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

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 $$\text{supp} \ \omega \subseteq W$$.

1. 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".

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?

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 $$\omega'={1\over{\int_M\omega}}\omega$$.
• ...and an exercise, where I think your $f$ comes from. (Not sure if same bump as in wikipedia.) In particular, if your $f$ does come from the exercise, then I better understand the role of $W$ nonempty for Lemma 10.17. – user636532 May 2 '19 at 8:54
• Tsemo Aristide, is $(g,U)$ a positively oriented chart or at least an oriented chart? See my other question please – user636532 May 2 '19 at 10:28