# Volume form is “always positive”

The Wikipedia article on volume forms states the following:

In mathematics, a volume form on a differentiable manifold is a nowhere-vanishing top-dimensional form (i.e., a differential form of top degree).

Later it is mentioned that:

Any oriented pseudo-Riemannian (including Riemannian) manifold has a natural volume form. In local coordinates, it can be expressed as $\omega ={\sqrt {|g|}}dx^{1}\wedge \dots \wedge dx^{n}$

Is this the only volume form that we can define on a Riemannian manifold or is it just the most natural (since it is constructed from the Jacobian) For example, could we prove that:

$$\omega = dx^1 \wedge \dots \wedge dx^n$$ may be vanishing ?

• Yes, this could be vanishing. For example a non-orientable manifold doesn't have any volume form. – R. Alexandre Apr 23 '17 at 11:25
• @R.Alexandre I believe the manifold is assumed to be orientable in the first place. – Kenny Wong Apr 23 '17 at 11:30
• @gertian: The form $\Bbb d x^1 \wedge \dots \wedge \Bbb d x^n$ exists only on the chart where the coordinates $x^1, \dots, x^n$ exist. You gave no way of extending it outside of this chart. Keep in mind that volume forms are global objects, i.e. defined on the whole manifold, not just on some fixed coordinate chart. – Alex M. Apr 23 '17 at 11:32

If $\omega$ is a volume form on $M$, then $f \omega$ is also a volume form for any smooth positive function $f$ on $M$. So if there is one volume form, then there are infinite many volume forms!
The volume form $\sqrt{g} \ dx_1 \wedge \dots \wedge dx_n$ is the natural volume form in the context of general relativity because it actually corresponds to the physical volume.
[For example, if $M$ is a spherical surface, so the metric is $ds^2 = d\theta^2 + \sin^2 \theta \ d\phi^2$, then $\sqrt{g} \ d\theta \wedge d\phi = \sin \theta \ d\theta \ d\phi$ is the familiar volume form in spherical coordinates that actually describes the real physical volume.]
Another nice thing about the "natural" volume form is that it is given by the expression $\sqrt{g} \ dx_1 \wedge \dots \wedge dx_n$ in all local coordinate charts. Your suggestion of $dx_1 \wedge \dots dx_n$ will look very different in a different coordinate frame - and then you would have to decide how to define this outside your chosen coordinate patch as well, which may not even be possible.
[Back to the sphere example, picking $d\theta \wedge d\phi$ as the volume form wouldn't actually work because there would be no way of smoothly extending this to the north and south poles (which are not covered by the $(\theta, \phi)$ coordinate system).]