Example for calculating the Lie derivative of a 2-form

Question

Problem

Let $F: \mathbb{R}^3 \to \mathbb{R}^3$, $F(x,y,z) = (z,y,-x)$ be a vectorfield and $\chi_{(x,y,z)} = (z^2 - x^2)(dx \wedge dy - dz \wedge dx)$ a 2-form over $\mathbb{R}^3$. Calculate the Lie derivative $\mathcal{L}_F\chi$.

My Approach

We have the formula $\mathcal{L}_F\chi = i_Fd\chi + di_F\chi$ for the Lie derivative. Simplifying $\chi = (z^2 - x^2)(dx \wedge dy) + (z^2 - x^2)(dx \wedge dz)$. Calculating the exterior derivative $d\chi = (2z)(dx \wedge dy \wedge dz)$. However, I don't know how to calculate $i_F \chi$, $i_Fd\chi$ and $di_F\chi$.

Questions

1. How do I calculate $i_F \chi$, $i_Fd\chi$ and $di_F\chi$?
2. Did I miss an easier way to calculate the Lie derivative?

Answer 1

You need to know how to calculate the exterior derivative and the interior product of a $p$-form. \begin{align} \chi = &\left(z^2 - x^2\right)(dx \wedge dy - dz \wedge dx)\\ = &\left(z^2 - x^2\right)(dx \wedge dy + dx \wedge dz)\\ = &\left(z^2 - x^2\right)dx \wedge dy + \left(z^2 - x^2\right)dx \wedge dz\\\\ i_F\chi = &\left(z^2 - x^2\right)\left( F_xdy - F_ydx \right) + \left(z^2 - x^2\right) \left( F_xdz - F_zdx \right)\\ = &\left(z^2 - x^2\right)\left( zdy - ydx \right) + \left(z^2 - x^2\right)\left( zdz + xdx \right)\\ = &\left( x - y \right)\left(z^2 - x^2\right)dx + z\left(z^2 - x^2\right)dy + z\left(z^2 - x^2\right)dz\\\\ di_F\chi = &\left(\frac{\partial}{\partial x} \left( z\left(z^2 - x^2\right) \right) - \frac{\partial}{\partial y} \left( \left( x - y \right)\left(z^2 - x^2\right) \right) \right) dx \wedge dy\\ &+ \left(\frac{\partial}{\partial x} \left( z\left(z^2 - x^2\right) \right) - \frac{\partial}{\partial z} \left( \left( x - y \right)\left(z^2 - x^2\right)\right)\right) dx \wedge dz\\ &+ \left( \frac{\partial}{\partial y} \left( z\left(z^2 - x^2\right) \right) - \frac{\partial}{\partial z} \left( z\left(z^2 - x^2\right) \right) \right) dy \wedge dz \\ = &\left( z^2 - 2xz - x^2 \right) dx \wedge dy + 2z\left(y-2x\right) dx \wedge dz + \left( x^2 - 3z^2 \right) dy \wedge dz\\\\ d\chi = &\left( - \frac{\partial}{\partial y} \left( z^2 - x^2 \right) + \frac{\partial}{\partial z} \left( z^2 - x^2 \right) \right) dx \wedge dy \wedge dz\\ = &\ 2zdx \wedge dy \wedge dz\\\\ i_Fd\chi =\ &2z \left( F_x dy \wedge dz - F_y dx \wedge dz + F_z dx \wedge dy \right)\\ =\ &2z \left( zdy \wedge dz - ydx \wedge dz - xdx \wedge dy \right)\\ =\ &-2xzdx\wedge dy - 2yzdx\wedge dz + 2z^2dy\wedge dz\\\\ \mathcal{L}_F\chi =\ &di_F\chi + i_Fd\chi\\ =\ &\left( z^2 - 4xz - x^2 \right) dx \wedge dy - 4xz dx \wedge dz + \left( x^2 - z^2 \right) dy \wedge dz \end{align}

Answer 2

An easier way to calculate this is to use the Leibniz law: $$\mathcal L_X(\omega \otimes \eta)=(\mathcal L_X \omega) \otimes \eta+\omega \otimes (\mathcal L_X\eta)$$ $$\mathcal L_X(f\omega)=(Xf)\omega+f(\mathcal L_X\omega)$$ Where $X$ is a vector field and $\omega,\eta$ are differencial forms. We could further exploit the fact that Lie derivative commutes with exterior derivative: $$d\mathcal L_X\omega=\mathcal L_Xd\omega$$ Now in your case, $X=z\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}-x\frac{\partial}{\partial z}$, so $$\mathcal L_X(2zdx \wedge dy \wedge dz)$$ $$=2(\mathcal L_Xz)dx \wedge dy \wedge dz+2z(\mathcal L_Xdx) \wedge dy \wedge dz+2zdx \wedge (\mathcal L_Xdy) \wedge dz+2zdx \wedge dy \wedge (\mathcal L_Xdz)$$ $$=-2xdx \wedge dy \wedge dz+2z(d\mathcal L_Xx) \wedge dy \wedge dz+2zdx \wedge (d\mathcal L_Xy) \wedge dz+2zdx \wedge dy \wedge (d\mathcal L_Xz)$$ $$=2(z-x)dx \wedge dy \wedge dz$$ If you want the expression of $i_X\omega$ for a given coordinate chart, just check the identity $$i_X(dx^{i_1}\wedge\cdots\wedge dx^{i_n})=\sum_r(-1)^rdx^{i_r}dx^{i-1}\wedge\cdots\wedge\hat{dx^{i_r}}\wedge\cdots\wedge dx^{i_n}$$ by inserting all arguments($\hat{dx^{i_r}}$indicating it's omitted).

Answer 3

If $\chi = dx \wedge dy,$ then $\chi (F,X)= dx(F)dy(X)-dy(F)dx(X)$. Hence, $i_F \chi = dx(F)dy-dy(F)dx $

So, given that $F = (z,y,-x)$ and $\chi = (z^2 - x^2)( dx \wedge dy - dz \wedge dx)$, $$i_F \chi = (z^2 - x^2)(dx(z,y,-x)dy - dy(z,y,-x)dx -dz(z,y,-x)dx + dx(z,y,-x)dz $$ $$ =(z^2 - x^2 )(zdy-ydx + xdx +zdz) $$ I think everything can be calculated now.