On $\mathbb{R}^{4}$ with coordinates $x_{1},y_{2},x_{2},y_{2}$ consider the 2-form $\omega=dx_{1}\wedge dy_{1}+dx_{2}\wedge dy_{2}$ . GIven a smooth $f$ on $\mathbb{R}^{4}$ , let $X$ be the vector field: $\vec{X}=f_{y_{1}}\partial_{x_{1}}-f_{x_{1}}\partial_{y_{1}}+f_{y_{2}}\partial_{x_{2}}-f_{x_{2}}\partial_{y_{2}}$
Compute the lie derivative $\mathcal{L}_{X}\omega$ of $\omega$ w.r.t. $\vec{X}$
My Answer: $\mathcal{L}_{X}\omega=i_{X}\left(\textrm{d}\omega\right)+\textrm{d}\left(i_{X}\omega\right)$.
Now, $\textrm{d}\omega=0$ , since the coefficient functions of the two elementary 2-forms in $\omega$ are both constants, and hence, have partial derivatives of $0$. Thus, the $i_{X}\left(\textrm{d}\omega\right)$ term vanishes.
As for the other term, since $\omega$ is a 2-form, letting $\vec{v}=\left(v_{1},v_{1}^{\prime},v_{2},v_{2}^{\prime}\right)$ in the given coordinates, $i_{X}\omega$ is the one-form: $\left(i_{X}\omega\right)\left(\vec{v}\right)\overset{\textrm{def}}{=}\omega\left(\vec{X},\vec{v}\right)=\left(dx_{1}\wedge dy_{1}\right)\left(\vec{X},\vec{v}\right)+\left(dx_{2}\wedge dy_{2}\right)\left(\vec{X},\vec{v}\right)$ $=\left|\begin{array}{cc} dx_{1}\left(\vec{X}\right) & dx_{1}\left(\vec{v}\right)\\ dy_{1}\left(\vec{X}\right) & dy_{1}\left(\vec{v}\right) \end{array}\right|+\left|\begin{array}{cc} dx_{2}\left(\vec{X}\right) & dx_{2}\left(\vec{v}\right)\\ dy_{2}\left(\vec{X}\right) & dy_{2}\left(\vec{v}\right) \end{array}\right|$
$=\left|\begin{array}{cc} f_{y_{1}} & dx_{1}\left(\vec{v}\right)\\ -f_{x_{1}} & dy_{1}\left(\vec{v}\right) \end{array}\right|+\left|\begin{array}{cc} f_{y_{2}} & dx_{2}\left(\vec{v}\right)\\ -f_{x_{2}} & dy_{2}\left(\vec{v}\right) \end{array}\right|$
$=f_{y_{1}}dy_{1}\left(\vec{v}\right)+f_{x_{1}}dx_{1}\left(\vec{v}\right)+f_{y_{2}}dy_{2}\left(\vec{v}\right)+f_{x_{2}}dx_{2}\left(\vec{v}\right)$ $=\textrm{d}f\left(\vec{v}\right)$
Hence, $\mathcal{L}_{X}\omega=\textrm{d}\left(\textrm{d}f\right)=0$ . Is this right? Your help is much appreciated.