Question about Interior product computation

Question

I need to evaluate $\omega=i_X(dx\wedge dy)$ where $X$ is a vector field in $\mathbb{C}^2$ (which means $p=2)$. If I write $X(x,y)=(X_1(x,y),X_2(x,y))$, or simply $X=(X_1, X_2)$, then the interior product $\omega=i_X(dx\wedge dy)$ is given by the formula $$ \omega=(i_Xdx)\wedge dy+(-1)^pdx\wedge (i_Xdy) $$ which in turn (and here's my doubt) equals to: $$ \dfrac{\partial}{\partial x}(X_1)dy-\dfrac{\partial}{\partial y}(X_2)dx $$ Alternatively, it could be equal to $$ X_1dy-X_2dx $$ I'm not sure which one of those two is the correct (I'm not even sure if any of them is the correct computation of the interior product). Can somebody please point the way?

Answer 1

We have that $$X = X_1(x,y) \frac{\partial}{\partial x} + X_2(x,y) \frac{\partial}{\partial y}$$ for some functions $$X_1, X_2: \Bbb R^2 \longrightarrow \Bbb R.$$ Now since $dx$ is a $1$-form, we have that \begin{align*} \iota_X(dx \wedge dy) & = (\iota_X dx) \wedge dy - dx \wedge (\iota_X dy) \\ & = dx\left(X_1(x,y) \frac{\partial}{\partial x} + X_2(x,y) \frac{\partial}{\partial y} \right) \wedge dy - dx \wedge dy\left( X_1(x,y) \frac{\partial}{\partial x} + X_2(x,y) \frac{\partial}{\partial y} \right) \\ & = \left( X_1(x,y) dx\left(\frac{\partial}{\partial x}\right) + X_2(x,y) dx\left(\frac{\partial}{\partial y} \right) \right) \, dy \\ & \phantom{=} - \left( X_1(x,y) dy\left(\frac{\partial}{\partial x}\right) + X_2(x,y) dy\left(\frac{\partial}{\partial y} \right)\right) \, dx \\ & = X_1(x,y) \, dy - X_2(x,y) \, dx. \end{align*} Here we used the fact that $$dx\left(\frac{\partial}{\partial x}\right) = dy\left(\frac{\partial}{\partial y}\right) = 1$$ and $$dx\left(\frac{\partial}{\partial y}\right) = dy\left(\frac{\partial}{\partial x}\right) = 0.$$