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If $M$ is a compact orientable $n$-manifold and $\omega$ is an $n$-form on $M$. If I let $\mu$ be a volume form on $M$, can I say that every other $n$-form on $M$ is a multiple of the volume form and write is as $\omega = f \mu $ where $f$ is a smooth function?

HK Lee
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Jack
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1 Answers1

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Yes; if $M$ has dimension $n$, then $\Omega^\bullet(M):=C^\infty(M)\otimes \wedge^\bullet T^*(M)$ and

$$\operatorname{dim}~\wedge^n T^*(M)=1.$$

The volume form is a basis of $\Omega^n(M)$ and the diff. graded algebra $\Omega^\bullet(M)$ is a $C^\infty(M)$-module.

Avitus
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  • Oh, I see. Thank you. By the way, what's the difference between volumn form and the volumn element? I am kind of confused here about them. @Avitus – Jack Dec 15 '13 at 22:10
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    I think you should distinguish between $\Lambda^n T^\ast M$ and its space of sections $\Omega^n(M)$. The dimension in your answer is a dimension as a $C^\infty(M)$-module. – Jeremy Daniel Dec 15 '13 at 22:13
  • But volumn form is defined for Riemann manifolds, can we define for a general manifold? @Jeremy Daniel – Jack Dec 15 '13 at 22:50
  • @Jeremy Daniel yes, notation is a bit sloppy: I modify it. Thank you! To the OP: volume forms are defined on orientable manifolds. On Riemannian manifolds there is a canonical way to define them. – Avitus Dec 16 '13 at 08:54