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Aug 9, 2014 at 22:46 comment added Yemon Choi It seems to me that these comments could be left as answers
Aug 9, 2014 at 14:46 review Close votes
Aug 10, 2014 at 4:07
Aug 9, 2014 at 14:34 comment added Olivier Benoist abx's comment immediately provides counterexamples when $G$ is not connected. There are also counterexamples when $G$ is connected and $X$ not normal, but none if $G$ is connected and $X$ is normal : see question mathoverflow.net/questions/109310/… , its answer and the references therein.
Aug 9, 2014 at 14:29 comment added abx You forgot an obvious necessary condition, namely that some power $L^m$ of $L$ should be $G$-invariant (i.e. $g^*L^m\cong L^m$ for all $g$ in $G$). That this is equivalent to your statement is Proposition 1.5 in Mumford's GIT.
Aug 9, 2014 at 14:25 history edited User3773 CC BY-SA 3.0
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Aug 9, 2014 at 13:58 history asked User3773 CC BY-SA 3.0