# Two volumes forms on a compact manifold

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?

• Do you know the rank of the vector bundle of differential $n$-forms on a $n$-dimensional manifold? Commented Jun 4, 2022 at 12:05
• @Didier it's equal to 1. I see what you imply. But can you write a complete answer using it? Commented Jun 4, 2022 at 12:30
• Related. Compactness is indeed not necessary. Commented Jun 4, 2022 at 12:31
• @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 Commented Jun 4, 2022 at 12:45

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 $$M$$.
• 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. Commented Jun 4, 2022 at 14:52