Compute $L_\mathbb{X}\beta$

Question

Given information: $$\alpha+(x+y)dy+(x^2-y^2)dz$$$$\beta=zdx\wedge dy+xzdx\wedge dz$$$\mathbb{X}$ is the vector field given by $$\mathbb{X}=(0,-x,-1)$$

I have found $i_\mathbb{X}\beta=2xzdx$

Question: Compute $L_\mathbb{X}\beta$

Attempt: I can use the equation: $L_\mathbb{X}\beta=i_\mathbb{X}d\beta-di_\mathbb{X}\beta$

$d\beta=d(zdx\wedge dy+xzdx\wedge dz)=dz\wedge dx\wedge dy$

Therefore $i_\mathbb{X}d\beta=i_\mathbb{X}(dz\wedge dx\wedge dy)$ How do I compute this???

I continue to compute $di_\mathbb{X}\beta=d(2xzdx)=2xdz\wedge dx$

Subtracting one from the other gives the solution.

Answer 1

We have

\begin{align}i_{\Bbb X}(dz\wedge dx\wedge dy) &= i_{\Bbb X}(dz\wedge dx) \wedge dy + (-1)^2 dz \wedge dx \wedge i_{\Bbb X} dy\\ &= i_{\Bbb X}(dz \wedge dx) \wedge dy + dz\wedge dx \wedge (-x)\\ &= [i_{\Bbb X}dz \wedge dx + (-1)^1 dz \wedge i_{\Bbb X} dx]\wedge dy - x\, dz\wedge dx\\ &= [(-1)\wedge dx - dz \wedge 0]\wedge dy - x\, dz\wedge dx\\ &= -dx\wedge dy - x\, dz\wedge dx \end{align}

Also note $L_{\Bbb X} \beta = i_{\Bbb X} d\beta \color{red}{+} di_{\Bbb X}\beta$.