fields

$$$\mathfrak{L}_{X}\langle\alpha,Y\rangle=\langle\mathfrak{L}_{X}\alpha,Y\rangle+\langle\alpha,\mathfrak{L}_{X}Y\rangle\label{eq:derivation}$$$

To see that the above expressions for the Lie derivative do indeed satisfy Eq.\ref{eq:derivation}

\begin{split}\begin{aligned} \mathfrak{L}_{X}\langle\alpha,Y\rangle & =\mathfrak{L}_{X}\left(\alpha_{i}Y^{i}\right)\\ & =X^{j}\frac{\partial}{\partial x^{j}}\left(\alpha_{i}Y^{i}\right)\\ & =X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}Y^{i}+X^{j}\alpha_{i}\frac{\partial Y^{i}}{\partial x^{j}}\end{aligned}\end{split}
\begin{split}\begin{aligned} \langle\mathfrak{L}_{X}\alpha,Y\rangle & =\langle X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}dx^{i}+\frac{\partial X^{i}}{\partial x^{j}}\alpha_{i}dx^{j},Y^{k}\frac{\partial}{\partial x^{k}}\rangle\\ & =X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}Y^{k}\langle dx^{i},\frac{\partial}{\partial x^{k}}\rangle+\frac{\partial X^{i}}{\partial x^{j}}\alpha_{i}Y^{k}\langle dx^{j},\frac{\partial}{\partial x^{k}}\rangle\\ & =X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}Y^{k}\delta_{k}^{i}+\frac{\partial X^{i}}{\partial x^{j}}\alpha_{i}Y^{k}\delta_{k}^{j}\\ & =X^{j}\frac{\partial\alpha_{i}}{\partial x^{j}}Y^{i}+\frac{\partial X^{i}}{\partial x^{j}}\alpha_{i}Y^{j}\end{aligned}\end{split}
\begin{split}\begin{aligned} \langle\alpha,\mathfrak{L}_{X}Y\rangle & =\langle\alpha_{k}dx^{k},X^{i}\frac{\partial Y^{j}}{\partial x^{i}}\frac{\partial}{\partial x^{j}}-Y^{i}\frac{\partial X^{j}}{\partial x^{i}}\frac{\partial}{\partial x^{j}}\rangle\\ & =\left(\alpha_{k}X^{i}\frac{\partial Y^{j}}{\partial x^{i}}-\alpha_{k}Y^{i}\frac{\partial X^{j}}{\partial x^{i}}\right)\langle dx^{k},\frac{\partial}{\partial x^{j}}\rangle\\ & =\left(\alpha_{k}X^{i}\frac{\partial Y^{j}}{\partial x^{i}}-\alpha_{k}Y^{i}\frac{\partial X^{j}}{\partial x^{i}}\right)\delta_{j}^{k}\\ & =\alpha_{j}X^{i}\frac{\partial Y^{j}}{\partial x^{i}}-\alpha_{j}Y^{i}\frac{\partial X^{j}}{\partial x^{i}}\end{aligned}\end{split}

After a simple relabeling of the dummy indices it can be shown that Eq.\ref{eq:derivation} is indeed satisfied.

Thus, in two dimensions it can be shown that

\begin{split}\begin{aligned} \mathfrak{L}_{X}f & =X^{x}\frac{\partial f}{\partial x}+X^{y}\frac{\partial f}{\partial y}\\ \mathfrak{L}_{X}Y & =[X^{x}\frac{\partial}{\partial x}+X^{y}\frac{\partial}{\partial y},Y^{x}\frac{\partial}{\partial x}+Y^{y}\frac{\partial}{\partial y}]\\ & =X^{x}\frac{\partial Y^{x}}{\partial x}\frac{\partial}{\partial x}+X^{y}\frac{\partial Y^{x}}{\partial y}\frac{\partial}{\partial x}+X^{x}\frac{\partial Y^{y}}{\partial x}\frac{\partial}{\partial y}+X^{y}\frac{\partial Y^{y}}{\partial y}\frac{\partial}{\partial y}-\\ & -Y^{x}\frac{\partial X^{x}}{\partial x}\frac{\partial}{\partial x}-Y^{y}\frac{\partial X^{x}}{\partial y}\frac{\partial}{\partial x}-Y^{x}\frac{\partial X^{y}}{\partial x}\frac{\partial}{\partial y}-Y^{y}\frac{\partial X^{y}}{\partial y}\frac{\partial}{\partial y}\\ & =\left(X^{x}\frac{\partial Y^{x}}{\partial x}+X^{y}\frac{\partial Y^{x}}{\partial y}-Y^{x}\frac{\partial X^{x}}{\partial x}-Y^{y}\frac{\partial X^{x}}{\partial y}\right)\frac{\partial}{\partial x}+\\ & +\left(X^{x}\frac{\partial Y^{y}}{\partial x}+X^{y}\frac{\partial Y^{y}}{\partial y}-Y^{x}\frac{\partial X^{y}}{\partial x}-Y^{y}\frac{\partial X^{y}}{\partial y}\right)\frac{\partial}{\partial y}\\ \mathfrak{L}_{X}\alpha & =X^{x}\frac{\partial\alpha_{x}}{\partial x}dx+X^{y}\frac{\partial\alpha_{x}}{\partial y}dx+X^{x}\frac{\partial\alpha_{y}}{\partial x}dy+X^{y}\frac{\partial\alpha_{y}}{\partial y}dy+\\ & +\frac{\partial X^{x}}{\partial x}\alpha_{x}dx+\frac{\partial X^{y}}{\partial x}\alpha_{y}dx+\frac{\partial X^{x}}{\partial y}\alpha_{x}dy+\frac{\partial X^{y}}{\partial y}\alpha_{y}dy\\ & =\left(X^{x}\frac{\partial\alpha_{x}}{\partial x}+X^{y}\frac{\partial\alpha_{x}}{\partial y}+\frac{\partial X^{x}}{\partial x}\alpha_{x}+\frac{\partial X^{y}}{\partial x}\alpha_{y}\right)dx+\\ & +\left(X^{x}\frac{\partial\alpha_{y}}{\partial x}+X^{y}\frac{\partial\alpha_{y}}{\partial y}+\frac{\partial X^{x}}{\partial y}\alpha_{x}+\frac{\partial X^{y}}{\partial y}\alpha_{y}\right)dy\end{aligned}\end{split}

Given a two form

$\Omega=\omega\,dx\wedge dy$

it can be shown that

\begin{split}\begin{aligned} \mathfrak{L}_{X}\Omega & =\mathbf{d}\mathbf{i}_{X}\Omega\\ & =\left(\frac{\partial}{\partial x}(\omega X^{x})+\frac{\partial}{\partial y}(\omega X^{y})\right)\,dx\wedge dy\end{aligned}\end{split}

Other quantities

\begin{split}\begin{aligned} \operatorname{div}(X) & =\frac{\partial X^{x}}{\partial x}+\frac{\partial X^{y}}{\partial y}\\ \operatorname{vort}(X) & =\frac{\partial X^{y}}{\partial x}-\frac{\partial X^{x}}{\partial y}\end{aligned}\end{split}