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Let $M$ be a smooth oriented manifold. Does there exist a smooth measure $m$ on $M$ which is not induced by the volume form of some Riemannian metric $g$ on $M$? I would say that the set of volume forms induced by Riemannian metrics is strictly contained in the set of all smooth measures on $M$...My interest would be to have some criteria for deciding whether a given measure on $M$ is induced by a Riemannian metric or not

edited Apr 4 2011 at 14:23

asked Apr 4 2011 at 14:01

Kikiriku
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Without some additional assumptions on $m$ the answer is trivially yes. For example, if $m$ is a point mass. – Mark Meckes Apr 4 2011 at 14:07

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As Mark indicates, this is easily answered using elementary facts from measure theory and Riemannian geometry. At best, it is suitable not here but on math.stackexchange.com – Deane Yang Apr 4 2011 at 14:13

Ok, that would be a degenerate case. But let us assume that $m$ is in some sense a smooth measure... – Kikiriku Apr 4 2011 at 14:15

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My comment above still stands. – Deane Yang Apr 4 2011 at 14:27

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To save you some time reasking this, here is an answer: Choose a reference metric $g$ on $M$. Clearly if there is any hope at all of $m$ coming from some metric, $m=\rho vol_M$ for $\rho \in C^\infty(M)$, $\rho>0$. Note that in coordinates the volume form is $\omega = \sqrt{g} dx^1\wedge\dots\wedge dx^n$, so we would like $\tilde g$ such that the associated volume form is $\rho \sqrt{g} dx^1\wedge\dots\wedge dx^n$. Thus $\rho \sqrt{g} = \sqrt{\tilde g}$. Thus, if we take $\tilde g_{ij} = \rho^{2/n} g_{ij}$, we clearly have that the volume form associated to $\tilde g$ gives the measure $m$. – Otis Chodosh Apr 4 2011 at 15:11

Corbennick · Answer 1 · 2011-04-04 19:35:31Z UTC

The only possible meaning for smoothness that comes to mind is the following: A measure on ${\mathbb R}^n$ is smooth, if it has a smooth density against the Lebesgue-measure. On a manifold, a measure is smooth if it transforms to smooth measures on every smooth chart. The question, whether a given smooth measure comes from a metric is equivalent to the question whether its density has a zero or not. If it has no zero, simply choose any Riemannian metric. Then your given measure has a nowhere vanishing density against the measure coming from the metric. Simply multiply the metric with a suitable power of the density to get a metric that induces the given measure. Since Radon-Nikodym densities are uniquely determined, this is an if and only if criterion.

S.A.A · Answer 2 · 2012-06-22 21:04:29Z UTC

I think the answer to your question is given at least within the class of k\"ahlerian manifolds. I would refer you to standard texts of complex geometry for the preliminaries. But in the case of k\"hlerian manifolds, any volume form, that is, a measure, whose Radon-Nikodym density with respect to a fixed reference metric $g$ is smooth enough, and has the same total volume as $d Vol_g$ can be realised as the volume form of a k\"ahlerian metric $g^{\phi}$ in the same cohomology class, ie. $g_{i \bar{j}}^{\phi} = g_{i \bar{j}} + \partial_{i} \partial_{\bar{j}}$. This was Calabi's volume conjecture.

Also, you can realise an -reasonably- arbitrary volume forms by conformal change of the metric. However, in the conformal case the total volume can change.

Measures on Riemannian manifolds which are not induced by the volume form of some Riemannian metric

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