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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 $\int_{T^2} \omega = \operatorname{vol}(T^2)?$

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    $\begingroup$ 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.) $\endgroup$
    – Potato
    Aug 8, 2013 at 22:56
  • $\begingroup$ 40 votes? @Potato $\endgroup$ Aug 8, 2013 at 22:59
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    $\begingroup$ What? $\textbf{}$ $\endgroup$
    – Potato
    Aug 8, 2013 at 22:59
  • $\begingroup$ Oh~ There is a user named 40 votes.... $\endgroup$ Aug 8, 2013 at 23:03

1 Answer 1

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The space of top forms is the space of sections of a real vector bundle of dimension 1 over your manifold. A volume form is a non-vanishing section of this bundle. So as Potato says, if there is one volume form (I.e. if your manifold is orientable), then there are many. You can, however, distinguish a unique volume form if your manifold is oriented and Riemannian: You then require the form to take the value 1 on oriented orthonormal bases at each point.

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  • $\begingroup$ So you mean top forms can't have coefficients, except for $\pm 1$ as orientation? But I've seen expressions like $4 \, dx_1 \wedge dx_2.$ $\endgroup$ Aug 9, 2013 at 4:47
  • $\begingroup$ You can multiply volume forms by nonzero real numbers and still get a volume form. The point is: without a metric and an orientation there is no natural choice for the form to the right of the coefficient. $\endgroup$ Aug 9, 2013 at 4:50
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    $\begingroup$ Lee's "introduction to smooth manifolds" is the best reference for this, IMO. $\endgroup$ Aug 9, 2013 at 4:52
  • $\begingroup$ 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 $\endgroup$ Aug 9, 2013 at 4:56
  • $\begingroup$ Thank you Tim, now I understand. :) $\endgroup$ Aug 9, 2013 at 5:30

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