Differential forms and simplification

Question

Let us suppose we have

$$(\partial_x \alpha - \partial_x \bar{\alpha} )(df+\lambda)\wedge dx+ (\partial_y \alpha - \partial_y \bar{\alpha} )(df+\lambda) \wedge dy + (\partial_z \alpha - \partial_z \bar{\alpha})(df+\lambda)\wedge dz$$

and then we decided to write those as

$$\partial_x (\alpha - \bar{\alpha} )(df+\lambda)\wedge dx+ \partial_y (\alpha - \bar{\alpha} )(df+\lambda) \wedge dy+\partial_z (\alpha - \bar{\alpha} )(df+\lambda)\wedge dz$$

where $\alpha$ is a complex function, $df$ is a one-form and so is $\lambda$.

Can we write this summation as $d[(\alpha -\bar{\alpha})(df+\lambda)]$?

Answer 1

It's the negative of your answer. To get $d\alpha\wedge \lambda$, for example, you need to switch the order in $d\lambda\wedge dx$, etc.

Answer 2

By the Leibniz rule,

\begin{align*} &\ d[(\alpha - \overline{\alpha})(df + \lambda)]\\ =&\ d(\alpha - \overline{\alpha})\wedge(df + \lambda) + (\alpha - \overline{\alpha})d(df + \lambda)\\ =&\ (\partial_x(\alpha - \overline{\alpha})dx + \partial_y(\alpha - \overline{\alpha})dy + \partial_z(\alpha - \overline{\alpha})dz)\wedge(df + \lambda) + (\alpha - \overline{\alpha})(d(df) + d\lambda)\\ =&\ \partial_x(\alpha - \overline{\alpha})dx\wedge(df + \lambda) + \partial_y(\alpha - \overline{\alpha})dy\wedge(df + \lambda) + \partial_z(\alpha - \overline{\alpha})dz\wedge(df + \lambda) + (\alpha - \overline{\alpha})d\lambda\\ =&\ (\alpha - \overline{\alpha})d\lambda-\partial_x(\alpha - \overline{\alpha})(df + \lambda)\wedge dx - \partial_y(\alpha - \overline{\alpha})(df + \lambda)\wedge dy - \partial_z(\alpha - \overline{\alpha})(df + \lambda)\wedge dz. \end{align*}

So your expression can be written as $(\alpha - \overline{\alpha})d\lambda - d[(\alpha-\overline{\alpha})(df + \lambda)]$. If $d\lambda = 0$ (i.e. $\lambda$ is closed), then your expression can be written as $-d[(\alpha - \overline{\alpha})(df + \lambda)]$.