Every smooth form is either closed or exact

Given: The smooth form $\alpha =xdy$ on ${\mathbb{R}}^2$.

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^k(M)=\ker (d:{\mathrm{\Omega }}^k(M)\to {\mathrm{\Omega }}^{k+1}(M))$ and $B^k(M)=\mathrm{im}(d:{\mathrm{\Omega }}^{k-1}(M)\to {\mathrm{\Omega }}^k(M))$. A form is closed if it belongs to $Z^k$ and exact if it belongs to $B^k$. If $\omega =d\eta$, then $d\omega =d^2\eta =0$ by thm-the-exterior-derivative-squares-to-zero, so $B^k\subseteq Z^k$. In particular $B^0=0$, since ${\mathrm{\Omega }}^{-1}=0$. The zero form is both closed and exact in every degree.

The local coordinate formula for the exterior derivative: Let $(U,x^1,\dots ,x^n)$ be a smooth chart on a smooth manifold and $\omega$ a smooth $k$-form on $U$, with $k\ge 0$. Summing over increasing $k$-tuples $I$, and writing $d{x}^I=d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$, if $\omega ={\sum }_I{\omega }_{I}d{x}^I$, then $d\omega ={\sum }_{I}d{\omega }_I\wedge d{x}^I\mathrm{.}$

The exterior derivative squares to zero: For every differential form $\omega$, $d(d\omega )=0$.