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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

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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 '11 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 '11 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 '11 at 14:15
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My comment above still stands. –  Deane Yang Apr 4 '11 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 '11 at 15:11
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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.

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There's one minor error in this. But again this belongs in math.stackexchange.com and not here. –  Deane Yang Apr 4 '11 at 15:12
    
It seems that the same is true for complete Riemannian metrics. –  Anton Petrunin Apr 4 '11 at 23:49
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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.

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