If a finite group acts $G$ on a variety $X$, consider the quotient $X/G$. I would like to understand which line bundle on $X$ descends to $X/G$. The action is not free. Can anyone direct me to some reference?
$\begingroup$
$\endgroup$
6
-
$\begingroup$ This may depend on your definition of the word "descend". If you want the original bundle to be $p^*(\text{line bundle})$, where $p\colon X\to X/G$ is the projection, the question reduces to whether $w_1$ (real bundles) or $c_1$ (complex bundles) is in the image of the appropriate induced homomorphism $p^*$. The latter, of course, depends on the action :) $\endgroup$– Alex DegtyarevCommented Mar 27, 2016 at 12:19
-
$\begingroup$ Cross posted from MSE: math.stackexchange.com/q/1715532/12885 $\endgroup$– BenCommented Mar 27, 2016 at 12:19
-
3$\begingroup$ See my answer to this MO question: a line bundle $L$ (or, more generally, a vector bundle) descends if and only if $L$ admits a $G$-linearization such that the stabilizer $G_x$ of each point $x$ acts trivially on $L_x$. $\endgroup$– abxCommented Mar 27, 2016 at 12:27
-
2$\begingroup$ @abx what you claim is fine for tame actions but may fail in positive characteristic. See the fantastic post of David Rydh : mathoverflow.net/questions/204701/… $\endgroup$– NielsCommented Mar 27, 2016 at 20:03
-
$\begingroup$ Mumford talks about this extensively for abelian varieties in his book of the same name. $\endgroup$– Avi SteinerCommented Mar 28, 2016 at 3:58
|
Show 1 more comment