Why do connected oriented manifolds have compactly supported forms with integral one but with support contained in a given open proper subset?

Question

My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave. It seems to be claimed

1. In the proof of Lemma 10.17:

For every proper open subset $W$ of $\mathbb R^n$, there is an $\omega \in \Omega_c^{n}(\mathbb R^n)$, namely the "$\omega_1$" in the proof, where $\int_{\mathbb R^n} \omega = 1$ and $\text{supp} \ \omega \subseteq W$.

2. In the proof of Theorem 11.9:

For a compact connected oriented smooth n-manifold $M$ and for every open subset $U$ of $M$, there is an $\omega \in \Omega_c^{n}(M) = \Omega^{n}(M)$, where $\int_M \omega = 1$ and $\text{supp} \ \omega \subseteq U$ (I don't believe this is dependent on the particular $U$ from Lemma 11.8).

Question: Why?

Please try to answer using the tools in the book such as Theorem 10.13 (or Corollary 10.14), Lemma 10.15, the last sentence of this or Lemma 10.3(ii).

• I think I'm missing something obvious about how such $\omega$ exists because the authors state it so naturally.

• Is the fact of the existence of such $\omega$ actually not obvious to the reader at this point in the text? For non-obvious facts, I think authors would usually say something like "First/next, observe that (insert fact), the proof of which is left to the reader/to the exercises".

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Here are my thoughts (assuming I understand right):

1. I think (2) follows from Theorem 10.13 (or Corollary 10.14).

   • I get $\tau \in \Omega_c^{n}(M)$ , where $\int_{M} \tau = 1$ and $\text{supp} \ \tau \subseteq M$. Choose $\omega$ to be the zero extension of $\tau|_{U}$ to $\tilde{\tau|_{U}} = \tau|_{U}1_{U} + 0 \ 1_{U^c}$. Similarly, (1) would follow from Lemma 10.15: I get $\tau \in \Omega_c^{n}(\mathbb R^n)$, $\int_{\mathbb R^n} \tau = 1$, $\text{supp} \ \tau \subseteq \mathbb R^n$, and then $\omega = \tilde{\tau|_{W}} = \tau|_{W}1_{W} + 0 \ 1_{W^c}$.

2. For (2), instead of applying Theorem 10.13 to $M$, I'll apply Theorem 10.13 to $U$, where the proof of Theorem 11.9 says we may actually be assume $U$ to be connected.

   • We get $\gamma \in \Omega_c^n(U)$, where $\int_{U} \gamma = 1$ and $\text{supp} \ \gamma \subseteq U$. Let $\psi = \gamma|_{\text{supp} \gamma}$, the restriction of $\gamma$ to its support. Denote the zero extensions of $\gamma$ and $\psi$ as $\tilde{\gamma} = \gamma 1_{U} + 0 \ 1_{M \setminus U}$ and $\tilde{\psi} = \psi 1_{\text{supp} \gamma} + 0 \ 1_{M \setminus \text{supp} \gamma}$.

   • Observe $\tilde{\psi} = \tilde{\gamma}$ (like here), and both are smooth by the last sentence of this.

   • Choose $\omega = \tilde{\psi} = \tilde{\gamma}$: As $\omega = \tilde{\psi}$, $\omega$ will satisfy $\text{supp} \ \omega \subseteq U$. As $\omega = \tilde{\gamma}$, $\omega$ will satisfy $\int_{M} \omega = 1$ because $\int_{M} \tilde{\gamma} = \int_{U} \gamma$ by Lemma 10.3(ii).

   • But with this method, how do we argue similarly for (1), where the proof of Lemma 10.17 does not say that we may assume $W$ is connected?

Answer 1

An open subset contain the domain of a chart $g:U\rightarrow\mathbb{R}^n$ which is diffeomorphic to an open subset of $\mathbb{R}^n$, let $f$ be a cut-off function defined on $g(U)$, (the support of $f$ is contained in $g(U)$, $f\neq 0$) and $\omega_0$ the standard volume form of $\mathbb{R}^n$. There exists an $n$-form $\omega$ defined on $M$ whose restriction to $U$ is $g^*(f\omega_0)$ and the restriction of $\omega$ on $M-U$ is zero. Define $\omega'={1\over{\int_M\omega}}\omega$.