Timeline for Measures on Riemannian manifolds which are not induced by the volume form of some Riemannian metric
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jun 22, 2012 at 21:04 | answer | added | S.A.A | timeline score: 1 | |
| Apr 4, 2011 at 15:11 | comment | added | Otis Chodosh | 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 $m$. | |
| Apr 4, 2011 at 15:10 | answer | added | user13945 | timeline score: 9 | |
| Apr 4, 2011 at 14:27 | comment | added | Deane Yang | My comment above still stands. | |
| Apr 4, 2011 at 14:23 | history | edited | Kikiriku | CC BY-SA 2.5 |
added 14 characters in body
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| Apr 4, 2011 at 14:15 | comment | added | Kikiriku | Ok, that would be a degenerate case. But let us assume that $m$ is in some sense a smooth measure... | |
| Apr 4, 2011 at 14:13 | comment | added | Deane Yang | 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 | |
| Apr 4, 2011 at 14:07 | comment | added | Mark Meckes | Without some additional assumptions on $m$ the answer is trivially yes. For example, if $m$ is a point mass. | |
| Apr 4, 2011 at 14:01 | history | asked | Kikiriku | CC BY-SA 2.5 |