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Let $D^2$ denote the closed unit disk in $\mathbb{R}^2$. Let $\omega = dx \wedge dy$ denote the standard area form on $\mathbb{R}^2$ (and on $D^2$ by restriction). Let $\phi$ be a diffeomorphism of $D^2$ which is equal to the identity in a neighborhood of $\partial D^2$, and which preserves area; i.e. $\phi^*\omega = \omega$. We denote the group of such diffeomorphisms by $\text{Diff}_\omega(D^2, \partial D^2)$. I know from here that there is a $1$-form $\alpha$ with $d\alpha = \omega$.

I have two questions.

  1. Is $\phi^* \alpha - \alpha$ exact?
  2. Is $\phi^*\alpha - \alpha$ equal to $df$ for some smooth function $f$?
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$\phi^*\alpha - \alpha$ is closed as $$\begin{split} d (\phi^* \alpha- \alpha) &= d\phi^*\alpha - d\alpha\\ & = \phi^* d\alpha - d\alpha \\ &= \phi^*\omega - \omega \\ &= \omega - \omega = 0 \end{split}$$

Thus $\phi^*\alpha - \alpha$ is also exact as $D^2$ is contractible, and equals to $df$ for some smooth function $f$.