I meet this question but have no idea...
Let $\alpha \in \Omega^{k}(M)$ be a closed form on a manifold, with $k>0$, and $p \in M$ any given point. Show that there exists a closed form $\alpha^{\prime} \in \Omega^{k}(M)$ such that $\alpha^{\prime}$ vanishes on a neighborhood of $p,$ and $\alpha-\alpha^{\prime}$ is exact.
I think I may use Poincare lemma, now $\alpha=d\beta$ locally, then how to proceed? Like, using some partition of unity, I can't see how to product $\alpha^\prime$.
Thanks in advance.
Choose a small enough neighborhood $U$ of $p$ so that $\alpha|_{U}$ is exact there and write $\alpha|_{U} = d\beta$ for a $k - 1$ form $\beta$ defined on $U$. Choose a smaller neighborhood $V$ of $p$ such that $\overline{V} \subseteq U$ and choose a smooth function $f \colon M \rightarrow \mathbb{R}$ which is $1$ on $\overline{V}$ and has support in $U$. Then $d(f\beta)$ (extended by zero outside $U$) gives you a global $k$-form. Set $\alpha' = \alpha - d(f\beta)$. Then, $\alpha'$ is closed and on $V$, we have
$$ \alpha'|_{V} = \alpha|_{V} - d(f\beta|_{V}) = \alpha|_{V} - d\beta \equiv 0.$$