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Let $M$ be a compact orientable manifold, $\alpha$ and $\beta$ are two volume forms (defined as nowhere vanishing (dim$M$)-forms) on $M$.

Is it true that there exists a smooth function $f:M\rightarrow\mathbb{R}$ such that $\alpha=f\beta$? If it's true, can you prove why? If not, can you please give a counterexample?

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    $\begingroup$ Do you know the rank of the vector bundle of differential $n$-forms on a $n$-dimensional manifold? $\endgroup$
    – Didier
    Commented Jun 4, 2022 at 12:05
  • $\begingroup$ @Didier it's equal to 1. I see what you imply. But can you write a complete answer using it? $\endgroup$ Commented Jun 4, 2022 at 12:30
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    $\begingroup$ Related. Compactness is indeed not necessary. $\endgroup$ Commented Jun 4, 2022 at 12:31
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    $\begingroup$ @randomexchanger The already existing answer is good. Here is a sketch of a proof using the rank: 1) show that if $\omega$ is a volume form, then $\{\omega\}$ is a basis of the set of $n$-forms as a $\mathcal{C}^{\infty}(M)$-module (e.g using the local coordinate expression) 2) In that case, $\alpha$ is a basis, so that $\beta = f\alpha$ for some function $f$. This does not rely on the fact that $\beta$ does not vanish $\endgroup$
    – Didier
    Commented Jun 4, 2022 at 12:45

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This is true, and here is one way of showing it. First consider a chart $(U, x)$ on $M$. There exist smooth functions $f, g : U \to \mathbb{R}$ such that

\begin{align*} \alpha &= g \, dx^1 \wedge \cdots \wedge dx^n \\ \beta &= h \, dx^1 \wedge \cdots \wedge dx^n. \end{align*}

Since, $\alpha, \beta$ are no-where vanishing, so are $g, h$. Thus, $f= \frac{g}{h}$ is well-defined and satisfies that $\alpha = f \beta$ on $U$. Now you can make an argument using a partition of unity and the fact that by compactness you may cover $M$ with finitely many chart domains, the details of which I leave to you, to show that you can make this work globally on $M$.

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    $\begingroup$ Another way to complete the argument without using a partition of unity is to note that your formulas show two things: (1) at each $p\in M$ there is a unique real number $f(p)$ such that $\alpha_p = f(p) \beta_p$; and (2) $f$ depends smoothly on $p$ in in a neighborhood of each point. A function that is smooth in a neighborhood of each point is smooth. $\endgroup$
    – Jack Lee
    Commented Jun 4, 2022 at 14:52
  • $\begingroup$ Yes, great point! Thanks for pointing it out. $\endgroup$ Commented Jun 4, 2022 at 16:13

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