Let $\bar{M}$ be a manifold with boundary and let $M$ be its interior. Is this statement correct?

A smooth k-form $\alpha$ on $\bar{M}$ is closed (exact) if and only if its restriction to $M$, i.e. $\alpha|_{M}$, is closed (exact).

I can also assume that $\bar{M}$ is a compact subset of $\mathbb{R}^{n}$ if necessary.



  • 1
    $\begingroup$ Let's start by getting definitions down. What is your definition of a smooth $k$-form on a manifold with boundary? I.e., what is a smooth $k$-form on the upper half-space $\mathbb H^n = \{x\in\mathbb R^n: x_n\ge 0\}$? $\endgroup$ – Ted Shifrin Jun 30 '13 at 19:28
  • $\begingroup$ A mapping $f:U\subset\mathbb{H}^{n}\rightarrow\mathbb{H}^{n}$ is smooth if it is a restriction of a smooth mapping $\bar{f}:\bar{U}\rightarrow\bar{V}$, where $\bar{U},\bar{V}$ are open subsets of $\mathbb{R}^{n}$. A smooth form has smooth coefficients in a local coordinate system. $\endgroup$ – Arzhang Jun 30 '13 at 19:49
  • $\begingroup$ OK, I agree. So clearly a form that's closed on $M$ has to be closed on $\bar M$. And, by the Poincaré lemma (since we can restrict to a star-shaped subset $U\subset\mathbb H^n$), I believe it follows immediately for exactness, too. $\endgroup$ – Ted Shifrin Jun 30 '13 at 20:30

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