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

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

2011-04-04T14:01:01Z

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

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

2011-04-04T15:10:26Z

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.

Answer by S.A.A for Measures on Riemannian manifolds which are not induced by the volume form of some Riemannian metric

2012-06-22T21:04:29Z

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.