most recent 30 from http://mathoverflow.net 2013-05-21T22:23:09Z http://mathoverflow.net/feeds/user/13875 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60554/measures-on-riemannian-manifolds-which-are-not-induced-by-the-volume-form-of-some Kikiriku 2011-04-04T14:01:01Z 2012-06-22T21:04:29Z

<p>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</p>

http://mathoverflow.net/questions/59422/cyclotomic-polynomials Kikiriku 2011-03-24T12:21:16Z 2011-03-25T07:07:00Z

<p>Let $\phi_{n}(x)$ be the $n$-th cyclotomic polynomial. What are the restrictions to $n$ (if any) to have $\phi_{n}(x)$ divides $\phi_{2n}(x)$ (where division is in $\mathbb{Z}[x]$)?Or is it true that $\frac{\phi_{2n}(x)}{\phi_{n}(x)}\in\mathbb{Z}[x]$ for all integers $n$?</p>

http://mathoverflow.net/questions/60554/measures-on-riemannian-manifolds-which-are-not-induced-by-the-volume-form-of-some Kikiriku 2011-04-04T14:15:43Z 2011-04-04T14:15:43Z

Ok, that would be a degenerate case. But let us assume that $m$ is in some sense a smooth measure...

http://mathoverflow.net/questions/59422/cyclotomic-polynomials/59426#59426 Kikiriku 2011-03-24T13:11:11Z 2011-03-24T13:11:11Z

Clear...thx....

http://mathoverflow.net/questions/59422/cyclotomic-polynomials Kikiriku 2011-03-24T13:04:03Z 2011-03-24T13:04:03Z

Oh I see...but is it still impossible to have &quot;$\phi_{n}(x)$ divides $\phi_{2n}(x)$&quot; (not necessarily over \mathbb{Z}[x])?