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Dec 21, 2011 at 20:19 answer added Niels timeline score: 2
Dec 21, 2011 at 18:36 comment added Keerthi Madapusi Yes, of course.
Dec 21, 2011 at 18:14 comment added Jack Huizenga I suppose we should also assume the action is faithful for this to be true.
Dec 21, 2011 at 17:45 comment added Keerthi Madapusi I will assume that you are working over an algebraically closed field. Let $\mathcal{O}$ be the completion of $X$ at $x$, and let $\mathcal{O}'$ be the completion of $X/G$ at the image of $x$. Then the fact is that $\mathcal{O}'$ is the ring of invariants in $\mathcal{O}$ under the stabilizer of $x$. In particular, $\mathcal{O}'=\mathcal{O}$ precisely when the stabilizer is trivial.
Dec 21, 2011 at 17:34 history asked user20107 CC BY-SA 3.0