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I'm reading Lee's Introduction to Smooth Manifolds, and his definition of the integral of a differential form is a bit hard for me to understand.

Suppose $M$ is an oriented smooth $n$-manifold with or without boundary, and $\omega$ is a compactly supported $n$-form on $M$. Let $\{U_i\}$ be a finite open cover of $\mathsf{supp}\omega$ by domains of positively or negatively oriented smooth charts, and let $\{\psi_i\}$ be a subordinate smooth partition of unity. Define the integral of $\omega$ over $M$ to be $$\int_M \omega=\sum_i\int_M \psi_i \omega.\tag{16.2}$$

I don't know exactly what guarantees the existence of $\{U_i\}$. More concretely, I haven't found any proposition in the previous sections that assures me of a smooth atlas $\{(U_\alpha,\varphi_\alpha)\}$ consisting of charts that are either positively-oriented or negatively-oriented. If such atlas existed, I would employ compactness of $\mathsf{supp}\omega$ to extract a finite subcover from the open cover $\{U_\alpha\}$. This subcover can then serve as $\{U_i\}$.

Does anyone have an idea? Thank you.

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3 Answers 3

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One way of seeing that this is true is by noting that on an oriented manifold $(M,\mu)$, every chart with connected domain is either positively or negatively oriented. To see this, note that in any chart $\varphi:U\to\mathbb{R}^n$ we have $\mu|_{U}=fdx^1\wedge\cdots\wedge dx^n$, where $f:U\to\mathbb{R}$ is continuous and nonvanishing.

From there you can either construct such a covering explicitly (e.g. out of coordinate balls), or note that given any cover by coordinate charts, we can subdivide the disconnected charts into their connected components and obtain a cover by connected coordinate charts.

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  • $\begingroup$ I'm not sure if I misunderstood your argument. Are you suggesting that every smooth manifold admits a smooth atlas consisting of smooth charts with connected domain? Thank you. $\endgroup$
    – Boar
    Feb 22, 2022 at 15:10
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    $\begingroup$ @Steve That's also true, but not strictly necessary here. To make sense of the definition in the question, all that's needed is to argue that $\operatorname{supp}(\omega)$ is covered by coordinate charts with connected domain. This can be done e.g. by taking a coordinate ball around each point in $\operatorname{supp}(\omega)$. $\endgroup$
    – Kajelad
    Feb 22, 2022 at 18:49
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Take a look at Proposition 15.6 in ISM:

Proposition 15.6 (The Orientation Determined by a Coordinate Atlas). Let $M$ be a smooth positive-dimensional manifold with or without boundary. Given any consistently oriented smooth atlas for $M$; there is a unique orientation for $M$ with the property that each chart in the given atlas is positively oriented. Conversely, if $M$ is oriented and either $\partial M=\varnothing$ or $\dim M >1$, then the collection of all oriented smooth charts is a consistently oriented atlas for $M$.

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The existence of a smooth atlas consisting of charts whose coordinate changes are positively (resp. negatively) oriented is equivalent to $M$ being orientable, which is a hypothesis we do have in the definition you cite. Since such an atlas does exist (again, because we're supposing $M$ is oriented), we can proceed as you say and use the compactness of $\operatorname{supp}(\omega)$ to extract the desired finite subcover.

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  • $\begingroup$ Thank you, but in Lee's book, a manifold is called orientable if we can endow it with a continuous pointwise orientation, whose definition does not refer to any positively-oriented charts or negatively-oriented charts. That's why I feel confused. $\endgroup$
    – Boar
    Feb 20, 2022 at 2:49

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