Lie derivative of a volume form

Question

Let $M$ be a $3-$dimensional orientable manifold. Suppose $\Omega = dx\wedge dy\wedge dz$ a volume form on it.

How can I compute explicitly the Lie derivative of it along the vector field $$ X = x\partial_x + y\partial_y+ xy\partial_{z}? $$

I see that by Cartan's formula we get $\mathcal{L}_X\Omega = di_X\Omega$, but how can I compute explicitly the contraction on the right?

By definition I think I should take 3 test vector fields $Y_1,Y_2$ so that I get $$ (i_X\Omega)(Y_1,Y_2)=\Omega(X,Y_1,Y_2), $$ but how can I compute this simply?

Answer 1

Hints:

$1).\ $ We have the Leibnitz rule

$L_X\Omega=L_Xdx\wedge dy\wedge dz+dx\wedge L_Xdy\wedge dz+dx\wedge dy\wedge L_Xdz$

$2).\ $ $L_X$ commutes with the exterior derivative; that is, $L_Xd\omega=dL_X\omega$ for any $k$-form, $\omega$

$3).\ $ The Lie derivative of a $C^\infty$ function $f$ is $L_Xf=Xf$

Answer 2

For $X=x\,\partial_x+y\,\partial_y+xy\,\partial_z$ and $\Omega=dx\wedge dy\wedge dz$ we get, using $$ dx(X)=x\,,\quad dy(X)=y\,,\quad dz(X)=xy\,, $$ that the contraction is $$ i_X\Omega=x\,dy\wedge dz-y\,dx\wedge dz+xy\,dx\wedge dy\,. $$ In this case, $$ d\,(i_X\Omega)=dx\wedge dy\wedge dz-dy\wedge dx\wedge dz+0=2\,\Omega\,. $$ The Lie derivative of $\Omega$ w.r.t. $X$ is therefore by, Cartan's formula, $$ L_X\Omega=d\,(i_X\Omega)+i_X\,\underbrace{d\Omega}_{0}=2\,\Omega\,. $$ Neat!