Timeline for When is the module of Kahler volume forms torsion-free?
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| Apr 16, 2013 at 23:13 | comment | added | Greg Muller | Long... thanks for the references! Your intuition in the last sentence is the opposite of what I expected, though. The one positive example I did completely was the cone $xz=y^2$, in which $\Omega^1$ is non-projective, but $\Omega^2$ is torsion-free. Hence, I was hoping for something like 'integrally-closed' implies 'torsion-free volume forms'. | |
| Apr 16, 2013 at 22:47 | comment | added | Mohan | If I remember correctly, there is a result of Donu Arapura which says that in addition to the above torsion-freeness, R is Cohen-Macaulay and smoothable, then R itself is smooth. | |
| Apr 16, 2013 at 20:14 | history | answered | Hailong Dao | CC BY-SA 3.0 |