# Is volume form equal to the top-dimensional form with coefficient $1$?

Wikipedia says a volume form on a differentiable manifold is a nowhere-vanishing top-dimensionial form.

And Guillemin and Pollack says $f: V \to U$ is a diffeomorphim of two open sets in $\mathbb{R}^k$ and $\omega = dx_1 \wedge \cdots \wedge dx_k$ is the volume form.

So, can I understand volume form as the top-dimensional form with coefficient $1$?

And as an exapmple, if $\omega$ is the volume form on $T^2$, then $\int_{T^2} \omega = \operatorname{vol}(T^2)?$

• It doesn't make sense to talk about "the top-dimensional form with coefficient 1." There isn't a canonical choice of top-dimensional form for most manifolds. (I think. It has been a while since I learned this stuff.) Aug 8, 2013 at 22:56
• 40 votes? @Potato Aug 8, 2013 at 22:59
• What? $\textbf{}$ Aug 8, 2013 at 22:59
• Oh~ There is a user named 40 votes.... Aug 8, 2013 at 23:03

• So you mean top forms can't have coefficients, except for $\pm 1$ as orientation? But I've seen expressions like $4 \, dx_1 \wedge dx_2.$ Aug 9, 2013 at 4:47
• When you write "$dx^1dx^2$" you probably have in mind $\mathbb{R}^2$ on which there is a natural Riemannian metric and orientation making that form the unique volume form with the property I mention in my answer Aug 9, 2013 at 4:56