1

I'm confused whether the following statement is true: Let $K$ be a compact, orientable manifold of dimension $n$ with boundary and $\omega$ be a smooth $n$-form on $K$. Say for example $K$ is the closed unit disc in $\mathbb{R}^2$ and $\omega$ is any $2$-form on $\mathbb{R}^2$ restricted to $K$. Is the following statement true?: $$ \int_{K} \omega = \int_{\overset{°}{K}} \omega $$ , where $\overset{°}{K}$ is the interior of $K$.

I.e. can the boundary of a set be ignored when integrating over the set ? I think it should be true, since the boundary should be a set of measure zero in $K$, but is there a flaw with this reasoning ? Thanks for your help.

Sebastian Hebold
  • 73
  • How do you define integrals on non-compact manifolds? – Exit path Dec 21 '19 at 15:42
  • The title differs significantly from the body. The integral over the boundary of a manifold is negligible. The boundary of a set need not be measure zero. https://math.stackexchange.com/questions/613408/when-does-the-boundary-have-measure-zero – Eric Towers Apr 05 '21 at 06:51

1 Answers1

1

Yes, this is correct. The integral of $\omega$ is defined by adding together its integrals in coordinate charts with a partition of unity. In each coordinate chart, the boundary $\partial K$ has measure $0$ (since the chart sends it to a hyperplane in $\mathbb{R}^n$ which has $n$-dimensional Lebesgue measure $0$), and so it does not affect the integral.

Eric Wofsey
  • 342,377