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I would like to prove the following:

Guess. Suppose M is an orientable smooth manifold without boundary and W is an open set in M. If $\omega$ is a smooth n-form on M such that $\operatorname{supp} \omega \subset W$ and $\operatorname{supp} \omega$ is compact, then $$ \int_M \omega = \int_W \omega. $$ To be precise, we here treat W as an n-dimensional submanifold with the orientation induced by M.

For a similar claim in $R^n$, see this question.

My attempt so far is the following (following J.M. Lee, Introduction to smooth manifolds). To apply the definition of $\int_M \omega$, let $U_1, \ldots, U_N$ be a finite cover of $\operatorname{supp} \omega$ with positively oriented charts $(U_i, \phi_i)$. Let $\psi_i$ be a smooth partition of unity subordinate to this cover. Then, by definition, \begin{eqnarray*} \int_M \omega &=& \sum_i \int_{\phi_i(U_i)} (\phi_i^{-1})^\ast(\psi_i \omega). \end{eqnarray*}

Furthermore, $U_1\cap W, \ldots U_N\cap W$ is an finite open cover for $\operatorname{supp} \omega$ and each $(U_i \cap W, \phi_i\vert_{U_i\cap W})$ is a positively oriented chart, so \begin{eqnarray*} \int_W \omega &=& \sum_i \int_{\phi_i(U_i\cap W)} ((\phi_i\vert_{U_i\cap W})^{-1})^\ast(\psi_i \omega). \end{eqnarray*}

Let us fix $i\in \{1, \ldots, N\}$, and let us drop the sub-indices. If $x^i$ are local coordinates for a chart $\phi: U\to R^d$ where $d = \dim M$, then $$ \int_{\phi(U)} (\phi^{-1})^\ast(\psi \omega) = \int_{\phi(U)} (\hat \psi \hat \omega)(x) dx^1 \wedge \cdots dx^d, $$ where $\hat \psi, \hat \omega: \phi(U)\to R$ are local functions representing $\psi$ and $\omega$ in $(U, \phi)$.

Any ideas on how to continue this? One would probably need some result to restrict the domain of a (Riemann) integral.

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Just take a cover of $supp \omega$ in $W$ and notice that it is a cover of $supp \omega$ also in $M$. So is you know that the integral does not depend on the choice of the cover, you are done.

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  • $\begingroup$ Yes. Thank you! $\endgroup$ – user117877 Mar 3 '13 at 15:32

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